Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
derive differential form $$\frac{l}{r^2}\frac{d}{d\theta}(\frac{l}{mr^2} \frac{dr}{d\theta})-\frac{l^2}{mr^3}=f(r)$$
I have to prove it is equal to
by taking $u=\dfrac{1}{r}$.
$$\frac{l}{r^2}\frac{l}{mr^2}\frac{d^2r}{d\theta^2}-\frac{l^2}{mr^3}=f(r)$$
$$\frac{d^2r}{d\theta^2}=\frac{mr^4}{l^2}f(r)+\frac{1}{r}$$
$$\frac{d^2r}{d\theta^2}-\frac{1}{r}=\frac{mr^4}{l^2}f(r)$$
I just can write that $du=-\frac{1}{r^2dr}$ I can't see how it will work for my equation.
|
We have $u=\dfrac{1}{r},$ so $r=\dfrac{1}{u}$ and $\dfrac{dr}{d\theta}=-\dfrac{1}{u^2}\dfrac{du}{d\theta}$ by the chain rule and thus substituting we have
$$\frac{l}{r^2}\frac{d}{d\theta}\left(\frac{l}{mr^2} \frac{dr}{d\theta}\right)-\frac{l^2}{mr^3}=f(r)$$
$$lu^2\frac{d}{d\theta}\left(\frac{lu^2}{m}\left(-\frac{1}{u^2}\frac{du}{d\theta}\right)\right)-\frac{l^2}{m}u^3=f\left(\frac{1}{u}\right)$$
$$-\frac{l^2u^{2}}{m}\frac{d^2u}{d\theta^2}-\frac{l^2}{m}u^3=f\left(\frac{1}{u}\right)$$
$$\frac{d^{2}u}{d\theta^2}+u=-\frac{m}{l^2}\frac{1}{u^2}f\left(\frac{1}{u}\right)=-\frac{m}{l^2}\frac{d}{du}V\left(\frac{1}{u}\right)$$
where $\dfrac{d}{du}V\left(\dfrac{1}{u}\right)=\dfrac{1}{u^2}f\left(\frac{1}{u}\right)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4258050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Proving $\frac{2\sin x+\sin 2x}{2\sin x-\sin 2x}=\csc^2x+2\csc x \cot x+\cot^2x$
Prove
$$\dfrac{2\sin x+\sin 2x}{2\sin x-\sin 2x}=\csc^2x+2\csc x \cot x+\cot^2x$$
Proving right hand side to left hand side:
$$\begin{align}\csc^2x+2\csc x \cot x+\cot^2x
&= \frac{1}{\sin^2x}+\dfrac{2\cos x}{\sin^2x}+\dfrac{\cos^2x}{\sin^2x} \tag1 \\[0.8em]
&=\frac{\cos^2x+2\cos x+1}{\sin^2x} \tag2 \\[0.8em]
&=\frac{1-\sin^2x+1+\dfrac{\sin 2x}{\sin x}}{\sin^2x} \tag3 \\[0.8em]
&=\frac{\;\dfrac{2\sin x-\sin^3x+\sin 2x}{\sin x}\;}{\sin^2x} \tag4 \\[0.8em]
&=\frac{2\sin x-\sin^3x+\sin2 x}{\sin^3x} \tag5
\end{align}$$
I could not prove further to the left hand side from here. I would need help. Thank you in advance.
|
We have that, using $\sin 2x=2\sin x\cos x$ and cancelling out $2\sin x\neq0$
$$\dfrac{2\sin x+\sin 2x}{2\sin x-\sin 2x}= \dfrac{2\sin x+2\sin x\cos x}{2\sin x-2\sin x\cos x}= \dfrac{1+\cos x}{1-\cos x}$$
then, assuming $\cos x \neq -1$ ( which holds since we need $\sin x\neq0$ for the original expression), multiply by $\frac{1+\cos x} {1+\cos x}$ and simplify to obtain the result using that $1-\cos^2x=\sin^2x$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4259836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Prove that $a^{2n+1}+(a-1)^{n+2}$ is divisible by $a^2-a+1$ for any $a \in\mathbb{Z}$. Prove that $a^{2n+1}+(a-1)^{n+2}$ is divisible by $a^2-a+1$ for any $a \in \mathbb{Z}$.
I tried to use induction here, but it seems like a dead end. Maybe I should somehow use binomial theorem?
I need a hint
|
For the divisibility, I am going to prove that $a^{2 n+1}+(a-1)^{n+2}\equiv 0 \quad \pmod{a^2-a+1}$
\begin{aligned}
& a^{2 n+1}+(a-1)^{n+2} \\
\equiv & a^{2 n+1}+\left(a^{2}\right)^{n+2} \quad\left(\bmod a^{2}-a+1\right) \\
\equiv & a^{2 n}\left(a+a^{4}\right) \quad\left(\bmod a^{2}-a+1\right)\\
\equiv & a^{2 n}\left[a+(1-a)^{2}\right] \quad\left(\bmod a^{2}-a+1\right)\\
\equiv & a^{2 n}\left(a^{2}-a+1\right) \quad\left(\bmod a^{2}-a+1\right)\\
\equiv & 0 \quad\left(\bmod a^{2}-a+1\right)\\
\end{aligned}
$\therefore a^{2 n+1}+(a-1)^{n+2} \text { is divisible by } a^{2}-a+1.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4261491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
$10$ circles ($2$ large of radius $R$, $6$ small of radius $r$ and 2 small of radius $t$) are enclosed in a square. How we find $r$ in terms of $t$? Let us embed $2$ large intersecting circles of radius $R$ into a square as depicted by the figure below. These two circles are highlighted green. Into these $2$ circles we embedd $6$ smaller ones of equal radius $r$ (highlighted orange).
Finally we have $2$ circles of the same radius $t$, which both touch the large (green) circles and the square.
How we can find a formula that calculates the radius $r$ of the $6$ small circles when inputting the radius $t$?
One idea to proceed might be to define a distance $d$ from the center of one of the large (green) circles to the point at which the large radius $R$ is touching one of the small (orange) circles. Then at least we would get the Pythagorean triangle equation $r^2+d^2=(R − r)^2$ as a possibly useful starting point:
|
The blue triangle is easy and has already been illustrated by the OP. Without loss of generality, suppose the square is the unit square in the coordinate plane. Then the common point of tangency of the two small circles is $(1/2,1/2)$, and the distance between this point and a center of the larger circle is simply $d_1 = \sqrt{2}(1/2 - R)^2$. Therefore, the blue triangle corresponds to the equation $$2 (1/2 - R)^2 + r^2 = (R - r)^2.$$
Now let $d_2$ be the distance from $(1/2,1/2)$ to the point of tangency of the small circle that is outside one of the larger circles, with the diagonal line. Then we must have the simultaneous conditions $$(d_2 + d_1)^2 + r^2 = (R + r)^2, \\ (d_2 - d_1)^2 + r^2 = (R - r)^2.$$ The solution to this system is left as an exercise for the reader; we have
$$\begin{align}
r &= \frac{10 - \sqrt{10}}{45}, \\
R &= \frac{10 - \sqrt{10}}{18}, \\
d_1 &= \frac{\sqrt{10} - 1}{9\sqrt{2}}, \\
d_2 &= \frac{2 \sqrt{5} - \sqrt{2}}{9}.
\end{align}$$
Note this uniquely fixes all of the circles in the square. This means the two externally tangent circles with radius $t$ are also uniquely determined, and their common radius is easily shown to obey the relationship $$(1-R-t)^2 + (R-t)^2 = (R+t)^2,$$ or $$t = 1 - 2 \sqrt{R} + R = \frac{28 - \sqrt{10} - 6\sqrt{20 - 2 \sqrt{10}}}{18}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4263323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Eliminating $\theta$ from $\cos^3\theta +a\cos\theta =b$ and $\sin^3\theta +a\sin\theta =c$
Eliminate $\theta$ from the equations.
$$\cos^3\theta +a\cos\theta =b$$
$$\sin^3\theta +a\sin\theta =c$$
Can anyone solve this question?
|
I have now my own answer developed out of conversation with David Quinn, so I thank him. I publish this answer to demonstrate the calculation. It seems to coincide with the calculation of Macavity in the comments above. Wow what a monster of a question.
We have
$$x^3+ax=b$$
$$y^3+ay=c$$
Where $x^2+y^2=1$.
Adding and subtracting,
$$x^3+y^3+a(x+y)=b+c$$
$$x^3-y^3+a(x-y)=b-c$$
$$(x+y)(x^2-xy+y^2)+a(x+y)=b+c$$
$$(x-y)(x^2+xy+y^2)+a(x-y)=b-c$$
$$(x+y)(1-xy)+a(x+y)=b+c$$
$$(x-y)(1+xy)+a(x-y)=b-c$$
So
$$(x+y)(1+a-xy)=b+c$$
$$(x-y)(1+a+xy)=b-c$$
squaring,
$$(1+2xy)((1+a)^2+x^2y^2-2(a+1)xy)=(b+c)^2$$
$$(1-2xy)((1+a)^2+x^2y^2+2(a+1)xy)=(b-c)^2$$
And adding,
$$2(1+a)^2+2x^2y^2-8(a+1)x^2y^2=2(b^2+c^2)$$
So
$$(8a+6)x^2y^2=2(1+a)^2-2(b^2+c^2)$$
$$(4a+3)x^2y^2=(1+a)^2-(b^2+c^2)$$
$$x^2y^2=\frac{(1+a)^2-(b^2+c^2)}{4a+3}$$
On the other hand if we multiply the two equations together we get
$$xy(x^2y^2 +a(1+a))=bc$$
and squaring,
$$x^2y^2(x^2y^2 +a(1+a))^2=b^2c^2$$
and substituting,
$$\frac{(1+a)^2-(b^2+c^2)}{4a+3}\left(\frac{(1+a)^2-(b^2+c^2)}
{4a+3} +a(1+a)\right)^2=b^2c^2$$
which becomes,
$$[(1+a)^2-(b^2+c^2)]\left((1+a)(2a+1)^2-(b^2+c^2)\right)^2=b^2c^2(4a+3)^3$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4265926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
}
|
Prove $\int_{0}^{\infty} \psi^{(2)} (1+x) \ln (x) \, dx = \zeta'(2) + \zeta(2)$ I'm looking for alternate methods of proving the following result:
$$\int_{0}^{\infty} \psi^{(2)} (1+x) \ln (x) \, dx = \frac{\pi^2}{6} \left( \gamma + \ln (2\pi)-12 \ln A +1\right) = \zeta'(2) + \zeta(2)$$
Where $\psi$ is the polygamma function, $A$ is the Glaisher-Kinkelin constant and $\gamma$ is the Euler-Mascheroni constant.
Here is how I originally proved it:
Begin with the known result $$\int_{0}^{\infty} \frac{x \ln x}{e^{2 \pi x}-1} \, dx = \frac{1}{2} \zeta'(-1)=\frac{1}{24}-\frac{1}{2} \ln A$$
then use the following well-known property of the Laplace transform
$$\int_{0}^{\infty} f(x) g(x) \, dx = \int_{0}^{\infty} \left( \mathcal{L} f\right) (y) \left( \mathcal{L}^{-1} g \right) (y) \, dy$$
Let $f(x) = \frac{x^2}{e^{2 \pi x} - 1}$ and $g(x) = \frac{\ln x}{x}$ then:
$$\left(\mathcal{L} f\right) (y) = -\frac{1}{8 \pi^3} \psi^{(2)} \left(1+\frac{y}{2\pi}\right)$$
$$\left(\mathcal{L}^{-1} g\right)(y) = -\gamma - \ln (y)$$
$$\implies \frac{1}{24} - \frac{1}{2} \ln A = \frac{1}{8\pi^3}\int_{0}^{\infty} \gamma \, \psi^{(2)} \left(1+\frac{y}{2\pi}\right) \, dy + \frac{1}{8\pi^3} \int_{0}^{\infty} \psi^{(2)} \left(1+\frac{y}{2\pi}\right) \ln (y) \, dy$$
Proceed with the substitution $y \mapsto 2\pi x \implies dy = 2 \pi\,dx$
$$\implies \frac{1+\gamma}{24} - \frac{1}{2} \ln A = \frac{1}{4\pi^2} \int_{0}^{\infty} \psi^{(2)} \left(1+x\right) \ln (2 \pi x) \, dx$$
Now separate $\ln (2\pi x) = \ln (2 \pi) + \ln (x)$ and the result quickly follows.
I was wondering if there are any alternative methods that don't rely on the initial $\frac{1}{2} \zeta'(-1)$ result.
Motivation:
I've been experimenting with the following identity:
$$\int_{0}^{\infty} \frac{x^{2n-1}}{e^{2\pi x}-1}\, dx = \frac{(-1)^{n-1} B_{2n}}{4n}\,\text{ for } n \in \mathbb{N}$$
I noticed that if one takes the partial derivative with respect to $n$ and then proceeds to take the limit as $n \to 1$, then we arrive at our $-\frac{1}{2} \zeta'(-1)$ result. This means it is possible for one to define the 'derivative' of the Bernoulli numbers at $n=1$.
More precisely:
$$B_{2}^{\prime} = \frac{1}{4} - 2 \ln A$$
is this result already known, and if so, are any other Bernoulli derivatives definable in a similar way using the Glaisher-Kinkelin constant? (I know it is possible to do it with Glaisher-Kinkelin-like constants such as the Bendersky-Adamchik constants due to their similar connection to hyperfactorials, $\zeta'(-2)$ and $\zeta(3)$ for example). I'm not sure what meaning one can give to the 'derivative' of the Bernoulli numbers, but figure 2 of this paper seems to roughly agree with my result.
|
For another solution using the complete beta function:
Using the same expansion
$$\psi^{(2)}(x) = - 2 \sum_{n=0}^{\infty} \frac{1}{(n+x)^{3}},$$
we have
$$ \begin{align} \int_{0}^{\infty} \psi^{(2)}(x+1) \ln (x) \, \mathrm dx &= -2\int_{0}^{\infty}\ln(x) \sum_{n=0}^{\infty} \frac{1}{(n+x+1)^{3}} \, \, dx \\ &= - 2 \sum_{n=0}^{\infty} \int_{0}^{\infty} \frac{\ln (x)}{(n+x+1)^{3}} \, dx.\\
=& - 2 \sum_{n=0}^{\infty} \frac{1}{(n+1)^3} \int_{0}^{\infty} \frac{\ln (x)}{(1+\frac{x}{(n+1)})^{3}} \, dx\\
=& - 2 \sum_{n=0}^{\infty} \frac{1}{(n+1)^3} \int_{0}^{\infty} \frac{\left(\frac{d}{dt}\Big|_{t=0+}x^t\right)}{(1+\frac{x}{(n+1)})^{3}} \, dx\\
=& \frac{d}{dt}\Big|_{t=0+} - 2 \sum_{n=0}^{\infty} \frac{1}{(n+1)^3} \int_{0}^{\infty} \frac{x^t}{(1+\frac{x}{(n+1)})^{3}}dx \\
=& \frac{d}{dt}\Big|_{t=0+} - 2 \sum_{n=0}^{\infty} \frac{(n+1)^t}{(n+1)^2} \int_{0}^{\infty} \frac{w^t}{(1+w)^{3}}dw \quad \\
=& \frac{d}{dt}\Big|_{t=0+} - 2 \sum_{n=0}^{\infty} \frac{(n+1)^t}{(n+1)^2}B(t+1,2-t)\\
=& \frac{d}{dt}\Big|_{t=0+} \sum_{n=0}^{\infty} \frac{(n+1)^t}{(n+1)^2}\pi (t-1)t \csc(\pi t)\\
=& \sum_{n=0}^{\infty} \frac{1}{(n+1)^2} - \sum_{n=0}^{\infty} \frac{\ln(n+1)}{(n+1)^2}\\
=& \zeta(2)+\zeta'(2)
\end{align}\\$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4269602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
}
|
the value of $\lim_{x\to0}\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}$ I want to compute this limit
$$\lim_{x\to0}\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x\sin(x^2)}.$$
I tried to apply Hopital rule, but I cannot compute it.
|
Various tricks can simplify the limit. Multiply by $\frac{\sin(x^2)}{x^2}$ which has a limit of one to reduce to
$$\frac{(x^2+1) \ln(x^2+1)-\sin(x^2)}{x^3}.$$Add $$\frac{\sin x^2 -x^2}{x^3}$$ to reduce to
$$\frac{(x^2+1) \ln(x^2+1)-x^2}{x^3}.$$Now break into pieces,
$$\frac{\ln(x^2+1)-1}{x}+\frac{\ln(x^2+1)}{x^3}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4277769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
What's the measure of the segment $LR$ in the figure below? For reference:In the picture shown we have an isosceles trapezoide,
The $QR$ side measures $b$ and the $PQ$ side measures $a$. find $LR$
(answer:$\frac{b(a+b)}{a-b}$ )
My progress:
I didn't get much on this question... a colleague indicated if PQRL form a harmonic range. we would have a quick solution. But it would be necessary to demonstrate that these points form a harmonic range ou or if someone sees another solution...
$LR = x\\
\frac{a+b+x}{x}=\frac{a}{b}\\
ax=ba+b.b+bx\\
ax - bx=b(a+b) \\
x(a-b) = b(a+b)\\ \therefore x = \frac{b(a+b)}{(a-b)}$
Edit: I found a way to demonstrate that points L, R, Q and P form a harmonic range:
$\angle DCP = 2\theta\\\angle DCB = 2\alpha\\\therefore \boxed{\alpha + \theta = 90^o}\\O ~is~ excenter~ \triangle ARL\\\angle LOR = \alpha\\O ~is~ excenter~ \triangle QCP\\\angle QOP = \theta\\\\ \angle ROQ = 90^o-\theta = \alpha \\
\triangle LOQ:\\
OR \text{ is angle bissector internal}\\
\text{OP is angle bissector external} $
|
Showing the pencil $O (LQ; RP)$ is harmonic definitely makes the work easier. Here is an alternate approach that requires solving a quadratic equation in the end.
$EF$ is perpendicular bisector of $AD$ and $BC$. In $EFRP$, $\angle FRP = 180^\circ - \angle EPR = 180^\circ - 2\theta$
As $F$ and $T$ are points of tangency on the circle from external point $R$, $OR$ bisects $\angle FRT$ and $\angle ORT = 90^\circ - \theta$
So, $\triangle OTR \sim \triangle PTO$. If $RT = c$ and $r$ is the radius of the circle, we have
$ \displaystyle \frac{c}{r} = \frac{r}{a+b-c} \implies r^2 = c (a+b-c) \tag1$
$\angle CQR = \angle PCQ + \angle CPQ = 180^\circ - 2 \alpha + 2\theta$
As $OQ$ bisects $\angle CQR$, $\angle OQT = 90^\circ + \theta - \alpha$ and it follows that $\angle ROQ = \alpha$
Using your work, $\angle LOR = \alpha$ or alternatively, $\angle OLT = \frac{180^\circ - (2 \alpha + 2\theta)}{2} = 90^\circ - (\alpha + \theta)$
SO, $\angle LOT = \alpha + \theta \implies \angle LOR = \alpha$. So $OR$ is bisector of $\angle LOQ$.
Therefore, $ \displaystyle \frac{OL^2}{OQ^2} = \frac{LR^2}{RQ^2}$
If $LR = x, ~ \displaystyle \frac{(x+c)^2 + r^2}{r^2 + (b-c)^2} = \frac{x^2}{b^2}$
Or, $b^2 c^2 + 2b^2cx + b^2 r^2 = x^2 ( r^2 + c^2 - 2 bc)$
Plugging in value of $r^2$ from $(1)$ and dividing both sides by $c$,
$b^2 (a+b) + 2b^2 x = (a-b) x^2$
$\implies (a-b) x^2 - 2b^2 x - b^2 (a+b) = 0$
$ ((a-b)x - b(a+b)) (x + b) = 0$
That leads to, $ \displaystyle x = \frac{b (a + b)}{a-b}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4277935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Maximum of a multivariable function with constraints I have to maximize the function
$$p(a,b,c)=\frac{1}{2} \left(\frac{1}{3} \sqrt{2 (\cos (a)+\cos (b)+\cos (c))+3}+1\right);$$
with the constarint that following function is a constant.
$$t(a,b,c)=\frac{1}{2} \left(\sqrt{2 (\cos (a)-\cos (b)-\cos (c))+3}+\sqrt{2 (-\cos (a)+\cos (b)-\cos (c))+3}+\sqrt{2 (-\cos (a)-\cos (b)+\cos (c))+3}+\sqrt{2 (\cos (a)+\cos (b)+\cos (c))+3}\right);$$
and
$$0\leq\{a,b,c\}\leq\pi/2,|a-b|\leq c \leq a+b $$
I believe the solution is when $a=b=c$ (see this answer), but I don't yet have a way to prove it. Is it possible to solve this using Lagrange multipliers? Kindly help any way you can.
EDIT
Can we use the symmetry in the objective and the constraint to prove this? As in, can we say that the critical points should satisfy certain symmetry and then proceed to show that the function is max at that point??
A related question on symmetry was answered here.
|
Remark: In the proof, we use the expression $\color{blue}{p=\frac{1}{12} \left(t+6 +\sqrt{3(12-t^2)}\right)}$ due to @Claude Leibovici which is simpler than @Domen's $p = \frac{1}{12}(\sqrt{2 t (\sqrt{36-3
t^2}-t)+36}+6) $ in the linked page, though they are the same (easy to prove).
Consider the following optimization problem:
\begin{align*}
&\max_{a, b, c} ~ p(a, b, c)\\
&\mathrm{s.t.} \quad t(a, b, c) = T, \\
&\qquad\ 0 \le a, b, c \le \pi/2, ~ |a - b| \le c \le a + b
\end{align*}
where $T$ is a constant such that the feasible region is not empty.
Denote its (global) maximum by $p(T)$.
As expected (to be proved), we have
$$p(T) = \frac{1}{12}\left(T + 6 + \sqrt{36 - 3T^2}\right).$$
(Note:
We have $t(a,b,c) = \frac12(x + y + z + w) \le \frac12\sqrt{4(x^2 + y^2 + z^2 + w^2)} = \sqrt{12}$ where $x, y, z, w$ denote $\sqrt{2 (\cos (a)-\cos (b)-\cos (c))+3}$ etc. Thus, $T \le \sqrt{12}$.)
Proof:
Fact 1: For all $0 \le a, b, c \le \pi/2, ~ |a - b| \le c \le a + b$,
$$p(a, b, c) \le \frac{1}{12}\left(t(a,b,c) + 6 + \sqrt{36 - 3[t(a,b,c)]^2}\right).$$
(The proof is given at the end.)
Using Fact 1, we have $p(T) \le \frac{1}{12}\left(T + 6 + \sqrt{36 - 3T^2}\right)$.
On the other hand, let
$$a = b = c =
\arccos \left(1 - \frac{1}{12}T^2 + \frac{1}{12}\sqrt{36 - 3T^2}\right)$$
which satisfy $0 \le a, b, c \le \pi/2, ~ |a - b| \le c \le a + b$; $~ t(a, b, c) = T$; and
$$p(a,b,c) = \frac{1}{12}\left(T + 6 + \sqrt{36 - 3T^2}\right).$$
As a result, we have
$$p(T) = \frac{1}{12}\left(T + 6 + \sqrt{36 - 3T^2}\right).$$
We are done.
Proof of Fact 1:
For convenience, use $p, t$ to denote $p(a,b,c), t(a,b,c)$.
It suffices to prove that
$$36 - 3t^2 \ge (12p - t - 6)^2$$
or
$$-144p^2 + 24pt - 4t^2 + 144p - 12t \ge 0. \tag{1}$$
Let
\begin{align*}
x &= \sqrt{2 (\cos (a)-\cos (b)-\cos (c))+3}, \\
y &= \sqrt{2 (-\cos (a)+\cos (b)-\cos (c))+3}, \\
z &= \sqrt{2 (-\cos (a)-\cos (b)+\cos (c))+3}, \\
w &= \sqrt{2 (\cos (a)+\cos (b)+\cos (c))+3}. \tag{2}
\end{align*}
We have
$$p = \frac12(w/3 + 1), \quad t = \frac12(x + y + z + w).$$
(1) becomes
$$36 - (x + y + z)^2 - 3w^2 \ge 0$$
which is true since
\begin{align}
36 - (x + y + z)^2 - 3w^2 &\ge
36 - 3(x^2 + y^2 + z^2) - 3w^2\\
&= 36 - 3(x^2 + y^2 + z^2 + w^2)\\
&= 0
\end{align}
where we have used $(x + y + z)^2 \le 3(x^2 + y^2 + z^2)$.
We are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4280771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Trigonometric equation in quadratic form If $\cos^4α+k$ and $\sin^4α+k$ are the roots of $x^2+\lambda \ (2x+1 )=0$ and $\sin^2α +\ell$ and $\cos^2 α +\ell$ are the roots of $x^2+8x+4=0$ , then the sum of the possible values of $λ$
is _________.
My approach is as follow
${\sin ^2}\alpha + \ell + {\cos ^2}\alpha + \ell = - 8;1 + 2\ell = - 8;\ell = - \frac{9}{2}$
$\left( {{{\sin }^2}\alpha + \ell } \right)\left( {{{\cos }^2}\alpha + \ell } \right) = 4 \\ \Rightarrow {\sin ^2}\alpha {\cos ^2}\alpha + \ell \left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) + {\ell ^2} = 4 \\ \Rightarrow {\sin ^2}\alpha {\cos ^2}\alpha + \ell + {\ell ^2} = 4$
$ \Rightarrow \dfrac{{4{{\sin }^2}\alpha {{\cos }^2}\alpha }}{4} + \ell + {\ell ^2} = 4 \\ \Rightarrow \dfrac{{{{\sin }^2}2\alpha }}{4} - \dfrac{9}{2} + {\left( { - \dfrac{9}{2}} \right)^2} = 4\\ \Rightarrow \dfrac{{{{\sin }^2}2\alpha }}{4} - \dfrac{9}{2} + \dfrac{{81}}{4} = 4 \\ \Rightarrow \dfrac{{{{\sin }^2}2\alpha }}{4} + \dfrac{{63}}{4} = 4\\ \Rightarrow {\sin ^2}2\alpha = - 47$
which is not possible hence how do I proceed?
|
For a quadratic equation, $Ax^2+Bx+C=0$, the difference of its roots can be shown to satisfy
$$(x_1-x_2)^2 = (x_1+x_2)^2-4x_1x_2 = \frac{B^2-4AC}{A^2} \tag{1}$$
Since $$\cos^4 \theta - \sin^4\theta = (\cos ^2 \theta - \sin ^2\theta)(\cos ^2 \theta + \sin ^2\theta)$$ $$=\cos ^2 \theta - \sin ^2\theta$$
the difference of two roots of given two qudratics are equal.
$$(\cos^4 \alpha + k)-(\sin^4 \alpha + k) = (\cos^2 \alpha + l) - (\sin^2 \alpha + l)$$
Then using $(1)$,
$$(2\lambda)^2-4\cdot\lambda\cdot 1 = 8^2-4\cdot 4\cdot 1$$
simplifies to $\lambda^2 - \lambda -12=0$.
Edit remark :
This is the method, but as mentioned in comment below the question, difference of two roots given is $\cos 2\alpha \le 1$ but for second quadratic we have
$$(x_1-x_2)^2 = 48 > 1$$
which is absurd.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4282120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Solve the diophantine equation $(n+1)^{2}X+n^2Y=1$ where $n\in\mathbb{N}$. I having difficulty applying Euclid's algorithm to find a particular solution.
|
Either $n=2k$ or $n+1=2k$ for some integer $k$.
In the first case, Euclid's algorithm gives:$$\begin{array}{c|c} (2k+1)^2&4k^2\\4k+1&-k\\1 \end{array}$$
Working backwards: \begin{eqnarray*}1&=&(4k+1) +4(-k)\\
&=& (4k+1)+4(4k^2-k(4k+1)\\ &=& (1-4k)(4k+1)+4(4k^2)\\&=&
(1-4k)((2k+1)^2-(4k^2))+4(4k^2)\\&=&
(1-4k)(2k+1)^2 +(3+4k)(4k^2)\\&=&(1-4k)(n+1)^2+(3+4k)n^2
\end{eqnarray*}
Thus $X=1-2n, Y=3+2n$.
The second case follows exactly the same method, though it is not necessary to repeat it, as substituting the above values for $X,Y$ we find the equation holds for all $n$.
In both cases $n^2, (n+1)^2$ are coprime, so lcm$(n^2, (n+1)^2)=n^2(n+1)^2$. Thus the solutions $$X=1-2n+tn^2,\qquad Y=3+2n-t(n+1)^2,\qquad t\in \mathbb{Z},$$
cover all cases.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4282728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Zero divisors in a Ring Let's say $a, b$ are zero divisors in a Ring, (i.e., there exist some $x,y \in R$ s.t. $ax=0, by=0$).
I feel that $a$ is a zero divisor of $xy$ (as $axy=0y=0$), but is $b$ a zero divisor of $xy$?
If I take a look at $bxy$, I know I can't commute $b$ and $x$, but can $b$ be a zero divisor of $(xy)$?
|
Your intuition is right. Indeed, $axy=0$, but we cannot say that $bxy=0$.
For example, consider $M_2(\mathbb{Z})$, the $2\times2$ matrices with integer coefficients. We have that
\begin{align*}
\begin{pmatrix}
1&0\\0&0
\end{pmatrix}
\begin{pmatrix}
0&0\\1&1
\end{pmatrix}=\begin{pmatrix}
0&0\\0&0
\end{pmatrix}
\end{align*}
and
\begin{align*}
\begin{pmatrix}
0&1\\0&0
\end{pmatrix}
\begin{pmatrix}
1&1\\0&0
\end{pmatrix}=\begin{pmatrix}
0&0\\0&0
\end{pmatrix}
\end{align*}
It is clear that
\begin{align*}
\begin{pmatrix}
1&0\\0&0
\end{pmatrix}
\begin{pmatrix}
0&0\\1&1
\end{pmatrix}
\begin{pmatrix}
1&1\\0&0
\end{pmatrix}=\begin{pmatrix}
0&0\\0&0
\end{pmatrix}
\end{align*}
but
\begin{align*}
\begin{pmatrix}
0&1\\0&0
\end{pmatrix}
\begin{pmatrix}
0&0\\1&1
\end{pmatrix}
\begin{pmatrix}
1&1\\0&0
\end{pmatrix}=\begin{pmatrix}
1&1\\0&0
\end{pmatrix}
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4283794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Approximate solution to a transcendental equation I'm working on a physics problem and stumbled upon the following equation:
$$h=\frac{2n\pi+\arctan\left(\frac{c}{b}\right)}{b}$$
where $n \in \mathbb{Z}$, $c \in [0,20]$ and $h \in \mathbb{R}^+$. This equation has to be solved for $b$, which can be done numerically ofcourse, but if possible, I would love to have a quite accurate expression in the given interval. Ive already did some work and obtained the following fair approximation for $b$:
$$b\approx\frac{(1+2n)\pi c}{2+ch}$$ However, this equation breaks down at low $c$. Thus, I was wondering if more accurate (analytical) expressions are possible.
|
Let $\frac c b=x$ to make the equation
$$2 n\pi x+x \tan ^{-1}(x)=k \qquad \text{with} \qquad k=ch$$ Expand the lhs as a Taylor series around $x=0$
$$2 n\pi x+x \tan ^{-1}(x)=2 n\pi x+x^2-\frac{x^4}{3}+\frac{x^6}{5}+O\left(x^8\right)$$ Use series reversion
$$x=\frac{k}{2 \pi n}-\frac{k^2}{8 \pi ^3 n^3}+\frac{k^3}{16 \pi ^5
n^5}+\frac{k^4 \left(4 \pi ^2 n^2-15\right)}{384 \pi ^7 n^7}+\frac{k^5 \left(7-4
\pi ^2 n^2\right)}{256 \pi ^9 n^9}+O\left(k^6\right)$$
In view of this results, we can also perform one single iteration iteration of Newton method
$$x_0=\frac k {2 n\pi}\implies x_1=\frac k {2 n\pi}-\frac{k \cot ^{-1}\left(\frac{2 \pi n}{k}\right)}{2 \pi n \left(2 \pi n
\left(\frac{k}{k^2+4 \pi ^2 n^2}+1\right)+\cot ^{-1}\left(\frac{2 \pi
n}{k}\right)\right)}$$
Edit (after @River Li's answer)
We can expand the function
$$f(b)=b-\frac{\tan ^{-1}\left(\frac{c}{b}\right)+2 \pi n}{h}$$ as series solution around $b=\frac {2n\pi}h$, use series reversion and have, as an approximation
$$b=\frac {2n\pi}h+t+\sum_{k=2}^p a_k \,t^k+ O(t^{p+1})$$ where $$t=\frac{c^2 h^2+4 \pi ^2 n^2}{h \left(c^2 h^2+c h+4 \pi ^2 n^2\right)}\,\tan ^{-1}\left(\frac{c h}{2 \pi n}\right)$$ The first coefficient is
$$a_2=\frac{2 \pi c h^2 n}{\left(c^2 h^2+4 \pi ^2 n^2\right) \left(c^2 h^2+c h+4 \pi ^2 n^2\right)}$$
Repeating @River Li's test cases, the second term is perfectly negligible.
