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How to simplify $\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}$ The answer is 2. But I want to learn how to simplify this expression without the use of calculator.
|
Frequently, numbers of this kind are obtained from Cardano's formula for the roots of a cubic equation $x^3+px+q=0$, that is,
$$
\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}
+
\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{q^2}{4}+\frac{p^3}{27}}}
$$
so you have
$$
-\frac{q}{2}=7,
\qquad
\frac{q^2}{4}+\frac{p^3}{27}=50,
$$
that easily gives
$$
q=-14,\qquad p=3.
$$
Therefore the equation is
$$
x^3+3x-14=0
$$
which has a single real root, precisely $2$.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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"answer_id": 4
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|
Trigonometric Equation Simplification $$3\sin x + 4\cos x = 2$$
To solve an equation like the one above, we were taught to use the double angle identity formula to get two equations in the form of $R\cos\alpha = y$ where $R$ is a coefficient and $\alpha$ is the second angle being added to $x$ when using the double angle identity.
Why can't we use the identity $\sin(x) = \cos(x-90)$ to get $3\cos(x-90) + 4\cos x = 2$? Is this equation difficult to simplify further?
Additionally, why was the relationship between $sinx$ and $cosx$ in the pythagorean theorem (modified for the unit circle) not put to use? I did the following:
$$\sin^2x = 1 - \cos^2x$$
$$\therefore \sin x = ±\sqrt{1-\cos^2x}$$
If:
$$\sin x = y$$
Then,
$$y = ±\sqrt{1-\cos^2x}$$
Meaning that,
$$\cos x = ±\sqrt{1-y^2}$$
Inputting this into the original equation,
$$3y + 4\sqrt{1-y^2} = 2$$
We see, $$3y-2=-4\sqrt{1-y^2}$$
So, $$(3y-2)^2=16-16y^2$$
Therefore, $$9y^2-12y+4=16-16y^2$$
Rearranging gives, $$25y^2-12y-12=0$$
And so the solutions are, $$y_0,y_1=\frac{12}{50}\pm\frac{1}{50}\sqrt{144+1200}$$
And simplifying yields, $$y_0,y_1=\frac{6\pm4\sqrt{21}}{25}$$
Checking these solutions will give us the unique solution: $$y_0=\frac{6-4\sqrt{21}}{25}$$
$$q.e.d.$$
Can the above method be generalised? Has it been generalised?
|
$$\begin{align}3\sin x + 4\cos x &= \sqrt{3^2 + 4^2 }\sin\left(x+\arctan\frac{4}{3}\right) \\
&= 5\sin(x+\arctan\frac{4}{3})\end{align}$$
In general, if $$a\sin x + b\cos x = c$$
$$\frac{a\sin x}{\sqrt{a^2 + b^2}}+\frac{b \cos x}{\sqrt{a^2 + b^2}} = \frac{c}{\sqrt{a^2 + b^2}}$$
Let $\cos\phi = \dfrac{a}{\sqrt{a^2 + b^2}} \ \ \therefore \sin\phi = \dfrac{b}{\sqrt{a^2 + b^2}}$
Hence $\tan\phi = \dfrac{b}{a}$. Substituting in above eqn,
$$\sin x \cos \phi + \cos x \sin\phi = \frac{c}{\sqrt{a^2 + b^2}}$$
$$\sin (x + \phi)\sqrt{a^2 + b^2} = c$$
$$\implies c = \sqrt{a^2 + b^2}\ \sin\left(x + \arctan\frac{b}{a}\right)$$
|
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|
Calculate $\sqrt{x^2+y^2+2x-4y+5} + \sqrt{x^2+y^2-6x+8y+25}$, if $3x+2y-1=0$ As the title says, given $x,y \in \mathbb{R}$ where $3x+2y-1=0$ and $x \in [-1, 3]$, calculate $A = \sqrt{x^2+y^2+2x-4y+5} + \sqrt{x^2+y^2-6x+8y+25}$.
I tried using the given condition to reduce the complexity of the roots, but couldn't get rid of them.
|
Hint: make a geometrical interpretation and draw a picture:
$x^2+y^2+2x-4y+5=(x+1)^2 +(y-2)^2$ and consider the point $P(-1,2)$.
$x^2+y^2-6x+8y+25=(x-3)^2+(y+4)^2$ and consider the point $Q(3,-4)$.
Observe that $P$ and $Q$ are on the line $3x+2y-1=0$.
Can you now see what $A$ means on the given interval?
The answer is here (mouse over) but try to find it yourself first:
$2\sqrt{13}$
|
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|
Area inside polar curve Find the area of the region inside
$r=7\sin\theta$ but outside $r=1$.
I have tried finding the area of both using $A=\frac{1}{2}\int_\alpha^\beta r^2 d\theta$, tried arc length...but I can't find the answer.
|
\begin{align*}
A &= \frac{1}{2}
\int_{\sin^{-1} \frac{1}{7}}^{\pi-\sin^{-1} \frac{1}{7}}
(r^{2}-1) \, d\theta \\
&= \frac{1}{2} \int_{\sin^{-1} \frac{1}{7}}^{\pi-\sin^{-1} \frac{1}{7}}
(49\sin^{2}-1) \, d\theta \\
&= \frac{1}{2} \int_{\sin^{-1} \frac{1}{7}}^{\pi-\sin^{-1} \frac{1}{7}}
\left[ 49\left( \frac{1-\cos 2\theta}{2} \right)-1 \right] \, d\theta \\
&= \left[
\frac{47}{4}\theta-\frac{49}{8} \sin 2\theta
\right]_{\sin^{-1} \frac{1}{7}}^{\pi-\sin^{-1} \frac{1}{7}} \\
&= \frac{47\pi}{4}-\frac{47\pi}{2}\sin^{-1} \frac{1}{7}+
\frac{49}{4} \sin \left( 2\sin^{-1} \frac{1}{7} \right) \\
&= \frac{47\pi}{4}-\frac{47\pi}{2}\sin^{-1} \frac{1}{7}+
\frac{49}{2} \times \frac{1}{7} \times \sqrt{\frac{48}{49}} \\
&= 2\sqrt{3}+\frac{47\pi}{2}-\frac{47}{2}\sin^{-1} \frac{1}{7} \\
&= 2\sqrt{3}+\frac{47}{2}\cos^{-1} \frac{1}{7}
\end{align*}
|
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|
$x^4 - y^4 = 2z^2$ intermediate step in proof I am ultimately trying to prove, for an Exercise in Burton's Elementary Number Theory, that $x^4 - y^4 = 2z^2$ has no solution in the positive integers.
I can establish that if there is a solution, the solution with the smallest value of x has gcd(x,y)=1
I see that $x^4 - y^4 = (x^2 + y^2)(x^2 - y^2) = (x^2 + y^2)(x + y)(x - y)$
I also have the fact that $uv = w^2$ and $gcd(u,v)=1$ implies $u$ and $v$ are each squares.
If I can establish (along the lines of the hint from the textbook) that
$(x^2 + y^2) = 2a^2$ , $(x + y) = 2b^2$ , and $(x - y)=2c^2$ for some integers a,b,c then I can derive a contradiction via a previous theorem.
However, I am stuck on how to establish the above equalities. Can anyone offer me some direction?
|
If you want to establish your equalities,
Note that $\gcd(x,y)=1$. This implies that
$$\gcd(x^2+y^2,x+y)=2,1$$$$\gcd(x^2+y^2,x-y)=2,1$$$$ \gcd(x-y,x+y)=2,1$$
This implies that either $x+y=2a^2, x-y=b^2,x^2+y^2=c^2$
Or that $x+y=a^2, x-y=2b^2, x^2+y^2=c^2$
Or that $x+y=a^2, x-y=b^2, x^2+y^2=2c^2$
Or that $x+y=2a^2, x-y=2b^2, x^2+y^2=2c^2$. I suggest you proceed from here.