What we can also do is one iteration of Halley method using $b_0=\frac {2n\pi}h$ and have as estimate
$$b=\frac {2n\pi}h+\frac{\left(c^2 h^2+4 \pi ^2 n^2\right) \left(c h (c h+1)+4 \pi ^2 n^2\right) \cot
^{-1}\left(\frac{2 \pi n}{c h}\right)}{h \left(\left(c h (c h+1)+4 \pi ^2
n^2\right)^2-2 \pi c h n \cot ^{-1}\left(\frac{2 \pi n}{c h}\right)\right)}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4284566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Should I ignore ± sign when integrating square roots? I was solving the following integral:
$$
\int \:\frac{x^2}{\sqrt{x^2+4}}dx
$$
$$
u=\sqrt{x^2+4}
$$
$$
\:du=\frac{2x}{2\sqrt{x^2+4}}dx=\frac{x}{u}dx
$$
$$
\int \:\frac{x^2}{\sqrt{x^2+4}}dx=\int \:\frac{x^2}{u}dx=\int \:xdu
$$
Now I only need to find what x means in terms of u:
$$
u^2=x^2+4,\:u^2-4=x^2
$$$$
x=\pm \sqrt{u^2-4}
$$
But now I have a problem, which is the plus minus sign, so my integral would be:
$$
\int \pm \sqrt{u^2-4}du
$$
To avoid this problem, I decided to use integration by parts instead:
$$
\int xdu\:=\:xu-\int \:udx\:=
$$
$$
x\sqrt{x^2+4}-\int \:\sqrt{x^2+4}dx
$$
But it looks like both equations yielded the same result and the plus minus sign was unnecessary.
$$\int \pm \sqrt{u^2-4}du$$
$$x\sqrt{x^2+4}-\int \:\sqrt{x^2+4}dx$$
$$u=2sect,\:t=arcsec\left(\frac{u}{2}\right),\:du=2sec\left(t\right)tan\left(t\right)dt$$
$$x=2tan\left(t\right),\:t=arctan\left(\frac{x}{2}\right),\:dx=2sec^2tdt$$
$$\int \:\sqrt{u^2-4}du=\int \:2tan\left(t\right)\cdot 2sec\left(t\right)tan\left(t\right)dt=$$
$$x\sqrt{x^2+4}-\int \:\sqrt{x^2+4}dx=\:x\sqrt{x^2+4}-\int \:2sec\left(t\right)\cdot 2sec^2tdt=$$
$$4\int \:sec\left(t\right)tan^2\left(t\right)dt=4\int \:\:sec\left(t\right)\left(sec^2\left(t\right)-1\right)dt=$$
$$\:x\sqrt{x^2+4}-4\int \:sec^3tdt$$
$$4\int \:\:sec^3tdt-4\int \:sec\left(t\right)dt$$
$$\int \:sec^3tdt=\frac{1}{2}\sec \:\left(t\right)\tan \:\left(t\right)+\frac{1}{2}\ln \:\left|\tan \:\left(t\right)+\sec \:\left(t\right)\right|+C$$
$$=4\left(\frac{1}{2}\sec \:\:\left(t\right)\tan \:\:\left(t\right)+\frac{1}{2}\ln \:\:\left|\tan \:\:\left(t\right)+\sec \:\:\left(t\right)\right|-ln\left|\tan \:\:\:\left(t\right)+\sec \:\:\:\left(t\right)\right|\right)$$
$$=x\sqrt{x^2+4}-4\left[\frac{1}{2}\sec \left(t\right)\tan \left(t\right)+\frac{1}{2}\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|\right]$$
$$=2\sec \left(t\right)\tan \left(t\right)-2\ln \left|\tan \:\:\left(t\right)+\sec \:\:\left(t\right)\right|$$
$$=x\sqrt{x^2+4}-2\sec \left(t\right)\tan \left(t\right)-2\ln \left|\tan \left(t\right)+\sec \left(t\right)\right|$$
$$=2\sec \left(sec^{-1}\left(\frac{u}{2}\right)\right)\tan \left(sec^{-1}\left(\frac{u}{2}\right)\right)-2\ln \left|\tan \:\:\left(sec^{-1}\left(\frac{u}{2}\right)\right)+\sec \:\:\left(sec^{-1}\left(\frac{u}{2}\right)\right)\right|$$
$$=x\sqrt{x^2+4}-2\sec \left(tan^{-1}\left(\frac{x}{2}\right)\right)\tan \left(tan^{-1}\left(\frac{x}{2}\right)\right)-2\ln \left|\tan \left(tan^{-1}\left(\frac{x}{2}\right)\right)+\sec \left(tan^{-1}\left(\frac{x}{2}\right)\right)\right|$$
$$sec=\frac{h}{a}=\frac{u}{2},\:o=\sqrt{u^2-2^2},\:tan=\frac{o}{a}=\frac{\sqrt{u^2-4}}{2}$$
$$tan=\frac{o}{a}=\frac{x}{2},\:h=\sqrt{x^2+2^2},\:sec=\frac{h}{a}=\frac{\sqrt{x^2+4}}{2}$$
$$=2\left(\frac{u}{2}\right)\frac{\sqrt{u^2-4}}{2}-2\ln \left(\left|\frac{\sqrt{u^2-4}}{2}+\frac{u}{2}\right|\right)$$
$$=x\sqrt{x^2+4}-2\frac{\sqrt{x^2+4}}{2}\left(\frac{x}{2}\right)-2\ln \:\left|\frac{x}{2}+\frac{\sqrt{x^2+4}}{2}\right|$$
$$=\frac{\sqrt{x^{2}+4}\sqrt{\left(\sqrt{x^{2}+4}\right)^{2}-4}}{2}-2\ln\left(\left|\frac{\sqrt{\left(\sqrt{x^{2}+4}\right)^{2}-4}}{2}+\frac{\sqrt{x^{2}+4}}{2}\right|\right)$$
$$=\frac{2x\sqrt{x^2+4}}{2}-\frac{x\sqrt{x^2+4}}{2}-2\ln \:\left|\frac{x}{2}+\frac{\sqrt{x^2+4}}{2}\right|$$
$$=\frac{x\sqrt{x^{2}+4}}{2}-2\ln\left|\frac{x+\sqrt{x^{2}+4}}{2}\right|$$
$$=\frac{x\sqrt{x^2+4}}{2}-2\ln \:\left|\frac{x+\sqrt{x^2+4}}{2}\right|$$
So, since both of them yield the exact same answer after simplification, I wonder if we can always assume that square roots are positive and omit the plus minus sign, or was my logic actually right that I should always try to avoid substitutions with plus minus square roots?
As you can see by the graph it seems to work for both positive and negative x. My only suspicion is that in cases where it is not possible to simplify the formations such as fractional angles. Then maybe we could be getting it wrong... for example when answer is like this... $$sin\left(\frac{1}{8}cos^{-1}x\right)$$
|
What the $\pm$ sign in $x=\pm \sqrt{u^2-4}$ implies is that you have a choice of how to make the substitution of $u$ for $x.$ You need this choice to be available in order that your evaluation of the integral can be valid for all values of $x.$
If we suppose that $x$ can have either positive or negative values, then there is an error in the last step of your evaluation of $\int \sqrt{u^2 - 4}\,\mathrm du.$
In fact,
$$
\sqrt{\left(\sqrt{x^2+4}\right)^2 - 4} = \sqrt{x^2} = \lvert x\rvert,
$$
not simply $x$ as you appear to have assumed in your work. So what you should have at the bottom of the left-hand column in your work is
$$
\frac{\lvert x\rvert\sqrt{x^2+4}}{2} - 2\ln\left\lvert\frac{\lvert x\rvert + \sqrt{x^2+4}}{2}\right\rvert,
$$
At first glance, this looks different from what you get at the bottom of the right-hand column. And it is different. You will find that if you plug in a negative value of $x,$ the expression in the left-hand column has a value exactly opposite the value in the right-hand column; that is, when $x < 0,$
\begin{multline}
\frac{\lvert x\rvert\sqrt{x^2+4}}{2} - 2\ln\left\lvert\frac{\lvert x\rvert + \sqrt{x^2+4}}{2}\right\rvert \\
= \frac{(-x)\sqrt{x^2+4}}{2} - 2\ln\left\lvert\frac{-x + \sqrt{x^2+4}}{2}\right\rvert \\
= -\left(\frac{x\sqrt{x^2+4}}{2} - 2\ln\left\lvert\frac{x + \sqrt{x^2+4}}{2}\right\rvert\right).
\end{multline}
The fact that
$$ \frac{(-x)\sqrt{x^2+4}}{2} = - \left(\frac{x\sqrt{x^2+4}}{2}\right) $$
is obvious; the fact that
$$
\ln\left\lvert\frac{-x + \sqrt{x^2+4}}{2}\right\rvert
= -\ln\left\lvert\frac{x + \sqrt{x^2+4}}{2}\right\rvert
$$
is less obvious, but it might be a little less surprising if you realize that
$$
\ln\left\lvert\frac{x + \sqrt{x^2+4}}{2}\right\rvert
= \ln\left(\frac{x + \sqrt{x^2+4}}{2}\right)
= \sinh^{-1}\left(\frac x2\right).
$$
Now you might think this is a disaster: the results agree for non-negative $x,$ but for all negative values of $x$ we have violent disagreement.
But this is where the $\pm$ sign saves you.
In fact, what you showed in the left-hand column is the result of the substitution
$x=\sqrt{u^2-4},$ $u=\sqrt{x^2+4},$
which results from choosing the $+$ part of the $\pm$ sign.
But if $x$ is negative that is not a valid substitution, because there is no real number $u$ that makes $\sqrt{u^2-4}$ negative.
For negative values of $x$ you can use $x=-\sqrt{u^2-4},$ $u=\sqrt{x^2+4}.$
That means you need to evaluate $\int -\sqrt{u^2 - 4}\,\mathrm du.$
Fortunately this is just $-\int \sqrt{u^2 - 4}\,\mathrm du$
and you have already evaluated $\int \sqrt{u^2 - 4}\,\mathrm du,$
so all you need to do is to take your result for $\int \sqrt{u^2 - 4}\,\mathrm du$
and reverse its sign.
And that makes it exactly equal to the result in the right-hand column
in the case where $x$ is negative.
Therefore you have the same results from both methods for non-negative $x$ and also for negative $x.$
But only if you acknowledge both the positive and negative square roots of
$u^2 - 4$ and use the correct root in every case.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4285670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
}
|
Area of a square on top of a quadrilateral
In quadrilateral $ABCD, \angle ABC + \angle DCB = 90^\circ$ and $ADEF$ is a square constructed on side $AD$ in the exterior of the quadrilateral $ABCD.$ If $BC = 10 $ cm, $AC = 9$ cm and $BD = 8$ cm, then the area of the square $ADEF$ lies between
(a) $70$ and $80$
(b) $60$ and $70$
(c) $50$ and $60$
(d) $40$ and $50$
I drew a rough sketch using Geogebra. Looks like the area of square should be between $40$ and $50.$
But I am not able to come with a proof. Any hints on how to proceed
|
Extend $\small CD$ to meet $\small AB$ at $\small E$.
Since $\small \measuredangle B+\measuredangle C=90^\circ$, we can see $\small \angle BEC=90^\circ.$
Now from Pythagorean theorem,
$\small BC^2=EB^2+EC^2\tag1$
$\small AD^2=EA^2+ED^2\tag2$
Adding both,
$\small \begin{align} BC^2+AD^2&=(EB^2+ED^2)+(EC^2+EA^2)\\ &=BD^2+AC^2\end{align}$
Now substituting,
$\small 10^2+AD^2=8^2+9^2\implies AD^2=\dots$
And $\small AD^2$ means ...
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4286027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
If $\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab}$, prove that $\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$.
Problem: Let $a, b, c, x, y, z$ be real numbers
such that $abc \ne 0$, $ ~ a + b + c \ne 0$, $~ (a^2 - bc)(b^2 - ca)(c^2 - ab)\ne 0$, $xyz\ne0$, $x+y+z\ne0$, $(x-y)^2+(y-z)^2+(z-x)^2\ne0$
and $$\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab} \ne 0.$$
Prove that $$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}.$$
The converse can be proved easily.
My Attempt:
$
\frac{x^2-yz}{a^2-bc}=\frac{y^2-zx}{b^2-ca}=\frac{z^2-xy}{c^2-ab}\\
\Rightarrow\frac{x^2-yz-y^2+zx}{a^2-bc-b^2+ca}=\frac{y^2-zx-z^2+xy}{b^2-ca-c^2+ab}=\frac{z^2-xy-x^2+yz}{c^2-ab-a^2+bc}\ [By\ Addendo]\\\Rightarrow\frac{x-y}{a-b}\cdot\frac{x+y+z}{a+b+c}=\frac{y-z}{b-c}\cdot\frac{x+y+z}{a+b+c}=\frac{z-x}{c-a}\cdot\frac{x+y+z}{a+b+c}\\\Rightarrow\frac{x-y}{a-b}=\frac{y-z}{b-c}=\frac{z-x}{c-a}\ \left[Considering\ (x+y+z)\ and\ (a+b+c)\ \neq0\right]
$
Known Exceptional Counter-examples:-
Macavity: $a=1, b=2, c=3, x=y=z$
River Li: $a = 1, b=2, c = -3, x = 7, y = 7, z = -14$
The conditions in the question have been edited to exclude exceptional counter-examples.
|
Proof:
First of all, clearly we have
$$ (a - b)^2 + (b - c)^2 + (c - a)^2 \ne 0. \tag{1}$$
Now, we split into two cases:
Case 1 $~ x = y$:
We have
$$\frac{x^2-xz}{a^2-bc}=\frac{x^2-zx}{b^2-ca}=\frac{z^2-x^2}{c^2-ab} \ne 0$$
which results in
\begin{align*}
a^2 - bc &= b^2 - ca, \tag{2}\\
x(c^2 - ab) &= - (z + x)(b^2 - ca). \tag{3}
\end{align*}
From (2), we have $(a - b)(a + b + c) = 0$
which results in $a = b$.
With this and (3), we have
$x(c^2 - a^2) = - (z + x)(a^2 - ca)$ or $(a - c)(az - cx) = 0$ which results in $az - cx = 0$ or $\frac{x}{a} = \frac{z}{c}$,
where we have used $a = b$ and (1) to get $a \ne c$.
Thus, we have
$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$.
Case 2 $~ x \ne y$:
Let
\begin{align*}
f &= (a^2 - bc)(y^2 - zx) - (b^2 - ca)(x^2 - yz) \\
&= Az + B,\\
g &= (a^2 - bc)(z^2 - xy) - (c^2 - ab)(x^2 - yz) \\
&= Cz^2 + Dz + E
\end{align*}
where
\begin{align*}
A &= -{a}^{2}x-acy+{b}^{2}y+bcx, \\
B &= {a}^{2}{y}^{2}+ac{x}^{2}-{b}^{2}{x}^{2}-bc{y}^{2}, \\
C &= a^2 - bc,\\
D &= -aby+{c}^{2}y, \\
E &= -{a}^{2}xy+ab{x}^{2}+bcxy-{c}^{2}{x}^{2}.
\end{align*}
We eliminate $z$ from the system of $f = g = 0$ as follows.
Let
$$h = Cz \cdot f - A\cdot g = (BC - AD)z - AE.$$
Then let
\begin{align*}
F &= (BC - AD)\cdot f - A \cdot h\\
&=
A^2 E - ABD + B^2C\\
&= - \frac18 (x - y) [3(x + y)^2 + (x - y)^2] ( ay - bx ) ( a + b + c ) \\
&\qquad \times ( a^2 - bc ) [(a - b)^2 + (b - c)^2 + (c - a)^2].
\end{align*}
From $F = 0$, using (1) and the given conditions, we have $ay - bx = 0$
or $\frac{x}{a} = \frac{y}{b}$.
Since $x \ne y$, we have $a \ne b$.
Using $x = \frac{a}{b}y$,
we have
$$f = - \frac {y (a - b) [ 3(a + b)^2 + (a - b)^2] ( bz - cy) }{4 b^2} = 0
$$
which results in $bz = cy$ or $\frac{y}{b} = \frac{z}{c}$.
Thus, we have
$\frac{x}{a}=\frac{y}{b}=\frac{z}{c}$.
We are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4286648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
Dealing more efficiently with fractional forms in system of equations As an example, suppose we have to solve the following system of two equations and two unknowns:
$$
\begin{cases}
-\frac{10}{x}-\frac{8}{y} &= \frac{8}{3} \\
-\frac{6}{x}+\frac{6}{y} &= -\frac{1}{3}
\end{cases}
$$
My approach and solution
I opted to solve it by combination, referring to the first equation as (1) and the second equation as (2), I started by eliminating the y's:
\begin{align}
\frac{6}{8}(1)+(2): -\frac{60}{8x}-\frac{6}{x} &= \frac{6}{3}-\frac{1}{3} \\
\frac{-60-48}{8x} &= \frac{5}{3} \tag{*}\\
40x &=3(-108) \tag{**} \\
x &=-\frac{81}{10} \tag{1'}
\end{align}
Substituting (1') into (2):
\begin{align}
\frac{100}{81}-\frac{8}{y}&=\frac{8}{3} \\
y &= -\frac{162}{29}
\end{align}
So I find the tuple of $(-\frac{81}{10};-\frac{162}{29})$ as solution.
Questions
*
*For such systems, should we not be concerned about the conditions of existence of the system? Namely, in this case both $x\neq 0$ and $y\neq 0,$ in the same way that we would do when solving an equation. If yes, how do we formally write the domain of existence for a system?
*Is my transition from step $(*)$ to $(**)$ allowed? My understanding is that, yes for all $x\neq 0.$
*I am really eager to learn whether there are simple ideas that simplify the system (and other similar systems) before we start solving it. As shown in my approach, it was slightly awkward dealing with the fractions throughout and the "large" numbers, which inherently may render the approach more prone to mistakes. Any alternative, quicker approach (I only know by combination and substitution) would be much appreciated.
|
We start by getting rid of the fractions.
\begin{align*}
\begin{cases}
-\frac{10}{x}-\frac{8}{y} &= \frac{8}{3}
&\implies -30y -24x=8xy \\
-\frac{6}{x}+\frac{6}{y} &= -\frac{1}{3}
&\implies -18y+18x=-xy
\end{cases}
\\\text{We mulitply the second equation by 8 and then add}
\\
\begin{cases}
-\space \space 30y -\space \space 24x &=\space \space\space 8xy \\
-144y+144x &=-8xy
&\implies 120 x - 174 y=0\\
&&\implies y=\dfrac{20x}{29}
\end{cases}
\end{align*}
By substitution
\begin{align*}
-30\cdot \dfrac{20x}{29} -24x&=8x\cdot \dfrac{20x}{29}\\
\implies
-1296 x &= 160 x^2
&\implies x=\frac{-81}{10}\space \space \\
\\
-30y -24\cdot \frac{-81}{10}
&= 8y\cdot \frac{-81}{10} \\
\implies -174y&=972
&\implies y= \dfrac{-162}{29}\\
\end{align*}
We confirm the calculations with WA
here
and
here
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4288341",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Limits of integration for theta in double integrals I have a question that asks me to find the volume lying inside both the sphere $x^2+y^2+z^2=a^2$ and the cylinder $x^2+y^2=ax$. The worked solution in the textbook goes like this
One quarter of the required volume lies in the first octant. In polar coordinates the cylinder $x^2 + y^2 = ax$ becomes $r = a cos θ$. Thus, the required volume is
$$
V=4\iint_D\sqrt{a^2-x^2-y^2}\,dA=4\int_0^{\pi/2}\int_0^{a\cos\theta}\sqrt{a^2-r^2}\,r\,dr\,d\theta=\frac{2\pi a^3}{3}-\frac{8a^3}{9}
$$
My question is why can't I do the integral from $0$ to $2\pi$ and get the same answer.
$$
V=\int_0^{2\pi}\int_0^{a\cos\theta}\sqrt{a^2-r^2}\,r\,dr\,d\theta=\frac{2\pi a^3}{3}
$$
Why do these answers differ? and how can you know when you have to split it up and when you can integrate all the way through?
|
In polar coordinates, $ \theta \in (0, 2\pi)$ is valid for the circle centered at the origin. But we have a circle that is centered on x-axis away from the origin and in fact the origin is a point on the circle.
The projection of the cylinder in XY-plane is circle $x^2 + y^2 = ax$.
Or, $ \displaystyle \left(x - \frac{a}{2}\right)^2 + y^2 = \frac{a^2}{4}$. This is a circle centered at $(a/2, 0)$ with radius $a/2$.
Now if we use the parametrization $x = r \cos\theta, y = r \sin\theta$ which measures radial distance from the origin, $x^2 + y^2 = ax$ simplifies to $r = a \cos\theta$. Also note that the circle is in the fourth and the first quadrant, and hence $ \theta \in (- \pi/2, \pi/2)$ covers the entire circle. Please see the below diagram for further explanation.
However if we parametrize the circle as $x = \frac{a}{2} + r \cos\theta, y = r \sin \theta$, which translates the origin to $(a/2, 0)$, the limits of $\theta$ will be $(0, 2\pi)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4288853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Compute $\int_{\gamma} \nabla f \cdot d\mathbf{x}$ Compute $\int_{\gamma} \nabla f \cdot d\mathbf{x}$ for the following choices of $f$ and $\gamma$.
(a) $f(x,y) = x^2+y^2; \gamma:g(t) = (1+t^2, 1-t^2), -1 \le t \le2$
What I have tried:
$$\int_{-1}^2(1+t^2)^2dx + \int_{-1}^2(1-t^2)^2dy$$
Where $dx = 2t$ and $dy = -2t$
So we have
$$\int_{-1}^2(1+t^2)^2(2t)dt + \int_{-1}^2(1-t^2)^2(-2t)dt$$
however this outputs the wrong result. How do I proceed from here>?
|
For $f(x,y) = x^2+y^2$ we have
$$ \nabla f. X = 2(x^2+y^2)$$
Thus,
$\begin{array}{ccl}
\displaystyle\int_{\gamma} \nabla f. X ds &=& 2 \displaystyle\int_{\gamma} (x^2+y^2) ds\\
&=& 2 \displaystyle\int_{-1}^{2} ((1+t^2)^2+(1-t^2)^2) \sqrt{8t^2} dt\\
&=& 2 \sqrt{8} \left( - \displaystyle\int_{-1}^{0} ((1+t^2)^2+(1-t^2)^2) t dt +\displaystyle\int_{0}^{2} ((1+t^2)^2+(1-t^2)^2) t dt \right) \\
&=& 8\sqrt{2} \left(-\displaystyle\int_{-1}^{0} t^5+tdt+\displaystyle\int_{0}^{2}t^5+tdt \right) \\
&=&96\sqrt{2}.
\end{array}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4292070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Multiplying power series I'm trying to understand how the expression
$$
\sum\limits_{n=0}^\infty x^n \cdot \sum\limits_{n=0}^\infty x^n
$$
results in the following expression:
$$
\sum\limits_{n=0}^\infty (1+n)x^n.
$$
Could you please walk me through it?
|
A simple observation might be enlightening:
$$(1+x+x^2+x^3)(1+x+x^2+x^3)\\
=1+x+x^2+x^3+x+x^2+x^3+x^4+x^2+x^3+x^4+x^5+x^3+x^4+x^5+x^6\\
=1+2x+3x^2+4x^3+3x^4+2x^5+x^6.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4295137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Infinite binomial series over a column of Pascal's triangle: $F_k(x)= \sum\limits_{n=0}^\infty \binom{2n+k}n x^{2n+k}$ Is there any closed form formula (or an equivalent) for this binomial infinite series :
$$F_k(x)= \sum_{n=0}^{\infty} \binom{2n+k}{ n } x^{2n+k} $$ in which $|x|<1$ and $k$ is a given integer?
This is a sum over a (odd or even) infinitely long column of Pascal's triangle :
For example, above : $$F_3(x)= x^3+ 5 x^5 + 21 x^7 + 84 x^9 +... $$
$F_k(x)$ appears in the Fourier Series development of $\frac{1}{1+a\,sin(x)}$ : the k-th harmonic would be of amplitude $F_k(a)$
|
We have
$$
S(k,x) = \sum\limits_{n = 0}^\infty {\left( \matrix{ 2n + k \cr n \cr} \right)x^{2n + k} }
= x^k \sum\limits_{n = 0}^\infty {\left( \matrix{ 2n + k \cr n \cr} \right)\left( {x^2 } \right)^n }
$$
Indicating the series coefficients as
$$
t_n = \left( \matrix{ 2n + k \cr n \cr} \right)
$$
then we have
$$
\eqalign{
& t_0 = 1 \cr & {{t_{n + 1} } \over {t_n }} =
{{\left( \matrix{ 2n + 2 + k \cr n + 1 \cr} \right)} \over {\left( \matrix{ 2n + k \cr n \cr} \right)}} =
{{n!\left( {n + k} \right)!\left( {2n + 2 + k} \right)!} \over {\left( {n + 1} \right)!\left( {n + 1 + k} \right)!\left( {2n + k} \right)!}}
= {{\left( {2n + 2 + k} \right)\left( {2n + 1 + k} \right)} \over {\left( {n + 1} \right)\left( {n + 1 + k} \right)}} = \cr
& = 4{{\left( {n + 1 + k/2} \right)\left( {n + 1/2 + k/2} \right)} \over {\left( {n + 1} \right)\left( {n + 1 + k} \right)}} \cr}
$$
which means that the series might be expressed through a
Hypergeometric function
$$
\eqalign{
& S(k,x) = x^k \sum\limits_{0\, \le \;k} {{{\left( {1/2 + k/2} \right)^{\,\overline {\,n\,} }
\left( {1 + k/2} \right)^{\,\overline {\,n\,} } } \over {\left( {1 + k} \right)^{\,\overline {\,n\,} } }}
{{\left( {4x^2 } \right)^{\,n} } \over {n!}}} = \cr
& = x^k {}_2F_{\,1} \left( {\left. {\matrix{ {1/2 + k/2,\;1 + k/2} \cr {1 + k} \cr } \;}
\right|\;4x^2 } \right) \cr}
$$
which converges for $|x|<1/2$
To explain the convergence range, it is known that
$$
S(0,x) = \sum\limits_{n = 0}^\infty {\left( \matrix{ 2n \cr n \cr} \right)x^{2n} } =
{1 \over {\sqrt {1 - 4x^2 } }}
$$
The duplication formula for Gamma in fact gives
$$
\eqalign{
& \left( \matrix{
2n \cr
n \cr} \right) = {{\Gamma \left( {2n + 1} \right)} \over {\Gamma \left( {n + 1} \right)^2 }} =
{{4^{\,n} } \over {\sqrt \pi }}{{\Gamma \left( {n + 1/2} \right)} \over {\Gamma \left( {n + 1} \right)}} = \cr
& = 4^{\,n} {{\Gamma \left( {n + 1/2} \right)} \over {\Gamma \left( {1/2} \right)\Gamma \left( {n + 1} \right)}} =
4^{\,n} \left( \matrix{ n - 1/2 \cr n \cr} \right) = \cr
& = \left( { - 4} \right)^{\,n} \left( \matrix{ - 1/2 \cr n \cr} \right) \cr}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4295731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
Prove that is a perfect square. Let $x,y$ and $z$ be positive integers such that $$\frac{1}{x}-\frac{1}{y}=\frac{1}{z}.$$ Let $h=(x,y,z).$ Prove that $hxyz$ is a perfect square.
Solution: Let $z=hc, x=ha,y=hb$ so $(a,b,c)=1$. Notice that:
\begin{align*}
hxyz=h^4abc
\end{align*}
We need to see that $abc$ is an square. Notice that from the equation,
\begin{align*}
z=\frac{xy}{y-x}&=\frac{h^2ab}{h(b-a)}=\frac{hab}{b-a}\\
c&=\frac{ab}{b-a}\\
\Rightarrow ab=c(b-a)
\end{align*}
So from the first equation:
$$hxyz= (h^4c)(c(b-a))=h^2c^2(b-a)$$
So we need to check that $b-a$ is an square. From the last equation, $c(b-a)=ab$, if $d\mid b-a$ is a prime, then $d\mid ab$, so $d\mid a$ or $b$, but any of them lets that $d$ divides the other, so $d^2\mid ab$. Since $(a,b,c)=1$, as we have $c(b-a)=ab$, $d^2\mid c(b-a)$, but $(d,c)=1$, so $d^2\mid (b-a)$, implying that for each prime divisor of $b-a$, there are two. If there is no $d$ prime such that $d\mid b-a$, then $b-a=1$, that is a square too. Then you have that
$$hxyz=h^4c^2(b-a)$$
Is really an square.
Question: Why $z=\frac{xy}{y-x}$?
|
Why is $z=\frac{xy}{y-x}$?
Initially, $x,y,z$ satisfy
$$\frac 1 x - \frac 1 y = \frac 1 z$$
Therefore, by solving for $z$,
$$z = \dfrac{1}{\left( \dfrac 1 x - \dfrac 1 y \right)} = \dfrac{1}{\left(\dfrac{y-x}{xy} \right)} = \dfrac{xy}{y-x}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4297140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Showing that $\arctan(\cos(\alpha)*\tan(x)) = \arccos(\dfrac{\cos(x)}{\cos(\arcsin(\sin(\alpha)*\sin(x)))})$ In trying to calculate an arc length of a right spherical triangle, I reached this $\arccos$ expression. However, I see the expression given in the manuals for the same problem is much simpler. checking against several values of $\alpha$ and $x$ (Let them both be between $0$ and $\pi/2$) shows those expressions are indeed equivalent.
$$\arctan(\cos(\alpha)*\tan(x)) = \arccos(\dfrac{\cos(x)}{\cos(\arcsin(\sin(\alpha)*\sin(x)))})$$
I tried to prove this equivalence using some trigonometric identities, but I'm quite far it seems.
|
Notice that $\cos(\arcsin(x)) = \sqrt{1 - x^2}$ by drawing a triangle where $\sin\theta = x$.
Similarly, $\tan(\arccos(x)) = \frac{\sqrt{1 - x^2}}{x} = \sqrt{\frac{1}{x^2} - 1}$.
Now \begin{align*}\tan(RHS) &= \tan\arccos\big(\cos x\cdot\frac{1}{\sqrt{1 - (\sin\alpha\sin x)^2}}\big) \\ &= \sqrt{\frac{1}{\big(\cos x\cdot\frac{1}{\sqrt{1 - (\sin\alpha\sin x)^2}}\big)^2} - 1} \\ &= \sqrt{\frac{1 - \sin^2\alpha\sin^2 x}{\cos^2 x} - 1} \\ &= \sqrt{\frac{\sin^2 x(1 - \sin^2\alpha)}{\cos^2 x}} \\ &= \tan x\cos\alpha\end{align*}
as desired.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4297536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Fibonacci numbers, sum of squares and divisibility Let $(F_n)$ be the Fibonacci sequence (i.e. $F_{n+2} = F_{n+1}+F_n$ with $F_{-1}=1$ and $F_0=0$).
Consider $a,b \in \mathbb{Z}_{\ge 1}$ with $a \le b$, and $D=1+a^2+b^2$, such that $a$ and $b$ divide $D$.
Question: Must the ordered pair $(a,b)$ be equal to $(F_{2n-1},F_{2n+1})$ for some $n \ge 0$?
Remark: It was checked (by SageMath) for $b<10^5$.
Note that $F_{2n\pm1}$ divides $1+ F_{2n-1}^2 + F_{2n+1}^2$ thanks to the following identity:
Proposition: $1+ F_{2n-1}^2 + F_{2n+1}^2 = 3F_{2n-1}F_{2n+1}$.
proof: Apply Cassini's identity (below) with $2n$, then $F_{2n-1}F_{2n+1} = 1 + F_{2n}^2$. Now by definition, $F_{2n} = F_{2n+1} - F_{2n-1}$. Then
$$ F_{2n-1}F_{2n+1} = 1 + (F_{2n+1} - F_{2n-1})^2 = 1 + F_{2n+1}^2 - 2F_{2n+1}F_{2n-1} + F_{2n-1}^2.$$
The result follows. $\square$
Cassini's identity: $F_{n-1}F_{n+1} = (-1)^n + F_n^2$.
proof: $F_{n-1}F_{n+1} - F_n^2 = \det \left(\begin{matrix} F_{n+1} & F_n \\ F_n & F_{n-1} \end{matrix} \right) = \det \left[\left(\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right)^n \right] = \left[\det \left(\begin{matrix} 1 & 1 \\ 1 & 0 \end{matrix} \right) \right]^n = (-1)^n$. $\square$
|
Yes. This uses Vieta Jumping. Note first that your $\gcd(a,b) = 1,$ so that $ab| a^2 + b^2 + 1.$ Then $a^2 + b^2 + 1 = 3ab.$ Furthermore, the only Ground Solution, in the sense of Hurwitz 1907, is $(1,1).$ Every solution derives as a finite number of jumps from $1,1.$ That all solutions are Fibonacci numbers, two indices apart, may be proved by induction.
My way of dealing with this is closer to Hurwitz, and consists in looking at the hyperbola branch $x^2 - kxy + y^2 = -1 \; \; \; $ ($x,y>0$) and the location of any GrundLosung, which lie between lines $ y = \frac{k}{2} x$ and $ y = \frac{2}{k} x,$ showing by inequalities that there are no such points when $k \geq 4.$ Indeed, the entire arc on which fundamental solutions lies within $0 \leq x \leq 2, 0 \leq y \leq 2. \; $ The endpoints are $(x_1, y_1)$ and $(y_1, x_1)$ with $x_1 = \frac{2}{\sqrt{k^2-4}}$ and $$ y_1 = \frac{k}{\sqrt{k^2 - 4}} = \sqrt{1 + \frac{4}{k-2}} - \frac{2}{\sqrt{k^2-4}}$$
which is between $1$ and $2 \; . \; \;$
Furthermore, the arc does not pass through $(1,1)$ when $k \geq 4.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4298283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
}
|
Putting 12 balls of different colors in 4 urns, 3 balls in each. 5 red, 4 green and 3 blue balls are put in 4 boxes so that every box has exactly 3 balls.
Find the probability that:
*
*At least 1 box has a ball of every color;
*Only 1 box has a ball of every color
I've tried to compute the probability of at least 1 box having an RGB as follows:
$$\mathbb{P}(\geq 1 RGB)=\frac{\binom{5}{1}\cdot\binom{4}{1}\cdot\binom{3}{1}\cdot\frac{9!}{(3!)^3}}{\frac{12!}{(3!)^4}}=\frac{3}{11}$$
since we choose an RGB and then put the remaining 9 balls into boxes at random.
The problem is, I don't know whether I need to account for the allocation of boxes. If, say, we choose one box to put the RGB into and the total number of ways to allocate boxes is $4!$, then the $\mathbb{P}(\geq 1 RGB)$ will multiply by $\frac{4}{4!}$ and be equal to $\frac{2}{11}$.
No progress with the other probability so far.
From the examples found online, this problem may be solved with the inclusion-exclusion formula, but I struggle to figure out what the events $A_1,\dots,A_n$ are in this model.
How do I apply the inclusion-exclusion formula here or is there any other way to obtain the probabilities in question?
|
Line up the urns from left to right. There are
$$\binom{12}{3}\binom{9}{3}\binom{6}{3}\binom{3}{3}$$
to distribute three balls to each of the four urns.
An urn with a ball of each color: There are four ways to select the urn which will receive a ball of each color, five ways to select a red ball, four ways to select a green ball, and three ways to select a blue ball for that urn. There are $\binom{9}{3}\binom{6}{3}\binom{3}{3}$ ways to distribute the remaining nine balls to the remaining three urns so that they each receive three balls. Hence, there are
$$\binom{4}{1}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{9}{3}\binom{6}{3}\binom{3}{3}$$
such distributions.
Two urns with a ball of each color: There are $\binom{4}{2}$ ways to select the two urns which will receive a ball of each color. There are five ways to select a red ball, four ways to select a green ball, and three ways to select a blue ball for the leftmost of those two urns. There are four ways to select a red ball, three ways to select a green ball, and two ways to select a blue ball for the rightmost of those two urns. There are $\binom{6}{3}\binom{3}{3}$ ways to distribute the remaining balls to the remaining two urns so that they each receive three balls. There are
$$\binom{4}{2}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{4}{1}\binom{3}{1}\binom{2}{1}\binom{6}{3}\binom{3}{3}$$
such distributions.
Three urns with a ball of each color: There are $\binom{4}{3}$ ways to select which three urns will receive a ball of each color. There are five ways to select a red ball, four ways to select a green ball, and three ways to select a blue ball for the leftmost of those three urns. There are four ways to select a red ball, three ways to select a green ball, and two ways to select a blue ball for the rightmost of those three urns. There are three ways to select a red ball, two ways to select a green ball, and one way to select a blue ball for the middle of those three urns. All three of the remaining balls must be placed in the other urn. Hence, there are
$$\binom{4}{3}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{4}{1}\binom{3}{1}\binom{2}{1}\binom{3}{1}\binom{2}{1}\binom{1}{1}\binom{3}{3}$$
such distributions.
It is not possible for all four urns to each receive a ball of each color since there are only three blue balls. By the Inclusion-Exclusion Principle, the probability that there is at least one ball of each color is
$$\frac{\binom{4}{1}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{9}{3}\binom{6}{3}\binom{3}{3} - \binom{4}{2}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{4}{1}\binom{3}{1}\binom{2}{1}\binom{6}{3}\binom{3}{3} + \binom{4}{3}\binom{5}{1}\binom{4}{1}\binom{3}{1}\binom{4}{1}\binom{3}{1}\binom{2}{1}\binom{3}{1}\binom{2}{1}\binom{1}{1}\binom{3}{3}}{\dbinom{12}{3}\dbinom{9}{3}\dbinom{6}{3}\dbinom{3}{3}}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4299628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Find all positive integers $x,y,z$ for which the expression is true. So I wish to find all the positive integers $x,y,z$ such that the following expression is true.
$$\frac{-1 + x + y + x y}{-1 - x - y + x y} = z$$
The problem is that the expression does not simplify any further as far as I can tell. The only constraint that I can think of is the denominator has to be greater than zero and so we get that $y > \frac{x+1}{x-1}$. We can also write the left-hand side as:
$$\frac{(x+1)(y+1)-2}{(x-1)(y-1)-2} = z$$
But I still can't see the algorithm that would let me find the integer solutions. Any ideas?
PS. Wolfram can easily find the solutions but I wish to know how it approaches the problem.
|
Let $x$, $y$ and $z$ be positive integers such that
$$\frac{-1+x+y+xy}{-1-x-y+xy}=z.\tag{1}$$
First, it should be clear that $x\geq2$ and $y\geq2$ because the ratio $z$ is positive, and that $z\geq2$ because the numerator is strictly greater than the denominator. If $x=2$ then $(1)$ simplifies to
$$z=\frac{1+3y}{-3+y}=3+\frac{10}{y-3},$$
which shows that $y-3$ divides $10$. This yields the following triplets $(x,y,z)$ as solutions:
$$(2,4,13),\quad(2,5,8),\quad(2,8,5),\quad(2,13,4).$$
Of course by symmetry in $x$ and $y$ we also get the following solutions:
$$(4,2,13),\quad (5,2,8),\quad(8,2,5),\quad(13,2,4).$$
For the remaining solutions we have $x,y\geq3$. Clearing the denominator in $(1)$ shows that
$$(z-1)xy-(z+1)x-(z+1)y-(z-1)=0,$$
and multiplying by $(z-1)$ we can express this as
$$\big((z-1)x-(z+1)\big)\big((z-1)y-(z+1)\big)=2(z^2+1).\tag{2}$$
Of course, because $x,y\geq3$ we see that
$$2(z^2+1)=\big((z-1)x-(z+1)\big)\big((z-1)y-(z+1)\big)\geq(2z-4)^2=4(z-2)^2,$$
which implies that $z\leq7$. Plugging these values of $z$ into $(2)$ we get the identities
\begin{eqnarray*}
(x-3)(y-3)&=&10,\\
(x-2)(y-2)&=&5,\\
(3x-5)(3y-5)&=&34,\\
(2x-3)(2y-3)&=&13,\\
(5x-7)(5y-7)&=&74,\\
(3x-4)(3y-4)&=&25,\\
\end{eqnarray*}
This yields the following triplets $(x,y,z)$ of solutions:
\begin{eqnarray*}
z=2:&\qquad (4,13,2),&&\quad (5,8,2),\quad (8,5,2),\quad
(13,4,2)\\
z=3:&\qquad (3,7,3),&&\quad (7,3,3),\\
z=4:&\qquad (2,13,4),&&\quad(13,2,4),\\
z=5:&\qquad (2,8,5),&&\quad(8,2,5),\\
z=6:&\qquad \text{None,}\\
z=7:&\qquad (3,3,7).
\end{eqnarray*}
These $15$ solutions are all positive integral solutions to $(1)$. Note that for $z=4$ and $z=5$ we have found some earlier solutions with $x=2$ or $y=2$ again.
Original sketch of answer:
Let $x$, $y$ and $z$ be positive integers such that
$$\frac{-1+x+y+xy}{-1-x-y+xy}=z.$$
Clearly the numerator is strictly greater than the denominator and so $z\geq2$. It follows that
$$-1+x+y+xy\geq2(-1-x-y+xy),$$
and a bit of rearranging shows that this is equivalent to
$$(x-3)(y-3)\leq10.$$
This greatly limits the possible values for $x$ and $y$. Without loss of generality $x\leq y$, and this inequality shows that then $x\leq6$. Moreover it is clear from this inequality, that for $x>3$ there are only a few values of $y$ to check. For $x=3$ identity $(1)$ reduces to
$$z=\frac{2+4y}{-4+2y}=2+\frac{5}{y-2},$$
which shows that either $y=3$ or $y=7$. Similarly, for $x=2$ identity $(1)$ reduces to
$$z=\frac{1+3y}{-3+y}=3+\frac{10}{y-3}.$$
which shows that $y\in\{4,5,8,13\}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4303464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
How to find following limit of sequence?
How to following limit
$$\lim _{n \rightarrow \infty}\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\cdots+\frac{(n-1)}{n^{2}}\right)\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right). $$
My attempt:
\begin{align*}
L_n&=\left(\frac{1}{n^{2}}+\frac{2}{n^{2}}+\cdots+\frac{(n-1)}{n^{2}}\right)\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right).\\
&=\left(\frac{n-1}{2n}\right)\left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right).\\
\end{align*}
By using camparison with
$$\frac{n}{2n}\le \left(\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}\right)\le \frac{n}{n+1}$$
I got $$\frac{1}{4}\le \lim_{n\to \infty}L_n\le \frac{1}{2}$$
But I need exact answer. But I could not get. Its answer given is $\ln \sqrt 2$.
Thank you very much for your valuable time.
|
Hint:
Rewrite the second factor as
$$\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{n+n}=\frac 1n\biggl(\frac 1{1+\dfrac1n}+\frac 1{1+\dfrac2n}+\dots+\frac 1{1+\dfrac nn}\biggr) $$
and observe this is a Riemann sum of the function $\dfrac1{1+x}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4305903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Integrate: $\frac{x^2+x-1}{x^2-1}$ with respect to $x$ Ok so here is where I am up to:
I can find some similarity with the numerator and denominator and managed to reduce to the following:
$\int\frac{x^2+x-1}{x^2-1}dx = \int\frac{x^2-1}{x^2-1}+\frac{x}{x^2-1}dx$
$=\int1+\frac{x}{x^2-1}dx.$
I tried to reduce it further by factoring out an $(x-1)$ but did not get anywhere after that.
Is there any other way to transpose this into a simpler format so I can integrate it?
|
$\int x^2 + x - 1\over x^2 - 1$ dx = $\int ({x^2 - 1\over x^2 - 1} + {x\over x^2 - 1})$dx
= $\int 1 dx + \int {x\over x^2 - 1}$ dx
Let $x^2 - 1$ = t $\implies$ 2x dx = dt
= x + $\int {dt\over 2t}$
= x + ${1\over 2} \int {1\over t}$ dt
since, integration of ${1\over x}$ = log x
= x + ${1\over 2}$ log t + C
Substituting the value of t here, we get
= x + $log(x^2 - 1)\over 2$ + C
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4307299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Number of real solutions of $\begin{array}{r} {\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]} =\frac{3 x-1}{2} \end{array}$ Solve for $x \in \mathbb{R}$ $$\begin{array}{r}
{\left[\frac{2 x+1}{3}\right]+\left[\frac{4 x+5}{6}\right]}
=\frac{3 x-1}{2}
\end{array}$$ where $[x]$ denotes greatest integer less than or equal to $x$.
My try:
Letting $a=\frac{2x+1}{3}$ we get
$$\left[a\right]+\left[a+\frac{1}{2}\right]=\frac{9 a-5}{4} \tag{1}$$
Now knowing that:
$$a-1<[a]\leq a$$ and $$a-\frac{1}{2}<\left[a+\frac{1}{2}\right]\leq a+\frac{1}{2}$$
Adding both the above inequalities and using $(1)$ we get
$$2a-\frac{3}{2}<\frac{9a-5}{4}\leq 2a+\frac{1}{2}$$
So we get $$a \in (-1, 7]$$
Any help here?
|
Letting $\displaystyle a=\frac{2x+1}{3}$ we get
$$\left[a\right]+\left[a+\frac{1}{2}\right]=\frac{9 a-5}{4} \tag{1}$$
The solution will involve identifying all satisfying values of $a$, and then translating these values into all satisfying values of $x$ via $\displaystyle a=\frac{2x+1}{3}$.
The problem may be attacked analytically, from scratch.
Let $a$ be represented by $P + r ~: P \in \Bbb{Z}, ~0 \leq r < 1.$
This implies that $[a] = P$.
From (1) above, the from scratch procedure is to consider the mutually exclusive cases:
*
*$\displaystyle 0 \leq r < \frac{1}{2}.$
*$\displaystyle \frac{1}{2} \leq r < 1.$
$\underline{\textbf{Case 1:} ~\displaystyle 0 \leq r < \frac{1}{2}}$
The LHS of (1) above is $2P.$
Therefore $\displaystyle 2P = \frac{9[P + r] - 5}{4} \implies $
$8P = 9P + 9r - 5 \implies $
$$ 5 = P + 9r.\tag2 $$
Combining the Case 1 constraint with (2) above, you have that $(P,r)$ must be an element in
$$\left\{(5,0), \left(4,\frac{1}{9}\right), \left(3,\frac{2}{9}\right), \left(2,\frac{3}{9}\right), \left(1 + \frac{4}{9}\right)\right\}.\tag3 $$
This means that the Case 1 candidate values for $a$ are
$$\left\{(5), \left(4 + \frac{1}{9}\right), \left(3 + \frac{2}{9}\right), \left(2 + \frac{3}{9}\right), \left(1 + \frac{4}{9}\right)\right\}.\tag4 $$
In fact, all five of the candidate values for $(a)$ shown in (4) above satisfy (1) above.
$\underline{\textbf{Case 2:} ~\displaystyle \frac{1}{2} \leq r < 1}$
The LHS of (1) above is $2P + 1.$
Therefore $\displaystyle (2P + 1) = \frac{9[P + r] - 5}{4} \implies $
$8P + 4 = 9P + 9r - 5 \implies $
$$ 9 = P + 9r.\tag5 $$
Combining the Case 2 constraint with (5) above, you have that $(P,r)$ must be an element in
$$\left\{\left(4,\frac{5}{9}\right), \left(3,\frac{6}{9}\right), \left(2,\frac{7}{9}\right), \left(1 + \frac{8}{9}\right)\right\}.\tag6 $$
This means that the Case 2 candidate values for $a$ are
$$\left\{\left(4 + \frac{5}{9}\right), \left(3 + \frac{6}{9}\right), \left(2 + \frac{7}{9}\right), \left(1 + \frac{8}{9}\right)\right\}.\tag7 $$
In fact, all four of the candidate values for $(a)$ shown in (7) above satisfy (1) above.
This leads to the following chart of solutions (in terms of $x$):
\begin{array}{| r | r |}
\hline
a & x = \displaystyle \frac{3a - 1}{2} \\
\hline
5 & 7 \\
\hline
\left(4 + \frac{1}{9}\right) & \left(5 + \frac{2}{3}\right) \\
\hline
\left(3 + \frac{2}{9}\right) & \left(4 + \frac{1}{3}\right) \\
\hline
\left(2 + \frac{3}{9}\right) & \left(3\right) \\
\hline
\left(1 + \frac{4}{9}\right) & \left(1 + \frac{2}{3}\right) \\
\hline
\left(4 + \frac{5}{9}\right) & \left(6 + \frac{1}{3}\right) \\
\hline
\left(3 + \frac{6}{9}\right) & \left(5\right) \\
\hline
\left(2 + \frac{7}{9}\right) & \left(3 + \frac{2}{3}\right) \\
\hline
\left(1 + \frac{8}{9}\right) & \left(2 + \frac{1}{3}\right) \\
\hline
\end{array}
Edit
Actually, there is a shortcut. Because the relationship between $a$ and $x$ is linear, each distinct value of $a$ gives a distinct value in $x$.
Further, the satisfying solutions for $a$ are all of the form $\displaystyle P + \frac{s}{9}$, where $s$ runs through each of the elements in $\{0,1,2,\cdots, 8\}$ exactly once.
Therefore, if all that is desired is a count of the distinct satisfying values of $x$, you can conclude, looking only at the fractional part of the confirmed Case 1 and Case 2 satisfying values for $a$, that there are exactly $9$ solutions.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4311152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
}
|
Use Induction to prove recurrence
if the above screenshot is not visible, here is the text format:
Question:
Solve the following recurrence and prove your result is correct using induction:
$a_1 = 0$
$a_n = 3(a_{n-1}) + 4^{n}$ for $n>=2$
Use induction to prove this recursive sequence.
So my approach was that, I plug in the $a_1 = 0$ into $a_2 = 3(0) + 4^{2} = 4^{2}$
and then
$a_3 = 3(4^{2}) + 4^{3}$
$a_4 = 3(3(4^{2}) + 4^{3}) + 4^{4} = 3^{2} *4^{2} + 3 * 4^{3} +4^{4}$
$a_5 = 3(3^{2} *4^{2} + 3 * 4^{3} +4^{4}) + 4^{5} = 3^{3} *4^{2} + 3^{2} * 4^{3} +3*4^{4} + 4^{5}$
$a_6 = 3^{4} *4^{2} + 3^{3} * 4^{3} +3^{2}*4^{4} + 3*4^{5} + 4^{6}$
...
$a_n = 3^{n-2} *4^{2} + 3^{n-3} * 4^{3} + ... + 3*4^{n-1} + 4^{n} = \sum_{k=2}^{n} 3^{n-k}*4^{k} $
now through induction, I'm not sure how to get there. So far what I know is that:
$a_{n+1} = 3(a_{n}) + 4^{n+1} = \sum_{k=2}^{n+1} 3^{n+1-k}*4^{k}$
= $3(\sum_{k=2}^{n} 3^{n-k}*4^{k}) + 4^{n+1} = \sum_{k=2}^{n+1} 3^{n+1-k}*4^{k}$
= $(\sum_{k=2}^{n} 3*3^{n-k}*4^{k}) + 4^{n+1} = \sum_{k=2}^{n+1} 3^{n+1-k}*4^{k}$
= $(\sum_{k=2}^{n} 3^{n+1-k}*4^{k}) + 4^{n+1} = \sum_{k=2}^{n+1} 3^{n+1-k}*4^{k}$
but then I'm stuck, because I cannot simplify it to make them equal to each other, or I'm in the wrong direction?
|
I assume you mean $a_0=0$ and $a_n = 3a_{n-1}+4^n$ for $n\ge 1$ (not $n\ge 2$). And by "prove this recursive sequence", you mean to find a closed formula.
Consider the generating function $f(x)=\sum_{n=0}^\infty a_nx^n$,
then by the initial condition and recursive relation, we have $$3xf(x)+\sum_{n=1}^\infty 4^nx^n = \sum_{n=1}^\infty a_nx^n = f(x)$$
$$3xf(x) + \frac{1}{1-4x}-1=f(x)$$
$$\begin{align} f(x) &= \frac{4x}{(1-4x)(1-3x)} =4x(\sum_{a=0}^\infty (4x)^a)(\sum_{b=0}^\infty (3x)^b) \\ &=\sum_{n=1}^\infty 4(\sum_{a+b=n-1}4^a3^b)x^n =\sum_{n=1}^\infty 4 \sum_{b=0}^{n-1}(\frac{3}{4})^b 4^{n-1} x^n\\&=\sum_{n=1}^\infty 4^{n+1}(1-(\frac{3}{4})^n)x^n\end{align}$$
Hence $a_n = 4^{n+1}(1-(\frac{3}{4})^n)=4(4^n-3^n)$.
The above reasoning is solid. But if you prefer, you may prove this by induction.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4314529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Optimal wager on two games given probability and odds Question. Suppose I have two games I would like to wager \$100 dollars on, but that I'm not required to wager the full \$100, i.e. I can bet \$50 if I want.
The games: $A$ vs. $B$ and $C$ vs. $D$, where the probabilities of each team winning are
$$P(A) = \frac{7}{10},\quad P(B) = \frac{3}{10}\quad\quad\text{and}\quad\quad P(C)=\frac{1}{4},\quad P(D) = \frac{3}{4}.$$
I am quoted the following odds for each game:
$$
\begin{array}{|c|c|}
\hline
\text{team} & \text{odds} \\ \hline
A & 2:7 \\ \hline
B & 3:1 \\ \hline
C & 7:2 \\ \hline
D & 1:3 \\ \hline
\end{array}.
$$
For example, if we bet on team $A$ we would need to wager \$7 to win \$2.
How should you bet?
So far. It seems to me that there may be several "right" answers, but many "wrong" answers. Initially, my idea was to just look at the individual expected payoffs:
$$
\mathbb{E}A = \frac{7}{10}(2) + \frac{3}{10}(-7) = \frac{-7}{10}\\
\mathbb{E}B = \frac{3}{10}(3) + \frac{7}{10}(-1) = \frac{1}{5} \\
\mathbb{E}C = \frac{1}{4}(7) + \frac{3}{4}(-2) = \frac{1}{4} \\
\mathbb{E}D = \frac{3}{4}(1) + \frac{1}{4}(-3) = 0$$
So we see that we should primarily wager on $B$ and $C$. As far as how much to allocate, I guess we could just scale the expected values of $B$ and $C$ such that we get
$$
\frac{1/5}{1/5 + 1/4} = \frac{4}{9}, \quad\quad \frac{1/4}{1/5 + 1/4} = \frac{5}{9}.
$$
So we wager about \$44 on $B$ and \$56 on $C$. Is this line of thinking correct, or is there actually an optimal wager that isn't what I have?
|
If the player is infinately risk averse she should bet $x$ on $C$ and $100-x$
on $D$ where $x$ is determined by the equation
\begin{eqnarray*}
\frac{x}{2}\ast 7+x &=&\frac{100-x}{3}+100-x\rightarrow \\
x &=&\frac{160}{7}
\end{eqnarray*}
Then, if $C$ wins she will get a net profit of $\frac{160}{7}\frac{7}{2}+%
\frac{160}{7}-100=2.8571.$
And if $D$ wins she will get a net profit of $\ \frac{100-\frac{160}{7}}{3}-(%
\frac{160}{7})=2.\,857\,1$
Hence, there is a possibility to make a guaranteed profit of $%
2.\,8571$ for every $100$ invested. It is thus a good strategy
if she had a credit card without limit.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4316422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Reduction formula for $\int x\tan^n(x)$ So I've been trying to derive a reduction formula for the following integral
$$\int x\tan^n(x)dx$$
I tried to use integration by parts by factoring out a $\tan^2(x)$ and taking the following:
$$u=x\tan^{n-2}(x)\implies du=\tan^{n-2}(x)+(n-2)x\tan^{n-3}(x)\sec^2(x)dx\\
dv=\tan^2(x)dx=(\sec^2(x)-1)dx\implies v=\tan(x)-x$$
which then gives
$$I_n=x\tan^{n-1}(x)-x^2\tan^{n-2}(x)-\int\left[\tan^{n-1}(x)+(n-3)x\tan^{n-2}(x)+(n-2)x\tan^n(x)\\-(n-2)x^2\tan^{n-3}(x)\sec^2(x)\right]dx\\
\implies(n-1)I_n=x\tan^{n-1}(x)-x^2\tan^{n-2}(x)-\int\tan^{n-1}(x)dx-(n-3)\int x\tan^{n-2}(x)dx\\+(n-2)\int x^2\tan^{n-3}(x)\sec^2(x)dx$$
The final integral evaluates to
$$\int x^2\tan^{n-3}(x)\sec^2(x)dx=\frac{1}{n-2}\left(x^2\tan^{n-2}(x)-2\int x\tan^{n-2}(x)dx\right)$$
so we end up with
$$(n-1)I_n=x\tan^{n-1}(x)-x^2\tan^{n-2}(x)-\int\tan^{n-1}(x)dx-(n-3)\int x\tan^{n-2}(x)dx\\+x^2\tan^{n-2}(x)-2\int x\tan^{n-2}(x)dx\\
\implies(n-1)I_n=x\tan^{n-1}(x)-\int\tan^{n-1}(x)dx-(n-1)\int x\tan^{n-2}(x)dx\\
=x\tan^{n-1}(x)-\int\tan^{n-1}(x)dx-(n-1)I_{n-2}$$
But this isn't entirely a reduction formula given that I'm unable to deal with the $\int\tan^{n-1}(x)dx$ term. Any help?
|
$$\begin{split}
I_{n+2} +I_n &= \int x \tan^n (x) (1+\tan^2 x)dx \\
&= \frac 1 {n+1}x \tan^{n+1}(x)-\frac 1 {n+1}\underbrace{\int \tan^n x dx}_{J_n} &\,\,\,\,\,\,\,\,(1)
\end{split}$$
where the last line comes by integrating by parts.
Now
$$\begin{split}
J_{n+2}+J_n &= \int \tan^n (x)(1+\tan^2x) dx\\
&= \frac 1 {n+1}\tan^{n+1}x
\end{split}$$
The above implies that
$$J_n = \frac{\tan^{n-1}x}{n-1}-J_{n-2}=\frac{\tan^{n-1}x}{n-1}-\left(\frac{\tan^{n-3}x}{n-3}-J_{n-4}\right)=...$$
In other words, using the fact that $J_0=x$ and $J_1=-\ln \cos x$,
$$\left\{
\begin{split}
J_{2p}&=\sum_{k=0}^{p-1} \frac{(-1)^k \tan^{2p-2k-1}x}{2p-2k-1}+(-1)^px\\
J_{2p+1}&=\sum_{k=0}^{p-1} \frac{(-1)^k \tan^{2p-2k}x}{2p-2k}+(-1)^{p+1}\ln\cos x
\end{split}
\right.
$$
With that done, we can go back to $I_n$. Equation $(1)$ can be rewritten as
$$I_n = \frac{x\tan^{n+1}x}{n+1} -\frac{J_n}{n+1}-I_{n-2}$$
Following a similar path as for the $J_n$'s, one can find expressions for $I_{2p}$ and $I_{2p+1}$ as a function of the $J_n$'s, as well as $I_0$ and $I_1$. The result is a bit ugly, especially for odd indices since $I_1$ involves a dilogarithm.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4318915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
For real numbers $z$ and $w$, $|(1+z)(1+w)-1| \leq (1+|z|)(1+|w|)-1$. I have written an attempted proof of the theorem on the title, and I need help verifying it.
I have used the following theorems to proof the theorem on the title.
Theorem 5.14)a) Let $x$ be a real number. $-|x| \leq x \leq |x|$.
Theorem 5.14)b) Let $a \geq 0$. $|x| \leq a$ if and only if $-a \leq x \leq a$.
Theorem 5.14)c) Let $x$ and $y$ be real numbers. $|x+y| \leq |x| + |y|$ (The Triangle Inequality).
For real numbers $z$ and $w$, $|(1+z)(1+w)-1| \leq (1+|z|)(1+|w|)-1$.
Proof. From Theorem 5.14)a), $(1+z) \leq |(1+z)|$ and $(1+w) \leq |(1+w)|$. Multiplying the $(1+z) \leq |(1+z)|$ by $|(1+w)|$, one obtains
\begin{align}
(1+z)|(1+w)| \leq |(1+z)||(1+w)|
\end{align}
Since $(1+w) \leq |(1+w)|$, it follows that
\begin{align}
(1+z)(1+w) \leq |(1+z)||(1+w)|
\end{align}
Observe that from The Triangle Inequality,
\begin{align}
|(1+z)| \leq 1 + |z|\\
|(1+w)| \leq 1 + |w|
\end{align}
Since $|(1+z)|(1 + |w|) \leq (1 + |z|)(1 + |w|)$ and $|(1+w)| \leq (1 + |w|)$, it follows that
\begin{align}
(1+z)(1+w) -1 \leq |(1+z)||(1+w)| -1 \leq (1 + |z|)(1 + |w|) -1
\end{align}
It is clear that $(1 + |z|)(1 + |w|) -1 \geq 0$.
In case that $(1+z)(1+w) -1 \geq 0$, $|(1+z)(1+w) -1| = (1+z)(1+w) -1$.
Hence,
\begin{align}
-[(1 + |z|)(1 + |w|) -1] \leq 0 \leq (1+z)(1+w) -1 \leq (1 + |z|)(1 + |w|) -1
\end{align}
From Theorem 5.14)b),
\begin{align}
|(1+z)(1+w) -1| \leq (1 + |z|)(1 + |w|) -1
\end{align}
establishing the result for this case.
On the other hand, in case that $(1+z)(1+w) -1 < 0$, $|(1+z)(1+w) -1| = -[(1+z)(1+w) -1]$.
Hence,
\begin{align}
-[(1 + |z|)(1 + |w|) -1] \leq (1+z)(1+w) -1 < 0
\end{align}
Since $(1 + |z|)(1 + |w|) -1 \geq 0$,
\begin{align}
-[(1 + |z|)(1 + |w|) -1] \leq (1+z)(1+w) -1 \leq (1 + |z|)(1 + |w|) -1
\end{align}
Multiplying the inequalities by $-1$, one obtains
\begin{align}
(1 + |z|)(1 + |w|) -1 \geq -[(1+z)(1+w) -1] \geq -[(1 + |z|)(1 + |w|) -1]
\end{align}
From Theorem 5.14)b),
\begin{align}
|(1+z)(1+w) -1| \leq (1 + |z|)(1 + |w|) -1
\end{align}
establishing the result for this case.
Because the result for all cases of $(1+z)(1+w) -1$ have been established, it is the case that $|(1+z)(1+w) -1| \leq (1 + |z|)(1 + |w|) -1$.
|
We have :
$(1+z)(1+w)-1 = z+w+zw$
Thus :
$|(1+z)(1+w)-1| = |z+w+zw| \le |z| + |w| + |zw| = (1+|z|)(1+|w|) -1$
QED
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4321116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
How many four-digit positive integers are there that contain the digit $3$ and are divisible by $5$?
How many four-digit positive integers are there that contain the digit $3$ and are divisible by $5$?
The answer is:
the number of four-digit integers that are divisible by $5\;-\;$ the number of four-digit integers that are divisible by $5$ and not contain the digit $3$
So, $9\cdot10\cdot10\cdot2-8\cdot 9\cdot 9\cdot 2=1800-1296=504\tag{*}$
I know that, but when I tried to solve this problem with an other way I got a different result.
Four digit integers : $\overline{xyzw}$
Suppose $\overline{xyzw}$ contain at least one digit equal $3$
So, $x = \{1, \ldots ,9\}, y = 3, z = \{0, \ldots, 9\}, w = \{0,5\}$
or $x = \{1, \ldots, 9\}, y = \{0, \ldots, 9\}, z = 3, w = \{0,5\}$
or $x = 3, y=\{0, \ldots, 9\}, z = \{0, \ldots, 9\}, w = \{0,5\}$
The number of all $\overline{xyzw}$ that must be divisible by $5$ and contain $3$ is: $9\cdot 10\cdot 2+9\cdot 10\cdot 2+10\cdot 10\cdot 2=180+180+200=560\tag{**}$
but (*) contradicts (**), so where is the mistake?
|
You double counted, for example $3335$ appears in all your sum.
You can use inclusion-exclusion to adjust your count.
\begin{align}
&|A|+|B|+|C|-|A\cap B|-|A \cap C|-|B \cap C| + |A \cap B \cap C|\\
&=560 - 18 - 20-20+2\\
&=504
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4323371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
$1^2 - 2^2 + 3^2 - 4^2 + \dots + 1999^2$ using closed form formula for sum of squares of first $n$ natural numbers The question is simple, the solution, not so much
Q.Find the sum of the given expression $1^2- 2^2 + 3^2 - 4^2 + \dots + 1999^2$
My idea is we know $1^2 + 2^2 + 3^2 + 4^2 + \dots + n^2 = \frac{n(n + 1)(2n + 1)}{6}$
So for $n=1999$ I get the sum as $2,66,46,67,000$
From this I need to subtract the squares of the even terms twice because subtracting once leaves with only the sum of the squares of the odd nos.
I observed something :
$2^2 + 4^2 + 6^2 + \dots + 1998^2 = (2 \cdot 1)^2 + (2 \cdot 2)^2 + \dots + (2 \cdot 999)^2$
Therefore to obtain the sum of the square of the even terms, I can take $4$ as common and use the aforementioned formula for $n=999$ and multiply it by $4$.
therefore sum of square of even terms = $1,33,13,34,000$
I need to subtract this sum twice to get the answer, because subtracting once simply leaves me with the sum of the squares of the odd numbers.
The answer is now $1999000$, which still doesn't match the answer key. Can someone explain where I am going wrong ?
|
Use the decomposition of squares into triangular numbers, $n^2=T_{n-1}+T_n$, and telescope:
$$1^2-2^2+3^2-\cdots+1999^2=1-(1+3)+(3+6)-(6+10)+\cdots+(T_{1998}+T_{1999})$$
$$=T_{1999}=\frac{1999×2000}2=1999000$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4335944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
}
|
Different solutions to the same system of linear equations I want to solve a real homogeneous system of linear equations represented by this matrix:
\begin{bmatrix}0&2&2&7&1&0\\1&0&1&3&1&0\\-1&2&1&4&0&0\end{bmatrix}
The reduced echelon form is:
\begin{bmatrix}1&0&1&3&1&0\\0&1&1&7/2&1/2&0\\0&0&0&0&0&0\end{bmatrix}
Since I have used Maple to verify my answer, I know that the above matrix is correct. Now comes my question. When deriving the parameterized solution from the reduced echelon form, I want to define the free variables as:
\begin{align*}
x3=t1
\end{align*}
\begin{align*}
x4=t2
\end{align*}
\begin{align*}
x5=t3
\end{align*}
Then the solution has to satisfy the following equations:
\begin{align*}
x1+t1+3t2+t3=0 \leftrightarrow\ x1=-t1-3t2-t3
\end{align*}
\begin{align*}
x2+t1+(7/2)t2+(1/2)t3=0 \leftrightarrow\ x2=-t1-(7/2)t2-(1/2)t3
\end{align*}
Thus the parameterized solution satisfy:
\begin{align*}
\begin{bmatrix}x1\\x2\\x3\\x4\\x5\end{bmatrix}=t1\begin{bmatrix}-1\\-1\\1\\0\\0\end{bmatrix}+t2\begin{bmatrix}-3\\-7/2\\0\\1\\0\end{bmatrix}+t3\begin{bmatrix}-1\\-1/2\\0\\0\\1\end{bmatrix}t1,t2,t3\in\mathbb{R}
\end{align*}
However, when I solve the system in Maple, I get this:
\begin{align*}
\begin{bmatrix}x1\\x2\\x3\\x4\\x5\end{bmatrix}=t1\begin{bmatrix}2\\1\\0\\0\\-2\end{bmatrix}+t2\begin{bmatrix}1\\0\\1\\0\\-2\end{bmatrix}+t3\begin{bmatrix}4\\0\\0\\1\\-7\end{bmatrix}t1,t2,t3\in\mathbb{R}
\end{align*}
Clearly, this is the solution I will get if I define the free variables as:
\begin{align*}
x2=t1
\end{align*}
\begin{align*}
x3=t2
\end{align*}
\begin{align*}
x4=t3
\end{align*}
Since the solution then has to satisfy:
\begin{align*}
t1+t2+(7/2)t3+(1/2)x5=0 \leftrightarrow\ x5=-2t1-2t2-7t3
\end{align*}
\begin{align*}
x1+t2+3t3+x5=0 \leftrightarrow\ x1=-t2-3t3-x5=-t2-3t3-(-2t1-2t2-7t3)=2t1+t2+4t3
\end{align*}
Are both solutions correct? If yes, does it mean that the three vectors in the two different solutions respectively spans the same vector space with respect to different bases? Also, I would really like to know why Maple does this? What is the advantage of choosing other free variables than those you normally choose?
|
It is easy to make Maple to use your choices of free variables, how? Let us rewrite what you are looking for mathematically. What you are asking can be written as a new linear system.
$$\begin{bmatrix}
0 & 0 & 1 & 0 & 0\\
0 & 0 & 0 & 1 & 0\\
0 & 0 & 0 & 0 & 1\\
1 & 0 & 1 & 3 & 1\\
0 & 1 & 1 & \frac{7}{2} & \frac{1}{2}\\
0 & 0 & 0 & 0 & 0
\end{bmatrix}\begin{bmatrix}
x_1\\
x_2\\
x_3\\
x_4\\
x_5
\end{bmatrix}=\begin{bmatrix}
t_1\\
t_2\\
t_3\\
0\\
0\\
0
\end{bmatrix}$$
Then the Maple code using the same LinearSolve command of the LinearAlgebra package that you have used is the following.