However, the shorter way would be to squaring both sides and adding $4x^2y^2$.
|
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|
Find the integral $\int \frac{1+x}{\sqrt{1-x^2}}\mathrm dx$ The integral can be represented as
$$
\int \frac{1+x}{\sqrt{1-x^2}}\mathrm dx=
\int \left(\frac{1+x}{1-x}\right)^{1/2}\mathrm dx
$$
Substitution $$t=\frac{1+x}{1-x}\Rightarrow x=\frac{t-1}{t+1}\Rightarrow dx=\frac{2}{(t+1)^2}dt\Rightarrow \int\limits \left(\frac{1+x}{1-x}\right)^{1/2}\mathrm dx=2\int\limits \frac{\sqrt{t}}{(t+1)^2}\mathrm dt$$
What substitution to use for solving the integral $\int\limits \frac{\sqrt{t}}{(t+1)^2}\mathrm dt$?
|
We can also use the substitution $u=\sqrt{1-x}$, then $\mathrm{d}u=-\frac{\mathrm{d}x}{2\sqrt{1-x}}$ and $\sqrt{1+x}=\sqrt{2-u^2}$. We will also use $u=\sqrt2\sin(\theta)$
$$
\begin{align}
\int\frac{\sqrt{1+x}}{\sqrt{1-x}}\,\mathrm{d}x
&=-2\int\sqrt{2-u^2}\,\mathrm{d}u\\
&=-2\int\sqrt2\cos(\theta)\cdot\sqrt2\cos(\theta)\,\mathrm{d}\theta\\
&=-2\int(1+\cos(2\theta))\,\mathrm{d}\theta\\[3pt]
&=-2\theta-\sin(2\theta)+C\\[3pt]
&=-2\arcsin\left(\sqrt{\frac{1-x}{2}}\right)-\sqrt{1-x^2}+C\\
&=-\arccos(x)-\sqrt{1-x^2}+C\\[6pt]
&=\arcsin(x)-\sqrt{1-x^2}+\left(C-\tfrac\pi2\right)
\end{align}
$$
Explanation of $\boldsymbol{2\arcsin\left(\sqrt{\frac{1-x}{2}}\right)=\arccos(x)}$
Let $\alpha=\arcsin\left(\sqrt{\frac{1-x}2}\right)$, then $x=1-2\sin^2(\alpha)=\cos(2\alpha)$. Thus, $\alpha=\frac12\arccos(x)$. Therefore,
$$
2\arcsin\left(\sqrt{\frac{1-x}2}\right)=\arccos(x)
$$
|
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|
Taylor expansion of $\cos{x}$ I found a pdf file on the internet which gives you known expansions of Taylor's.
There is something I cant understand :
Why is the remainder of $\cos x$ is written like this?
$$\frac{\cos ^{(2n+2)}(c)x^{2n+2}}{(2n+2)!}$$
And not like this:
$$\frac{\cos^{(2n+1)} (c)x^{2n+1}}{(2n+1)!}$$
when the last element was :
$$\frac{(-1)^nx^{2n}} {(2n)!} $$
|
I can't guess what a random, unlinked PDF on the Internet says. However, this is not the behaviour of the Taylor series of cosine. For instance, the expansion of $\cos x$ around $x=1$ is $$\cos (1)-(x-1) \sin (1)-\frac{1}{2} (x-1)^2 \cos (1)+\frac{1}{6} (x-1)^3
\sin (1)+\frac{1}{24} (x-1)^4 \cos (1)-\frac{1}{120} (x-1)^5 \sin
(1)-\frac{1}{720} (x-1)^6 \cos (1)+\frac{(x-1)^7 \sin
(1)}{5040}+\frac{(x-1)^8 \cos (1)}{40320}-\frac{(x-1)^9 \sin
(1)}{362880}-\frac{(x-1)^{10} \cos (1)}{3628800}+\cdots, $$ which you might notice does have the error behaviour you seem to expect: the degree of the error term is equal to the degree of the first omitted term.
The Maclaurin series for cosine, $$1-\frac{x^2}{2}+\frac{x^4}{24}-\frac{x^6}{720}+\frac{x^8}{40320}-\frac{x^{10}}{3628800}+\cdots$$ does have the form you cite from the paper and does have the error term you mention. In the expansion I just showed, the first omitted term has degree $12$, not $11$. This behaviour is predictable: cosine is even (i.e., is unchanged on reflection through the line $x=0$). We would expect similar behaviour for Taylor series expansion around multiples of $\pi$ as well, due to the same reflection symmetry, which we see. For example, expanding around $x=\pi$, $$-1+\frac{1}{2} (x-\pi )^2-\frac{1}{24} (x-\pi )^4+\frac{1}{720} (x-\pi)^6-\frac{(x-\pi )^8}{40320}+\frac{(x-\pi )^{10}}{3628800}+\cdots.$$
Further, much like sine, cosine has odd symmetry around odd multiples of $\pi/2$, so we would expect the Taylor expansion to contain only odd degree terms,which it does.$$-\left(x-\frac{\pi }{2}\right)+\frac{1}{6} \left(x-\frac{\pi}{2}\right)^3-\frac{1}{120} \left(x-\frac{\pi}{2}\right)^5+\frac{\left(x-\frac{\pi}{2}\right)^7}{5040}-\frac{\left(x-\frac{\pi}{2}\right)^9}{362880}+\cdots$$ Again, the error term has degree matching that of the first omitted term, which here is $11$, not $10$.
|
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|
If $A+B+C=π$, verify the given If $A+B+C=π$, prove that
$$\cos A \sin B \sin C + \cos B \sin C \sin A + \cos C \sin A \sin B=1+\cos A \cos B\cos C$$
ATTEMPT:
Here,
$$A+B+C=π$$
Now,
\begin{align*}
\text{L.H.S} &= \cos A \sin B \sin C + \cos B \sin C \sin A + \cos C \sin A \sin B \\
&=\sin C(\sin A \cos B + \cos A \sin B) + \cos C \sin A \sin B \\
&= \sin C \sin(A+B) + \sin A \sin B \cos C
\end{align*}
What should I do next?
|
take the trig term fro the RHs to the left and combine to get
$$
\sin C \sin(A+B) - \cos C \cos (A+B) = -\cos (A+B+C)
$$
and now $A+B+C=\pi$...
|
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|
Evaluation of $\lim_{n\rightarrow \infty}\sum_{k=1}^n\sin \left(\frac{n}{n^2+k^2}\right)$
Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\sin \left(\frac{n}{n^2+1}\right)+\sin \left(\frac{n}{n^2+2^2}\right)+\cdots+\sin \left(\frac{n}{n^2+n^2}\right)$
$\bf{My Try::}$ We can write the Sum as $$\lim_{n\rightarrow \infty}\sum^{n}_{r=1}\sin\left(\frac{n}{n^2+r^2}\right)$$
Now how can I convert into Riemann Sum, Help me
Thanks
|
As a consequence of the fact that $\,\,\int_0^1 f(x)\,dx=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n f\big(\frac{k}{n}\big)$, we obtain
$$
\sum_{k=1}^n\frac{n}{n^2+k^2}=\frac{1}{n}\sum_{k=1}^n\frac{1}{1+\left(\frac{k}{n}\right)^2}\to\int_0^1\frac{dx}{1+x^2}=\tan^{-1}1=\frac{\pi}{4}
$$
Next, Taylor expansion of $\sin x$, provides that $0\le x-\sin x\le \frac{x^3}{6} $, for $0\le x\le 1$.
Thus
$$
0\le \frac{n}{n^2+k^2}-\sin\left(\frac{n}{n^2+k^2}\right)\le \frac{1}{6n^3},
$$
and thus
$$
0\le \sum_{k=1}^n\left(\frac{n}{n^2+k^2}-\sin\left(\frac{n}{n^2+k^2}\right)\right)\le \frac{1}{6n^2}.
$$
Thus
$$
\lim_{n\to\infty}\sum_{k=1}^n\sin\left(\frac{n}{n^2+k^2}\right)=
\lim_{n\to\infty}\sum_{k=1}^n\frac{n}{n^2+k^2}=\frac{\pi}{4}.
$$
|
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|
How many different integer solutions are there to the equation $x_1 + x_2 + x_3 + x_4 = 21$ with restrictions So i was Given this question. How many different integer solutions are there to
the equation
$x_1 + x_2 + x_3 + x_4 = 21$
$0 \leq x_i \leq 9$?
I just assumed it would be
${21+4-4-1 \choose 4-1} = {20 \choose 3}$
I'm confused by the $0 \leq x_i \leq 9$.
|
All solutions
Split $(x_{1}+x_{2})+(x_{3}+x_{4})=21$. There are 16 different pairs $(x_{1}+x_{2},x_{3}+x_{4})$ (if you count $3+18$ different from $18+3$) simply $(3,18),(4,17),(5,16),...,(10,11),...,(18,3)$
For $(10,11)$ and $(11,10)$ you have $9$ combinations for $10$ and $8$ combinations for $11$. That makes $2\cdot 9 \cdot 8=144$ combinations.
For $(9,12)$ and $(12,9)$ you have $10$ combinations for $9$ and $7$ combinations for $12$.
For $(8,13)$ and $(13,8)$ you have $9$ combinations for $8$ and $6$ combinations for $13$.
...
The pattern emerges $2m(m-3)$. So the total number of solutions is then
$$144+ \sum\limits_{m=4}^{10} 2m(m-3) = 592$$
|
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|
$\sqrt{13a^2+b^2}$ and $\sqrt{a^2+13b^2}$ cannot be simultaneously rational Let $a,b \in \mathbb{N}^{*}$. Prove that $\sqrt{13a^2+b^2}$ and $\sqrt{a^2+13b^2}$ cannot be simultaneously rational.