A := Matrix([ [0,0,1,0,0], [0,0,0,1,0], [0,0,0,0,1], [1,0,1,3,1], [0,1,1,7/2,1/2], [0,0,0,0,0] ]);
b := < t__1, t__2, t__3, 0, 0, 0 >;
LinearAlgebra:-LinearSolve( A, b );
The output is exactly what you found manually.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4336818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solving $x^2 \equiv 106234249 \bmod{12^2}$ I have the following congruence:
$$x^2 \equiv 106234249 \bmod{12^2}$$
When I tried to solve it, the equation becames in 2 equations:
$x^2 \equiv 4 \bmod{9}$
$x^2 \equiv 9 \bmod{16}$
Because $12^2 = 9 \times 16$ and, 9 and 16 are coprimes.
Then I applied the method of this video, but I obtained 4 results instead of 8.
The following page give me 8 results: https://www.alpertron.com.ar/QUADMOD.HTM
The results that I obtained are:
$x \equiv 83 \bmod{12^2}$
$x \equiv 29 \bmod{12^2}$
$x \equiv 115 \bmod{12^2}$
$x \equiv 61 \bmod{12^2}$
There are 4 results missing.
Where am I wrong? How to solve it?
|
When you solved $x^2 \equiv 4 \pmod 9$, the only solutions are $x \equiv ±2 \pmod 9$. This is because if $x = 3k ± 2, x^2 = 9k^2 ± 12k + 4$ and $±3k + 4$ is never a multiple of $9$.
However, solving $x^2 \equiv 9 \pmod {16}$ is different. When $x = 16k ± 3$, $x^2 \equiv ±(2 \cdot 16k \cdot 3) + 9 \equiv 9 \pmod {16}$. This leads to the fact that $x = 8k ± 3$ is also a solution as $2 \cdot 8k \cdot 3$ is still divisible by $16$.
Each possible combination gives a distinct solution using CRT, hence there are $8$ solutions in total.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4339506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Is there a polynomial $g(x)$ such that $p \circ f=g \circ p$, where $f(x)=9x+30x^3+27x^5+9x^7+{{x}^{9}}$? Let us consider a polynomial $f(x)=9x+30x^3+27x^5+9x^7+{{x}^{9}}$ over a field of characteristic $0$.
Set $p(x)=x^nh(x), n \geq 1$, where $h(x)$ can be any rational function.
I am trying to find out another polynomials $g(x)$ such that $p \circ f=g \circ p$. One more information, for any zero $\alpha$ of $f(x)$, $~~p(\alpha)$ is a root of $g(x)$.
If $h(x)=1$, then we need find a $g(x)$ such that $(f(x))^n=g(x^n)$.
I exclude the case $f = g, n=1$.
I have found the roots of $f(x)$ here in decimal form.
I thing this gives a hints about $g(x)$ because $f(\alpha)=0 \Rightarrow f(p(\alpha))=0$.
Edit: sorry for multiple edits but I hope this is the final one
|
$$
\begin{array}{rcl}
p(x)&=&9x,\\
g(x)&=&\displaystyle \frac{x^9}{43046721}+\frac{x^7}{59049}+\frac{x^5}{243}+\frac{10 x^3}{27}+9 x.
\end{array}
$$
Upd. For the new version
$$
\begin{array}{rcl}
p(x)&=&x^2,\\
g(x)&=&x (x+3)^2 \left(x (x+3)^2+3\right)^2;
\end{array}
$$
$$
\begin{array}{rcl}
p(x)&=&x^4+4 x^2,\\
g(x)&=&x (x+3)^2 \left(x (x+3)^2+3\right)^2.
\end{array}
$$
$$
\begin{array}{rcl}
p(x)&=&x^2 \left(x^2+3\right)^2,\\
g(x)&=&x (x+3)^2 \left(x (x+3)^2+3\right)^2.
\end{array}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4339675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solving the system $x^2+y^2+x+y=12$, $xy+x+y=-7$ I've been trying to solve this system for well over the past hour.$$x^2+y^2+x+y=12$$
$$xy+x+y=-7$$
I've tried declaring $x$ using $y$ ($x=\frac{-7-y}{y+1}$) and solving from there, but I've gotten to $$y^4+3y^3-9y^2-45y+30=0$$ and I don't see how we can get $y$ from here.
If anyone sees it, I'd appreciate the help. But if you see a simpler solution please do not hesitate to leave a comment and notify me of such.
|
Let $S = x+y$ and $P=xy$. You have $S^2 - 2P +S = 12$ and $P+S=-7$. Therefore
$$S^2 - 2P +S = S^2-2(-7-S) + S=S^2+3S+14=12$$ or $$S^2+3S+2=0.$$ The roots of this last equation are $-1,-2$. Therefore $(S,P) \in \{(-1, -6) , (-2,-5)\}$. Which implies that $x,y$ are the roots of either $u^2 +u -6 = 0$ or of $u^2 +2u -5=0$. Which are quadratic equations that you can solve... You'll get
$$(x,y) \in \{(2,-3),(-3,2),(-1- \sqrt 6, -1 + \sqrt 6),(-1+ \sqrt 6, -1 - \sqrt 6)\}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4342256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Can one integral can give more than one answer one with natural log and other with tan inverse question is $$\int\frac{1}{\sin^6(x) + \cos^6(x)}\,dx$$
My method :
$$\sin^6\left(x\right)+\cos^6\left(x\right)$$
$$=\left(\sin^2\left(x\right)+\cos^2\left(x\right)\right)\left(\sin^4\left(x\right)-\cos^2\left(x\right)\sin^2\left(x\right)+\cos^4\left(x\right)\right)$$
$$=1-3\cos^2\left(x\right)\sin^2\left(x\right)$$
so integration is
$$\int\dfrac{1}{1-3\cos^2\left(x\right)\sin^2\left(x\right)}\,dx$$
divide numerator and denominator by $$cos^2x$$
$$\int\dfrac{\sec^2\left(x\right)}{\sec^2\left(x\right)-3\tan^2\left(x\right)} \, dx$$
Now put $$t=\tan x$$
So $$dt = \sec^2x\,dx$$
So
$$\int\dfrac{t}{1-2t^2} \, dt$$
$$=\frac{1}{2\sqrt{2}}\ln\left(\left|\frac{\sqrt{2}\tan x+1}{\sqrt{2}\tan x-1}\right|\right)$$
But answer is $$\arctan\left(\tan\left(x\right)-\cot\left(x\right)\right)$$
So can there be two answers of different forms of same integration?
Thanks!
|
You made a mistake : \begin{aligned}\int{\frac{\mathrm{d}x}{1-3\cos^{2}{x}\sin^{2}{x}}}&=\int{\frac{\sec^{2}{x}}{\sec^{2}{x}-3\color{red}{\sin^{2}{x}}}\,\mathrm{d}x}\\ &=\int{\frac{\mathrm{d}y}{1+y^{2}-3\left(1-\frac{1}{1+y^{2}}\right)}}\\ &=\int{\frac{y^{2}+1}{y^{4}-y^{2}+1}\,\mathrm{d}y}\\ &=\int{\frac{1+\frac{1}{y^{2}}}{\left(y-\frac{1}{y}\right)^{2}+1}\,\mathrm{d}y}\\ &=\int{\frac{\mathrm{d}u}{u^{2}+1}}\\ &=\arctan{u}+C\\ &=\arctan{\left(y-\frac{1}{y}\right)}+C\\ \int{\frac{\mathrm{d}x}{1-3\cos^{2}{x}\sin^{2}{x}}}&=\arctan{\left(\tan{x}-\cot{x}\right)}+C\end{aligned}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4342540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Find parameters $a,b$ such that $x^6-2 x^5+2 x^4+2 x^3-x^2-2 x+1-\left(x^3-x^2+a x+b\right)^2>0$ The probrem is to prove that
$$x^6-2 x^5+2 x^4+2 x^3-x^2-2 x+1>0.$$
(the minimum value is about 0.02, tested by wolframalpha.)
I use sos(sum of squares) method, my idea is to reduce the degree of the polynomial gradually.
First I need to find $a,b$ so that
$$x^6-2 x^5+2 x^4+2 x^3-x^2-2 x+1-\left(x^3-x^2+a x+b\right)^2>0.$$
It's equivalent to
$$(1-2 a) x^4+(2 a-2 b+2)x^3 +\left(-a^2+2 b-1\right)x^2 +(-2 a b-2)x -b^2+1>0,$$ or
$$
\left( (1-2\!\:a)\!\:x^4+(2+2\!\:a-2\!\:b)\!\:x^3+(-1-a^2+2\!\:b)\!\:x^2+(-2-2\!\:a\!\:b)\!\:x+1-b^2 \right) _{\min}>0.$$
Let
$$f(x)=
(1-2\!\:a)\!\:x^4+(2+2\!\:a-2\!\:b)\!\:x^3+(-1-a^2+2\!\:b)\!\:x^2+(-2-2\!\:a\!\:b)\!\:x+1-b^2$$
and $f'(x)=0$, I get
$$
-2-2\!\:a\!\:b+2\!\:(-1-a^2+2\!\:b)\!\:x+3\!\:(2+2\!\:a-2\!\:b)\!\:x^2+4\!\:(1-2\!\:a)\!\:x^3=0.
$$ However, it's difficult and ugly to solve the equation. And it's impossible to solve the extreme point when the degree of the polynomial increases, for example, prove that $$x^{12}-x^{11}+x^8-x^7+x^6+x^3-2 x+1>0,$$ we need to find 5 parameters $a,b,c,d,e$ such that $$
x^{12}-x^{11}+x^8-x^7+x^6+x^3-2x+1-\left( x^6-\frac{x^5}{2}+ax^4+bx^3+cx^2+dx+e \right) ^2>0.
$$
How can I solve the problem? Is there any general method?
|
For the original problem, observe that
$$x^6 - 2x^5 + 2x^4 + 2x^3 - x^2 - 2x + 1 = (x^3 - x^2)^ 2 + (x^2 +x - 1)^2.$$
Since both squares cannot be 0 at the same time, hence it is strictly positive.
For your stated problem, $ a, b = 0 $ is still not a solution to your strict inequality because the expression does hit 0 when $x^2 + x - 1 = 0 $.
Testing values near that point indicates that $ a = -0.01, b = 0.01$ works.
The resulting quartic has 4 complex roots, hence is strictly positive.
Proof by Wolfram Alpha since it's not that interesting.
Given the above, I'm not motivated to figure out the rest of the SOS.
Notes
*
*You do not always have to "reduce the degree of the polynomial gradually".
*
*EG We could have (say) $ f(x) = (0.6x^3 + g(x) ) ^2 + (0.8 x^3 + h(x) ) ^2 $ for some quadratics $g(x), h(x)$.
*EG Had you tried to do so, you'd likely have ended up with $ (x^3 - x^2 + 0.5 x + 1.5)^2$ in order to remove the $x^4$ and then $x^3 $ term. However, this leaves us with $1.75x^2 - 3.5x - 1.25$ which has distinct real roots, so we can't SOS further.
*I chanced upon this identity, in part from recognizing some of the terms. You guessed that $(x^3-x^2)^2$ might be involved, so looking at $ f(x) - (x^3 -x^2)^2$ would have been a good next step.
*The minimum of $f(x)$ occurs near the positive root of $x^2 + x - 1 = 0$. This shouldn't be too surprising based on the identity.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4342792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
How far can I go with the integral $\int \frac{\sin ^{n} x \cos ^{n} x}{1-\sin x \cos x} d x, $ where $n\in N$? Latest Edit
By the aid of my recent post, a closed form for its definite integral is obtained as below:
$$
\int_{0}^{\frac{\pi}{2}} \sin ^{k} \theta d \theta= \frac{\sqrt{\pi} \Gamma\left(\frac{k+1}{2}\right)}{k \Gamma\left(\frac{k}{2}\right)}
$$
Hence \begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{\sin ^{n} x \cos ^{n} x}{1-\sin x \cos x} d x&=\frac{2}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x\right]_{0}^{\frac{\pi}{2}} -\sum_{k=1}^{n-1} \frac{1}{2^{k}} \cdot \frac{\sqrt{\pi} \Gamma\left(\frac{k+1}{2}\right)}{k \Gamma\left(\frac{k}{2}\right)}\\&=\frac{\pi}{3\sqrt3}- \sum_{k=1}^{n-1} \frac{1}{2^{k}} \cdot \frac{\sqrt{\pi} \Gamma\left(\frac{k+1}{2}\right)}{k \Gamma\left(\frac{k}{2}\right)}
\end{aligned}
In my answer, I have found the integral
$$\int \frac{d x}{1-\sin x \cos x} =\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)+C_0
$$
Next,
$$
\begin{aligned}
& \int \frac{\sin x \cos x}{1-\sin x \cos x} d x \\
=& \int \frac{d x}{1-\sin x \cos x}-\int 1 d x \\
=& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x+C_{1}
\end{aligned}
$$
and
$$
\begin{aligned}
& \int \frac{\sin ^{2} x \cos ^{2} x}{1-\sin x \cos x} d x \\
=& \int \frac{1-\left(1-\sin ^{2} x \cos ^{2} x\right)}{1-\sin x \cos x} d x
\\
=& \int \frac{d x}{1-\sin x \cos x}-\int(1+\sin x \cos x) d x \\
=& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x+\frac{\cos 2 x}{4}+C_2
\end{aligned}
$$
Now I want to go further, $$
\begin{aligned}
& \int \frac{\sin ^{3} x \cos ^{3} x}{1-\sin x \cos x} d x \\
=& \int \frac{1-\left(1-\sin ^{3} x \cos ^{3} x\right)}{1-\sin x \cos x} d x \\
=& \int \frac{d x}{1-\sin x \cos x}-\int\left(1+\sin x \cos x+\sin ^{2} x \cos ^{2} x\right) d x \\
=& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x+\frac{\cos 2 x}{4}-\int \frac{\sin ^{2} 2 x}{4} d x \\
=& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x+\frac{\cos 2 x}{4}-\frac{1}{4}\int\frac{1-\cos 4 x}{2} d x\\
=& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x+\frac{\cos 2 x}{4}-\frac{1}{8}\left(x-\frac{\sin 4 x}{4}\right) +C\\=& \frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-\frac{9}{8} x+\frac{\cos 2 x}{4}+\frac{\sin 4 x}{32} +C_3
\end{aligned}
$$
Then I discovered that the integral
$$
I(n)=\int \frac{\sin ^{n} x \cos ^{n} x}{1-\sin x \cos x} d x
$$
has a telescoping series
$$I(k+1)-I(k)=-\int \sin ^{k} x \cos ^{k} x d x$$
Hence $$
I(n)-I(1)=-\sum_{k=1}^{n-1} \frac{1}{2^{k}} \int\sin ^{k}(2 x) d x
$$
We can conclude that
$$
I(n)=\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2 \tan x-1}{\sqrt{3}}\right)-x-\sum_{k=1}^{n-1} \frac{1}{2^{k}}\int\sin ^{k}(2 x) d x
$$
Then I was stuck with the last sum.
My question is whether we can find a closed form for the last sum.
|
Just for your curiosity.
As said in comments, the result is not very pretty.
Using
$$I_k=\int_0^{\frac \pi 2} \sin^k(x)\,dx=\frac{\sqrt{\pi }}{2}\,\,\frac{\Gamma \left(\frac{k+1}{2}\right)}{\Gamma \left(\frac{k+2}{2}\right)}$$
$$S_n=\sum_{k=1}^{n-1} \frac{1}{2^{k+1}} \int_{0}^{\frac{\pi}{2}} \sin ^{k}( x)\, d x$$
$$S_n=\frac{8 \sqrt{3}-9}{36} \pi-\frac{\sqrt{\pi }}{ 2^{n+3}}\, T_n$$
$$T_n=2\frac{ \Gamma
\left(\frac{n+1}{2}\right)}{\Gamma \left(\frac{n+2}{2}\right)}\, _2F_1\left(1,\frac{n+1}{2};\frac{n+2}{2};\frac{1}{4}\right)+\frac{ \Gamma
\left(\frac{n+2}{2}\right)}{\Gamma \left(\frac{n+3}{2}\right)}\,
_2F_1\left(1,\frac{n+2}{2};\frac{n+3}{2};\frac{1}{4}\right)$$ where appears the gaussian hypergeometric function.
However, the individual values of the $S_n$ are not bad. They write
$$S_n=a_n+ \pi b_n$$
The $a_n$ form the sequence
$$\left\{0,\frac{1}{4},\frac{1}{4},\frac{7}{24},\frac{7}{24},\frac{3}{10},\frac{3}{10},
\frac{169}{560},\frac{169}{560},\frac{1523}{5040},\frac{1523}{5040},\frac{133}{440},\cdots\right\}$$ and the $b_n$ form the sequence
$$\left\{0,0,\frac{1}{32},\frac{1}{32},\frac{19}{512},\frac{19}{512},\frac{157}{4096},
\frac{157}{4096},\frac{5059}{131072},\frac{5059}{131072},\frac{40535}{1048576},\frac{40535}{1048576},\cdots\right\}$$
Edit
If you plan to integrate in the range $0\le x \le \frac \pi 2$
$$I_{k+1}-I_k=\frac{\sec ^{-(k+1)}(x) }{k+1}\,\, _2F_1\left(\frac{1-k}{2},\frac{k+1}{2};\frac{k+3}{2};\cos^2(x)\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4344864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
}
|
A 3d line intersecting 2 other 3d lines The equation of the line parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and intersecting the lines $9x + y + z + 4 = 0 = 5x + y + 3z$ & $x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3$
My solution is as follow
$\frac{{x - a}}{2} = \frac{{y - b}}{3} = \frac{{z - c}}{4}$ represent the line parallel to $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and intersecting the line and intersecting the line ${L_1}:9x + y + z + 4 = 0 = 5x + y + 3z$ & ${L_2}:x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3$
Let $\frac{{x - a}}{2} = \frac{{y - b}}{3} = \frac{{z - c}}{4} = {\lambda _i}$ where $\left( {x,y,z} \right) = \left( {a + 2{\lambda _i},b + 3{\lambda _i},c + 4{\lambda _i}} \right)$
$ \Rightarrow 9a + 18{\lambda _1} + b + 3{\lambda _1} + c + 4{\lambda _1} + 4 = 0 \Rightarrow \frac{{9a + b + c + 4}}{{ - 25}} = {\lambda _1}$ after putting the values in $L_1$
$5a + 10{\lambda _1} + b + 3{\lambda _1} + 3c + 12{\lambda _1} = 0 \Rightarrow \frac{{5a + b + 3c}}{{ - 25}} = {\lambda _1}$
$9a + b + c + 4 = 5a + b + 3c \Rightarrow 4a - 2c + 4 = 0$
${L_2}:x + 2y - 3z - 3 = 0 = 2x - 5y + 3z + 3$
$\left( {x,y,z} \right) = \left( {a + 2{\lambda _2},b + 3{\lambda _2},c + 4{\lambda _2}} \right)$
$a + 2{\lambda _2} + 2b + 6{\lambda _2} - 3c - 12{\lambda _2} - 3 = 0 = 2a + 4{\lambda _2} - 5b - 15{\lambda _2} + 3c + 12{\lambda _2} + 3$
$ \Rightarrow a + 2b - 3c - 3 - 4{\lambda _2} = 0 = 2a - 5b + 3c + 3 + {\lambda _2} \Rightarrow \frac{{a + 2b - 3c - 3}}{4} = {\lambda _2} = \frac{{2a - 5b + 3c + 3}}{{ - 1}}$
$ \Rightarrow a + 2b - 3c - 3 = - 8a + 20b - 12c - 12 \Rightarrow 9a - 18b + 9c + 9 = 0 \Rightarrow a - 2b + c + 1 = 0$
$ \Rightarrow a - 2b + c + 1 = 0\& 2a - c + 2 = 0$
Let $b = - 1 \Rightarrow a + c + 3 = 0\& 2a - c + 2 = 0 \Rightarrow c = - \frac{4}{3}\& a = - \frac{5}{3}$
$\frac{{x + \frac{5}{3}}}{2} = \frac{{y + 1}}{3} = \frac{{z + \frac{4}{3}}}{4} = \lambda \Rightarrow \left( {x,y,z} \right) = \left( { - \frac{5}{3}\hat i - \hat j - \frac{4}{3}\hat k} \right) + \lambda \left( {2\hat i + 3\hat j + 4\hat k} \right)$
But my answer is not matching. Can you tell me the error that I have made. Each steps is elaborated
|
To find a line parametric equation, we have to find a point on the line, and then use the cross product of the normal vectors of the the two planes given.
The two planes for the first line are
$9x + y + z = -4 $
$5 x + y + 3 z = 0 $
To find a point on both these planes, set $x = 0$ and solve the system
$ y + z = -4 $
$ y + 3 z = 0 $
The solution is $y = -6, z = 2 $ so our point is $(0, -6, 2) $
Next the direction vector of the line is along the cross product of the first normal and second normal
$d = (9, 1, 1) \times (5, 1, 3) = (2, -22, 4) = 2 (1, -11, 2) $
So the parametric equation of the first line is
$ P_1(t) = (0, -6, 2) + t (1, -11, 2) $
Similarly, for the second line, the planes are
$ x + 2 y - 3 z = 3 $
$ 2 x - 5 y + 3 z = -3 $
To find a point of the line, it must satisfy both equations, so set $y = 0$ and solve the system
$ x - 3 z = 3 $
$ 2 x + 3 z = -3 $
The solution is $x = 0 , z = -1 $, so our point is $(0, 0, -1) $
And the direction vector is
$ (1, 2, -3) \times (2, -5, 3) = -9 (1, 1, 1) $
So the parametric equation of the second line is
$ P_2 (s)= (0, 0, -1) + s (1, 1, 1) $
Now, since the line whose equation is requested is parallel to $(2,3,4)$ and it intersects both the first and second lines obtained above, then the setup is as follows:
$ P_1(t) - P_2(s) = \lambda (2, 3, 4) $
For some $t, s, \lambda$. Writing the coordinates as column vectors, this last equation becomes the linear system
$ \begin{bmatrix} 1 && -1 && -2 \\ -11 && -1 && -3 \\ 2 && -1 && - 4 \end{bmatrix} \begin{bmatrix} t \\ s\\ \lambda \end{bmatrix} = \begin{bmatrix} 0 \\ 6 \\ -3 \end{bmatrix} $
Its solution is $ (t, s, \lambda ) = ( - \dfrac{3}{5} , - 3, \dfrac{6}{5} ) $
Thus the first point of intersection on the first line is
$ P_1^* = (0, -6, 2) +\left( -\dfrac{3}{5} \right)(1, -11, 2) = (-\dfrac{3}{5}, \dfrac{3}{5}, \dfrac{4}{5} ) $
And the intersection with the second line is
$P_2^* = (0, 0, -1) +(- 3) (1, 1, 1) = (-3, -3, -4) $
We can use either point for the parametric equation of our line. Using the second point, we get
$ Q(t) = (-3, -3, -4) + t (2, 3, 4) $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4346271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Is this series known? I encountered the following series and was wondering if it can be expressed using a known value?
$$\sum_{a,b=1}^{\infty} \frac{(-1)^a+1}{a^3-a^2+ab^2+b^2},$$
where all terms in the sum for which $a$ is odd are interpreted as zero.
|
Since $a$ can only contribute nontrivially when even, we may as well replace $a\to 2a$
$$\sum_{(a,b)\in\Bbb{Z}^+\times\Bbb{Z}^+} \frac{2}{8a^3-4a^2+(2a+1)b^2} = \sum_{a=1}^\infty \frac{2}{2a+1}\sum_{b=1}^\infty\frac{1}{4a^2\left(\frac{2a-1}{2a+1}\right)+b^2}$$
The sum on the interior is the famous result
$$\sum_{n=1}^\infty \frac{1}{x^2+n^2} = \frac{\pi x \coth \pi x - 1}{2x^2}$$
leaving us with
$$\sum_{a=1}^\infty \frac{2}{2a+1}\left[\frac{\pi}{4a}\sqrt{\frac{2a+1}{2a-1}}\coth\left(2\pi a\sqrt{\frac{2a-1}{2a+1}}\right) - \frac{1}{8a^2}\left(\frac{2a+1}{2a-1}\right)\right]$$
$$ = \sum_{a=1}^\infty \frac{\pi}{2a}\frac{1}{\sqrt{4a^2-1}}\coth\left(2\pi a\sqrt{\frac{2a-1}{2a+1}}\right) - \frac{1}{4a^2}\frac{1}{2a-1}$$
Both terms converge by comparison test so the series converges, and the term on the right has an easy closed form
$$\sum_{a=1}^\infty \frac{-1}{4a^2(2a-1)} = \frac{1}{4}\sum_{a=1}^\infty\frac{1}{a^2} + \sum_{a=1}^\infty \frac{1}{2a}-\frac{1}{2a-1} = \frac{\pi^2}{24} - \log 2$$
However the term on the left with the $\coth $ would be tricky, and it is unlikely a nice closed form would exist.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4346417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
show that $(x,y)\to \ln(\sqrt{x^2+y^2})$ is harmonique over $\mathbb{R}^2\backslash \{(0,0)\} $ by use of following result:
If $f$ is a holomorphic function over the open $O$ of $\mathbb{C}$ then the real part of $f$ is harmonic over the open $O$
show that $(x,y)\to \ln(\sqrt{x^2+y^2})$ is harmonique over $\mathbb{R}^2\backslash \{(0,0)\} $?
we notice that $\ln(| z | ) $ is the real part of any determination of $\ln (z)$, of which it suffices to choose the cut so that $\ln(z)$ is holomorphic on an open neighborhood of the point $(x, y) \in \mathbb{R}^2\backslash \{(0,0)\}$. But how can I choose the cut?
|
We may also show $u(x, y)$ harmonic by direct differentiation as follows:
$u(x, y) = \ln(x^2 + y^2)^{1/2} = \dfrac{1}{2} \ln(x^2 + y^2); \tag 1$
$u_x = \dfrac{1}{2} \dfrac{1}{x^2 + y^2} (2x) = \dfrac{x}{x^2 + y^2}; \tag 2$
$u_{xx} = \dfrac{x^2 + y^2 - 2x^2}{(x^2 + y^2)^2} = \dfrac{y^2 - x^2}{(x^2 + y^2)^2}; \tag 3$
$u_y = \dfrac{y}{x^2 + y^2}; \tag 4$
$u_{yy} = \dfrac{x^2 + y^2 - 2y^2}{(x^2 + y^2)^2} = \dfrac{x^2 - y^2}{(x^2 + y^2)^2}; \tag 5$
comparing (3) and (5) we see that
$u_{yy} = -u_{xx}, \tag 6$
whence
$u_{xx} + u_{yy} = 0, \tag 7$
and thus we see that $u(x, y)$ is harmonic.
Notice we haven't introduced any cuts in obtaining this result.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4354683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Area of a circle passing through two vertices of a parallelogram touching one edge. For reference:
Let $ABCD$ be a parallelogram, $AB = 6, BC= 10$ and $AC = 14$ ; traces a circle passing through the vertices $C$ and $D$ with $AD$ being tangent and this circle is $BC$ a secant. Calculate the area of the circle. (Answer: $12\pi$)
My progress:
$AD^2 =AI\cdot AC \implies 10^2=AI\cdot14$
$ \implies AI = \dfrac{50}{7} ,\ IC = \dfrac{48}{7}$
In $\triangle ABC$
$14^2=6^2+10^2-2\cdot6\cdot10\cdot\cos \angle B\implies
\cos \angle B =-\dfrac{1}{2}\therefore \angle B =120^o $
....I can't see many options????
|
Drop the perp from $A$ to $BC$ and call the foot $E$.
Let $BE=x$, then using Pythagoras' theorem, $$6^2-x^2=14^2-(10+x)^2\implies x=3.$$
Therefore in right triangle $ABE$, $\angle BAE=30^\circ$ and so is $\angle CDO$.
Let $M$ be the midpoint of side $CD$, then considering $\triangle ODM$, $OD$, or, radius of the circle is $2\sqrt3$.
Hence, the area of circle is $12\pi$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4359958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Prove that if $a^5-a^3+a=3$, then $a^6\geq 5$ The problem, which I encountered in a highschool book, goes as following:
Prove that if $a^5-a^3+a=3,$ then $a^6\geq 5$ must hold.
Now, obviously, I have tried a lot of things such as:
$a^6=a^4-a^2+3a\Rightarrow a^4-a^2+3a\geq 5$ or multiplying that again by $a^2$ to create a new expression for $a^6$ and simplify some terms, but higher powers keep appearing and it just doesnt seem to end
|
The equation $a^5-a^3+1-3=0$ has at least one real solution.
As
$$3=a(a^4-a^2+1)=a\bigl((a^2-1/2)^2+3/4\bigr)$$
we have $a>0$ assuming $a$ is real.
Now
$$a^4-a^2=\frac3a-1,$$
hence
$$a^6=a^4-a^2+3a=\frac3a+3a-1
=3(\sqrt{a}-1/\sqrt{a})^2+6-1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4363244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Solve the system $3^x-2^{y^2}=77$, $3^{\frac{x}{3}}-2^{\frac{y^2}{2}}=7$ in $\mathbb{R}$ I had to solve the similar system $3^x-2^{y^2}=77,\; 3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7$ before, which can be solved like this:
$$3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7 \implies 2^{\frac{y^2}{2}}=3^{\frac{x}{2}}-7$$
$$2^{\frac{y^2}{2}}=3^{\frac{x}{2}}-7 \implies 2^{y^2}=(3^{\frac{x}{2}}-7)^2$$
$$2^{y^2}=(3^{\frac{x}{2}}-7)^2 \implies 2^{y^2}=3^{x}-14\cdot3^{\frac{x}{2}}+49$$
So putting that in the first equation:
$$3^x-2^{y^2}=77 \implies 3^x-3^{x}+14\cdot3^{\frac{x}{2}}-49=77$$
$$\mathbf{3^x-3^{x}+14\cdot3^{\frac{x}{2}}-49=77 \implies 14\cdot3^{\frac{x}{2}}=126}$$
$$14\cdot3^{\frac{x}{2}}=126 \implies 3^{\frac{x}{2}}=9$$
$$3^{\frac{x}{2}}=9 \implies x = 4$$
And putting the value of $x$ in the second equation we get $y$:
$$3^{\frac{x}{2}}-2^{\frac{y^2}{2}}=7 \implies 3^{2}-2^{\frac{y^2}{2}}=7$$
$$3^{2}-2^{\frac{y^2}{2}}=7 \implies 2^{\frac{y^2}{2}}=2$$
$$2^{\frac{y^2}{2}}=2 \implies y^2 = 2$$
$$y^2 = 2 \implies y = \sqrt{2}$$
So I tried solving the system in the question title with the same strategy until the step in bold, which gave me $3^x - 3^{\frac{2x}{3}}+14\cdot3^{\frac{x}{3}}=126$, and I have no idea how to progress further.
|
If $u=3^{\frac x 2}$ and $v=2^{\frac {y^2}2}$ then
$$\left \{
\begin{split}
u^3-v^2&=77\\
u-v&=7
\end{split}
\right.$$
So $v=u-7$ and thus $u^3-u^2+14u-126=0$.
Solving the latter gives you 3 roots (1 real and 2 complex), and you can find $u$ and $v$, and get back to $x$ and $y$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4367073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Prove that $x_n = (-1)^n \frac {3n + 2} {n + 1}$ is divergent Prove that the sequence {${x_n}$} to be defined by $x_n = (-1)^n \frac {3n + 2} {n + 1}$ is divergent.
Also, you can assume without proof that $\lim{n\to\infty}$ $\frac{3n+2}{n+1} = 3 $
Proposed Solution: (by contradiction)
Given that there were a limit $x$, then for $\varepsilon = \frac{1}{2}$ we expect an $M$ to satisfy the divergence definition. Suppose such an $M$ exists. Then for $n\geq M$ we need to compute:
$\frac{1}{2} > |x_n - x| = |1 - x|$ and $\frac{1}{2} > |x_{n+1} - x| = |-1 - x|$
*
*Do I do the following next? : plug in $x_{2n} = \frac{6n}{2n+1}$ and $x_{2n+1} = -\frac{6n+3}{2n+2} $
*Then conclude this? : $\lim{n\to\infty}$ $x_{2n} = 3$ and $\lim{n\to\infty}$ $x_{2n+1} = -3$
*Therefore, $\frac{1}{2} \ngtr |1 - 3|$ and $\frac{1}{2} \ngtr |-1 - (-3) |$ ?
|
Suppose the limit exists and call it $x$. Thus, for $\epsilon = \dfrac{1}{2}$, there is a $n$ such that $|x_n - x| < \dfrac{1}{2}, |x_{n+1} - x| < \dfrac{1}{2}\implies |x_{n+1} - x_n| < 1$. If $n$ is even, then $|x_n - x_{n+1}| = \left|\dfrac{3n+2}{n+1} + \dfrac{3n+5}{n+2}\right| > 2$, contradiction. The case $n$ is odd is similar. So $x$ can't exist.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4370323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Describe the locus of $w$ if $w=\frac{1-z}{1+z}$ and $z=1+iy$ (i.e $z$ is a complex number that moves along the line $x=1$) So I'm trying to solve the following problem: If $z=x+iy$, express $w=\frac{1-z}{1+z}$ in the form $a+bi$ and hence find the equation of the locus of $w$ if $z$ moves along the line $x=1$.
My attempt is:
$w=\frac{1-z}{1+z}\times\frac{1+\bar{z}}{1+\bar{z}}$
$=\frac{1-2i\Im{(z)}-|z|^2}{1+2\Re{(z)}+|z|^2}$
$=\frac{1-2yi-x^2-y^2}{1+2x+x^2+y^2}$
$\therefore a+bi=\frac{1-x^2-y^2}{(x+1)^2+y^2}-\frac{2y}{(x+1)^2+y^2}i$
Substituting $x=1$ gives the parametric equations $a=\frac{-y^2}{4+y^2}$ and $b=\frac{-2y}{4+y^2}$
My problem is that I can't reduce these two to only an equation in terms of $a$ and $b$ (best I can do is $b=2ya$, when I try to find $y$ in terms of $b$ it goes $4+y^2=-\frac{2y}{b}$ which is $y^2-\frac{2y}{b}+\left(\frac{y}{b}\right)^2=-4+\left(\frac{y}{b}\right)^2$ and so $\left(y-\frac{y}{b}\right)^2=-4+\left(\frac{y}{b}\right)^2$ but clearly this is not working) so I'm not sure if my steps are right into deriving the parametric equations or if the question just requires me to be more capable of reducing parametric equations. If they can be reduced into equations of just $a$ and $b$, could someone show me how?
|
$\,w=\frac{1-z}{1+z} \;\;(z \ne -1)\, \iff z=\frac{1-w}{1+w} \;\; (w \ne -1)\,$, then the condition $\,\text{Re}(z) = 1\,$ translates to:
$$
\require{cancel}
2 = 2\text{Re}(z)=z + \bar z = \frac{1-w}{1+w}+\frac{1-\bar w}{1+\bar w}
= \frac{\left(1-|w|^2-\cancel{w}+\bcancel{\bar w}\right)+\left(1-|w|^2+\cancel{w}-\bcancel{\bar w}\right)}{1+|w|^2+w+\bar w}
\\ \iff \cancel{2} + 2|w|^2+2w+2\bar w=\cancel{2} - 2|w|^2 \\ \iff 4 |w|^2+2w+2\bar w=0 \\ \iff (2w+1)(2\bar w+1)=1
$$
The last equation can be written as $\,\left|w+\frac{1}{2}\right|^2=\frac{1}{4}\,$ which is a circle of radius $\,\frac{1}{2}\,$ centered at $\,- \frac{1}{2}\,$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4372102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Prove that $\sum \frac{a^3}{a^2+b^2}\le \frac12 \sum \frac{b^2}{a}$
Let $a,b,c>0$. Prove that
$$ \frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}\le \frac12 \left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right).\tag{1}$$
A idea is to cancel the denominators, but in this case Muirhead don't work because the inequality is only cyclic, not symmetric.
Another idea would be to apply Cauchy reverse technique:
$$(1)\iff\sum \left(a-\frac{a^3}{a^2+b^2}\right)\ge \frac12 \sum (2a-b^2/a)\iff \sum\frac{ab^2}{a^2+b^2}\ge \frac12\sum\frac{2a^2-b^2}{a}$$
$$\iff \sum \frac{(ab)^2}{a^3+ab^2}\ge \frac12\sum \frac{2a^2-b^2}a.$$
Now we can apply Cauchy-Schwarz, and the problem reduces to
$$\frac{(\sum ab)^2}{\sum a^3+\sum ab^2}\ge \frac12\sum \frac{2a^2-b^2}{a},$$
and at this point I am stuck. Here the only idea is to cancel the denominators, but as I say above it can't work.
|
Let me continue your approach. We will use the following simple observation:
Lemma. For positive $x$ and $y$ the following inequality holds
$$
\frac{x^2}{y}\ge 2x-y.
$$
Proof. It is equivalent to $(x-y)^2\ge 0$.
We need to prove that
$$
\sum_{cyc}\frac{(ab)^2}{a^3+ab^2}\ge \frac{1}{2}\sum_{cyc}\frac{2a^2-b^2}{a}.
$$
Note that
$$
\frac{(ab)^2}{a^3+ab^2}=\frac{b^2}{a+\frac{b^2}{a}}=\frac{1}{4}\cdot\frac{(2b)^2}{a+\frac{b^2}{a}}\ge\frac{1}{4}\left(4b-a-\frac{b^2}{a}\right).
$$
Thus, summing up similar inequalities we obtain
$$
\sum_{cyc}\frac{(ab)^2}{a^3+ab^2}\ge\frac{3}{4}(a+b+c)-\frac{1}{4}\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right).
$$
It remains to check that
$$
\frac{3}{4}(a+b+c)-\frac{1}{4}\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right)\ge\frac{1}{2}\sum_{cyc}\frac{2a^2-b^2}{a}.
$$
Since $\sum_{cyc}\frac{2a^2-b^2}{a}=2(a+b+c)-\left(\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\right)$, the last inequality is equivalent to
$$
\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\ge a+b+c,
$$
which is again a consequence of the lemma:
$$
\frac{b^2}{a}+\frac{c^2}{b}+\frac{a^2}{c}\ge (2b-a)+(2c-b)+(2a-c)=a+b+c.
$$
Comment. Here is a more general form of the lemma: $\frac{a^x}{b^y}\ge\frac{xa^{x-y}-yb^{x-y}}{x-y}$ for any $a,b,x,y>0$ such that $x\neq y$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4372219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
}
|
Six people (half are female, half are male) for seven chairs. Problem:
Suppose there are $7$ chairs in a row. There are $6$ people that are going to randomly
sit in the chairs. There are $3$ females and $3$ males. What is the probability that
the first and last chairs have females sitting in them?
Answer:
Let $p$ be the probability we seek. Out of $3$ females, only $2$ can be sitting at the end of the row. I consider the first and last chairs to be at the end of the row.
\begin{align*}
p &= \dfrac{ {3 \choose 2 } 3(2) (4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ 3(3)(2) (4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ 3(4)(3)(2) } { 7(5)(4)(3)(2) } = \dfrac{ 3(3)(2) } { 7(5)(3)(2) } \\
p &= \dfrac{ 18 } { 35(3)(2) } \\
p &= \dfrac{ 3 } { 35 }
\end{align*}
Am I right?
Here is an updated solution.
Let $p$ be the probability we seek. Out of $3$ females, only $2$ can be sitting at the end of the row. I consider the first and last chairs to be at the end of the row.
\begin{align*}
p &= \dfrac{ 3(2) (5)(4)(3)(2) } { 7(6)(5)(4)(3)(2) } \\
p &= \dfrac{ (5)(4)(3)(2) } { 7(5)(4)(3)(2) } \\
p &= \dfrac{1}{7}
\end{align*}
Now is my answer right?
|
Here is an approach that is closest to your thought process. The three females and the three males are all indistinguishable, so we use combinations instead of permutations. Picking any one sex first gives:
$${7 \choose 3} \cdot {4 \choose 3}$$
total possibilities.
Now if two women are already sitting at the ends, then there is one woman and three men left to fill $5$ seats. Choosing the woman first we have:
$$5 \cdot {4 \choose 3}$$
ways and here we can see another simpler approach: where 3 men sit in 4 seats and the women sit in the remaining 3 seats.
Thus the probability is $\frac{5}{7 \choose 3} = \frac{1}{7}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4374307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
$A+B+C+D=\pi$, and $0\leq A,B,C,D \leq \frac{\pi}{2}$. Prove that $\sin^4(A)+\sin^4(B)+\sin^4(C)+\sin^4(D)\leq 2$ I tried to simplify it from some ways.
(1).$\sin^4(A)+\sin^4(B)+\sin^4(C)+\sin^4(D)$
$=\left(\frac{1-\cos(2A)}{2}\right)^2+\left(\frac{1-\cos(2B)}{2}\right)^2+\left(\frac{1-\cos(2C)}{2}\right)^2+\left(\frac{1-\cos(2D)}{2}\right)^2$
$=1-\frac{1}{2}\left(\cos(2A)+\cos(2B)+\cos(2C)+\cos(2D)\right)+\frac{1}{4}\cdot\left(2+\frac{\cos(4A)+\cos(4B)+\cos(4C)+\cos(4D)}{2}\right)$
$=\frac{3}{2}-\frac{1}{2}\left(\cos(2A)+\cos(2B)+\cos(2C)+\cos(2D)\right)+\frac{1}{8}\left(\cos(4A)+\cos(4B)+\cos(4C)+\cos(4D)\right)$
(2).$\cos(2A)+\cos(2B)+\cos(2C)+\cos(2D)$
$=2\cos(A+B)\cos(A-B)+2\cos(C+D)\cos(C-D)$
since $A+B=\pi-(C+D)$, thus $\cos(A+B)=-\cos(C+D)$. It follows that
$=2\cos(A+B)\left(\cos(A-B)-\cos(C-D)\right)$
$=2\cos(A+B)\cdot\left(-2\sin(\frac{A-B+C-D}{2})\sin(\frac{A-B-C+D}{2})\right)$
$=2\cos(A+B)\cdot\left(-2\sin\left(\frac{\pi}{2}-(B+D)\right)\sin\left(\frac{\pi}{2}-(B+C)\right)\right)$
$=2\cos(A+B)\cdot\left(-2\cos(B+D)\cos(B+C)\right)$
$=-4\cos(A+B)\cos(A+C)\cos(A+D)$
(3).Similarly,$\cos(4A)+\cos(4B)+\cos(4C)+\cos(4D)$
$=2\cos(2A+2B)\cos(2A-2B)+2\cos(2C+2D)\cos(2C-2D)$
$=2\cos(2A+2B)\cos(2A+2C)\cos(2A+2D)$
Am I on the right track? Maybe it is a Jensen's-inequality problem but it seems not always concave up or down in the interval $[0,\frac{\pi}{2}]$. I am stuck here. Please help, and thank you.
|
A bit of an ugly solution and would love a clever one.
First, let's assume wlog $D \le C \le B \le A$ and then $D+C \le \pi/2$ and we show that $\sin^4C+\sin^4D \le \sin^4(C+D)$ so we reduce the problem to three angles where we use calculus.
Noting that $\sin^2x-\sin^2y=\sin(x-y)\sin(x+y)$, one has $\sin^4(C+D)-\sin^4C=(\sin^2(C+D)+\sin^2C)\sin D \sin (2C+D)$ and since $C+D \le \pi/2$ then $D \le \pi-(2C+D)$ and $D \le 2C+D$ so $\sin D \le \sin (2C+D)=\sin (\pi-(2C+D))$ since $\sin$ is increasing on $[0, \pi/2]$. Since $\sin^2D \le \sin^2C$ by the same reason, one gets that $\sin^4D \le \sin^2C \sin D\sin (2C+D) \le (\sin^2(C+D)+\sin^2C)\sin D \sin (2C+D)$ and we are done with the reduction.
Now with $A+B+C =\pi, 0 \le A,B,C \le \pi/2$ we consider the function $f(x,y)=\sin^4x+\sin^4y+\sin^4(x+y)$ and show that it has only one critical point (at $x=y=\pi/3, (f(x,y)=27/16$) inside the triangle $0 \le x,y, \pi-x-y \le \pi/2$ while on the boundary at least one of $x,y,x+y=\pi/2$ and then if wlog $x+y=\pi/2$ we can apply the result in the first paragraph, hence $\sin^4x+\sin^4y \le \sin^4(x+y)=1$ (or directly as $\sin^4x+\sin^4y=\sin^4x+\cos^4x\le \sin^2x+\cos^2x=1$) so the maximum of $f$ on the closed triangle (which exists as $f$ continous) is $2$ so we are done!
So looking at $f_x=f_y=0$ we get $\sin^3x\cos x+\sin^3(x+y)\cos (x+y)=\sin^3y\cos y+\sin^3(x+y)\cos (x+y)=0$ or with the original notation $\sin^3A\cos A=\sin^3B\cos B=\sin^3C\cos C$
But now $g(x)=\sin^3x \cos x$ on $(0,\pi/2)$ has $g'(x)=3\sin^2x\cos^2x-\sin^4x=3\sin^2x-4\sin^4x$, so $g'$ has clearly only one zero at $x=\pi/3$ and $g'>0$ on $(0, \pi/3)$, while $g'<0$ on $(\pi/3,\pi/2)$ hence $g$ monotonic (so injective) on each of the two intervals, which implies that at least two of $A,B,C$ must be equal and wlog let $A=B=x$ hence from $f_x=f_y=0$ we get $\sin^3x\cos x=-\sin^3(2x)\cos(2x)$ which (since $2x \ge \pi/2$) gives $\cos x= -8\cos^3 x\cos (2x)$; but now we are inside $(0, \pi/2)$ so $\cos x \ne 0$ either, hence $1=-4(1+\cos 2x)\cos 2x$ or $\cos 2x=-1/2, x=\pi/3$ and our claim about critical points is proven and the problem solved!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4375138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Prove $\sum_{k=1}^{n-1} k \cos\frac{3 \pi k}{n} \sin\frac{\pi k}{n}=\frac n4\csc\frac{2\pi}n$ I would like to evaluate the sum
$$\sum_{k=1}^{n-1} k \cos\frac{3 \pi k}{n} \sin\frac{\pi k}{n}
$$
which is said to reduce to the simple close-form $\frac n4\csc\frac{2\pi}n$. I have verified it numerically for a large number of $n$’s. However, I have struggled to prove it by using familiar trigonometric identities. The index term $k$ in front of the sequence is problematic. I am not sure how to get around it. Appreciate any hint or proof.
|
Noting that
$$
\begin{aligned}
\sum_{k=1}^{n-1} k \sin (k \theta) &=-\sum_{k=1}^{n-1} \frac{d}{d \theta}(\cos (k \theta))=-\frac{d}{d \theta}\left(\sum_{k=1}^{n-1} \cos k \theta\right)
\end{aligned}
$$
Using the identity
$$\sum_{k=1}^{n-1} \cos k \theta=
\frac{1}{2}\left[-\cos (n \theta)+\cot \left(\frac{\theta}{2}\right) \sin (n \theta)-1\right]$$
Hence $$\sum_{k=1}^{n-1} k \sin (k \theta)= \frac{1}{2}\left[-n \sin (n \theta)-n \cot \left(\frac{\theta}{2}\right) \cos (n\theta)+\frac{1}{2} \csc ^{2}\left(\frac{\theta}{2}\right) \sin (n \theta)\right] \cdots(*) $$
Converting the product to sum yields
$$ \begin{aligned}\sum_{k=1}^{n-1} k \cos\frac{3 \pi k}{n} \sin\frac{\pi k}{n} &= \frac 1 2\sum_{k=1}^{n-1} k \left[\sin \left (\frac {4\pi k}{n}\right) -\sin \left (\frac {2\pi k}{n}\right)\right]\end{aligned}$$
Putting $\theta=\frac{4 \pi}{n}$ and $\frac{2 \pi}{n}$ in the identity (*), simplifies the sum as
$$\begin{aligned}\sum_{k=1}^{n-1} k \cos\frac{3 \pi k}{n} \sin\frac{\pi k}{n}=
&=\frac{n}{4}\left[\cot \left(\frac{\pi}{n}\right)-\cot \left(\frac{2 \pi}{n}\right)\right]=\frac{n}{4} \csc \left(\frac{2 \pi}{n}\right)
\end{aligned} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4375626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Using Lagrange Mult. to find maximum of $2x + y^2$ with constraints: $x^2 + y^2 + z^2 = 2$ and $x + y^2 + z = 0$ I was trying to maximize the function $2x + y^2$ with the following constraints: $x^2 + y^2 + z^2 = 2$ and $x + y^2 + z = 0$ using Lagrange multipliers. I already did all the work of proving that the maximum does exist and showing that we can apply the technique of Lagrange multiplies. After that I arrived at the following systems of equations:
$$\begin{cases}2 = 2\alpha x + \beta \\
2y = 2\alpha y + 2 \beta y \\
0 = 2\alpha z + \beta \\
x^2 +y^2 + z^2 =2 \\
x + y^2 + z = 0\end{cases}$$
Where $\alpha$ and $\beta$ are the Lagrange multipliers. The problem is that I'm having a lot of trouble solving this system for $(x,y,z)$.
My plan was trying to write $x,y,z$ in terms of $\alpha,\beta$ and then using the equations $x^2 + y^2 + z^2 = 2$ and $x + y^2 + z = 0$ find the values for $\alpha $ and $\beta$, but I wasn't able to find an expression for $y$ in terms of $\alpha$ and $\beta$.
How can this system be solved?
|
$x^2+y^2+z^2=2$.
$x+y^2+z=0.$
Minimize: $2x+y^2$
$z=-(x+y^2)\implies z^2=x^2+2xy^2+y^4.$
$x^2+y^2+(x^2+2xy^2+y^4)=2\implies 2x^2+(2x+1)y^2+y^4=2$
$4x+2y^2=2\lambda\implies 2x+y^2=\lambda$
$2y(2x+1)+4y^3=2y\lambda$
$y=0$ or $(2x+1)+2y^2=\lambda\implies 1+y^2=0$
Case 1: $y=0$.
$x^2+z^2=2$ and $x+z=0$. Then $x=\pm1$ and $z=-x$.
Minimum occurs if $x=-1$. Minimum = $-2$.
Case 2: $y^2=-1$
$x^2+z^2=3$ and $x+z=1$.
Then: $x^2+x^2-2x+1=3\implies 2x^2-2x-2=0\implies x=\frac{1\pm\sqrt{5}}{2}$, $z=\frac{1-\mp\sqrt{5}}{2}$.
Minimum is $-\sqrt{5}$. While that's lower, complex values aren't allowed, so minimum is $-2$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4379156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Functional analysis problem. Find the smallest value of $f(x) = \frac{x^2}{x-9}$ on the interval $(9, +\infty)$.
We should basically find the biggest $a$ such that $\frac{x^2}{x-9} \geq a$. We can multiply both sides by $x-9$ since it's positive and than we get $x^2-ax+9a \geq 0.$ I don't know how to proceed.
|
Rephrased:
$y:=x-9,$ $y\in (0,\infty)$;
$f(y)=\dfrac{(y+9)^2}{y} =$
$y+18+\dfrac{81}{y}=$
$(y^{1/2}-\dfrac{9}{y^{1/2}})^2+18+18 \ge 36;$
Equality for
$y^{1/2}=\dfrac{9}{y^{1/2}}$,
$y=9.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4385475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Evaluate $\int_0^\infty\frac{x^2\ln x}{x^4+x^2+1}dx$ by the residue theorem The result should be $\frac{\pi^2}{12}$.
Edit: I have tried to reproduce the image, but limitations of MathJax required some reformatting. Here is the original image.
$$
\begin{align}
\int_0^\infty\frac{x^2\ln x\,\mathrm{d}x}{x^4+x^2+1}&=\int_0^\infty\frac{x^2\ln x}{\left(x^2+x+1\right)\left(x^2-x+1\right)}\\
&\text{poles}\left\{\begin{array}{}
\boxed{\textstyle\frac12+\frac{i\sqrt3}2}&\raise{5pt}{\text{lies inside}\\\text{contour}}\\
\frac12-\frac{i\sqrt3}2\\
\boxed{\textstyle-\frac12+\frac{i\sqrt3}2}&\raise{5pt}{\text{lies inside}\\\text{contour}}\\
-\frac12-\frac{i\sqrt3}2\\
\end{array}\right.
\end{align}
$$
$$
\begin{align}
\int_{-\infty}^0\frac{x^2\ln x\,\mathrm{d}x}{x^4+x^2+1}+\int_0^\infty\frac{x^2\ln x\,\mathrm{d}x}{x^4+x^2+1}&=2\pi i\sum\text{Res}\\
\int_0^\infty\frac{x^2\ln(-x)\,\mathrm{d}x}{x^4+x^2+1}+\int_0^\infty\frac{x^2\ln(x)\,\mathrm{d}x}{x^4+x^2+1}&=\qquad"\\
\int_0^\infty\frac{x^2(i\pi+\ln x)\,\mathrm{d}x}{x^4+x^2+1}+\int_0^\infty\frac{x^2\ln(x)\,\mathrm{d}x}{x^4+x^2+1}&=\qquad"\\
\int_0^\infty\frac{x^2(i\pi)\,\mathrm{d}x}{x^4+x^2+1}+2\int_0^\infty\frac{x^2\ln(x)\,\mathrm{d}x}{x^4+x^2+1}&=\qquad"\\
\end{align}
$$
We need to find the residue
Then we can compare RHS & LHS side
|
The Residues
$\newcommand{\Res}{\operatorname*{Res}}$
If we have $g(z)$ with a simple zero at $z=z_0$, then
$$
\begin{align}
\Res_{z=z_0}\left(\frac{f(z)}{g(z)}\right)
&=\lim_{z\to z_0}\frac{(z-z_0)f(z)}{g(z)}\tag{1a}\\
&=\frac{f(z_0)}{g'(z_0)}\tag{1b}
\end{align}
$$
Applying $(1)$, we get
$$
\begin{align}
\Res_{z=e^{\pi i/3}}\left(\frac{z^2}{z^4+z^2+1}\right)
&=\frac1{4z+2/z}\tag{2a}\\
&=\frac{3-i\sqrt3}{12}\tag{2b}
\end{align}
$$
$$
\begin{align}
\Res_{z=e^{\pi i/3}}\left(\frac{z^2\color{#C00}{\log(z)}}{z^4+z^2+1}\right)
&=\frac{3-i\sqrt3}{12}\color{#C00}{\frac{\pi i}3}\tag{3a}\\
&=\pi\frac{\sqrt3+3i}{36}\tag{3b}
\end{align}
$$
$$
\begin{align}
\Res_{z=e^{2\pi i/3}}\left(\frac{z^2}{z^4+z^2+1}\right)
&=\frac1{4z+2/z}\tag{4a}\\
&=\frac{-3-i\sqrt3}{12}\tag{4b}
\end{align}
$$
$$
\begin{align}
\Res_{z=e^{2\pi i/3}}\left(\frac{z^2\color{#C00}{\log(z)}}{z^4+z^2+1}\right)
&=\frac{-3-i\sqrt3}{12}\color{#C00}{\frac{2\pi i}3}\tag{5a}\\
&=\pi\frac{2\sqrt3-6i}{36}\tag{5b}
\end{align}
$$
Applying the Residues
You had gotten
$$
\int_{-\infty}^\infty\frac{z^2\log(z)\,\mathrm{d}z}{z^4+z^2+1}
=2\int_0^\infty\frac{z^2\log(z)\,\mathrm{d}z}{z^4+z^2+1}
+\frac{\pi i}2\int_{-\infty}^\infty\frac{z^2\,\mathrm{d}z}{z^4+z^2+1}\tag6
$$
which gives
$$
\begin{align}
\int_0^\infty\frac{z^2\log(z)\,\mathrm{d}z}{z^4+z^2+1}
&=\frac12\int_{-\infty}^\infty\frac{z^2\log(z)\,\mathrm{d}z}{z^4+z^2+1}-\frac{\pi i}4\int_{-\infty}^\infty\frac{z^2\,\mathrm{d}z}{z^4+z^2+1}\tag{7a}\\
&=\pi i\left(\pi\frac{\sqrt3-i}{12}\right)+\frac{\pi^2}2\left(-\frac{i\sqrt3}6\right)\tag{7b}\\
&=\frac{\pi^2}{12}\tag{7c}
\end{align}
$$
Explanation:
$\text{(7a)}$: algebraic manipulation of $(6)$
$\text{(7b)}$: apply $(3)$ and $(5)$ to the first integral
$\phantom{\text{(7b):}}$ and $(2)$ and $(4)$ to the second integral
$\text{(7c)}$: evaluate
We used an upper half-plane semi-circular contour in both integrals on the right-hand side of $\text{(7a)}$. Therefore, we included the residues from the poles at $e^{\pi i/3}$ and $e^{2\pi i/3}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4387831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
$y'=2-\frac{3}{x}y+\frac{2}{x^2}y^2$ (Riccati) $y'=2-\frac{3}{x}y+\frac{2}{x^2}y^2$ (Riccati)
(a) Find the solutions.
(b) $y(x_0)=y_0$, prove two cases:
$$0<y_0<x_0 \implies \text{solution's domain is} [x_0,\infty) $$
$$0<x_0<y_0 \implies \text{solution's domain is} [x_0,x_0+\alpha) , \alpha \in \mathbb {R}.$$
I will be grateful for help in $(b)$.
My solution for (a):
$y_1(x)=x$
Denote $y=x+z$
$$(x+z)'=2-\frac{3}{x}(x+z)+\frac{2}{x^2}(x+z)^2$$
$$1+z'=-1-\frac{3z}{x}+\frac{2}{x^2}(x^2+2xz+z^2)$$
$$z'=\frac{z}{x}+\frac{2z^2}{x^2}$$
This is a Bernoulli differential Equation.
Denote $u=z^{-1} \implies -\frac{z'}{z^2} \implies z'=-u'z^2$
Then,
$$u'=\frac{u}{x}-\frac{2}{x^2}$$
Integration factor is $\mu=x$
$$\int(xu)'=\int-\frac{2}{x} \implies xu=-2\ln|x|+c \implies u=\frac{-2\ln|x|+c}{x} \implies z=\frac{x}{-2\ln|x|+c}$$
The solution is $y=x+\frac{x}{-2\ln|x|+c}$
Is my solution correct ?
Thanks !
|
The equation is homogeneous, so you can use $v=\frac{y}{x}$ so that $y=vx$ and $y'=v+xv'$. Then
$$v+xv'= 2 -3v +2v^2$$
$$xv' = 2(1-2v +v^2)$$
Then separating variables
$$\int \frac{dv}{(v-1)^2} = \int\frac{2dx}{x}$$
$$-\frac{1}{v-1} = \ln(x^2) -C$$
$$ v-1 = \frac{1}{C-\ln(x^2)}$$
$$ v = 1+\frac{1}{C-\ln(x^2)}$$
Then
$$y= x +\frac{x}{C-\ln(x^2)}$$
So your solution is correct and confirmed by another method.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4389227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Calculating $\int \frac{dx}{\sqrt[3]{(x+1)^2(x-1)^7}}$ I am trying to calculate the integral
$$
\int \frac{dx}{\sqrt[3]{(x+1)^2(x-1)^7}}.
$$
I know that the answer is
$$
\frac3{16}(3x-5)\sqrt[3]{\frac{x+1}{(x-1)^4}}+C;
$$
this led me to the idea of introducing the new variable
$$
t^3=\frac{x+1}{(x-1)^4}.
$$
But that got me nowhere, because I can't express $x$ in terms of $t$.
|
Substitute $t^3=\frac{x+1}{x-1}$. Then, $x=\frac{t^3+1}{t^3-1}$, $dx= -\frac{6t^2}{(t^3-1)^2}dt$ and
\begin{align}
&\int \frac{1}{\sqrt[3]{(x+1)^2(x-1)^7}}dx\\
=&-\frac34 \int (t^3-1)dt = -\frac3{16}t^4 +\frac34t +C\\
= & -\frac3{16}\left(\frac{x+1}{x-1} \right)^{\frac43} +\frac34 \left(\frac{x+1}{x-1} \right)^{\frac13} +C\\
=& \frac3{16}(3x-5)\sqrt[3]{\frac{x+1}{(x-1)^4}}+C
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4392730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
How can I show that $\underset{\left(x,y\right)\rightarrow\left(0,0\right)}{\lim}{\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}}$ exists, using limit def? I am trying to solve an exercise to show that this function $$f(x,y)=\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}$$ has a limit as $(x,y)$ approaches $(0,0)$:
$$\underset{\left(x,y\right)\rightarrow\left(0,0\right)}{\lim}{\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}}$$
To do so, I have to use the limit precise definition:
$$\forall\epsilon>0\ \exists\delta>0∶\ \left|\left(x,y\right)-\left(a,b\right)\right|<\delta\ {\Rightarrow}\left|f\left(x,y\right)-L\right|<\ \epsilon$$
I've found out that the the potential limit $L=2$ by evaluating the limits across the x-axis, y-axis, arbitrary line and x^2 parabola and y^2 parabola.
I have plugged all the necessary values and ended up with this:
$$\forall\epsilon>0\ \exists\delta>0∶\ \sqrt{x^2+y^2}<\delta\ {\Rightarrow}\left|\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}-2\right|<\ \epsilon$$
Now, I am in trouble to construct inequalities that lead to prove that the following is true:
$$\left|\frac{x^2+y^2}{\sqrt{x^2+y^2+1}-1}-2\right|<\sqrt{x^2+y^2}\ $$
Can someone help me out here?
|
A proof involving geometry:
Set $r^2:=x^2+y^2$, where $r \gt 0$.
Need to show: $(r^2+1)^{1/2}-1 \lt r$.
Consider a right triangle with the $2$
legs $r$ and $1$.
Pythagoren theorem:
The hypothenuse is given by $(r^2+1)^{1/2}.$
In any triangle:
The sum of $2$ sides is greater than the $3$rd side.
Hence
$r+1> (r^2+1)^{1/2}$,
$(r^2+1)^{1/2}-1<r$, and we are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4395898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Confusions in Holder's Inequality Holder's Inequality states that for nonnegative real numbers $a_1,...,a_n$ and $b_1,...,b_n$ we have $$\left(\sum_{i=1}^na_i\right)^p\left(\sum_{i=1}^nb_i\right)^q\ge \left(\sum_{i=1}\sqrt[p+q]{a_i^pb_i^q}\right)^{p+q}$$
Where $p$ and $q$ are positive real numbers.
Here is my problem : $a,b,c$ are positive reals, prove that $$(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3)\ge (a+b+c)^3$$
But for $x>0$ we have $x^5-x^2+3\ge x^3+2$ so we only need to prove $$(a^3+2)(b^3+2)(c^3+2)\ge (a+b+c)^3$$
But when I read the solution, it says ''From Holder's Inequality, it follows '' $$(a^3+2)(b^3+2)(c^3+2)\ge (a+b+c)^3$$
But I don't see that.
|
It should be $a^5 - a^2 + 3$ etc. rather than $a^5 - a^2 + 2$ etc.
Problem: Let $a, b, c $ be positive reals. Prove that
$$(a^5-a^2+3)(b^5-b^2+3)(c^5-c^2+3)\ge (a+b+c)^3.$$
Proof: It is easy to prove that $x^5 - x^2 + 3 \ge x^3 + 2$ for all $x \ge 0$.
It suffices to prove that
$$(a^3+2)(b^3+2)(c^3+2)\ge (a+b+c)^3.$$
Using Holder's inequality, we have
\begin{align*}
(a^3+2)(b^3+2)(c^3+2)
&= (a^3 + 1 + 1)(1 + b^3 + 1)(1 + 1 + c^3)\\
&\ge (\sqrt[3]{a^3 \cdot 1 \cdot 1} + \sqrt[3]{1 \cdot b^3 \cdot 1} + \sqrt[3]{1\cdot 1 \cdot c^3})^3\\
& = (a + b + c)^3.
\end{align*}
For Holder's inequality, see:
https://artofproblemsolving.com/wiki/index.php/H%C3%B6lder%27s_Inequality
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4400473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
If $ax^2+bx+c$ is irreductible, then exists constants $k_1,k_2,k_3$ such that $ax^2+bx+c = k_1((k_2x+k_3)^2+1).$ I want to prove that:
If $ax^2+bx+c$ is irreductible, then exists constants $k_1,k_2,k_3$ such that
$$ax^2+bx+c = k_1((k_2x+k_3)^2+1)$$
We note that
\begin{align*}
&ax^2+bx+c = \frac{1}{4a}(4a^2x^2+4abx+4ac)= \frac{1}{4a}[(4a^2x^2+4abx+b^2)-b^2+4ac]=\\
&= \frac{1}{4a}[(2ax+b)^2-(b^2-4ac)]\\
\therefore\ ax^2+bx+c &= \frac{1}{4a}[(2ax+b)^2-(b^2-4ac)].
\end{align*}
Now, $ax^2+bx+c$ is irreductible implies that $b^2-4ac<0$, since otherwise we have that $ax^2+bx+c$ has roots and in consequence it can be expressed as something of the form $(x-r_1)(x-r_2)$, where $r_1,r_2\in\mathbb{R}$, and in consequence it is not irreductible.
I think thta is necesary that $b^2-4ac=-1$ but I can't see why, can anyone help me please?
|
$$
q(x) = \frac{1}{4a} ((2ax+b)^2 - (b^2 - 4ac) )\\
D = b^2 - 4ac < 0\\
r \equiv \sqrt{-D}\\
q(x) = \frac{1}{4a} ((2ax+b)^2 + r^2 )\\
= \frac{r^2}{4a} (\frac{(2ax+b)^2}{r^2} + 1 )\\
= \frac{r^2}{4a} ( (\frac{2a}{r} x + \frac{b}{r})^2 + 1)\\
k_1 = \frac{r^2}{4a}\\
k_2 = \frac{2a}{r}\\
k_3 = \frac{b}{r}\\
$$
Note that you had to take a square root of a positive number. This is allowed if you just wanted to have $k_i$ be real numbers. But if you said for example, $a,b,c$ were rational and you wanted $k_i$ to also be rational this would not work. But since this is about precalculus, I am guessing you only care about numbers as real numbers not whether they are more specifically rational, integers, etc.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4400587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Solving $\frac{\ 3x}{x^2+x+1}+\frac{2x}{x^2-x+1}=3$ with other approaches
$$\frac{\ 3x}{x^2+x+1}+\frac{2x}{x^2-x+1}=3$$
$$x=?$$
I solved this problem as follow:
$x=0$ is not a root, we can divide numerator and denominator of each fraction by $x$:
$$\frac{3}{t+1}+\frac{2}{t-1}=3\quad\quad\text{where $t=x+\frac1x$}$$
$$5t-1=3t^2-3\Rightarrow t=2 , \frac{-1}6$$
Only $x+\frac1x=2$ is acceptable and $x=1$.
I'm looking for other elegant methods to solve this equation.
|
We have $(x-1)^2\ge 0$ with equality if and only of $x=1$.