If $(a,b)=(k,k\cdot6)$, then $\sqrt{13a^2+b^2}$ is rational, but I do not know if those are the only solutions.
|
$\sqrt{x}$ (where $x\in\mathbb Z^+$) is rational if and only if $\sqrt{x}$ is a positive integer (see this question).
Therefore, your problem is equivalent to proving the system $13a^2+b^2=k^2$ with $a^2+13b^2=m^2$ has no positive integer solutions.
$$a^2+13\left(k^2-13a^2\right)=m^2\iff m^2+168a^2=13k^2$$
I'll prove the last equation has no positive integer solutions.
$m^2\equiv 5k^2\pmod{8}$. But we have $y^2\equiv \{0,1,4\}\pmod{8}$ for all $y\in\mathbb Z$.
Therefore $m,k$ are both even. Let $m=2m_1$, $k=2k_1$. Then $m_1^2+42a^2=13k_1^2$.
$m_1,k_1$ have the same parity. If $m_1,k_1$ are both odd, then $m_1^2\equiv k_1^2\equiv 1\pmod{8}$, so $42a^2\equiv 4\pmod{8}$, i.e. $a^2\equiv 2\pmod{4}$, contradiction ($2$ is not a quadratic residue mod $4$). Therefore $m_1,k_1$ are both even, so $a$ is also even.
Therefore $(m,a,k)$ is a positive integer solution if and only if $(m/2,a/2,k/2)$ is also a positive integer solution. By Infinite Descent we get $m=a=k=0$.
|
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|
What is the procedure to solve for the $\lim\limits_{x\rightarrow 0} \frac{\cos(3x)-1}{x^2}$? What is the procedure to solve for the $\lim\limits_{x\rightarrow 0} \frac{\cos(3x)-1}{x^2}$?
My calculator tells me the answer is -9/2, but I don't know how to solve without substituting values of x.
I suspect there is some trigonometric identity involved.
|
From the trigonometric identity $\;\color{blue}{\cos(2\theta)=1-2\sin^2\theta}\;$ we have
\begin{align}
\lim_{x\to 0}\frac{\cos(3x)-1}{x^2}&=\lim_{x\to 0}\frac{1-2\sin^2(3x/2)-1}{x^2}\\
&=\lim_{x\to 0}\frac{-2\sin^2(3x/2)}{x^2}\\
&=-2\cdot\frac{9}{4}\cdot\lim_{x\to 0}\frac{\sin^2(3x/2)}{\frac{9}{4}x^2}\\
&=-2\cdot\frac{9}{4}\cdot\left(\lim_{x\to 0}\frac{\sin(3x/2)}{3x/2}\right)^2\\
&=-2\cdot\frac{9}{4}\cdot\left(1\right)^2\\
&=\boxed{\color{blue}{-\frac{9}{2}}}
\end{align}
|
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|
How to prove $ \left(\sum\limits_{cyc}{xy}\right)^2 \ge3xyz(x+y+z)$ with $x,y,z$ being positive real numbers I have tried to improve the inequality by AM-GM. Here is what I have done:
Since
$$\sum_{cyc}{xy}\ge3\sqrt[3]{x^2y^2z^2}$$
$$\Rightarrow \left(\sum_{cyc}{xy}\right)^2 \ge3xyz\sqrt[3]{xyz}$$ That means we got to prove$\sqrt[3]{xyz} \ge x+y+z$.
However that is false in the case $x=1$,$y=2$,$z=3$.
Any help?
|
Simplify the inequality to
$$x^2y^2+y^2z^2+x^2z^2\ge xyz(x+y+z)$$
If some of $x,y,z$ equals $0$, then the inequality is obvious.
So assume $xyz\neq 0$ and divide by $x^2y^2z^2>0$ both sides to get
$$\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge \frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}$$
This is true by the power of Cauchy Schwarz $$\frac{1}{y}\cdot\frac{1}{z}+\frac{1}{z}\cdot\frac{1}{x}+\frac{1}{x}\cdot\frac{1}{y}\leq \sqrt{\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{x^2}}\cdot\sqrt{\frac{1}{z^2}+\frac{1}{x^2}+\frac{1}{y^2}}$$
|
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|
Sticks and stones problem Ten squares in a row are labelled 1, 2, 3, . . . 10, in order. A counter starts at square 1. At every step,
the counter can move ahead 1, 2, or 3 squares. However, beginning with the second step, the counter
cannot move the
same
number of squares as it did in the previous step. For example, in the first step,
the counter can move from square 1 to square 3. Then in the next step, the counter can move to square
4 or square 6, but not square 5. Find the number of possible sequences of steps that take the counter
from square 1 to square 10. (In the last step, the counter must land exactly on square 10.)
Hi,
I am thinking of using sticks and stones for this problem, but cannot think of a way to use it. Can anyone help me?
Thanks!
|
I get $21$.
$2$ with no $3$'s:
$1,2,1,2,1,2$
$2,1,2,1,2,1$
$12$ with one $3$:
$3,1,2,1,2$
$3,2,1,2,1$
$1,3,2,1,2$
$2,3,1,2,1$
$1,2,3,1,2$
$1,2,3,2,1$
$2,1,3,1,2$
$2,1,3,2,1$
$1,2,1,3,2$
$2,1,2,3,1$
$1,2,1,2,3$
$2,1,2,1,3$
$7$ with two $3$'s:
$3,1,2,3$
$3,2,1,3$
$1,3,2,3$
$3,2,3,1$
$1,3,1,3,1$
$2,3,1,3$
$3,1,3,2$
$2+12+7=21.$
|
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"timestamp": "2023-03-29T00:00:00",
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|
A factory makes three types of biased coins with probability of getting head when tossed is $\frac{1}{2}$,$\frac{1}{3}$,$\frac{1}{4}$ respectively
A factory makes three types of biased coins with probability of getting head when tossed is $\frac{1}{2}$,$\frac{1}{3}$,$\frac{1}{4}$ respectively. Each coin is tossed before packing and the face that appears is marked.
(1) A coin is selected from a bag containing one coin of each type and the coin is tossed and head appears. Find the probability that the face that appeared is marked.
(2) If a coin is selected from the bag and the 'heads' side is marked, the probability that the tossed coin will show heads is?
In the first case, the probability of choosing any type of coin is $\frac{1}{3}$
The probability of 'head' being marked in each type of coin is, $\frac{1}{2}$,$\frac{1}{3}$,$\frac{1}{4}$
The probability of getting heads is $\frac{1}{2}$,$\frac{1}{3}$,$\frac{1}{4}$
There are three possible cases. The total probability is $$\frac{1}{3}\times\frac{1}{2}\times\frac{1}{2}+\frac{1}{3}\times\frac{1}{3}\times\frac{1}{3}+\frac{1}{3}\times\frac{1}{4}\times\frac{1}{4}$$
But the answer I got was wrong. The answer given is $\frac{61}{12\times 13}$.
(I can get the answer if I divide the answer I got, by the probability of getting a head irrespective of the mark.)
|
Your last remark is indeed correct. The probability you calculated is $P(H \cap H_M)$, the probability that a coin from those 3 has its head marked and comes up $H$.
$P(H) = \frac{1}{3}(\frac{1}{2} + \frac{1}{3} + \frac{1}{4}) = \frac{13}{36}$, by conditioning again.
The asked for probability is (head is marked given that head came up!): $P(H_M | H) = {P(H \cap H_M) \over P(H)}$ and then we get the answer $\frac{61}{3 \cdot 12 \cdot 12} \cdot \frac{36}{13} = \frac{61}{12 \cdot 13}$ as required.
|
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|
Why is $\left ( \frac{x}{j}+1 \right )^{-1}\left ( \frac{1}{j}+1 \right )^{x}=1+\frac{x(x-1)}{2j^2}+O(j^{-3})$? On this page, it says
$$ \left ( \frac{x}{j}+1 \right )^{-1}\left ( \frac{1}{j}+1 \right )^{x}=1+\frac{x(x-1)}{2j^2}+O(j^{-3}). $$
Could anyone please enlighten me how it is deduced? I tried to find an approximation of $j\mapsto \left ( \frac{1}{j}+1 \right )^{x}$, and then multiplying it by $\left ( \frac{x}{j}+1 \right )^{-1}$. But there is no way I could find a such approximation.
|
As already said, this is the binomial theorem.