But
\begin{align}
(x-1)^2 \ge 0 &\iff x^2-2x+1\ge 0\\
&\iff x^2+x+1\ge 3x\\
&\iff \frac{3x}{x^2+x+1}\le 1
\end{align}
and
\begin{align}
(x-1)^2 \ge 0 &\iff x^2-2x+1\ge 0\\
&\iff x^2-x+1\ge x\\
&\iff \frac{x}{x^2-x+1}\le 1\\
&\iff \frac{2x}{x^2-x+1}\le 2
\end{align}
Then $$\frac{3x}{x^2+x+1} + \frac{2x}{x^2-x+1} \le 3$$
with equality if and only if $x=1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4401595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
How to obtain the asymptotic behavior of this integral? Consider the integral
$$ I(x) = \int_{0}^{\infty} \exp \bigl( - s^2/2 \bigr) \cos \bigl( xs + \lambda s^3 \bigr) \, \text{d} s,$$
where $\lambda >0$ is a constant. I would like to know the asymptotic behavior of this integral in the limit $x \rightarrow \infty$. Any ideas on how to proceed?
|
This is an elaboration on Maxim's comment. Your integral is
$$
I(x) = \frac{1}{2}\int_{ - \infty }^{ + \infty } {\exp \left( { - \frac{1}{2}s^2 + i\lambda s^3 + ixs} \right)ds} .
$$
With the change of variables $s = \frac{t}{{(3\lambda )^{1/3} }} + \frac{1}{{6i\lambda }}$, we find
$$
I(x) \!=\! \frac{1}{2}\frac{1}{{(3\lambda )^{1/3} }}\exp \left( {\frac{x}{{6\lambda }} \!+\! \frac{1}{{108\lambda ^2 }}} \right)\int_{ - \infty + i/(6\lambda )}^{ + \infty + i/(6\lambda )} {\exp \left( {i\left( {\frac{{t^3 }}{3} \!+\! \frac{1}{{(3\lambda )^{1/3} }}\!\left( {x \!+\! \frac{1}{{12\lambda }}} \right)t} \right)} \right)dt}.
$$
Pushing the contour downwards and using the know integral representation of the Airy function, we deduce
\begin{align*}
I(x) & = \frac{1}{2}\frac{1}{{(3\lambda )^{1/3} }}\exp \left( {\frac{x}{{6\lambda }} + \frac{1}{{108\lambda ^2 }}} \right)\int_{ - \infty }^{ + \infty } {\exp \left( {i\left( {\frac{{t^3 }}{3} + \frac{1}{{(3\lambda )^{1/3} }}\left( {x + \frac{1}{{12\lambda }}} \right)t} \right)} \right)dt}
\\ & = \frac{1}{{(3\lambda )^{1/3} }}\exp \left( {\frac{x}{{6\lambda }} + \frac{1}{{108\lambda ^2 }}} \right)\int_{ - \infty }^{ + \infty } {\cos \left( {\frac{{t^3 }}{3} + \frac{1}{{(3\lambda )^{1/3} }}\left( {x + \frac{1}{{12\lambda }}} \right)t} \right)dt}
\\ & = \frac{\pi }{{(3\lambda )^{1/3} }}\exp \left( {\frac{x}{{6\lambda }} + \frac{1}{{108\lambda ^2 }}} \right)\operatorname{Ai}\left(\frac{1}{{(3\lambda )^{1/3} }}\left( {x + \frac{1}{{12\lambda }}} \right) \right).
\end{align*}
Now we know that
$$
\operatorname{Ai}(z) = \frac{1}{{2\sqrt \pi z^{1/4} }}\exp \left( {-\frac{{2z^{3/2} }}{3}} \right)\left( {1 + \mathcal{O}\!\left( {\frac{1}{{z^{3/2} }}} \right)} \right)
$$
as $z\to \infty$ in the sector $|\arg z|\leq \pi-\delta$ ($<\pi$). Since
$$
\frac{2}{3}\frac{1}{{\sqrt {3\lambda } }}\left( {x + \frac{1}{{12\lambda }}} \right)^{3/2} = \frac{{2x^{3/2} }}{{3\sqrt {3\lambda } }} + \frac{{x^{1/2} }}{{12\sqrt 3 \lambda ^{3/2} }} + \frac{1}{{576\sqrt 3 \lambda ^{5/2} x^{1/2} }} - \cdots
$$
and
$$
(3\lambda )^{1/12} \left( {x + \frac{1}{{12\lambda }}} \right)^{ - 1/4} = (3\lambda )^{1/12} x^{ - 1/4} \left( {1 - \frac{1}{{48\lambda x}} + \cdots } \right),
$$
we obtain
$$
I(x) = \frac{{\sqrt \pi }}{{2(3\lambda x)^{1/4} }}\exp \left( {-\frac{{2x^{3/2} }}{{3\sqrt {3\lambda } }} + \frac{x}{{6\lambda }} - \frac{{x^{1/2} }}{{12\sqrt 3 \lambda ^{3/2} }} + \frac{1}{{108\lambda ^2 }}} \right)\left( {1 + \mathcal{O}\!\left( {\frac{1}{{x^{1/2} }}} \right)} \right)
$$
as $x\to \infty$ in the sector $|\arg x|\leq \pi-\delta$ ($<\pi$). More terms in the expansion can be obtained if necessary.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4403218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Prove $\left(\frac{a+b+c}{3}\right)^p\leq \frac{a^p+b^p+c^p}{3}$. Prove:
Let $p$ be an integer greater than $1$. Suppose $a,b,c$ be positive real numbers. Then $\left(\frac{a+b+c}{3}\right)^p\leq \frac{a^p+b^p+c^p}{3}$.
By AM-GM, I get $\frac{a+b+c}{3}\geq (abc)^{1/3}$ and $\frac{a^p+b^p+c^p}{3}\geq (a^pb^pc^p)^{1/3}$. Then $\left(\frac{a+b+c}{3}\right)^p\geq (abc)^{p/3}$.
I get stuck in how to relate $\left(\frac{a+b+c}{3}\right)^p$ and $\frac{a^p+b^p+c^p}{3}$. What should I do? I also face the same problem on proving $\left(\frac{a+b+c+d}{4}\right)^p\leq \frac{a^p+b^p+c^p+d^p}{4}$.
My tutor suggests me to let $u=\frac{a+b}{2}$ and $v=\frac{c+d}{2}$.
I know $\left(\frac{u+v}{2}\right)^p\leq \frac{u^p+v^p}{2}$. So I substitute $u$ and $v$ into the inequality, and get $\left(\frac{a+b+c+d}{4}\right)^p\leq \frac{{(a+b)}^p+{(c+d)}^p}{2^{p+1}}$.
I haven't learnt Jensen's inequality. I suppose I should make use of AMGM to solve the problem(?
|
To make the problem less tedious, let $x = \dfrac{a}{a+b+c}, y = \dfrac{b}{a+b+c}, z = \dfrac{c}{a+b+c}$, then you prove: $x^p+y^p+z^p \ge \dfrac{1}{3^{p-1}}$, with $x+y+z=1$. Let $u = 3x, v = 3y, w = 3z$, then you again prove: $u^p+v^p+w^p \ge 3$, with $u+v+w = 3$. Observe that by AM-GM inequality: $u^p + 1(p-1) \ge p\sqrt[p]{u^p}=pu$. Repeat this twice for $v,w$ we have: $v^p + 1(p-1) \ge p\sqrt[p]{v^p}=pv$, and $w^p+1(p-1) \ge p\sqrt[p]{w^p}=pw$. Adding these inequalities we obtain: $u^p+v^p+w^p+3p-3 \ge p(u+v+w) = 3p\implies u^p+v^p+w^p \ge 3$. Equality occurs when $u = v = w = 1 \implies x = y = z \implies a = b = c$. Done !
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4409037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find an orthonormal basis and the signature of the quadratic form
Consider the quadratic form given by the matrix below (in the canonical basis)
\begin{pmatrix}
1 & 1 & -1\\
1 & 1 & 3\\
-1 & 3 & 1
\end{pmatrix}
Find an orthonormal basis of it and find its signature.
First I calculated the eigenvalues, which are $4, \frac{-1+ \sqrt{17}}{2}, \frac{-1-\sqrt{17}}{2}$. Then I calculated the eigenvectors associated to $4$ and $\frac{-1+ \sqrt{17}}{2}$ and normalized them, which gave me
\begin{align}
e_1 &= \frac{1}{\sqrt{2}}\begin{bmatrix}
0 \\
1 \\
1
\end{bmatrix}, \quad
e_2 = \frac{\sqrt{2}}{\sqrt{17+3\sqrt{17}}}\begin{bmatrix}
-\frac{3+\sqrt{17}}{2} \\
-1 \\
1
\end{bmatrix}
\end{align}
And the third vector of the basis I want to be orthogonal to $e_1$ and $e_2$, so
$$e_3 = \frac{1}{\sqrt{17+3\sqrt{17}}} e_1 \wedge e_2 =\frac{1}{\sqrt{17+3\sqrt{17}}} \begin{bmatrix}
2 \\
-\frac{3+\sqrt{17}}{2} \\
\frac{3+\sqrt{17}}{2}
\end{bmatrix} $$
I can't detail the calculations because they are very big. Perhaps someone can confirm the results. For the signature I know that the two possibilities are $(0,3)$ and $(2,1)$ but I don't know how to find the right one.
|
Using SymPy:
>>> from sympy import *
>>> A = Matrix([[ 1, 1,-1],
[ 1, 1, 3],
[-1, 3, 1]])
>>> Q, D = A.diagonalize(normalize=True)
>>> D
Matrix([
[4, 0, 0],
[0, -1/2 + sqrt(17)/2, 0],
[0, 0, -sqrt(17)/2 - 1/2]])
>>> simplify(Q)
Matrix([
[ 0, -sqrt(2)*(3 + sqrt(17))/(2*sqrt(3*sqrt(17) + 17)), sqrt(2)*(-3 + sqrt(17))/(2*sqrt(17 - 3*sqrt(17)))],
[sqrt(2)/2, -sqrt(2)/sqrt(3*sqrt(17) + 17), -sqrt(2)/sqrt(17 - 3*sqrt(17))],
[sqrt(2)/2, sqrt(2)/sqrt(3*sqrt(17) + 17), sqrt(2)/sqrt(17 - 3*sqrt(17))]])
Using function latex, we obtain
$$Q = \left[\begin{matrix}0 & - \frac{\sqrt{2} \left(3 + \sqrt{17}\right)}{2 \sqrt{3 \sqrt{17} + 17}} & \frac{\sqrt{2} \left(-3 + \sqrt{17}\right)}{2 \sqrt{17 - 3 \sqrt{17}}}\\\frac{\sqrt{2}}{2} & - \frac{\sqrt{2}}{\sqrt{3 \sqrt{17} + 17}} & - \frac{\sqrt{2}}{\sqrt{17 - 3 \sqrt{17}}}\\\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{\sqrt{3 \sqrt{17} + 17}} & \frac{\sqrt{2}}{\sqrt{17 - 3 \sqrt{17}}}\end{matrix}\right]$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4411687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Rhombus rotates in a circle. Two vertexes of the rhombus is on the circle, rotate the rhombus $ABCD$ clockwise around the point A to the rhombus $AB'C'D'$, where the point $B'$ falls on the circle, link $B'D,C'C$.
If $B'D:CC'=4:3$, then value of $tan\angle BAD$ will be?
It seems $B',B,D'$ or $D,D',C'$ are in a straight line on the image, but I can't proof it.By the way, there are also many things that seem right but hard to proof.It confuses me a lot.
|
If the angle of rotation is $\phi$ (clockwise) then
$ CC' = 2 \overline{AC} \sin(\dfrac{\phi}{2}) $
and
$ BD' = 2 \overline{AD} \sin(\angle BAC + \dfrac{\phi}{2} ) $
Now, $ \dfrac{\overline{AD}}{\overline{AC}} = \dfrac{1}{2} \sec(\angle BAC) $
Also, from the figure, we have $\phi = \pi - 4 \angle BAC $
Let $ x = \angle BAC $ , then
$\dfrac{4}{3} = \dfrac{1}{2} \sec(x) \dfrac{ \sin( \dfrac{\pi}{2} - x ) }{\sin( \dfrac{\pi}{2} - 2 x)} $
Hence,
$8 \cos(2x) = 3 $
Therefore
$x = \angle BAC = \dfrac{1}{2} \cos^{-1}\left(\dfrac{3}{8}\right) $
Thus
$ \angle BAD = 2 x = \cos^{-1} \left( \dfrac{3}{8} \right ) $
Thus $ \cos(\angle BAD) = \dfrac{3}{8} $
From which
$\tan(\angle BAD) = \sqrt{\dfrac{64}{9} - 1} = \dfrac{\sqrt{55}}{3}$
The figure below shows the original rhombus (blue) and its rotated image (red).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4423086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
}
|
Find $k\in\mathbb{R}$ given $w = k+i$ and $z=-4+5ki$ and $\arg(w+z)$ I am working on problem 2B-11 from the book Core Pure Mathematics, Edexcel 1/AS. The question is:
The complex numbers $w$ and $z$ are given by $w = k + i$ and $z = -4 + 5ki$ where $k$ is a real constant. Given at $\arg(w + z) = \frac{2\pi}{3}$, find the exact value of $k$.
I have attempted this question, but by answer is different from the given one and I can't figure out why. My answer:
Let $w + z = a + bi$.
$a = k - 4 \\
b = 1 + 5k$
The angle of the triangle formed by the vector $w + z$ (on Argand diagram) and the x-axis is $\pi -\frac{2\pi}{3} = \frac{\pi}{3}$
$\tan\frac{\pi}{3} = \frac{b}{a} \\
\tan\frac{\pi}{3} = \frac{1 + 5k}{k - 4} \\
k\tan\frac{\pi}{3} - 4\tan\frac{\pi}{3} = 1 + 5k \\
k\tan\frac{\pi}{3} - 5k = 1 + 4\tan\frac{\pi}{3} \\
k = \frac{1 + 4\tan\frac{\pi}{3}}{\tan\frac{\pi}{3} - 5} = -2.42$
However, the given answer is:
$\frac{4\sqrt{3} - 1}{5 + \sqrt{3}} = 0.88$ where $\tan\frac{\pi}{3} = \sqrt{3}$
I have just noticed that assuming the angle is $-\tan\frac{\pi}{3}$, my answer is correct. How can you tell the quadrant of $w + z$ when the constant $k$ appears in both $a$ and $b$?
|
It seems to me that one can simply say
$$
w+z=k+4+i(1+5k)\\
\arg(w+z)=\tan^{-1}\frac{1+5k}{k+4}=\frac{2\pi}{3}\\
\frac{1+5k}{k+4}=\tan\frac{2\pi}{3}=-\sqrt{3}\\
k=\frac{4\sqrt{3}-1}{5+\sqrt{3}}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4423608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
If $f(x)=\frac {x^ 2 -2x +4}{ x^ 2 +2x+4}$ for $x \in \mathbb{R}$, prove that the range of $f(x)$ is $[1/3, 3]$ One method to solve it would by putting $y = f(x)$ then multiplying the denominator with $y$ hence making a quadratic equation in x then we can just use the inequalities for $x$ being real to prove it.
For an alternative what I did is
$y = \frac{x^2 + 2x + 4}{x^2 - 2x + 4}$
Then$ \frac{1+y}{1-y} =\frac{x^2+4}{2x} $
Therefore$ \frac{1+y}{1-y} = {x/2+2/x}$
If we go further putting rhs range to lhs the answer gets altered . Is this method wrong if so why if not where did the process go wrong.
|
Notice that $f(-x) = \frac{1}{f(x)}$, so the analysis we do for one side will apply to the other (aka take $x\geq0$ WLOG).
We have
$$\frac{x^2-2x+4}{x^2+2x+4} \leq 1$$
with equality achieved at least at $x=0$. Then rewriting the function we have
$$\frac{(x-2)^2+2x}{(x-2)^2+6x} = 1 - \frac{4x}{(x-2)^2+6x} \geq 1 - \frac{4x}{6x} = \frac{1}{3}$$
and in fact this minimum is reached at $x=2$. Therefore
$$1 \geq f(x) \geq \frac{1}{3}$$
on $x\in[0,\infty)$ which implies
$$3 \geq f(x) \geq \frac{1}{3}$$
on $x\in\Bbb{R}$, with max and min achieved at least at $x=\pm2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4424161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Spivak Calculus, Ch 10, problem 32b: Can we use Leibniz's formula to calculate $f^{(k)}(x)$ if $f(x)=\frac{1}{x^2-1}$? Spivak's Calculus, Chapter 10 on Differentiation, problem 32:
*What is $f^{(k)}(x)$ if
a)$f(x) = \frac{1}{(x-1)^n}$
*b) $f(x)=\frac{1}{x^2-1}$
My question regards part $b)$.
$a)$ can be solved quite easily if you use the conclusion of a previous problem (30), which shows
If $f(x)=x^{-n}$ for $n \in \mathbb{N}$ then
$$f^{(k)}(x)=(-1)^k\frac{(n+k-1)!}{(n-1)!}x^{-n-k}\tag{1}$$
$$g(x)=f(x+1)=x^{-n}$$
$$g^{(k)}(x)=f^{(k)}(x+1)=(-1)^k\frac{(n+k-1)!}{(n-1)!}x^{-n-k}$$
$$g^{(k)}(x-1)=f^{(k)}(x)=\frac{d^{k} \frac{1}{(x-1)^n}}{dx^k}=(-1)^k\frac{(n+k-1)!}{(n-1)!}(x-1)^{-n-k}\tag{2}$$
Similarly, we can show that
$$\frac{d^k \frac{1}{(x+1)^n}}{dx^k}=(-1)^k\frac{(n+k-1)!}{(n-1)!}(x+1)^{-n-k}\tag{3}$$
Now for part b).
We can rewrite $f(x)=\frac{1}{x^2-1}$ in two different ways:
$$f(x)=\frac{1}{x+1}\cdot \frac{1}{x-1}\tag{4}$$
$$f(x)=\frac{1}{2} \left ( \frac{1}{x-1}-\frac{1}{x+1} \right )\tag{5}$$
Spivak's solution manual differentiates $(5)$.
$$f^{(k)}(x)=\frac{1}{2} \left ( \frac{d^k \frac{1}{(x-1)^n}}{dx^k} - \frac{d^k \frac{1}{(x+1)^n}}{dx^k} \right )$$
So we can just sub in $(2)$ and $(3)$:
$$f^{(k)}(x)=\frac{1}{2}(-1)^k\frac{(n+k-1)!}{(n-1)!}((x-1)^{-n-k}-(x+1)^{-n-k})\tag{6}$$
My question is: can we apply Leibniz's formula to differentiate $(4)$?
Leibniz's formula tells us that
$$(f \cdot g)^{(n)}(x)=\sum_{k=0}^n \binom{n}{k}f^{(k)}(x) \cdot g^{(n-k)}(x)$$
If we apply this to $f(x)=\frac{1}{x+1}\cdot \frac{1}{x-1}$ then we get
$$f^{(n)}(x)=\sum_{k=0}^n \left [(-1)^k \frac{k!}{0!}(x-1)^{-1-k} \right ] \left [(-1)^{n-k} \frac{(n+k)!}{0!}(x+1)^{-1-k} \right ]$$
$$= \sum_{k=0}^n \binom{n}{k}(-1)^nk! (n+k)! (x^2-1)^{-1-k}\tag{7}$$
Differentiating the sum in $(5)$ was the easier route. But is the calculation of differentiating $(4)$ correct? Is one route better than the other?
|
The calculation can be done in both ways. On the one hand we have
\begin{align*}
\color{blue}{f^{(k)}(x)}&=\left(\frac{1}{x^2-1}\right)^{(k)}\\
&=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)^{(k)}\\
&=\frac{1}{2}\left(\left((x-1)^{-1}\right)^{(k)}-\left((x+1)^{-1}\right)^{(k)}\right)\\
&=\frac{1}{2}\left((-1)(-2)\cdots(-1-(k-1))(x-1)^{-1-k}\right.\\
&\qquad\qquad\left.-(-1)(-2)\cdots(-1-(k-1))(x+1)^{-1-k}\right)\\
&=\frac{1}{2}\left((-1)^kk!(x-1)^{-1-k}-(-1)^kk!(x+1)^{k+1}\right)\\
&\,\,\color{blue}{=\frac{(-1)^kk!}{2}\left(\frac{1}{(x-1)^{k+1}}-\frac{1}{(x+1)^{k+1}}\right)}\tag{1}
\end{align*}
On the other hand using the general Leibniz rule we obtain
\begin{align*}
\color{blue}{f^{(k)}(x)}&=\left(\frac{1}{x^2-1}\right)^{(k)}\\
&=\left(\frac{1}{x-1}\cdot\frac{1}{x+1}\right)^{(k)}\\
&=\sum_{j=0}^k\binom{k}{j}\left(\frac{1}{x-1}\right)^{(j)}\left(\frac{1}{x+1}\right)^{(k-j)}\\
&=\sum_{j=0}^k\binom{k}{j}\frac{(-1)^jj!}{(x-1)^{j+1}}\,\frac{(-1)^{k-j}(k-j)!}{(x+1)^{k-j+1}}\\
&=\frac{(-1)^kk!}{(x+1)^{k-1}}\,\frac{1}{x-1}\sum_{j=0}^k\left(\frac{x+1}{x-1}\right)^j\\
&=\frac{(-1)^kk!}{(x+1)^{k-1}}\,\frac{1}{x-1}\,\frac{1-\left(\frac{x+1}{x-1}\right)^{k+1}}{1-\frac{x+1}{x-1}}\\
&=\frac{(-1)^kk!}{(x+1)^{k-1}}\,\frac{1-\left(\frac{x+1}{x-1}\right)^{k+1}}{-2}\\
&\,\,\color{blue}{=\frac{(-1)^kk!}{2}\left(\frac{1}{(x-1)^{k+1}}-\frac{1}{(x+1)^{k+1}}\right)}\tag{2}
\end{align*}
in accordance with (1).
Obviously the first way is more convenient, since we obtain immediately the closed form (1). In (2) we have to cope with a sum and to apply the formula for finite geometric sums till we finally obtain the same form as in (1).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4426875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Elegant solutions to $ \int_{0}^{\frac{\pi}{2}} x\tan 2 x \ln (\tan x) d x $ . Letting $x\mapsto \frac{\pi}{2} -x$ converts the integral
$\displaystyle I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \tan 2 x \ln (\tan x) d x \tag*{} $
Using the identity $ \displaystyle \tan x=\frac{\sin 2 x}{1+\cos 2 x} $ , we get
$$\displaystyle I=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}}\left[\frac{\sin 2 x}{\cos 2 x} \ln (\sin 2 x)-\frac{\sin 2 x}{\cos 2 x} \ln (1+\cos 2 x)\right] d x \tag*{} $$
Letting $2x\mapsto x$ yields
$\displaystyle I=\frac{\pi}{8}\left[\underbrace{\int_{0}^{\pi} \frac{\sin x}{\cos x} \ln (\sin x)}_{J} d x-\underbrace{\int_{0}^{\pi} \frac{\sin x}{\cos x} \ln (1+\cos x) d x}_{K}\right]\tag*{} $
As $ \displaystyle J \stackrel{x \rightarrow \pi-x}{=}-J \Rightarrow J=0 $ , therefore $\displaystyle I=-\frac{\pi}{8} K. $
By my post ,
$\displaystyle \begin{aligned}K \stackrel{y=\cos x}{=}& \int_{-1}^{1} \frac{\ln (1+y)}{y} d y=\frac{\pi^{2}}{4}\end{aligned}\tag*{} $
Now we can conclude that
$\displaystyle \boxed{I=-\frac{\pi^{3}}{32}}\tag*{} $
Request for elegant solutions. Your suggestion and alternative methods are warmly welcome!
|
Substitute $t=\tan^2x$, along with $\tan 2x =\frac{2\tan x}{1-\tan^2x}$\begin{align}
& I
= \frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \tan 2 x \ln (\tan x) d x = \frac\pi8 \int_0^\infty\frac{\ln t}{1-t^2}dt
=\frac\pi8\left(-\frac{\pi^2}4\right)=-\frac{\pi^3}{32}
\end{align}
where $\int_0^\infty \frac {\ln t}{1-t^2}dt=-\frac{\pi^2}4 $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4427985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
Find $\angle{CAD}$ in equilateral triangle $\triangle{ABC}$ for $D$ inside with $\angle{ABD}=18^{\circ}, \angle{BCD}=12^{\circ}$
This problem is feasible to solve using trigonometric functions. I am looking for pure geometric solution as usual. Thanks...
|
Let $E$ be the circumcenter of $\triangle{BCD}$, so
$BE=CE=DE, \angle{BED}=2\angle{BCD}=2*12^{\circ}=24^{\circ} \\
\implies \angle{EBD}=\angle{BDE}=\dfrac{180^{\circ}-24^{\circ}}{2}=78^{\circ} \\
\implies \angle{EBC}=\angle{ECB}=78^{\circ}-42^{\circ}=36^{\circ} \\
\implies \angle{BEC}=108^{\circ}$
Find $F$ so that $BE=BF=CE=DE, \angle{FBE}=108^{\circ}$, connect $AF, CF, DF, EF$. So
*
*$\angle{FBC}=108^{\circ}-36^{\circ}=72^{\circ}$
*$\triangle{BCE} \cong \triangle{BEF} \implies EF=BC=AB, \angle{BFE}=\angle{BEF}=36^{\circ}=\angle{BCE}=\angle{EBC} \\
\implies \angle{DEF}=36^{\circ}-24^{\circ}=12^{\circ} $
*$\angle{ABF}=108^{\circ}-78^{\circ}-18^{\circ}=12^{\circ}=\angle{DEF}$
$\implies \triangle{ABF} \cong \triangle{FED} \text{ [SAS]} \\
\implies AF=DF, \angle{FAB}=\angle{DFE} \\
\implies \angle{FDA}=\angle{FAD} \\
\text{Also we have }\angle{BFE}=\angle{BCE} \implies BECF \text{ are cyclic} \\
\implies \angle{FCB}=\angle{BEF}=36^{\circ} \\
\implies \angle{BFC}=180^{\circ}-\angle{FCB}-\angle{FBC}=72^{\circ}=\angle{FBC} \\
\implies BC=FC=AC \\
\implies \text{C is circumcenter of } \triangle{ABF} \\
\implies \angle{AFB}=180^{\circ}-\dfrac{\angle{ACB}}{2}=180^{\circ}-30^{\circ}=150^{\circ} \\
\implies \angle{FAB}=\angle{DFE}=180^{\circ}-150^{\circ}-12^{\circ}=18^{\circ} \\
\implies \angle{BFD}=36^{\circ}-18^{\circ}=18^{\circ} \\
\implies \angle{DFA}=150^{\circ}-18^{\circ}=132^{\circ} \\
\implies \angle{FDA}=\angle{FAD}=\dfrac{180^{\circ}-132^{\circ}}{2}=24^{\circ} \\
\implies \angle{BAD}=\angle{FAD}-\angle{FAB}=24^{\circ}-18^{\circ}=6^{\circ} \\
\implies \angle{DAC}=\angle{BAC}-\angle{BAD}=60^{\circ}-6^{\circ}=\boxed{54^{\circ}}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4428191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
An inequality $2(ab+ac+ad+bc+bd+cd)\le (a+b+c+d) +2(abc+abd+acd+bcd)$
Let $a, b, c, d \ge 0$ such that $a+b, a+c, a+d, b+c, b+d, c+d \le 1$. Show that
$$2(ab+ac+ad+bc+bd+cd)\le (a+b+c+d) +2(abc+abd+acd+bcd).$$
I am trying to maximise
$$f(a,b,c,d)=(a+b+c+d) /(1-(1-a)(1-b)(1-c)(1-d) +abcd) $$ over the set of nonnegative real numbers $a, b, c, d$ subject to $a+b, a+c, a+d, b+c, b+d, c+d \le 1$.
When $a=b=c=d=0$, set $f(0,0,0,0)=1$. Some massage with Wolfram Alpha gives $2$ as a local maximum, which, if it were the global maximum, would then be equivalent to the above inequality.
It seems one standard way to solve inequalities of this class is the pqrs, uvwt method.
|
Remark: Here is a proof which is similar to my answer
in An inequality $16(ab + ac + ad + bc + bd + cd) \le 5(a + b + c + d) + 16(abc + abd + acd + bcd)$
Let $x = a + b, \, y = c + d$.
We have $x, y \in [0, 1]$.
We have
\begin{align*}
\mathrm{RHS} - \mathrm{LHS} &= (x + y) + 2(ab y + cdx) - 2(ab + cd + xy)\\
&= (x + y) - 2xy - 2(1 - y)ab - 2(1 - x)cd\\
&\ge (x + y) - 2xy - 2(1 - y)\cdot \frac{x^2}{4} - 2(1 - x)\cdot \frac{y^2}{4} \tag{1}\\
&= (x + y) - 2xy - x^2/2 - y^2/2 + x^2y/2 + xy^2/2 \\
&= (x + y) - xy - (x + y)^2/2 + xy(x + y)/2 \\
&= (x + y) - (x + y)^2/2 -\frac{1}{2}(2 - x - y)xy\\
&\ge (x + y) - (x + y)^2/2 -\frac{1}{2}(2 - x - y)\cdot \frac{(x + y)^2}{4}\\
&= \frac18(x + y)(2 - x - y)(4 - x - y)\\
&\ge 0
\end{align*}
where we have used $ab \le (a + b)^2/4 = x^2/4$ and $cd \le (c + d)^2/4 = y^2/4$ in (1).
We are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4435584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Turning a pole/ladder horizontally at a corner I've been attempting to solve the following problem:
A lane runs perpendicular to a road 64 ft wide. If it is just possible to carry a pole 125 ft long from the road into the lane, keeping it horizontal, then what is the minimum width of the lane?
At first, I tried to solve this by thinking that the mid-point of the pole must be at the intersection point of the lane and the road, but that doesn't give the answer, and I have no ideas on how to even attempt this. I have been thinking about it for a while, and finally decided to ask for help.
Any hints/ clues would also be appreciated. Thanks.
|
From the figure attached, we'll relate the maximum possible pole length $L$ to $w$, the width of the lane.
From the figure
$L = w \csc(\theta) + 64 \sec(\theta)$
for a given $w$, we want to find the minimum attainable length $L$ and this critical value will be our maximum possible pole length $L$
$\dfrac{dL}{d\theta} = - w \csc(\theta) \cot(\theta) + 64 \sec(\theta) \tan (\theta) = 0 $
multiply through by $\sin^2(\theta) \cos^2(\theta) $, then
$0 = - w \cos^3( \theta) + 64 \sin^3(\theta) $
Hence, the critical value for $\theta$ is determined by
$\tan(\theta) = \left( \dfrac{w}{ 64} \right)^{\frac{1}{3}}$
Let $u = \dfrac{w}{64} $ then
$\cos(\theta) = \dfrac{1}{ \sqrt{ 1 + u^\frac{2}{3} }}$
$\sin(\theta) = \tan(\theta) \cos(\theta) = \dfrac{u^\frac{1}{3} } {\sqrt{ 1 + u^\frac{2}{3} } }$
therefore, the maximum $L$ is
$L = (1 + u^\frac{2}{3} )^(\frac{1}{2}) \left( 64 u^\frac{2}{3} + 64 \right) = 64 ( 1 + u^\frac{2}{3} )^\frac{3}{2} $
Since we are given that the maximum $L$ is $125$ then
$ 125 = 64 ( 1 + u^\frac{2}{3} )^\frac{3}{2} $
so that
$ \left(\dfrac{ 125}{64} \right )^{2}{3} = 1 + u^\frac{2}{3} $
This reduces to
$ \dfrac{25}{16} = 1 + u^\frac{2}{3} $
From which,
$ u = \left( \dfrac{9}{16} \right)^\frac{3}{2} = \dfrac{ 27 }{64 } $
But $ u = \dfrac{ w }{64 } $
Therefore
$ w = 27 $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4435751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Find integral of $\sqrt{x}$ using Riemann sum definition Let $a > 1$ be a real number. Evaluate the definite integral
\begin{equation}
\int_{1}^{a} \sqrt{x} \,dx
\end{equation}
from the Riemann sum definition.
My approach I know a Riemann sum consists of a sigma notation with a width and function. However, I am confused and not sure where to start. Any hints/answers are appreciated. Thanks.
|
Using the hints and tips provided, I successfully proved the integral from the Riemann sum definition:
The function $f(x)=\sqrt{x}$ is continuous on $[1,a]$, hence integrable on $[1,a]$. For every positive integer $n$, we consider the left Riemann sum of $f$ with respect to the partition $[1,a^{1/n},a^{2/n},a^{1/n}...a^{n/n}]$ of $[1,a]$ into $n$ subintervals. Then,
\begin{align*}
\int_{1}^{a} \sqrt{x} \,dx &= \lim_{n\to\infty} \sum_{k=0}^{n-1} \sqrt{a^{k/n}}(a^{\frac{k+1}{n}}-a^{k/n}) \\
&= \lim_{n\to\infty} (a^{\frac{1}{n}}-1) \sum_{k=0}^{n-1} (a^{\frac{3}{2n}})^k \\
&= \lim_{n\to\infty} (a^{\frac{1}{n}}-1) \frac{(a^{\frac{3}{2n}})^n-1}{a^{\frac{3}{2n}}-1} \\
&= (a^{\frac{3}{2}}-1) \lim_{n\to\infty} \frac{a^{\frac{1}{n}}-1}{a^{\frac{3}{2n}}-1} \\
&= (a^{\frac{3}{2}}-1) \lim_{n\to\infty} \frac{(a^{\frac{1}{2n}}-1)(a^{\frac{1}{2n}}+1)}{(a^{\frac{1}{2n}}-1)(a^{\frac{1}{n}}+a^{\frac{1}{2n}}+1)} \\
&= (a^{\frac{3}{2}}-1) \lim_{n\to\infty} \frac{a^{\frac{1}{2n}}+1}{a^{\frac{1}{n}}+a^{\frac{1}{2n}}+1} \\
&= (a^{\frac{3}{2}}-1) \frac{a^0+1}{a^0+a^0+1} \\
&= \frac{2}{3}(a^{\frac{3}{2}}-1)
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4439680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Show that $p^k \mid \mid (x-y)$ iff $p^k \mid \mid (x^6-y^6)$. Definition. Let $n>1$ be an integer and $p$ be a prime. We say that $p^k$ fully divides $n$ and write $p^k \mid \mid n$ if $k$ is the greatest positive integer such that $p^k \mid n$.
Let $p>6$ be a prime. Let $x$ and $y$ be two distinct integers such that $p\nmid x, p\nmid y,$ and $p\mid (x-y)$. Show that $p^k \mid \mid (x-y)$ iff $p^k \mid \mid (x^6-y^6)$.
Attempt:
$(\implies)$ Let $p^k \mid \mid (x-y)$. Then $k$ is the greatest positive integer such that $p^k \mid (x-y)$. Write $x-y=p^km$ for some $m \in \Bbb Z$. Notice that
$$x^6-y^6 = (x-y)(x^5+x^4y+\cdots+y^5) = p^km(x^5+x^4y+\cdots+y^5)=p^k(m(x^5+x^4y+\cdots+y^5)).$$
Hence, $p^k \mid (x^6-y^6)$ and then $p^k \mid \mid (x^6-y^6)$.
Does this approach correct? If not, how to approach it? And how to approach the reverse direction?
Any ideas? Thanks in advanced.
|
$(\implies)$
Let $p^k \mid \mid (x-y)$. Then $p^k \mid (x-y)$ with $k$ is the greatest positive integer such that $x-y = p^km$ for some nonzero integer $m$. Notice that we must have $p \nmid m$, because otherwise we would have $m=pc$ for some nonzero integer $c$ such that $x-y=p^{k+1}c$, which implies $p^{k+1} \mid (x-y)$, contradicting the maximality of $k$.
For the convenience, let's write $A=x^5+\cdots+y^5$.
Now, notice that
$$x^6-y^6 = (x-y)A= p^kmA = p^k(mA).$$
Hence, $p^k \mid (x^6-y^6)$. The goal is to show that $p^k \mid \mid (x^6-y^6)$. In other words, we want to show that $p^{k+1} \nmid (x^6-y^6)$.
If $p \mid A$, then $A=ps$ for some nonzero integer $s$. Hence,
$$x^6-y^6=(x-y)A=p^km(ps)=p^{k+1}ms,$$ which gives $p^{k+1} \mid (x^6-y^6)$, which is not being our goal. So, $p \nmid A$. It follows that $p^{k+1} \nmid (x^6-y^6)$. To this end, suppose for contradiction that $p^{k+1} \mid (x^6-y^6)$. Then $$x^6-y^6 = p^{k+1}s = p^kps = (x-y)A = p^kmA,$$ for some nonzero integer $s$. Hence, $p^kps=p^kmA$, which implies $ps=mA$. It follows that $p \mid m$ or $p \mid A$. Since $p \nmid m$, then we must have $p \mid A$, contradicting the assumption that $p \nmid A$. Thus,
$$p^{k+1} \nmid (x^6-y^6).$$
This means that $p^k \mid \mid (x^6-y^6)$, and so we're done. $\Box$
$(\impliedby)$
Let $p^k \mid \mid (x^6-y^6)$, i.e.,
$k$ is the greatest positive integer such that
$p^{k} \mid (x^6-y^6)$. We want to show that $p^k \mid \mid (x-y)$, i.e., $p^{k+1} \nmid (x-y)$.
First way: Suppose for contradiction that $p^{k+1} \mid (x-y)$. Then $x-y=p^{k+1}s$ for some nonzero integer $s$.
Now, let $A=x^5+\cdots+y^5$.
Hence,
$$x^6-y^6 = (x-y)A = p^{k+1}sA.$$
Clearly, $A$ is an integer. Thus, $sA$ also being an integer. Therefore, $p^{k+1} \mid (x^6-y^6)$, a contradiction to our assumption that $p^{k+1} \nmid (x^6-y^6)$. Hence, $p^{k+1} \nmid (x-y)$. Thus, $p^k \mid \mid (x-y). \qquad \Box$
Second way: Let $p^k \mid \mid (x^6-y^6)$. Then $x^6-y^6=p^km$ for some nonzero integer $m$. Notice that we must have $p \nmid m$, because otherwise we would have $m=pj$ such that $x^6-y^6=p^{k+1}j$ for some nonzero integer $j$, contradicting the maximality of $k$. Now, let $A=x^5+\cdots+y^5$. We have
$$x^6-y^6 = p^km = (x-y)A.$$
Hence $p^k \mid (x-y)$ or $p^k \mid A$.
If $p^k \mid A$, then $A=p^ks$ for some nonzero integer $s$. Since $p \mid (x-y)$, then $x-y=pt$ for some nonzero integer $t$. Hence
$$x^6-y^6=(x-y)A=pt(p^ks)=p^{k+1}st.$$
Thus, $p^{k+1} \mid x^6-y^6$, contradicting our assumption that $p^k \mid \mid (x^6-y^6)$. Hence, $p^k \nmid A$. Therefore, $p^k \mid (x-y)$. If $p^{k+1} \mid (x-y)$, then $x-y=p^{k+1}r$ for some nonzero integer $r$. Hence
$$x^6-y^6=p^km=p^{k+1}rA,$$ which gives $p \mid m$, which contradicts the maximality of $k$. Thus, $p^{k+1} \nmid (x-y)$. Therefore, $p^k \mid \mid (x-y)$, and we are done. $\Box$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4439907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
}
|
Is my construction of one-point compactification of $\mathbb{R}$ correct? (+ Clarifying questions) Could you please check if this construction makes sense and answer to questions to the parts in bold?
My construction:
The construction can be given explicitly as an inverse stereographic projection. Consider the map
$s: \mathbb{R} \rightarrow S^1$ given by
$x \rightarrow (\frac{1-x^2}{1 + x^2},\frac{2x}{1 + x^2})$
Then $s$ is a homeomorphism between $\mathbb{R}$ and $S^1 \setminus \{(-1,0)\}$. So, since $S^1 \setminus \{(-1,0)\}$ is dense in $S^1$ and $S^1 \setminus\left(S^1 \setminus\{(-1,0)\}\right)$ consists of a single point, $S^1$ is the one-point compactification of $\mathbb{R}$.
Question 1: Why is s a homeomorphism? (How to prove that?) I take this as a known fact from math. analysis, but I am not sure.
Question 2: How do we know that $S^1 \setminus \{(-1,0)\}$ is dense in $S^1$?
|
Stereographic projection from $p = (-1,0)$ is given by
$$\sigma : S^1 \setminus \{p\} \to \mathbb R, \sigma(x,y) = \frac{y}{1+x}.$$
Your map $s$ has the property $p \notin s(\mathbb R)$ (because $s(x)$ has second coordinate $0$ only for $x = 0$, and $s(0) = (1,0)$). Thus we can regard $s$ as a map $s : \mathbb R \to S^1 \setminus \{p\}$. Noting $x^2 + y^2 = 1$ for $(x,y) \in S^1$ we get
$$s(\sigma(x,y)) = s\left(\frac{y}{1+x}\right) = \left(\frac{1- (\frac{y}{1+x})^2}{1+ (\frac{y}{1+x})^2}, \frac{2\frac{y}{1+x}}{1+ (\frac{y}{1+x})^2}\right) = \left(\frac{(1+x)^2 -y^2}{(1+x)^2 +y^2}, \frac{2(1+x)y}{(1+x)^2 +y^2}\right) \\=\left(\frac{(1+2x +x^2 -y^2}{1+2x +x^2 +y^2}, \frac{2(1+x)y}{1+2x +x^2 +y^2}\right) = \left(\frac{(2x +2x^2}{2+2x}, \frac{2(1+x)y}{2+2x}\right) =(x,y), $$
$$\sigma(s(x)) = \sigma\left(\frac{1-x^2}{1+x^2},\frac{2x}{1+x^2} \right) = \frac{\frac{2x}{1+x^2}}{1+\frac{1-x^2}{1+x^2}} = \frac{2x}{1+x^2 + 1 -x^2} = x .$$
This shows that $s$ and $\sigma$ are inverse to each other, i.e. both maps are homeomorphisms.
To see that $S^1 \setminus \{p\}$ is dense in $S^1$, observe that the points $p_n = \left(-\sqrt{1 -\frac{1}{n^2}},\frac 1 n \right)$ are in $ S^1 \setminus \{p\}$ and $p_n \to p$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4440057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Is there any closed form for $\frac{d^{2 n}( \cot z)}{d z^{2 n}}\big|_{z=\frac{\pi}{4}}$? Latest Edit
Thanks to Mr Ali Shadhar who gave a beautiful closed form of the derivative which finish the problem as
$$\boxed{S_n = \frac { \pi ^ { 2 n + 1 } } { 4 ^ { n + 1 } ( 2 n ) ! } | E _ { 2 n } | }, $$
where $E_{2n}$ is an even Euler Number.
In the post, I had found the sum
$$
\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{3}}= \frac{\pi^{3}}{32},
$$
and want to investigate it in a more general manner, $$
S_{n}=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{2 n+1}}
$$
where $n\in N.$
$$
\begin{aligned}
S_{n}&= \lim _{N \rightarrow \infty} \sum_{k=0}^{N} \frac{1}{(4k+1)^{2 n+1}}-\sum_{k=0}^{N} \frac{1}{(4 k+3)^{2 n+1}} \\
&=\frac{1}{4^{2 n+1}} \lim _{N \rightarrow \infty} \left[\sum_{k=0}^{N} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}-\sum_{k=0}^{N} \frac{1}{\left(k+\frac{3}{4}\right)^{2 n+1}}\right] \\
&=\frac{1}{4^{2 n+1}} \lim _{N \rightarrow \infty} \left[\sum_{k=0}^{N} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}+\sum_{k=0}^{N} \frac{1}{\left(-k-\frac{3}{4}\right)^{2 n+1}}\right] \\
&=\frac{1}{4^{2 n+1}} \lim _{N \rightarrow \infty} \left[\sum_{k=0}^{N} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}+\sum_{k=-N}^{-1} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}\right] \\
&=\frac{1}{4^{2 n+1}}\left[\lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{1}{\left(k+\frac{1}{4}\right)^{2 n+1}}\right]
\end{aligned}
$$
Using the Theorem:
$$(*):\pi \cot (\pi z)=\lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{1}{k+z} ,\quad \forall z \not \in Z.$$
Differentiating (*) w.r.t. $z$ by $2 n$ times yields
$$
\begin{aligned}
& \lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{(-1)^{2 n}(2 n) !}{(k+z)^{2 n+1}}=\frac{d^{2 n}}{d z^{2 n}}[\pi \cot (\pi z)] \\
\Rightarrow & \lim _{N \rightarrow \infty} \sum_{k=-N}^{N} \frac{1}{(k+z)^{2 n+1}}=\frac{\pi}{(2 n) !} \frac{d^{2 n}}{d z^{2 n}}[\cot (\pi z)]
\end{aligned}
$$
Now we can conclude that
$$\boxed{\sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2 k+1)^{2n+1}}=\left.\frac{\pi^{2n+1}}{4^{2 n+1}(2 n) !} \frac{d^{2 n}}{d z^{2 n}}[\cot z]\right|_{z=\frac{\pi}{4}}}$$
My Question:
Is there any closed form for $\displaystyle \left.\frac{d^{2 n} (\cot z)}{d z^{2 n}}\right|_{z=\frac{\pi}{4}}$?
|
$$\begin{align}\cot(x+\pi/4)&=\sec2x-\tan2x\\&=\sum_{n\ge0}\frac{(-1)^n}{(2n)!}E_{2n}(2x)^{2n}+\sum_{n\ge1}\frac{(-1)^n}{(2n)!}\cdot2^{2n}(2^{2n}-1)B_{2n}(2x)^{2n-1}\\&=1+\sum_{n\ge1}\frac{(-1)^n2^{2n}}{(2n)!}E_{2n}x^{2n}+\sum_{n\ge1}\frac{(-1)^n}{(2n)!}2^{4n-1}(2^{2n}-1)B_{2n}x^{2n-1}\\\implies\cot z&=1+\sum_{n\ge1}\frac{(-1)^n2^{2n}}{(2n)!}E_{2n}(z-\pi/4)^{2n}\\&+\sum_{n\ge1}\frac{(-1)^n}{(2n)!}2^{4n-1}(2^{2n}-1)B_{2n}(z-\pi/4)^{2n-1}\end{align}$$
You may accordingly extract the even and odd derivatives.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4440406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
How to factorize $2ab^2+2b^2c-ac^2-bc^2-a^2b-a^2c$? If $a,b,c$ are in AP $\implies \frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}$ are in AP $\implies
\frac{ab+bc+ca}{bc}, \frac{ab+bc+ca}{ca}, \frac{ab+bc+ca}{ab}$ are in AP $\implies \frac{bc+ca}{bc},\frac{ab+bc}{ca},\frac{bc+ca}{ab}$ are in AP $\implies \frac{bc}{bc+ca}, \frac{ca}{ab+bc}, \frac{ab}{bc+ca}$
are in HP.
If $P,Q,R$ are in HP $\implies Q=\frac{2PR}{P+R}.$ By this method proving that $\frac{bc}{bc+ca}, \frac{ca}{ab+bc}, \frac{ab}{bc+ca}$, if $a,b,c$ are in AP; is not straight forward.
Curiously, then we need to factorize $(2ab^2+2b^2c-ac^2-bc^2-a^2b-a^2c)$.
The question is: How do you factorize the last expression?
|
Given expression is $2(a+c)b^2-(a^2+c^2)b-ac(a+c)$
=$2(a+c)b^2-(a+c)^2b+2acb - ac(a+c)=(a+c)b\left[2b-(a+c) \right]+ac\left[2b-(a+c)\right]$
$=(ab+bc+ca)(2a-b-c)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4441291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Determining if a die is fair or not by rolling six times and observing only the sum Suppose you have two bags and each has 6 dice. In one bag all the die are fair. In the other each die has a bias - die “1” has a probability of returning 1 of $\frac{1}{6} +\epsilon$ and $\frac{1}{6} -\frac{\epsilon}{5}$ for any of the other numbers. Die 2 has a probability of returning 2 of $\frac{1}{6} +\epsilon$ and $\frac{1}{6} - \frac{\epsilon}{5}$ for any of the other numbers and so on. Someone chooses a bag at random and one of the die out of that bag at random. They roll the same die six times and give you the sum. What is the probability that the die came from the biased bag. You can denote the number of ways to get sum S from r rolls as $N^{S}_{r}$.
This is what I got so far:
From Bayes,
P(bag|sum) = P(sum|bag)P(bag) / P(sum) $\propto$ P(sum|bag)
For the fair bag P(sum|bag is fair) = $\frac{N^{S}_{r}}{6^{6}}$
For the unfair bag P(sum|bag is not fair) = $\frac{1}{6}$ P(sum|die biased to return 1) + $\frac{1}{6}$ P(sum|die biased to return 2) + ... + $\frac{1}{6}$ P(sum|die biased to return 6)
Thanks!
|
So let's look at $P(\mathrm{sum}=S \mid \mathrm{die\ biased\ toward\ } k)$.
Each roll of the biased die is equivalent to a process where we first choose between case $A$ with probability $\frac{6}{5} \epsilon$ and case $B$ with probability $1 - \frac{6}{5} \epsilon$. In case $A$, the result of the process is $k$. In case $B$, we'll roll a fair die and use that as the result. The net probability works out to the stated $\frac{1}{6} + \epsilon$ for result $k$ and $\frac{1}{6} - \frac{1}{5} \epsilon$ for any result other than $k$. (I came up with the probabilities for $A$ and $B$ by requiring $P(B) \frac{1}{6}$ must be the probability the result is a particular number other than $k$.)
Pretending the biased die value instead comes from this process, let $M$ be the number of times case $B$ happens, so case $A$ happens $6-M$ times. The probability that case $B$ happens $m$ times is
$$ P(M=m) = {6 \choose m} \left(\frac{6}{5} \epsilon\right)^{6-m} \left(1 - \frac{6}{5} \epsilon\right)^m $$
If case $B$ happens $m$ times, then we have $6-m$ results of exactly $k$, and $m$ results equally likely to be any number from $1$ to $6$. If all $6$ add up to $S$, then the case $B$ values must add up to $S - (6-m)k$. The number of ways this could happen is $N_m^{S-(6-m)k}$. ($N_r^s$ is zero for any impossible combinations of $r$ and $s$, or if $r$ or $s$ is negative. But $N_0^0 = 1$.) So
$$ P(\mathrm{sum} = S \mid (\mathrm{die\ biased\ toward\ } k) \land (M=m)) =
\frac{N_m^{S-(6-m)k}}{6^m} $$
$$ \begin{align*}
P((M=m) \land (\mathrm{sum} = S) \mid \mathrm{die\ biased\ toward\ } k) &= {6 \choose m} \left(\frac{6}{5} \epsilon\right)^{6-m} \left(1 - \frac{6}{5} \epsilon\right)^m \frac{N_m^{S-(6-m)k}}{6^m} \\
&= {6 \choose m} \left( \frac{6}{5} \epsilon \right)^{6-m} \left(\frac{1}{6} - \frac{\epsilon}{5} \right)^m N_m^{S-(6-m)k}
\end{align*} $$
Combine all possible values of $m$ to get:
$$ P(\mathrm{sum} = S \mid \mathrm{die\ biased\ toward\ } k) = \sum_{m=0}^6 {6 \choose m} \left( \frac{6}{5} \epsilon \right)^{6-m} \left(\frac{1}{6} - \frac{\epsilon}{5} \right)^m N_m^{S-(6-m)k} $$
Then as you already described, you can combine all possible values of $k$ to get:
$$ P(\mathrm{sum = S} \mid \mathrm{unfair\ bag}) =
\frac{1}{6} \sum_{k=1}^6 \sum_{m=0}^6 {6 \choose m} \left( \frac{6}{5} \epsilon \right)^{6-m} \left(\frac{1}{6} - \frac{\epsilon}{5} \right)^m N_m^{S-(6-m)k} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4441640",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
Question from isi previous years (a) Show that $\left(\begin{array}{l}n \\ k\end{array}\right)=\sum_{m=k}^{n}\left(\begin{array}{c}m-1 \\ k-1\end{array}\right)$.
(b) Prove that
$$
\left(\begin{array}{l}
n \\
1
\end{array}\right)-\frac{1}{2}\left(\begin{array}{l}
n \\
2
\end{array}\right)+\frac{1}{3}\left(\begin{array}{l}
n \\
3
\end{array}\right)-\cdots+(-1)^{n-1} \frac{1}{n}\left(\begin{array}{l}
n \\
n
\end{array}\right)=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}
$$
I was thinking mathematical induction for the second part that is,,,
Let
$$
\begin{aligned}
P(n):\left(\begin{array}{l}
n \\
1
\end{array}\right)-\left(\frac{1}{2}\right)\left(\begin{array}{l}
n \\
2
\end{array}\right) &+\frac{1}{3}\left(\begin{array}{c}
n \\
3
\end{array}\right) \cdots+(-1)^{n-1} \frac{1}{n}\left(\begin{array}{l}
n \\
n
\end{array}\right) \\
&=1+\frac{1}{2}+\cdots \frac{1}{n}
\end{aligned}
$$
$P(1):\left(\begin{array}{l}1 \\ 1\end{array}\right)=1$.
$P(1)$ true.
$$
\begin{aligned}
P(2):\left(\begin{array}{l}
2 \\
1
\end{array}\right)-\frac{1}{2}\left(\begin{array}{l}
2 \\
2
\end{array}\right) \\
=& 2-\frac{1}{2}=1+\frac{1}{2}
\end{aligned}
$$
$P(2)$ is also true.
Let $p(k)$ tque $\Rightarrow\left(\begin{array}{l}k \\ 1\end{array}\right)-\frac{1}{2}\left(\begin{array}{l}k \\ 2\end{array}\right)+\frac{1}{3}\left(\begin{array}{l}k \\ 3\end{array}\right) \cdots+(-1)^{k+\frac{1}{k}}$ Now we try to show $p(k+1)$ will be true $1+\frac{1}{2}+\cdots \frac{1}{k}$ $P(k+1)=\left(\begin{array}{c}k+1 \\ 1\end{array}\right)-\frac{1}{2}\left(\begin{array}{c}k+1 \\ 2\end{array}\right)+\cdots(-1)^{k} \frac{1}{k+1}\left(\begin{array}{l}k+1 \\ k+1\end{array}\right)$
I cannot argue from here,,, please help me for both part. Thank you.
|
Suppose we seek to evaluate
$$S_n = \sum_{k=1}^n \frac{(-1)^{k-1}}{k}{n\choose k}.$$
We introduce the function
$$f(z) = n! (-1)^{n-1} \frac{1}{z} \prod_{q=0}^n \frac{1}{z-q}.$$
We have for $1\le k\le n$ that
$$\mathrm{Res}_{z=k} f(z) =
n! (-1)^{n-1} \frac{1}{k}
\prod_{q=0}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^n \frac{1}{k-q}
\\ = n! (-1)^{n-1} \frac{1}{k}
\frac{1}{k!} \frac{(-1)^{n-k}}{(n-k)!}
= \frac{(-1)^{k-1}}{k}{n\choose k}.$$
It follows that the desired sum is given by
$$S_n = \sum_{k=1}^n \mathrm{Res}_{z=k} f(z).$$
Now residues sum to zero and the residue at infinity is zero by
inspection. Therefore the sum must be equal to
$$S_n = -\mathrm{Res}_{z=0} f(z)
= - n! (-1)^{n-1} \mathrm{Res}_{z=0} \frac{1}{z^2}
\prod_{q=1}^n \frac{1}{z-q}
\\ = n! (-1)^n
\left.\left(\prod_{q=1}^n \frac{1}{z-q}\right)'\right|_{z=0}
\\ = n! (-1)^n
\left. \prod_{q=1}^n \frac{1}{z-q}
\sum_{q=1}^n \frac{1}{q-z} \right|_{z=0}
= n! (-1)^n \frac{(-1)^n}{n!} H_n
\\ = H_n = 1+ \frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}.$$
This is the claim.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4443034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
What would be the Mean shortest distance from random points in the right angled triangle to the Hypotenuse. The problem is to find the average shortest distance of uniformly random points from the hypotenuse in a right angled rectangle.
The distance d shows the shortest distance to the hypotenuse from a random point N (x1,y1). I want to find the average (mean) shortest distance from N random points inside the triangle.
What I have in mind is to integrate the distance formula of point distance to the hypotenuse.
Let P be (0,0) => Q (a,0) and R (a,b)
Then slope of hypotenuse will be $m = \frac{b-0}{a-0} = \frac{b}{a}$
The equation of hypotenuse be $bx -ay = 0$.
The distance d would be $$d = \frac{|b*x_1 -a*y_1|}{\sqrt{a^2 +b^2}}$$
I thought integrating it over x and y for the given range would provide me with the mean shortest distance -
$$D_{mean} = \int_{0}^{a} \int_{0}^{b} d.dx.dy = \int_{0}^{a} \int_{0}^{b} \frac{|b*x_1-a*y_1|}{\sqrt{a^2+b^2}} dx.dy$$
I am stuck in solving it... (given that my approach is the correct one.)
|
Take $P$ to be the origin. Then the equation of the hypotenuse is
$ y = \dfrac{b}{a} x , \hspace{15pt} x \in [0, a] $
Now pick a point $ (x_1, y_1) $ such that $ y_1 \in [0, \dfrac{b}{a} x_1 ], then its perpendicular distance from the hypotenuse is
$d (x_1) = \dfrac{ \dfrac{b}{a} x_1 - y_1 } {\sqrt{1 + \left(\dfrac{b}{a}\right)^2} } = \dfrac{ b x_1 - a y_1 }{\sqrt{a^2 + b^2}}$
So the average distance is
$ \overline{d} = \displaystyle \dfrac{1}{A} \int_{x_1=0}^{a} \int_{y_1 = 0 }^{\dfrac{b}{a} x_1} \dfrac{ b x_1 - a y_1}{\sqrt{a^2 + b^2}} dy_1 dx_1 $
where $A = \displaystyle \int_{x_1=0}^{a} \int_{y_1 = 0 }^{\dfrac{b}{a} x_1} dy_1 dx_1 = \dfrac{1}{2} a b $
Integrating with respect to $y_1$,
$\overline{d} = \dfrac{2}{ab \sqrt{a^2 + b^2}} \displaystyle \int_{x_1=0}^{a} \dfrac{b^2}{2a} x_1^2 dx_1 = \left(\dfrac{b}{a^2 \sqrt{a^2 + b^2}}\right) \left(\dfrac{a^3}{3}\right) =\dfrac{ab}{3 \sqrt{a^2 + b^2}} $
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4443762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Determine all solutions of the linear equation system $A \vec{x} = \vec{b} = \begin{bmatrix} 2 \\ 4 \\ 18\\ -10 \end{bmatrix}$ Could you give me your feedback ?
Determine **all** solutions of the linear equation system $A \vec{x} = \vec{b}$ for
$$A = \begin {bmatrix}
1 & -2 & 4 & 3 & 2 \\
-7 & 14 & -28 & -12 & -23 \\
4 & -8 & 12 & 12 & 8 \\
-3 & 6 & -4 & -1 & -14
\end {bmatrix} , \vec{b} = \begin{bmatrix} 2 \\ 4 \\ 18\\ -10 \end{bmatrix} $$
Add 3 first row to fourth row, add -4 first row to third row,add 7 first row to second row. Then divide second row by 9, divide third row by -4, divide fourth row by 8. Then add -1 second row to fourth row, add -1 third row to fourth row [EDIT: and add -4 third row to first row]. Then add 2 second row to first row.
$$A = \begin {bmatrix}
1 & -2 & 4 & 3 & 2 \\
0 & 0 & 0 & 9 & -9 \\
0 & 0 & -4 & 0 & 0 \\
0 & 0 & 8 & 8 & -8
\end {bmatrix} = \begin {bmatrix}
1 & -2 & 4 & 3 & 2 \\
0 & 0 & 0 & 1 & -1 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 1 & 1 & -1
\end {bmatrix} = \begin {bmatrix}
1 & -2 & 0 & 3 & 2 \\
0 & 0 & 0 & 1 & -1 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end {bmatrix} = \begin {bmatrix}
1 & -2 & 0 & 5 & 0 \\
0 & 0 & 0 & 1 & -1 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 0 & 0
\end {bmatrix} $$
$\vec{b} = \begin{bmatrix} 2 \\ 18 \\ 10\\ -4 \end{bmatrix} = \begin{bmatrix} 2 \\ 2 \\ -5/2\\ -1/2 \end{bmatrix} = $[EDIT]$ \begin{bmatrix} 12 \\ 2 \\ -5/2\\ 0 \end{bmatrix} = \begin{bmatrix} 16 \\ 2 \\ -5/2\\ 0 \end{bmatrix} $
So we get $x_3 = -5/2, x_4 = x_5 + 2, $ [EDIT] $ x_1 = 16 + 2x_2 -5x_4 = 16 - 10 + 2x_2 - 5x_5 = 6 + 2x_2 -5x_5$ with $x_2, x_5$ free
[EDIT: forgot to add -4 third row to first row in step 3. Problem solved now]
Is this correct ? When I plug that into a calculator https://matrix.reshish.com/gaussSolution.php they get the same result except $x_1 = 6 + 2x_2 -5x_5$, is this something to worry about ?
Thanks for your help !
|
I took these steps
*
*$(01)~ R_2: R_2 + 7 R_1$
*$(02)~ R_3: R_3 - 4 R_1$
*$(03)~ R_4: R_4 + 3 R_1$
*$(04)~$ Swap $R_2$ with $R_4$
*$(05)~ R_3: R_3 + \dfrac{1}{2} R_2$
*$(06)~$ Swap $R_3$ with $R_4$
*$(07)~ R_4: R_4 -\dfrac{4}{9} R_3$
*$(08)~ R_3: \dfrac{1}{9} R_3$
*$(09)~ R_2: R_2 - 8 R_3$
*$(10)~ R_1: R_1 - 3 R_3$
*$(11)~ R_2: \dfrac{1}{8} R_2$
*$(12)~ R_1: R_1 - 4 R_2$
The final result is
$$\begin{pmatrix}
1 & -2 & 0 & 0 & 5 & 6 \\
0 & 0 & 1 & 0 & 0 & -\dfrac{5}{2} \\
0 & 0 & 0 & 1 & -1 & 2 \\
0 & 0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}$$
This means
$$x_4 = 2 + x_5, x_3 = -\dfrac{5}{2}, x_1 = 6 + 2 x_2 -5 x_5$$
You answer has an issue. You should get the same as the book.
You can test it by plugging it back in and seeing that it does not work.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4444826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Does this proof of the binomial expansion (a+b)^2 work? I was rereading Terence Tao's Analysis 1 and found this question in the section:
Exercise $2.3.4.$ Prove the identity $(a + b)^2 = a^2 + 2ab + b^2$ for all natural
numbers a, b.
Prior to this we already have proved:
$1.a\cdot b=b\cdot a\\2.a\cdot b=0\implies a=0\lor b=0\\3.a\cdot (b\cdot c)=(a\cdot b)\cdot c\\4. a\cdot (b+c)=a\cdot b+a\cdot c\\$
So I wrote this down:
$(x+y)^2\\=(x+y)(x+y)\\=x(x+y)+y(x+y)\\=x^2+xy+xy+y^2\\=x^2+2xy+y^2$
Is this a valid proof?
|
Here is the modified version of your proof with few skipped steps added and properties mentioned.
\begin{align}
(x+y)^2 &=(x+y).(x+y) & \\
&=(x(x+y))+(y(x+y))&\text{(using property 4 and 1)} \\
&=(x.x +x.y )+(y.x +y.y) &\text{(using property 4 again)}\\&=x^2+(x.y+y.x)+y^2 &\text{(using associativity of addition)}\\ &=x^2+(x.y+x.y)+y^2 &\text{(since $y.x=x.y$, property 1)}\\&=x^2+(1.xy+1.xy)+ y^2 &\text{(using property: $1.a=a$ for all natural numbers $a$)} \\
&=x^2 + (1+1). xy + y^2 &\text{(using property 4)}\\
&=x^2+2xy+y^2 &\text{(using 1+1=2)}\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4447307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Solve for x: $\lfloor x\rfloor \{\sqrt{x}\}=1$ where $\{x\}=x-\lfloor x\rfloor$ Solve for x:
$\lfloor x\rfloor \{\sqrt{x}\}=1$
where $\{x\}=x-\lfloor x\rfloor$
My Attempt:
I took intervals of $x$ as $x\in (2,3)$ so $\sqrt x\in (1,2)$. Due to which $\lfloor x\rfloor=2$ and $\{\sqrt x\}=\sqrt x-\lfloor \sqrt x\rfloor=\sqrt x-1$.
Substituting in the equation
$\lfloor x\rfloor \{\sqrt{x}\}=1$
I got
$2(\sqrt x-1)=1$
$x=\frac{9}{4}$
which satisfies the equation.
But when I apply the same process to $x\in (3,4)$ I end up getting value of $x$ that does not satisfy the given equation. But when I plotted the graph a solution clearly exists between two consecutive integers.
What am I missing
|
We can proceed similar as you did. Let $n=\lfloor x\rfloor$. From the assumption $\lfloor x\rfloor\{\sqrt x\}=1$ one concludes $\{\sqrt x\}=\frac1n$. Since $\sqrt x$ changes its integral part at integral values, we can conclude that $\lfloor\sqrt n\rfloor=\lfloor\sqrt x\rfloor$, so that
$$\sqrt x=\lfloor\sqrt x\rfloor+\{\sqrt x\}=\lfloor\sqrt n\rfloor+\frac1n\Longrightarrow x=\lfloor\sqrt n\rfloor^2+\frac{2\lfloor\sqrt n\rfloor}{n}+\frac1{n^2}.$$
Now we have to check under which conditions this value for $x$ indeed satisfies the assumption $\lfloor x\rfloor=n$. For this, let $n=k^2+m$ with $k,m\in\mathbb N_0$ and $k$ maximal, i.e. $\lfloor\sqrt n\rfloor=k$. Now if $n\geq5$, then
$$x=k^2+\frac{2k}{n}+\frac1{n^2}=k^2+\frac{2kn+1}{n^2}<k^2+1,$$
so in this case, we must have $m=0$. Then we have $n=k^2<x<k^2+1$, so indeed $\lfloor x\rfloor=n$, hence we found solutions. The cases $n=0,\ldots,4$ are easily checked, and $\frac94$, which you found, is the only solution in these cases.
To summarize, the solutions are $\frac94$ and $x=\lfloor\sqrt n\rfloor^2+\frac{2\lfloor\sqrt n\rfloor}{n}+\frac1{n^2}$ with $n$ a perfect square other than $1$ or $4$, or letting $n=k^2$, $$x=\frac94\quad\text{and}\quad x=k^2+\frac2k+\frac1{k^4},\quad k\in\mathbb N,\ k>2.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4450325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Calculate: $\Delta=\left|\begin{array}{ccc} b c & c a & a b \\ a(b+c) & b(c+a) & c(a+b) \\ a^{2} & b^{2} & c^{2} \end{array}\right|$ Calculate:
$$\Delta=\left|\begin{array}{ccc}
b c & c a & a b \\
a(b+c) & b(c+a) & c(a+b) \\
a^{2} & b^{2} & c^{2}
\end{array}\right|$$
Does anyone know any easy way to calculate this determinant?
I tried the classic way but I was wondering if anyone knows an easier method.
|
\begin{gathered}
\left|\begin{array}{ccc}
b c & a c & a b \\
a b+a c & a b+b c & a c+b c \\
a^{2} & b^{2} & c^{2}
\end{array}\right|=b c \cdot(a b+b c) \cdot c^{2}+a c \cdot(a c+b c) \cdot a^{2}+a b \cdot(a b+a c) \cdot b^{2}-a^{2} \cdot(a b+b c) \cdot(a b)-b^{2} \cdot(a c+b c) \cdot(b c)-c^{2} \cdot(a b+a c) \cdot(a c) \\
\equiv \\
=a^{2} b^{4}-a^{4} b^{2}-a^{2} c^{4}+b^{2} c^{4}+a b^{2} c^{3}-a^{2} b c^{3}+a^{4} c^{2}-b^{4} c^{2}-a b^{3} c^{2}+a^{3} b c^{2}+a^{2} b^{3} c-a^{3} b^{2} c
\end{gathered}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4450509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
Solving $\tan ^{-1}(\frac{1-x}{1+x})=\frac{1}{2}\tan ^{-1}(x)$ I was solving the following equation,$$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\tan ^{-1}\left(x\right)$$
But I missed a solution (don't know where's the mistake in my work).