You can also do it with Taylor series. Consider first $$A=\left ( \frac{1}{j}+1 \right )^{x}$$ $$\log(A)=x\, \log\left( \frac{1}{j}+1 \right)=x\left(\frac{1}{j}-\frac{1}{2 j^2}+\frac{1}{3 j^3}+O\left(\frac{1}{j^4}\right)\right)$$ $$A=e^{\log(A)}=1+\frac{x}{j}+\frac{x(x-1)}{2 j^2}+\frac{x(x-1)(x-2)}{6
j^3}+O\left(\frac{1}{j^4}\right)$$ $$B=\left ( \frac{x}{j}+1 \right )^{-1}=\frac{1}{\frac{x}{j}+1}=1-\frac{x}{j}+\frac{x^2}{j^2}-\frac{x^3}{j^3}+O\left(\frac{1}{j^4}\right)$$ All of the above makes $$\left ( \frac{x}{j}+1 \right )^{-1}\left ( \frac{1}{j}+1 \right )^{x}=A \times B=1+\frac{x(x-1) }{2 j^2}-\frac{(x+1)x(x-1)}{3 j^3}+O\left(\frac{1}{j^4}\right)$$
|
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|
Does $\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta\dots}}}}=\sqrt{1 \sqrt{1 \sqrt{1\dots}}} \implies \cos{\theta}=1$? I was solving this equation:
$$\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta\dots}}}}=1$$
I solved it like this:
The given equation can be written as:
\begin{align*}
\sqrt{\cos{\theta} \sqrt{\cos{\theta} \sqrt{\cos{\theta\dots}}}}&=\sqrt{1 \sqrt{1 \sqrt{1\dots}}} \\
\cos{\theta}&=1 \\
\theta&=\arccos {1}
\end{align*}
So the solution is $2n\pi, n \in \mathbb Z$.
Have I solved it the wrong way?
(The title originally contained a more general question: Does $\sqrt{a \sqrt{a \sqrt{a\dots}}}=\sqrt{b \sqrt{b \sqrt{b\dots}}} \implies a=b$? The current title is consistent with the body and the accepted answer.)
|
To answer the question in the title, let $A=\sqrt{a \sqrt{a \sqrt{a\cdots}}}$ and $B=\sqrt{b \sqrt{b \sqrt{b\cdots}}}$.
Then $A=B$ implies $aA=A^2=B^2=bB$ and so $a=b$.
|
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|
Infinity Series to Approximate a fraction I have observed that the following series is a good approximation for
$\frac{1}{10}$.
$\frac{1}{8}- \frac{1}{16} + \frac{1}{32} + \frac{1}{64} - \frac{1}{128} - \frac{1}{256} + \frac{1}{512} + \frac{1}{1024} - \frac{1}{2048} -
\frac{1}{(2\cdot2048)} + \frac{1}{(4\cdot2048)} + \frac{1}{(8\cdot2048)} - \frac{1}{(16\cdot2048)} - \frac{1}{(32\cdot2048)} + ...$
I believe there is a pattern to the series. However, I cannot prove that pattern holds. Does it? Can it be proved? I am thinking some Taylor series might be the way to go but that is just a hunch.
Bob
|
The series is similar to $\sum \limits_{n=3}^{\infty} \frac{1}{2^n}=2-1/1-1/2-1/4=1/4$.
Note that $\sum \limits_{n=0}^{\infty} x^n=\frac{1}{1-x}$ for $|x|<1$.
EDIT, since I misread the question:
$$1/8 - 1/16 + 1/32 + 1/64 - 1/128 - 1/256 + 1/512 + 1/1024 - 1/2048 -
1/4096 + 1/8192 + 1/16384 - 1/32768 - 1
/65536 + ...=$$
$$\frac{1}{8}-\frac{1}{16}+\sum \limits_{n=2}^{\infty} \frac{(-1)^n}{2^{2n+1}}+\sum \limits_{n=2}^{\infty} \frac{(-1)^n}{2^{2n+2}}=$$
$$\frac{1}{8}-\frac{1}{16}+\frac{1}{40}+\frac{1}{80}=\frac{1}{10}$$
This expression can be evaluated using the formula given above
|
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|
Prove that $f(x)=8$ for all natural numbers $x\ge{8}$ A function $f$ is such that $$f(a+b)=f(ab)$$ for all natural numbers $a,b\ge{4}$ and $f(8)=8$. Prove that $f(x)=8$ for all natural numbers $x\ge{8}$
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Yes. It is true that $f(x) = 8 \quad \forall\;\;\; x \in N $
Manually, we can prove this for $x \le 20$.
Now, let $x$ be even. $x = 2y$ for some $y$.
$$f(2y)=f((2y-4) +(4))=f(4(2y-4))=f(8(y-2))=f(8+y-2)=f(y+6)$$
Note: This is true only if the $y-2$ factor is greater than $4$, so let $y \ge 6$.
Similarly, if $x$ is odd, $x = 2y + 1$ for some $y$.
$$f(2y+1)=f((2y-4)+5)=f(5(2y-4))=f(10(y-2))=f(10 + y-2)=f(y+8)$$
Note: Similarly, this has the same condition $y \ge 6$.
And we can see that $2y > y+6$ and $2y+1 > y+8$ for $y\ge6$. ($y > 7$ for the second case). Therefore, for any $f(m)$, we can find $f(n)=f(m)$ for $n < m
$.
Thus after reducing, we get a number lesser than 20 which can be proved manually equal to $8$.
Therefore $f(x) = 8\;\;\; \forall \;\;\; x \in N$
|
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|
How can I solve $7^{77}\mod 221$ How is it possible to solve this without calculator etc.: $$7^{77} \mod 221$$
I started with:
\begin{align}
7&\equiv 7 \mod 221 \\
7^2 &\equiv 49 \mod 221 \\
7^4 &\equiv \ ? \mod 221
\end{align}
Sure i can calculate this by my own, but is there a trick to calclulate this with any tools?
|
Use the Chinese Remainder Theorem. Note that $221=13\times17$. Modulo $13$ we have
$$7^2\equiv-3\ ,\quad 7^6\equiv(-3)^3\equiv-1$$
and so
$$7^{77}=(7^6)^{12}7^5\equiv7^5=(7^2)(7^2)7\equiv(-3)(-3)7=63\equiv-2\ .$$
Modulo $17$ we have
$$7^2\equiv-2\ ,\quad 7^8\equiv(-2)^4\equiv-1$$
and so
$$7^{77}=(7^8)^97^5\equiv-7^5\equiv-(-2)(-2)7\equiv6\ .$$
So you have to solve simultaneously
$$x\equiv-2\pmod{13}\ ,\quad x\equiv6\pmod{17}\ .$$
Standard methods (look up Chinese Remainder Theorem) give $x\equiv193\pmod{221}$.
|
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|
Solving recurrence relation? Consider the recurrence relation $a_1=8,a_n=6n^2+2n+a_{n−1}$. Let $a_{99}=K\times10^4$ .The value of $K$ is______ .
My attempt :
$a_n=6n^2+2n+a_{n−1}$
$=6n^2+2n+6(n−1)^2+2(n−1)+a_{n−2}$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+a_1$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+6.1^2+2.1$
$=6(n^2+(n−1)^2+...+2^2+1^2)+2(n+(n−1)+...+2+1)$
$=6×n(n+1)(2n+1)/6+2×n(n+1)/2$
$=n(n+1)(2n+1+1)$
$=2n^3+2n^2+2n^2+2n$
$=2n(n^2+n+n+1)$
$=2n(n^2+2n+1)$
$a_n=2n(n+1)^2$
for $n=99, a_{99}=2×99×(99+1)^2=198×10^4$
I'm looking for short trick or alternative way, can you explain please?
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As $a_n-a_{n-1}$ is a quadratic polynomial, $a_n$ must be a cubic polynomial. By integration, close to $\dfrac63n^3=2n^3$.
Then the factor $10^4$ is very likely to be $(99+1)^2$, and we can gamble on the expression $2n(n+1)^2$. It works for $n=1$ ($a_1=8$) and $n=2$ ($a_2=36$). This is convincing enough.
$$2n=198.$$
By better knowledge of Faulhaber's formula, (the first coefficient is always $\dfrac1d$, the second coefficient is always $\dfrac12$), we can predict that the cubic term is exactly $\dfrac63n^3=2n^3$ and the quadratic term is exactly $\dfrac62n^2+n^2=4n^2$. This fits with $2n(n+1)^2$.
|
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|
How could I find the interval where the solution is valide for the initial value problem : $y'=\frac{1+3x^2}{3y^2-6y}$ with $y(0)=1$? How could I find the interval where the solution is valide for the initial value problem : $y'=\frac{1+3x^2}{3y^2-6y}$ with $y(0)=1$?
I think I have to consider the points where the curve-solution admit a vertical tangent.
I got $y^3-3y^2-x-x^3+2=0$ for the curve solution. According to the solution sheet, this interval should be $|x|<1$. Any help?
EDIT: I found a solution on https://www.math.psu.edu/shen_w/250/Notes/only_ch2.pdf. when we put it into google. However I don't understand why we work on interval instead of discrete points.
Thanks!
|
See my comment:
The differential equation is only valid on the following continuous intervals: 1) $-\infty < y < 0$, 2) $0 < y < 2$, and 3) $2 < y < \infty$. It might be valid on more but since $1 + 3x^2 = 0$ permits no real solutions, there's no need to worry about possible removeable discontinuities.
Find the "solution" first (which you did):
\begin{align}
(3y^2 - 6y)dy =&\ (1 + 3x^2)dx \\
y^3 - 3y^2 =&\ x + x^3 + C
\end{align}
Plug in $y = 1$ and $x = 0$ to find:
$$
1 - 3 = C \rightarrow C = -2
$$
This gives the following:
$$
y^3 - 3y^2 = x + x^3 - 2
$$
(this is what you found)
Since $y(0) = 1$ we know that we must be on the interval $0 < y < 2$. Find the min and max on this interval: max of $f(y) = y^3 - 3y^2$ on the interval $0 < y < 2$. We already know that the critical points will be $y = 0$ and $y = 2$--the end points. So just input those two numbers. This tells us that the right side ranges from $y = 0$, $0^3 - 3\cdot0^3 = 0$ to $y = 2$, $2^3 - 3\cdot2^2 = 8 - 12 = -4$. So the $x$ values, $x + x^3 - 2$, range from $-4$ to $0$.
We need to find the continuous interval of $f(x) = x + x^3 - 2$ which is bounded by $-4 < f(x) < 0$. You can verify that there are no (real) extrema (meaning it's either monotonically increasing or decreasing) of the function $f(x) = x + x^3 - 2$ by showing that $f'(x) = 1 + 3x^2 = 0$ has no real solutions. This means that the interval is simply finding the two following solutions:
$$
x + x^3 - 2 = -4 \\
x + x^3 - 2 = 0
$$
Clearly $x = 1$ solves $x + x^3 - 2 = 0$. Also it's clear that $x = -1$ solves $x + x^3 - 2 = -4$. This means the $x$ interval is $-1 < x < 1$ or $|x| <1$. There is no "algebraic" way (that I'm aware of) to solve those two equations other than Rational Zeros Theorem (which means your particular problem is very special--i.e. it admits rational solutions, which will not generally be the case). Otherwise, it requires numerical methods.
|
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|
Looking for a nonrecursive formula for the general derivatives of the quotient of functions I want to prove that the $k$-th derivative $h^{(k)}(x)$ of the function $h(x)=\frac{1}{1+x^2}$ is zero at $x=0$ for all integer values $k>0$.
My only idea was to go the stubborn way applying iteratively the elementary formula for the derivative of a quotient of functions. Alas I didnt find a general formula similar to the Leibniz formula for the derivatives of a product of functions neither in Wikipedia nor else in the web so far.
This puzzles me. It wouldnt surprise me if a non-recursive closed expression (using iterated binomial coefficient sums) would be existing.
I tackled the problem so far in using first the Leibniz formula on $h(x)=f(x)\frac{1}{g(x)}$ in the way
$$h^{(k)}(x)=\sum_{r=0}^k f^{(k)}(x)\left( \frac{1}{g(x)}\right)^{(k-r)}$$
So I am in front of the problem calculating $\left( \frac{1}{g(x)}\right)^{(s)}$ which in the first step tackled by decreasing iteratively by $1$
$$\left( \frac{1}{g(x)}\right)^{(s)}=\left( -\frac{g^{(1)}(x)}{g^2(x)}\right)^{(s-1)}$$
If one applies to that quotient iteratively the Leibniz product rule
the next calculation problem comes up for
$$\left( \frac{1}{g^2(x)}\right)^{(t)}=\left( -\frac{(g^2)^{(1)}(x)}{g^4(x)}\right)^{(t-1)}$$
So I arrive at the problem calculating
$$\left( -\frac {g^{{2^{m-1}}^{(1)}}(x)} {g^{2^m}(x)} \right)^{(1)}$$
Then I tried to go on by de-exponentiating and getting the square
$$(g^{2^{m-1}})^{(1)}(x)=\left((g^{2^{m-2}}(x))^2\right)^{(1)}(x)=2g^{2^{m-2}}(x)g^{2^{m-2}})^{(1)}(x)$$
This leads me iteratively to the ( surprising/erroneous(?) ) result
$$\left( -\frac {g^{{2^{m-1}}^{(1)}}(x)} {g^{2^m}(x)} \right)^{(1)}=2^{m-1}\frac{g^{\prime}}{g}$$
where the exponent at the $2$ is in doubt.
Now I am overwhelmed at putting this all together and especially simplifying the nested iterative Leibniz sums with the binomials.
|
In order to answer OPs second part of the question, we provide an expression of the $n$-th derivative of $\frac{f}{g}$ in terms of derivatives of $f$ and $g$.
Let $D_x$ represent differentiation with respect to $x$. Hence $D^n_x f(x)$ is the $n$-th derivative of $f$ with respect to $x$. The following holds true for $n$ times differentiable functions $f$ and $g$
\begin{align*}
D_x^n\left(\frac{f}{g}\right)=\sum_{k=0}^n\sum_{j=0}^{k} (-1)^j\binom{n}{k}\binom{k+1}{j+1}\frac{1}{g^{j+1}}
D_x^{n-k}\left(f\right) D_x^{k}\left( g^j\right)
\end{align*}
The following formula for the $n$-th derivative of the composite of two functions is stated as identity (3.56) in H.W. Gould's Tables of Combinatorial Identities, Vol. I and called:
Hoppe Form of Generalized Chain Rule
Let $D_z$ represent differentiation with respect to $z$ and $z=z(x)$. Hence $D^n_x f(z)$ is the $n$-th derivative of $f$ with respect to $x$. The following is valid
\begin{align*}
D_x^n f(z)=\sum_{k=0}^nD_z^kf(z)\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}z^{k-j}D_x^nz^j
\end{align*}
We consider $z=g$
\begin{align*}
f(g)=\frac{1}{g}\qquad\qquad g=g(x)
\end{align*}
and obtain
\begin{align*}
D_x^n\left(\frac{1}{g}\right)=\sum_{k=0}^nD_g^k\left(\frac{1}{g}\right)\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}g^{k-j}D_x^ng^j\tag{1}
\end{align*}
Since
\begin{align*}
D_g^k\left(\frac{1}{g}\right)=(-1)^kk!\frac{1}{g^{k+1}}
\end{align*}
We obtain from (1)
\begin{align*}
D_x^n\left(\frac{1}{g}\right)&=\sum_{k=0}^n\sum_{j=0}^k(-1)^j\binom{k}{j}\frac{1}{g^{j+1}}D_x^ng^j\\
&=\sum_{j=0}^n(-1)^j\frac{1}{g^{j+1}}D_x^ng^j\sum_{k=j}^n\binom{k}{j}\tag{2}\\
\end{align*}
Since the following identity holds,
\begin{align*}
\sum_{k=j}^n\binom{k}{j}=\binom{n+1}{j+1}
\end{align*}
we finally obtain from (2)
\begin{align*}
D_x^n\left(\frac{1}{g}\right)=\sum_{j=0}^n(-1)^j\binom{n+1}{j+1}\frac{1}{g^{j+1}}D_x^ng^j\tag{3}\\
\end{align*}
This identity is also stated as (3.63) in H.W. Goulds book, Vol. I.
Next we consider the $n$-th derivative of $f$ with $\frac{1}{g}$ and use the Leibniz formula. We derive
\begin{align*}
D_x^n\left(\frac{f}{g}\right)&=\sum_{k=0}^n \binom{n}{k}D_x^k\left(f\right)D_x^{n-k}\left(\frac{1}{g}\right)\\
&=\sum_{k=0}^n\sum_{j=0}^{n-k} (-1)^j\binom{n}{k}\binom{n-k+1}{j+1}\frac{1}{g^{j+1}}
D_x^k\left(f\right)D_x^{n-k}\left(g^j\right)\tag{4}\\
&=\sum_{k=0}^n\sum_{j=0}^{k} (-1)^j\binom{n}{k}\binom{k+1}{j+1}\frac{1}{g^{j+1}}
D_x^{n-k}\left(f\right)D_x^{k}\left(g^j\right)\tag{5}
\end{align*}
and the claim follows.
Comment:
*
*In (4) we apply the formula (3)
*In (5) we exchange $k$ with $n-k$.