Here's my work:
$$\tan ^{-1}\left(\frac{1-x}{1+x}\right)=\frac{1}{2}\tan ^{-1}\left(x\right)$$
Putting $x = \tan(\theta)$
$$\begin{align} \tan ^{-1}\left(\frac{1-\tan\theta}{1+\tan\theta}\right)&=\frac{1}{2}\tan ^{-1}\left(\tan\theta\right)\\
\tan ^{-1}\left(\tan\left(\frac\pi 4 - \theta\right)\right)&=\frac{1}{2}\theta\\
\frac\pi 4 - \theta&=\frac{1}{2}\theta\\
\frac\pi 4&=\frac{1}{2}\theta + \theta\\
\frac\pi 4&=\frac{3}{2}\theta\\
\frac\pi 6&=\theta\\
\frac\pi 6&=\tan^{-1}(x)\\
\tan\frac\pi 6&=(x)\\
\frac1{\sqrt{3}}&=x\end{align}$$
I got only one solution, but the answer in my textbook is $\pm\frac{1}{\sqrt{3}}$. Where is the mistake?
|
Thank you to @Robin'sPremiumCoffee for a nice algebraic solution. Now the problem with your original method is that although:
$$\frac{1-\tan \theta}{1+\tan \theta} = \tan(\frac \pi 4 -\theta)$$
It is also equal to:
$$\frac{1-\tan \theta}{1+\tan \theta} = \tan(\frac \pi 4 -\theta)=\tan(\pi +\frac \pi 4 -\theta)$$
This is because $\tan(\theta)$ is $\pi$ periodic. From there we may continue your steps to get:
\begin{align}
\pi + \frac\pi 4 - \theta&=\frac{1}{2}\theta\\
\frac{5\pi} 4&=\frac{1}{2}\theta + \theta\\
\frac{5\pi} 4&=\frac{3}{2}\theta\\
\frac{5\pi} 6&=\theta\\
\frac{5\pi}6&=\tan^{-1}(x)\\
\tan\frac{5\pi} 6&=(x)\\
-\frac1{\sqrt{3}}&=x\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4450671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
How to find $\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}$? By factorization:
$$\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}\tag{1}$$
$$=\lim_{x\to-\infty} \frac{x\sqrt{1+\frac{2}{x}}}{-x}$$
$$=\lim_{x\to-\infty}-\sqrt{1+\frac{2}{x}}$$
If I input $x=-\infty$, the limiting value seems to be $-1$. But according to desmos, the limiting value should be $1$.
By L'Hopital's rule:
$$\lim_{x\to-\infty} \frac{\sqrt{x^2+2x}}{-x}$$
$$=\lim_{x\to-\infty} \frac{\dfrac{x+1}{\sqrt{x^2+2x}}}{-1}$$
$$=-\lim_{x\to-\infty} \dfrac{x+1}{\sqrt{x^2+2x}}$$
$$=-\lim_{x\to-\infty} \frac{1}{\dfrac{x+1}{\sqrt{x^2+2x}}}$$
$$=-\lim_{x\to-\infty} \dfrac{\sqrt{x^2+2x}}{x+1}$$
$$=-\lim_{x\to-\infty} \frac{\dfrac{x+1}{\sqrt{x^2+2x}}}{1}$$
$$=-\lim_{x\to-\infty} \dfrac{x+1}{\sqrt{x^2+2x}}$$
I can't get a determinate form.
My questions:
*
*How do I find $(1)$ using factorization?
*How do I find $(1)$ using L'Hopital's rule?
Related
|
\begin{eqnarray}
\frac{\sqrt{x^2+2x}}{-x}&=&\frac{\sqrt{x^2}\sqrt{1+\frac{2}{x}}}{-x}\\
&=&\frac{|x|\sqrt{1+\frac{2}{x}}}{-x}\\
&=&\frac{-x\sqrt{1+\frac{2}{x}}}{-x} \text{ since }x<0\\
&=&\sqrt{1+\frac{2}{x}}
\end{eqnarray}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4451437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
How many different whole numbers are factors of number $2 \times 3 \times 5 \times 7 \times 11 \times 13$? The question is:
How many different whole numbers are factors of number $2 \times 3 \times 5 \times 7 \times 11 \times 13$?
My answer to this question is $63$ but the right answer is $64$. I don't know why it is $64$? I need some assistance.
|
You can obtain the solution as well by omitting the factor 1. One unique whole number is if you multiply all 6 primes. The corresponding sequence is $2,3,5,7,11,13$. So we have one sequence. The corresponding binomial coefficient is $\binom{6}{0}=1$
Then we can remove one of the 6 factors. This can be done with all 6 prime factors. The number of sequences is $\binom{6}{1}=5$.
Next we remove two factors. The resulting sequences are: $ 5,7,11,13\quad 3,7,11,13\quad 3,5,11,13\quad 3,5,7,13\quad 3,5,7,11$
$2,7,11,13\quad 2,5,11,13\quad 2,5,7,13\quad 2,5,7,11\quad 2,3,11,13$
$ 2,3,7,13\quad 2,3,7,11\quad 2,3,5,13\quad 2,3,9,11\quad 2,3,7,9$
Basically we two kinds of elements: The removed elements (x) and the remaining (a) elements. Then the number of such sequences is $\binom{6}{2}=15$.
We can go on like this by removing $3, 4$ and $5$ factors. Then the sum is
$$\sum_{i=1}^6 \binom{6}{i}=2^6-1=63$$
The result is easy to evaluate due the binomial theorem, see here.
One point of view is that $i=6$ is not possible since we would remove all prime factors. On the other hand they don't talk about prime factors, but factors only. Then the number 1 is valid and we obtain $63+1=64$ different numbers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4453808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Find $\sum_{k=1}^\infty\frac{1}{x_k^2-1}$ where $x_1=2$ and $x_{n+1}=\frac{x_n+1+\sqrt{x_n^2+2x_n+5}}{2}$ for $n \ge 2$ Given $x_1=2$
and $x_{n+1}=\frac{x_n+1+\sqrt{x_n^2+2x_n+5}}{2}, n\geq 2$
Prove that $y_n=\sum_{k=1}^{n}\frac{1}{x_k^2-1}, n\geq 1$ converges and find its limit.
*
*To prove a convergence we can just estimate $x_n > n$, therefore $y_n<z_n$, where $z_n=\sum_{k=1}^{n}\frac{1}{k^2-1}$ and $z_n$ converges, then $y_n$ converges too.
*We can notice that $x_n^2+2x_n+5=(x_n+1)^2+4$. So $x_{n+1}$ is one of the roots of the equation:
$x_{n+1}^2-(x_n+1)x_{n+1}-1=0$
So $x_{n+1}^2-1=(x_n+1)x_{n+1}$ and therefore: $y_n=\sum_{k=1}^n \frac{1}{(x_{n-1}+1)x_{n}}$
I'm stuck here.
|
Making use of the following equality, and standard partial fraction decompositions: $$x_{k+1}^2-(x_k+1)x_{k+1}-1=0\implies(x_k+1)=\frac{x_{k+1}^2-1}{x_{k+1}}$$
$$\begin{align}y_n&=\frac{1}{2}\sum_{k=1}^n\left(\frac{1}{x_k-1}-\frac{1}{x_k+1}\right)\\&=\frac{1}{2}\sum_{k=1}^n\left(\frac{1}{x_k-1}-\frac{x_{k+1}}{x_{k+1}^2-1}\right)\\&=\frac{1}{2}\sum_{k=1}^n\left(\frac{1}{x_k-1}-\frac{1}{2}\left(\frac{1}{x_{k+1}-1}+\frac{1}{x_{k+1}+1}\right)\right)\\&=\frac{1}{4}+\frac{1}{12}-\frac{1}{4}\frac{1}{x_{n+1}-1}-\frac{1}{4}\frac{1}{x_{n+1}+1}+\frac{1}{4}\sum_{k=1}^n\left(\frac{1}{x_k-1}-\frac{1}{x_k+1}\right)\\&=\frac{1}{3}-\frac{1}{4}\frac{1}{x_{n+1}-1}-\frac{1}{4}\frac{1}{x_{n+1}+1}+\frac{1}{2}y_n\\\implies y_n&=\frac{2}{3}-\frac{1}{2}\frac{1}{x_{n+1}-1}-\frac{1}{2}\frac{1}{x_{n+1}-1}\\\implies y_n&\to\frac{2}{3},\quad n\to\infty\end{align}$$
Convergence is slow, but my Python script indeed evaluates this as $0.66665666$ at $n=100,000$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4455721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
If $2\arccos(\frac45)-\arcsin(\frac45)=\arctan(y)$ then find the value of $y$ If $2\arccos(\frac45)-\arcsin(\frac45)=\arctan(y)$ then find the value of $y$
My Attempt:
Using $2\arccos(x)=\arccos(2x^2-1)$, I get
$$2\arccos(\frac45)=\arccos(2\times\frac{16}{25}-1)=\arccos(\frac7{25})$$
Also,
$$\arcsin(\frac45)=\arccos(\frac35)$$
Now, using $\arccos(x)-\arccos(y)=\arccos(xy+\sqrt{1-x^2}\sqrt{1-y^2})$, I get
$$\arccos(\frac7{25})-\arccos(\frac35)=\arccos(\frac{21}{125}+\frac{24}{25}\cdot\frac45)=\arccos(\frac{217}{125})$$
But $\arccos$ input can't be greater than $1$
What am I doing wrong?
|
Since $$7^2 + 24^2 = 25^2$$ and $$3^2 + 4^2 = 5^2,$$ it follows that
$$\frac{24}{25} = \sqrt{ 1 - \frac{7^2}{25^2}},$$ and $$\frac{4}{5} = \sqrt{1 - \frac{3^2}{5^2}}.$$
Hence
$$\frac{7}{25} \cdot \frac{3}{5} + \sqrt{ 1 - \frac{7^2}{25^2}} \sqrt{1 - \frac{3^2}{5^2}} = \frac{21}{125} + \frac{96}{125} = \frac{117}{125}.$$
In particular, $$21 + 24 \cdot 4 = 21 + 96 = 117.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4459661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Show $a^3+b^3+c^3+d^3 \le 27$ Given $a, b, c, d$ are real numbers and $a^2+b^2+c^2+d^2 = 9$ show that
$$a^3+b^3+c^3+d^3\le27$$
I have tried to do the following:
$$(a^2+b^2+c^2+d^2)^{\frac{3}{2}} = 27 \\ \text{after simplifying} \\ (a^3+b^3+c^3+d^3) + \frac{3}{2}(a^2b^2+(a^2+b^2)(c^2+d^2)+c^2d^2)=27$$
By setting the following:
$$(a^3+b^3+c^3+d^3) + \frac{3}{2}(a^2b^2+(a^2+b^2)(c^2+d^2)+c^2d^2)=(a^3+b^3+c^3+d^3) \\ \frac{3}{2}(a^2b^2+(a^2+b^2)(c^2+d^2)+c^2d^2)=0 \\
(a^2+b^2)(c^2+d^2)=-(a^2b^2+c^2d^2) \\ (a^3+b^3+c^3+d^3)+\frac{3}{2}(a^2b^2+-(a^2b^2+c^2d^2)+c^2d^2)=27 \\ (a^3+b^3+c^3+d^3)=27$$
and so we have shown the inequality to hold - does this proof work?
|
Sorry didn't really read your solution, but this might work:
Clearly $|a|,|b|,|c|,|d|\leq 3$ and thus $a^3\leq 3a^2$ and so on... and we are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4460740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
what is the solution of the ordinary differential equation? would have any complete solution to the equation plus complete below?
$$
y'= \frac{4y-{3x}}{2x-y}
$$
a part of the solution:$$
y'=xv'+v
$$
$$
y=xv
$$
not complete solution
$$
xv'+v= \frac{4xv-{3x}}{2x-xv}
$$
Continuation
$$
xv'+v= \frac{2v-{3}}{1-v}
$$
Continuation
$$
xv'= \frac{2v-3-v+{v^2}}{1-v}
$$
Continuation
$$
x\frac{dy}{dx}= \frac{2v-3-v+{v^2}}{1-v}
$$
Continuation
$$
xv'= \frac{1-v}{v^2-v-3}
$$
Answer $$|y-x|=C|y+x|^3$$
|
That's a homogeneous differential equation. I'd solve it like this:
divide by x $$y' = \frac{4\left(\frac{y}{x}\right) - 3}{2 - \left(\frac{y}{x}\right)}$$
Set $v = \frac{y}{x}$, so $y = xv$ and $y' = v + xv'$
Substitute: $$v + xv' = \frac{4v - 3}{2 - v}$$
$$xv' = \frac{4v - 3}{2 - v} - v$$
$$xv' = \frac{4v - 3 - 2v + v^2}{2 - v}$$
$$xv' = \frac{v^2 + 2v - 3}{2 - v}$$
$$xv' = \frac{(v + 3)(v - 1)}{-(v - 2)}$$
$$v' = \frac{(v + 3)(v - 1)}{v - 2}\left(-\frac{1}{x}\right)$$
Solve by separation of variables, where $B(v) = \frac{(v + 3)(v - 1)}{v - 2}$ and $A(x) = \left(-\frac{1}{x}\right)$. First $B(v) = 0$:
$$(v + 3)(v - 1) = 0$$
Solutions are $v = -3$ and $v = 1$, so $y(x) = -3x$ and $y(x) = x$.
Secondly, you solve $$\int_{}^{}\frac{v - 2}{(v + 3)(v - 1)} \,\mathrm dv = \int_{}^{}\left(-\frac{1}{x}\right) \,\mathrm dx$$
The one on the right is $-ln|x| + C$. The other can be solved using residues: $$\int_{}^{}\left(\frac{A}{v + 3} + \frac{B}{v - 1}\right) \,\mathrm dv$$
$$A = \lim\limits_{v\rightarrow -3} \frac{v - 2}{v - 1} = \frac{5}{4}$$
$$B = \lim\limits_{v\rightarrow 1} \frac{v - 2}{v + 3} = -\frac{1}{4}$$
You put these values and find that the integral is $\frac{5}{4}ln|v + 3| -\frac{1}{4}ln|v - 1| + C$
so: $$\frac{5}{4}ln|v + 3| -\frac{1}{4}ln|v - 1| = -ln|x|+ C /*4$$
$$5ln|v + 3| -ln|v - 1| = -ln|x^4|+ C$$
You go on and find $$\frac{(v + 3)^5}{v - 1} = \frac{C}{x^4}$$ with C $\neq$ 0.
Replace v with $\frac{y}{x}$ and finally find $$(y + 3x)^5 = C(y - x)$$
If C = 0, we find that $y = -3x$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4461580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Determine the image of the unit circle $S^1$ by the action of the matrix $e^A$. We have:
$$e^{ \begin{pmatrix}
-5 & 9\\
-4 & 7
\end{pmatrix} }$$
I need to determine the image of the unit circle $S^1$ by the action of the matrix $e^A$.
I think that I know how to calculate $e^A$:
I get the Jordan decomposition:
$$A = \begin{pmatrix}
-5 & 9\\
-4 & 7
\end{pmatrix} =
\begin{pmatrix}
-6 & 1\\
-4 & 0
\end{pmatrix}
\cdot
\begin{pmatrix}
1 & 1\\
0 & 1
\end{pmatrix}
\cdot
\frac{1}{4}
\begin{pmatrix}
0 & -1\\
1 & -6
\end{pmatrix}
$$
With eigenvalues: $\lambda$ = 1, algebraic multiplicity = 2, eigenvecotrs: $\left\{ \begin{pmatrix}
1\\
0
\end{pmatrix}, \begin{pmatrix}
0\\
1
\end{pmatrix} \right\}$
$$ \displaystyle e^A = \sum^{\infty}_{i = 0} \frac{A^i}{i!}$$
$$e^A =
\begin{pmatrix}
-6 & 1\\
-4 & 0
\end{pmatrix}
\cdot
\left(
\begin{pmatrix}
1 & 0\\
0 & 1
\end{pmatrix}
+
\displaystyle \sum^{\infty}_{i = 1} \frac{ \begin{pmatrix}
1 & 1\\
0 & 1
\end{pmatrix}}{i!}
\right)
\cdot
\frac{1}{4}
\begin{pmatrix}
0 & -1\\
1 & -6
\end{pmatrix}$$
$$e^A = \begin{pmatrix}
-5 & 9\\
-4 & 7
\end{pmatrix} =
\begin{pmatrix}
-6 & 1\\
-4 & 0
\end{pmatrix}
\cdot
\begin{pmatrix}
\displaystyle \sum^{\infty}_{i = 1} \frac{1}{i!}& \displaystyle \sum^{\infty}_{i = 1} \frac{2^{i-1}}{i!}\\
0 & \displaystyle \sum^{\infty}_{i = 1} \frac{1}{i!}
\end{pmatrix}
\cdot
\frac{1}{4}
\begin{pmatrix}
0 & -1\\
1 & -6
\end{pmatrix}$$
Where: $$\displaystyle \sum^{\infty}_{i = 1} \frac{2^{i-1}}{i!} = \frac{1}{2} \sum^{\infty}_{i = 1} \frac{2^{i}}{i!} = \frac{1}{2}(e^2 - 1)
$$
So:
$$e^A =
\begin{pmatrix}
-6 & 1\\
-4 & 0
\end{pmatrix}
\cdot
\begin{pmatrix}
e & \displaystyle \frac{e^2}{2} - \displaystyle \frac{1}{2}\\
0 & e
\end{pmatrix}
\cdot
\frac{1}{4}
\begin{pmatrix}
0 & -1\\
1 & -6
\end{pmatrix} =
\begin{pmatrix}
\displaystyle \frac{-3e^2 + e + 3}{4} & \displaystyle \frac{9e^2 - 9}{2}\\
\displaystyle \frac{-e^2 + 1}{2} & 3e^2 + e - 3
\end{pmatrix}
$$
Now, I don't know if I did it correctly up to this point and what I should do next - to operate on my unit circle.
Solution:
Because of @Oscar Lanzi we know that:
$$e^{\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}}=e\begin{pmatrix}-5 & 9\\-4 & 7\end{pmatrix}$$
Then because of that:
Equation of unit circle under linear transformation - can't understand role of inverse matrix
(answer by @Prototank)
We know that the image of unit circle in action of the matrix $A$ is given by: $$65x^{2}-166xy+106y^{2}=1$$
Now we need to scale by $e$ and we get the image of unit circle in action of the matrix $e^A$:
$$65x^{2}-166xy+106y^{2}=e^2$$
|
This address the general question of shape of image of an unit circle under the action of $e^A$. We can think of a linear dynamic system $\dot x=Ax$, the solution of which will be $x=e^{tA}$. Then what you are seeking is the ending position of points starting from the unit circle, after following linear dynamics for $t=1$.
This animation has credit to Ella Batty. I TAed her class last semester and used this as a demo for 2d dynamic systems.
https://twitter.com/i/status/1319061743679799297
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4467835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
To prove an Inequality: $ ( x^2 +2x)e^x + (x^2-2 x)e^{-x} \ge 0$
$ \left(x^2 +2x\right)e^x + \left(x^2-2 x\right)e^{-x} \ge 0$.
I used photomath to plot its graph: $y=(x^{2}+2x))e^{x} + \frac{{x}^{2}-2x}{{e}^{x}}$
But how do I prove it without an image? Should I take the derivative of it and reason, please tell me the solution.
|
HINT:
$f(x) =(x^2 + 2 x) e^x$ has a Taylor series with all coefficients positive. So $f(x) + f(-x)$ will have a Taylor series with all coefficients $\ge 0$.
$\bf{Note}$: the function $g(x) = x e^x$ has a positive Taylor expansion at $0$, so $g(x) + g(-x) \ge 0$. We get stronger inequalities
$(x^2 +a x) e^x + (x^2 - a x) e^{-x}\ge 0$
by decreasing $a$. Now the Taylor expansion of the above is:
$$(x^2 + a x) e^x + (x^2 - a x) e^{-x}= 2(1+a) x^2 + \frac{3+a}{3}x^4 + \cdots + \frac{2(2n-1+ a)}{ (2n-1)!} x^{2n} + \cdots$$
For $a=-1$ we get
$$(x^2 - x)e^x + (x^2 + x) e^{-x} \ge 0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4472916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
}
|
Is it possible to expand $\arctan x \ln(1-x^2)$ in series? I am not sure if its possible to find the Taylor series of $\arctan x \ln(1-x^2)$ so I am giving it a try:
By using the Cauchy product
$$\left(\sum_{n=1}^\infty a_n x^n\right)\left(\sum_{n=1}^\infty b_n x^n\right)=\sum_{n=1}^\infty \left(\sum_{k=1}^n a_k b_{n-k+1}\right)x^{n+1}$$
we have
$$\arctan x \ln(1-x^2)=\left(\sum_{n=1}^\infty \frac{(-1)^{n-1} x^{2n-1}}{2n-1}\right)\left(-\sum_{n=1}^\infty\frac{x^{2n}}{n}\right)$$
$$=\frac1x\left(\sum_{n=1}^\infty \underbrace{\frac{(-1)^{n}}{2n-1}}_{a_n}(x^2)^n\right)\left(\sum_{n=1}^\infty\underbrace{\frac{1}{n}}_{b_n}(x^2)^n\right)$$
$$=\frac1x\sum_{n=1}^\infty\left(\sum_{k=1}^n \frac{(-1)^k}{2k-1}\cdot\frac{1}{n-k+1}\right)(x^2)^{n+1}$$
$$\left\{\text{write $\frac{1}{(2k-1)(n-k+1)}=\frac{1}{2n+1}\left(\frac{2}{2k-1}+\frac{1}{n-k+1}\right)$}\right\}$$
$$=\sum_{n=1}^\infty \frac{2}{2n+1}\left(\sum_{k=1}^n \frac{(-1)^k}{2k-1}\right)x^{2n+1}+\sum_{n=1}^\infty \frac{1}{2n+1}\left(\sum_{k=1}^n \frac{(-1)^k}{n-k+1}\right)x^{2n+1}$$
$$\left\{\text{reverse the order of the terms in the second sum}\right\}$$
$$=\sum_{n=1}^\infty \frac{2}{2n+1}\left(\sum_{k=1}^n \frac{(-1)^k}{2k-1}\right)x^{2n+1}+\sum_{n=1}^\infty \frac{(-1)^{n+1}}{2n+1}\left(\sum_{k=1}^n \frac{(-1)^k}{k}\right)x^{2n+1}$$
$$=\sum_{n=1}^\infty\frac{2}{2n+1}\left(\sum_{k=1}^n\frac{(-1)^k}{2k-1}\right)x^{2n+1}+\sum_{n=1}^\infty \frac{(-1)^{n}\overline{H}_n}{2n+1}x^{2n+1}.$$
Mathematica gives
$$\sum_{k=1}^n\frac{(-1)^k}{2k-1}=\int_0^1\frac{(-x^2)^n-1}{1+x^2}dx=-\frac{\pi}{4}+\frac{(-1)^n}{4}\left(\psi\left(\frac{2n+3}{4}\right)-\psi\left(\frac{2n+1}{4}\right)\right)$$
but this form makes the problem harder I guess. Any idea how to find a better form for $\sum_{k=1}^n\frac{(-1)^n}{2k-1}$ or how to find the whole $\sum_{n=1}^\infty\frac{2}{2n+1}\left(\sum_{k=1}^n\frac{(-1)^k}{2k-1}\right)x^{2n+1}$ or maybe a different way to find the Taylor series of $\arctan x \ln(1-x^2)$?
|
A bit tricky but it makes things simpler.
Rewrite
$$\tan ^{-1}(x)\log \left(1-x^2\right) = x \frac{\tan ^{-1}(x)}{x}\log \left(1-x^2\right)$$ and let $x=\sqrt t$
$$\frac{\tan ^{-1}(\sqrt t)}{\sqrt t}=\sum_{n=0}^\infty \frac{(-1)^n }{2 n+1}t^n$$
$$\log \left(1-t\right)=-\sum_{n=1}^\infty \frac {t^n} n$$
Where is see a problem is that the coefficients of the product
$$\frac{\tan ^{-1}(\sqrt t)}{\sqrt t}\,\log \left(1-t\right)=-\sum_{n=1}^\infty a_n\,t^n$$ being
$$\left\{1,\frac{1}{6},\frac{11}{30},\frac{121}{1260},\frac{281}{1260},\frac{929}{1
3860},\frac{432683}{2702700},\frac{18527}{360360},\frac{84971}{680680},\frac{203
75981}{488864376},\cdots\right\}$$ nothing has been identified by $OEIS$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4473283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Study stability of a linear time-invariant system by its 5x5 A matrix I need to study the Lyapunov stability of these matrices.
Is there a criterion that does not need to calculate the eigenvalues?
For example if the trace of a matrix is > 0 then it is unstable (like the third one).
$$
\begin{bmatrix}
-1 & 2 & 0 & -5 & 3 \\
0 & -1 & 1 & 6 & -4 \\
0 & 0 & 0 & 1 & -1 \\
0 & 0 & 0 & 0 & 2 \\
0 & 0 & 0 & -1 & -2 \\
\end{bmatrix}
$$
$$
\begin{bmatrix}
1 & 2 & 0 & 0 & 0 \\
-5 & -8 & 0 & 0 & 0 \\
-1 & 3 & -12 & 1 & 1 \\
1 & -3 & 0 & -3 & 0 \\
-11 & 12 & 0 & 3 & -2 \\
\end{bmatrix}
$$
$$
\begin{bmatrix}
-1 & 0 & 0 & 0 & 0 \\
-5 & -1 & 0 & 0 & 0 \\
-1 & 3 & -0.8 & 0 & 0 \\
1 & -3 & 0 & 3 & -2.1 \\
-11 & 12 & 0 & 3 & -2 \\
\end{bmatrix}
$$
$$
\begin{bmatrix}
-1 & 0 & 0 & 0 & 0 \\
-5 & -2 & 0 & 0 & 0 \\
-1 & 3 & -0.8 & 0 & 0 \\
1 & -3 & 0 & 3 & -2.1 \\
-11 & 12 & 0 & 3 & -2 \\
\end{bmatrix}
$$
|
Note that these matrices are block triangular. Thus, each matrix is stable iff its blocks on the diagonal are each (separately) stable.
For example, the second matrix can be partitioned as
$$
A =
\left[\begin{array}{cc|ccc}
1 & 2 & 0 & 0 & 0 \\
-5 & -8 & 0 & 0 & 0 \\
\hline
-1 & 3 & -12 & 1 & 1 \\
1 & -3 & 0 & -3 & 0 \\
-11 & 12 & 0 & 3 & -2 \\
\end{array}
\right]
$$
Thus, $A$ is stable iff each of the matrices
$$
A_1 = \pmatrix{1&2\\-5&-8}, \quad A_2 = \pmatrix{-12&1&1\\0&-3&0\\0&3&-2}
$$
is stable. From there, we can see that $A_2$ is itself block triangular, since it can be partitioned as
$$
A_2 = \left[\begin{array}{c|cc}-12&1&1\\ \hline 0&-3&0\\0&3&-2 \end{array}\right].
$$
Thus, $A_2$ is stable iff the $1\times1$ matrix $-12$ and the $2\times2$ matrix $\pmatrix{-3&0\\3&-2}$ are stable, and it is easy to see that they are (the $2 \times 2$ matrix is upper triangular, so its diagonal entries are its eigenvalues). All that remains is to decide whether $A_1$ is stable. We can confirm that $A_1$ is stable either by computing its eigenvalues or by finding its characteristic polynomial and noting that all coefficients are positive (this trick is necessary and sufficient for $2 \times 2$ matrices; larger matrices would require the Routh-Hurwitz criterion).
Thus, the matrix $A$ is stable.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4473841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Find the minimum value of a trigonometric function If the minimum value of $f\left(x\right)=\left(1+\frac{1}{\sin ^6\left(x\right)}\right)\left(1+\frac{1}{\cos ^6\left(x\right)}\right),\:x\:∈\:\left(0,\:\frac{\pi }{2}\right)$ is $m$, find $\sqrt m$.
How do I differentiate this function without making the problem unnecessarily complicated? If there are any other methods to finding the minimum value I am open to those too.
|
Let $s\equiv\sin x$ and $c\equiv\cos x$ then
$$
\begin{align}
f(x)
&=
\left(1+{1\over s^6}\right)
\left(1+{1 \over c^6}\right)
\\
&=
{s^6+1\over s^6}{c^6+1 \over c^6}
\\
&=
{(sc)^6+s^6+c^6+1\over (sc)^6}
\\
\end{align}
$$
Now $s^6+c^6=(s^2+c^2)^3-3(sc)^2(s^2+c^2)=1-3(sc)^2$
so
$$
f(x)=1+{2\over (sc)^6}-{3\over (sc)^4}\\
=1+{2\times2^6\over \sin^6 2x}
-{3\times 2^4\over\sin^42x}
$$
and
$$
f'(x)=\cos 2x\times(96/\sin^5 2x-384/\sin^72x)
$$
Since $-1\leq\sin \leq 1$ there are no solutions to $96\sin^2 2x-384=0$, the stationary point must be at $\cos 2x=0$, hence $x=\pi/4$. We can easily see it's a minimum because $f(x)$ consists of positive powers of $1/\sin 2x$,
$$
f(x)=1+{32-24\sin^2 2x\over\sin^7 2x}
$$
and
$\sin^7\leq\sin^2$, and $24\sin^2<32$, and $\sin 2x$ is a maximum at this point.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4474533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
}
|
How to solve $\sin(x) = \pm a$ for $a \not = 0$? I was solving the below equation: $\left|\sqrt{2\sin^2x + 18 \cos^2x} - \sqrt{2\cos^2x + 18 \sin^2x} \right| = 1$ for $x \in [0, 2\pi]$.
My attempt:
$$\begin{align}&\left|\sqrt{2\sin^2x + 18 \cos^2x} - \sqrt{2\cos^2x + 18 \sin^2x} \right| = 1\\\implies& \left|\sqrt{2\sin^2x + 2 \cos^2x + 16\cos^2x} - \sqrt{2\cos^2x + 2 \sin^2x + 16\sin^2x} \right| = 1
\\\implies& \left|\sqrt{2(\sin^2x + \cos^2x) + 16\cos^2x} - \sqrt{2(\cos^2x + \sin^2x) + 16\sin^2x} \right| = 1
\\\implies& \left|\sqrt{2+ 16\cos^2x} - \sqrt{2 + 16\sin^2x} \right| = 1\end{align}$$
Squaring both sides,
$$\begin{align}\implies& (2+ 16\cos^2x) + (2 + 16\sin^2x) - 2\sqrt{(2+ 16\cos^2x) (2 + 16\sin^2x)} = 1
\\\implies& 4 + 16(\sin^2x + \cos^2x) - 2\sqrt{2\cdot 2 \cdot (1+ 8\cos^2x) (1+ 8\sin^2x)} = 1
\\\implies &20 - 4\sqrt{1+8\cos^2x + 8\sin^2x + 8^2 \sin^2x \cos^2x } = 1
\\\implies &19 = 4\sqrt{1+8(\cos^2x + \sin^2x) + 16 \cdot (2\sin x \cos x)^2 }
\\\implies &\frac{19}{4} = \sqrt{9 + 16 \cdot (2\sin x \cos x)^2 }
\\\implies &\frac{19}{4} = \sqrt{9 + 16 \sin^2(2x) }\end{align}$$
Again squaring both sides,
$$\begin{align}\implies& \frac{361}{16} = 9 + 16 \sin^2(2x)
\\\implies& \frac{361}{16} - 9 = 16 \sin^2(2x)
\\\implies &\frac{361 - 144}{16} = 16 \sin^2(2x)
\\\implies& \frac{217}{256} = \sin^2(2x)
\\\implies& \pm\frac{\sqrt{217}}{16} = \sin(2x)\end{align}$$
Now I'm not getting any way to solve this equation. Although the question states to find only the solutions in the interval $[0, 2\pi]$, can we find the general form for all the solutions in $\mathbb{R}$ i.e. for all real numbers?
Desmos shows that there are $8$ solutions over the interval $[0, 2\pi]$.
|
In order to solve $\sin(x)=a$; such that $a\in[-1,1]$ and $x\in [-\pi/2,\pi/2]$, one can use the $\arcsin$ function indeed for all $a\in[-1,1]$, $\arcsin(a)=x$ where $x\in[-\pi/2,\pi/2]$, in order to find all solutions one can use the periodicity of $\sin$, note that as long as $a\in [-1,1]$ there will be an infinite amount of solutions in $\mathbb{R}$. Note that there is no algebraic way to solve this in general since $\sin$ is a transcendental function.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4476529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
help me to evaluate these integrals Find the value of $$\int\frac{x}{x^2-x+1} dx$$ and $$\int\frac{1}{x^2-x+1} dx$$ This was not the original question. Original question was tougher and I have simplified that to these two integrals. I am having a hard time evaluating these two integrals but what I know is that, in these two integrals there is arctan occurring in a term of both integrals. Maybe we have to do some operations in the numerators?
Any help will be greatly appreciated.
EDIT: The original question was: $$\int\frac{3x}{x^3+1}dx$$
EDIT: $$\begin{align}\int\frac{3x}{x^3+1}dx&=3\int\frac{x}{x^3+1}dx\\&=3\int\frac{x+1}{3x^2-3x+3}dx - 3\int\frac{1}{3x+3}dx\\&=\int\frac{x+1}{x^2-x+1}dx - \int\frac{1}{x+1}dx\end{align}$$
The value of last integral is $\ln|x+1|$ so we will consider it later. We will only consider the first integral now and we can write that as:
$$\int\frac{x}{x^2-x+1}dx + \int\frac{1}{x^2-x+1}dx$$ after this I am stuck.
|
Integral No. 1
$$\int \frac{dx}{x^2-x+1}= \int \frac{dx}{(x-\frac12)^2+\frac34}$$ Substitute $x-\frac12=t$ so $du = dt$. The integral becomes $$\int \frac{dt}{(t^2+\frac34)}=\frac{2}{\sqrt3}\arctan\frac{t}{\frac{\sqrt3}{2}}+c= \frac{2}{\sqrt3}\arctan\frac{2(x-\frac12)}{\sqrt3}+c = \frac{2}{\sqrt3}\arctan\frac{2x-1}{\sqrt3}+c .$$
Integral No. 2
$$\int \frac{xdx}{x^2-x+1}= \int \frac{x-\frac12+\frac12}{x^2-x+1}dx= \int \frac{x-\frac12+\frac12}{(x-\frac12)^2+\frac34}dx$$$$= \int \frac{x-\frac12}{(x-\frac12)^2+\frac34}dx+ \int \frac{\frac12}{(x-\frac12)^2+\frac34} dx $$ Now the second part is half of the first integral itself. For the first part, substitute $(x-\frac12)^2+\frac34=m$ and you should be able to finish it.
EDIT: After the m substitution, $2(x-\frac12)dx=dm$ so $(x-\frac12)dx=\frac{dm}{2}$. So the integral now is : $$\int \frac{dm}{2m}=\frac 12 \ln m + c=\frac12\ln(x^2-x+1)+c.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/4478207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.