Let's look at a small example in order to see the formula in action
Example: $D^2_x\left(\tan x\right)$
\begin{align*}
D_x^2\left(\frac{\sin x}{\cos x}\right)&=\sum_{k=0}^2\binom{2}{k}\left(D_x^{2-k}\sin x\right)
\sum_{j=0}^k(-1)^j\binom{k+1}{j+1}\frac{1}{\cos^{j+1} x}\left(D_x^k\cos ^jx\right)\\
&=\binom{2}{0}\left(D_x^2\sin x\right)\left((-1)^0\binom{1}{1}\frac{1}{\cos x}D_x^0(1)\right)\\
&\qquad+\binom{2}{1}\left(D_x\sin x\right)\left((-1)\binom{2}{2}\frac{1}{\cos ^2x}\left(D_x \cos x\right)\right)\\
&\qquad+\binom{2}{2}\left(D_x^0 \sin x\right)\left((-1)\binom{3}{2}\frac{1}{\cos ^2x}\left(D_x^2 \cos x\right)\right.\\
&\qquad\qquad\qquad\qquad\qquad\qquad +\left.\binom{3}{3}\frac{1}{\cos^3x}\left(D_x^2\cos^2x\right)\right)\\
&=-\frac{\sin x}{\cos x}+2\cos x\left(\frac{\sin x}{\cos ^2x}\right)
+\sin x\left(3\frac{1}{\cos x}+\frac{2\sin^2 x - 2\cos ^2 x}{\cos ^3x}\right)\\
&=2\frac{\sin x}{\cos x}+2\frac{\sin ^3x}{\cos ^3x}\\
&=2\frac{\sin x}{\cos ^3x}\\
\end{align*}
in accordance with Wolfram Alpha
|
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|
Integral $\int \frac{\sqrt{x}}{(x+1)^2}dx$ First, i used substitution $x=t^2$ then $dx=2tdt$ so this integral becomes $I=\int \frac{2t^2}{(1+t^2)^2}dt$ then i used partial fraction decomposition the following way:
$$\frac{2t^2}{(1+t^2)^2}= \frac{At+B}{(1+t^2)} + \frac{Ct + D}{(1+t^2)^2} \Rightarrow 2t^2=(At+B)(1+t^2)+Ct+D=At+At^3+B+Bt^2 +Ct+D$$
for this I have that $A=0, B=2, C=0, D=-2$
so now I have
$I=\int \frac{2t^2}{(1+t^2)^2}dt= \int\frac{2}{1+t^2}dt - \int\frac{2}{(1+t^2)^2}dt$
Now,
$$ \int\frac{2}{1+t^2}dt = 2\arctan t$$
and
$$\int\frac{2}{(1+t^2)^2}dt$$
using partial integration we have:
$$u=\frac{1}{(1+t^2)^2} \Rightarrow du= \frac{-4t}{1+t^2}$$
and $$dt=dv \Rightarrow t=v$$
so now we have:
$$\int\frac{2}{(1+t^2)^2}dt =\frac{t}{(1+t^2)^2} + 4\int\frac{t^2}{1+t^2}dt = \frac{t}{(1+t^2)^2} + 4\int\frac{t^2 + 1 -1}{1+t^2}dt = \frac{t}{(1+t^2)^2} + 4t -4\arctan t$$
so, the final solution should be:
$$I=2\arctan t - \frac{t}{(1+t^2)^2} - 4t +4\arctan t$$
since the original variable was $x$ we have
$$I= 6\arctan \sqrt{x} - \frac{\sqrt{x}}{(1+x)^2} - 4\sqrt{x} $$
But, the problem is that the solution to this in my workbook is different, it says that solution to this integral is $$I=\arctan \sqrt{x} - \frac{\sqrt{x}}{x+1}$$
I checked my work and I couldn't find any mistakes, so i am wondering which solution is correct?
|
You have mistake here:
$u=\frac{1}{(1+t^2)^2} \Rightarrow du= \frac{-4t}{\color{red}(1+t^2\color{red}{)^3}}$
|
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|
Partition of $\{1,2,3,\cdots,3n\}$ into $n$ subsets, each with $3$ numbers, which have equal sum I want to show, that for every odd $n$ $(n\ge3)$, there exists a partition of $\{1,2,3,\cdots,3n\}$ into $n$ disjoint subsets, where each one has $3$ elements and equal sum.
The first such number is $3$. For $3$ it is obvious. $\{1,6,8\}, \{2,4,9\}, \{3,5,7\}$. I tried to show this using induction, but it seems I have some trouble with it. Please help me, if you can.
|
Let $S$ be the equal sum of each partition, then $nS=1+\ldots+3n$.
i.e. $\: nS=\frac{3n(3n+1)}{2} \implies S=\frac{3(3n+1)}{2}$
Now, $n=1$ is trivial and $n$ should be odd since the sum is an integer.
Note that the median of the sequence is $\frac{3n+1}{2}$.
Observing $1+\frac{3n+1}{2}+3n=S$
Take $a+b+c=0$:
$(1+a)+(\frac{3n+1}{2}+b)+(3n+c)=S$
By trial and error, one possible set of $n=5$ is
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
a & 0 & 1 & 2 & 3 & 4 \\
\hline
b & 0 & 1 & 2 & -2 & -1 \\
\hline
c & 0 & -2 & -4 & -1 & -3 \\
\hline
\end{array}
$$
That is,
$$
\begin{array}{|c|c|c|c|c|c|}
\hline
p & 1 & 2 & 3 & 4 & 5 \\
\hline
q & 8 & 9 & 10 & 6 & 7 \\
\hline
r & 15 & 13 & 11 & 14 & 12 \\
\hline
\end{array}
$$
Also see the links of similar question and magic rectangle
Snapshots one, two and three from Thomas R. Hagedorn, Magic rectangles revisited, Discrete Mathematics 207 (1999), 65-72.
|
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|
Interval of convergence of $\sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)x^n}{n!}$
Given the series
$$ \sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)x^n}{n!} \quad \quad k \geq 1 $$
Find the interval of convergence.
I started by applying the Ratio test
$$
\lim_{n\to \infty}\left|\frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)(k+n)x^{n+1}}{(n+1)!}\cdot \frac{n!}{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)x^n}\right|$$
$$\lim_{n\to \infty}\left|\frac{(k+n)x}{(n+1)}\right|$$
to show that the series converges when $|x| \lt 1$.
However, when I test the end points of $(-1,1)$ for convergence, I end up with two series whose convergence I am unable to show. Namely,
$$
\sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)}{n!}
$$
and
$$
\sum_{n=1}^{\infty} \frac{k(k+1)(k+2)\cdot \cdot \cdot (k + n - 1)(-1)^n}{n!}
$$
How can I show that these two series converge or diverge?
|
Your series is
$$
\sum_{n=0}^\infty\binom{-k}{n}(-x)^n=(1-x)^{-k}
$$
This alone should show that there is no convergence at $x=1$ for positive $k$.
For the series at $x=-1$ consider that
$$
\binom{-k}{n}(-1)^n=\binom{n+k-1}{n}=\binom{n+k-1}{k-1}=\frac{(n+1)(n+2)···(n+k-1)}{(k-1)!}
$$
is a polynomial in $n$ of degree $k-1$.
|
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|
integrate $\iint\limits_D \frac{2a}{\sqrt{4a^2-x^2-y^2}}dxdy, D: (x-a)^2+y^2 \le a^2$ $$I=\iint\limits_D \frac{2a}{\sqrt{4a^2-x^2-y^2}}dxdy, D: (x-a)^2+y^2 \le a^2(a>0)$$
The difficulty is to find a proper and easy way to solve this double integrals.
If do it like this, $ 0\le x \le 2a, -\sqrt{2ax-x^2} \le y \le \sqrt{2ax-x^2}$, wolfram calculation exceeds the standard time.
Maybe using polar coordinate is easier? $x-a=r\cos\theta, y=r\sin\theta(-\frac\pi2 \le \theta \le \frac\pi2, 0\le r \le 2a)$, then $I=\int_{-\pi/2}^{\pi/2} d\theta\int 2a/\sqrt{4a^2-(a+r\cos\theta)^2-(r\sin\theta)^2}rdr$ which looks rather complex too.
Am I doing it wrong? How to integrate this $I$ ?
|
Believe it or not, it's actually easier without polar coordinates.
$\int_{0}^{2a}\int_{-\sqrt{2ax-x^2}}^{\sqrt{2ax-x^2}}\frac{2a}{\sqrt{4a^2-x^2-y^2}}dydx = \int_{0}^{2a}2a\arctan{\frac{y}{\sqrt{4a^2-x^2-y^2}}}\Big\vert_{-\sqrt{2ax-x^2}}^{\sqrt{2ax-x^2}}dx$
$= 4a\int_{0}^{2a}\arctan{\frac{\sqrt{-2x(x-2a)}}{2\sqrt{-a(x-2a)}}} = 4a\int_{0}^{2a}\arctan{\sqrt{\frac{x}{2a}}}dx = 4ax\arctan{\sqrt{\frac{x}{2a}}}+8a^2\arctan{\sqrt{\frac{x}{2a}}}-4a^2\sqrt{\frac{2x}{a}}\Big\vert_{0}^{2a} = 4a^2(\pi-2)$
|
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|
Integration of rational functions by partial fractions I have the following integral :
$$\int \frac{x^2+2x+3}{x^2-5x+6}~dx$$
Although I tried to solve it by partial fractions, I could not come up with an appropriate answer. My question is more about the technique which I should use here in order to evalue the integral. I understand that the denominator could be expressed as
$$(x-2)(x-3)$$
But then how to proceed with such a complicated numerator.
|
First,
$$
x^2+2x+3=1\cdot (x^2-5x+6)+7x-3.
$$
Thus
$$
\frac{x^2+2x+3}{x^2-5x+6}=1+\frac{7x-3}{x^2-5x+6}.
$$
You know $x^2-5x+6=(x-2)(x-3)$. It is known that there exists $A,B$ such that
$$
\frac{7x-3}{x^2-5x+6}=\frac{A}{x-2}+\frac{B}{x-3}
$$
and so
$$
\frac{7x-3}{x^2-5x+6}=\frac{(A+B)x-(3A+2B)}{x^2-5x+6}.
$$
Thus we get $A=-11$ and $B=18$. Therefore
$$
\int\frac{x^2+2x+3}{x^2-5x+6}dx=x-11\ln|x-2|+18\ln|x-3|+C.
$$
|
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|
Nested radical $\sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}$ I am studying the $f(x) = \sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}$ for $x \in (0,\infty)$ and I am trying to get closed form formula for this, or at least some useful series/expansion. Any ideas how to get there?
So far I've got only trivial values, which are
$$f(1)=\sqrt{1+\sqrt{1+\cdots\sqrt{1}}}=\frac{\sqrt{5}+1}{2}$$
$$f(4) = 3$$
Second one follows from
$$2^n+1 = \sqrt{4^n+(2^{n+1}+1)} = \sqrt{4^n+\sqrt{4^{n+1}+(2^{n+2}+1)}} = \sqrt{4^n+\sqrt{4^{n+1}+\cdots}}$$
I have managed to compute several derivatives in $x_0=1$ by using chain rule recursively on $f_n(x) = \sqrt{x^n + f_{n+1}(x)}$, namely:
\begin{align*}
f^{(1)}(1) &= \frac{\sqrt{5}+1}{5}\\
f^{(2)}(1) &= -\frac{2\sqrt{5}}{25}\\
f^{(3)}(1) &= \frac{6\sqrt{5}-150}{625}\\
f^{(4)}(1) &= \frac{1464\sqrt{5}+5376}{3125}\\
\end{align*}
These gave me Taylor expansion around $x_0=1$
\begin{align*}
T_4(x) &= \frac{\sqrt{5}+1}{2} + \frac{\sqrt{5}+1}{5} (x-1) - \frac{\sqrt{5}}{25} (x-1)^2 + \frac{6\sqrt{5}-150}{3750} (x-1)^3 \\
&\ \ \ \ \ + \frac{61\sqrt{5}+224}{3125} (x-1)^4
\end{align*}
However this approach seems to be useful only very closely to the $x=1$. I am looking for something more general in terms of any $x$, but with my limited arsenal I could not get much further than this. Any ideas?
This was inspiring but kind of stopped where I did
http://integralsandseries.prophpbb.com/topic168.html
Edit:
Thanks for the answers, i will need to go through them, looks like the main idea is to divide by $\sqrt{2x}$, so then I am getting
$$\frac{\sqrt{x+\sqrt{x^2+\cdots\sqrt{x^n+\cdots}}}}{\sqrt{2x}} = \sqrt{\frac{1}{2}+\sqrt{\frac{1}{4}+\sqrt{\frac{1}{16x}+\sqrt{\frac{1}{256x^4}+...}}}}$$
Then to make expansion from this. This is where I am not yet following how to get from this to final expansion.
|
In the same spirit as Mark Fischler's answer, setting $x=\frac{y^2}2$, for large values of $x$, Taylor expansion is $$y+\frac{1}{4 \sqrt{2}}-\frac{5}{64 }\frac 1 {y}+\frac{85}{256 \sqrt{2} }\frac 1 {y^2}-\frac{1709}{8192
}\frac 1 {y^3}+\frac{6399}{32768 \sqrt{2} }\frac 1 {y^4}+O\left(\frac{1}{y^5}\right)$$ which would converge quite fast.
|
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|
Finding Maximum Area of a Rectangle in an Ellipse
Question: A rectangle and an ellipse are both centred at $(0,0)$.
The vertices of the rectangle are concurrent with the ellipse as shown
Prove that the maximum possible area of the rectangle occurs when the x coordinate of
point $P$ is $x = \frac{a}{\sqrt{2}} $
What I have done
Let equation of ellipse be
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
Solving for y
$$ y = \sqrt{ b^2 - \frac{b^2x^2}{a^2}} $$
Let area of a rectangle be $4xy$
$$ A = 4xy $$
$$ A = 4x(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) $$
$$ A'(x) = 4(\sqrt{ b^2 - \frac{b^2x^2}{a^2}}) + 4x\left( (b^2 - \frac{b^2x^2}{a^2})^{\frac{-1}{2}} \times \frac{-2b^2x}{a^2} \right) $$
$$ A'(x) = 4\sqrt{ b^2 - \frac{b^2x^2}{a^2}} + \frac{-8x^2b^2}{\sqrt{ b^2 - \frac{b^2x^2}{a^2}}a^2} = 0 $$
$$ 4a^2\left(b^2 - \frac{b^2x^2}{a^2} \right) - 8x^2b^2 = 0 , \sqrt{ b^2 - \frac{b^2x^2}{a^2}a^2} \neq 0 $$
$$ 4a^2\left(b^2 - \frac{b^2x^2}{a^2} \right) - 8x^2b^2 = 0 $$
$$ 4a^2b^2 - 4b^2x^2 - 8x^2b^2 = 0 $$
$$ 4a^2b^2 - 12x^2b^2 = 0 $$
$$ 12x^2b^2 = 4a^2b^2 $$
$$ x^2 = \frac{a^2}{3} $$
$$ x = \frac{a}{\sqrt{3}} , x>0 $$
Where did I go wrong?
edit:The duplicate question is the same but both posts have different approaches on how to solve it so I don't think it should be marked as a duplicate..
|
Its easier to solve this question using parametric points.
Let one vertex of the rectangle be $(a\cos\theta,b\sin\theta)$.
The other vertices are $(a\cos\theta,-b\sin\theta)$, $(-a\cos\theta,b\sin\theta)$, $(-a\cos\theta,-b\sin\theta)$
The area of rectangle formed is $$A(\theta)=4ab\cos\theta\sin\theta=2ab\sin2\theta$$
Maximum area is $2ab$ and it occurs when $\theta=\frac{\pi}{4}$ (or when $\sin2\theta$ is maximum).
When $\theta=\frac{\pi}{4}$, $x$-coordinate $=a\cos\frac{\pi}{4}=\frac{a}{\sqrt{2}}$
|
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|
How many tokens would person A have under these conditions? Persons A and B each have a positive integer number of tokens, and the number of tokens B has is a square number less than 100. B says to A, "If you give me all of your tokens, my total number of tokens will still be a square number." A says, "Yes - if on the other hand, you give me the same number of tokens that I already have, your total number of tokens will also be a square number." How many tokens does A have?
|
Let's start by listing all the square numbers between $0$ and $100$ $$1,4,9,16,25,36,49,64,81$$ Note we do not include $100$. We know that $$B \in [1,4,9,16,25,36,49,64,81]\\B+A \in [1,4,9,16,25,36,49,64,81,100,121,144,...]\\ B-A \in [1,4,9,16,25,36,49,64,81,100,121,144,...]$$ So what we need to find in a square equidistant from 2 other squares. Note that the last 2 sets aren't limited to be less than 100. We can derive some more equalities $$B= b^2, A=a, B+A=c^2, B-A=d^2\\b^2+a=c^2\\b^2-a=d^2\\b^2+b^2=c^2+d^2\\2b^2=c^2+d^2$$ This tells us that the sum of our two new squares can't be more than twice $b^2$. At this point we can almost brute force it as we only have $9$ choices for $b$. going through the list, $b$ cannot be $1,2,3,4$. When we get to $b=5$ however, we have $$2(5)^2= c^2 + d^2\\50 = c^2+d^2\\50=1^2+7^2\\50=1+49\\50=50$$So the solution to our problem is $$A=24$$
|
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|
Does $L^2 = P^2$ implies that $L = P$? I have encountered a problem when solving this problem:
Assume that $\alpha \in (\pi, \frac{3}{2} \pi)$, then prove $\sqrt{\frac{1 + \sin \alpha}{1 - \sin \alpha}} - \sqrt{\frac{1 - \sin \alpha}{1 + \sin \alpha}} = -2 \tan \alpha$
The most popular way to solve this kind of problems is to take left-hand side of the equation and prove that it is equal to the right-hand side. But this time, it is not so easy, because I can't see any way to transform LHS into RHS. My question is - can I take the square of LHS and prove that it equals to the RHS?
In the language of mathematics: does $L^2 = P^2$ implies that $L = P$?
|
(I made it a little more general)
Note that
$\sqrt{x}-\sqrt{\frac1{x}}
=\sqrt{x}-\frac1{\sqrt{x}}
=\frac{x-1}{\sqrt{x}}
$.
If
$x = \frac{1+y}{1-y}
$,
this is
$\begin{array}\\
\frac{x-1}{\sqrt{x}}
&=\frac{\frac{1+y}{1-y}-1}{\sqrt{\frac{1+y}{1-y}}}\\
&=\frac{1+y-(1-y)}{(1-y)\sqrt{\frac{1+y}{1-y}}}\\
&=\frac{2y}{\sqrt{(1+y)(1-y)}}\\
&=\frac{2y}{\sqrt{1-y^2}}\\
\end{array}
$
In your case,
with
$y = \sin a$,
this becomes
$\frac{2\sin a}{\sqrt{1-\sin^2a}}
=\pm\frac{2\sin a}{\cos a}
=\pm 2\tan a
$.
The restriction of $a$
then decides the sign.
|
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|
$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$ Calculate:
$$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$$
I don't know how to use L'Hôpital's Rule.
I tried to make $\tan x =\frac{\sin x}{\cos x}$ for the term ${\sqrt{\tan x}}$.
|
$$\frac{\tan x\sqrt{\tan x}-\sin x\sqrt{\sin x}}{x^3\sqrt x}=\left(\frac{\sin x} x\right)^{3/2}\cdot\frac{\frac1{\cos^{3/2}x}-1}{x^2}=$$
$$=\left(\frac{\sin x} x\right)^{3/2}\frac{1-\cos^{3/2}x}{x^2\cos^{3/2}x}=\left(\frac{\sin x} x\right)^{3/2}\frac{1-\cos^2x+\cos^{3/2}x(\cos^{1/2}x-1)}{x^2\cos^{3/2}x}=$$
$$=\left(\frac{\sin x} x\right)^{3/2}\left[\frac1{\cos^{3/2}x}\left(\frac{\sin x}x\right)^2+\frac{\cos^{1/2}x-1}{x^2}\right]=$$
$$=\left(\frac{\sin x} x\right)^{3/2}\left[\frac1{\cos^{3/2}x}\left(\frac{\sin x}x\right)^2+\frac{\cos x-1}{x^2(\cos^{1/2}x+1)}\right]=$$
$$=\left(\frac{\sin x} x\right)^{3/2}\left[\frac1{\cos^{3/2}x}\left(\frac{\sin x}x\right)^2-\frac2{\cos^{1/2}x+1}\frac{\sin^2\frac x2}{x^2}\right]=$$
$$=\left(\frac{\sin x} x\right)^{3/2}\left[\frac1{\cos^{3/2}x}\left(\frac{\sin x}x\right)^2-\frac2{\cos^{1/2}x+1}\cdot\frac14\cdot\left(\frac{\sin\frac x2}{\frac x2}\right)^2\right]\xrightarrow[x\to0^+]{}$$
$$\rightarrow1\left[1\cdot1-\frac22\cdot\frac14\cdot1\right]=1-\frac14=\frac34$$
|
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|
$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$ Solve this equation :
$$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$$
Such that $a+b+c=\pi$
I don't have any idea. I can't try anything.
|
$$x^{2}+2x \cdot \cos b \cdot \cos c+\cos^2{b}+\cos^2{c}-1=0$$
$$\frac D4= \cos^2 b \cdot \cos^2c-\cos^2b -\cos^2c+1= \left( 1-\cos^2b\right)\left(1-\cos^2 c \right)=\sin^2b \cdot \sin^2c$$
$$x_{1,2}=-\cos b \cdot \cos c \pm \sin b \cdot \sin c=-\cos (b \mp c)$$
$$x_1=-\cos (b+c)=-\cos (\pi - a)=\cos a$$
$$x_2=-\cos(b-c)$$
|
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|
Solve in positive integers the equation $a^3+b^3=9ab$ Solve in positive integers the equation:
$$a^3+b^3=9ab$$
I try to:
$$\dfrac{a^2}{b}+\dfrac{b^2}{a}=9\Longrightarrow a^2<9b,b^2<9a$$
Of course, I can't solve it. Can anyone help?
|
HINT: $x^3+y^3=9xy$ is the Cartesian equation of a Folium of Descartes $x^3+y^3=3axy$ with $a=3$. This is a rational cubic with a double point and its parametric equations are $x=\frac{9t}{1+t^3}$ and $y=tx$ The loop of the curve occurs for the values $t\gt 0$ so the integer points we are looking for belong to this loop.
For $t\ge 3$ one has $x\lt 1$ hence possible integers values $x, y$ correspond to values of the parameter $t$ such that $0\lt t \lt 3$.
When $t=1$ there is not integer solution $(a,b)$ but when $t=2$ there is the solution $\color {red}{(a,b)=(2,4)}$ (and trivially $(b,a)$ from symmetry over $y=x$).
Now,to discard fractional values of $t$, taking derivatives, we have
$$x’=\frac{9-18t^3}{(1+t^3)^2}$$
$$y’=\frac{18t-9t^4}{(1+t^3)^2}$$ it follows
$$\max x=\frac{9\sqrt[3]{2}}{1+2}=3\sqrt[3]{2}\approx 3.7797$$
$$\max y=\frac{9\sqrt[3]{4}}{1+2}=3\sqrt[3]{4}\approx 4.7622$$
Consequently there are just $12$ points to verify (in the “lattice” $\{1,2,3\}\text {x}\space \{1,2,3,4\}$). Thus the only solution in positive integers is the given one $(2,4)$.
|
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|
Criticize my math when I attempt to find the coefficient of $x^2y^6$ in the expansion of $(x+2y^2)^5$ So I look around this site and my textbook (Richmond&Richmond, discrete math) and I know I'm in the right direction but I'm also sure I am doing it wrong.
Original Question: find the coefficients of $x^2y^6$ in the expansion of
$$(x + 2y^2)^5$$
So far I have:
$$(x+2 y^2)^5 \implies x^2+2^3×y^6 \implies (5!)/(2!×6!)×2^3 = 2/3$$
But that definitely does not seem right.
I'm also avoiding learning combinations by drawing out Pascal's Triangle since on a time sensitive exam its not practical.
Comments, suggestions, snide remarks; anything to help is appreciated ahead of time.
|
There are two aspects in your derivation which should be corrected.
When selecting the terms with $x^2$ and $\left(2y^2\right)^3$ you have to multiply them instead of adding them.
\begin{align*}
(x+2 y^2)^5 \implies x^2\left(2y^2\right)^3=2^3x^2y^6
\end{align*}
When identifying the corresponding binomial coefficient, note that it is
\begin{align*}
(5!)/(2!3!)=\binom{5}{2}=\binom{5}{3}=10
\end{align*}
You derivation could therefore be
\begin{align*}
(x+2 y^2)^5 \implies 2^3x^2y^6 \implies \frac{5!}{2!3!}\cdot 2^3=\frac{120}{2\cdot6}\cdot8=80
\end{align*}
Another variation:
In case you already know the binomial theorem you can use
\begin{align*}
\left(x+2y^2\right)^5=\sum_{k=0}^5\binom{5}{k}x^k\left(2y^2\right)^{5-k}\tag{1}
\end{align*}
In order to find the coefficient of $x^2$ we see there is only the term with $k=2$ a candidate. The summand with $k=2$ gives
\begin{align*}
\binom{5}{2}x^2\left(2y^2\right)^{5-2}=2^3\binom{5}{2}x^2y^6=80x^2y^6
\end{align*}
and we conclude the coefficient of $x^2y^6$ is $80$.
|
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|
How to evaluate $\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x$ How to evaluate
$$\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x$$
I completely have no idea how to find the result.Mathematic gave me the following answer part of the integral
$$\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( 1+x^{2} \right )\mathrm{d}x=-\frac{1}{4}\mathbf{G}\pi +\frac{\pi ^{2}}{16}\ln 2+\frac{21}{64}\zeta \left ( 3 \right )$$
where $\mathbf{G}$ donates the Catalan's Constant.
But it can't evaluate the other part.So I'd like to know how to evaluate the original integral or the above integral.
|
\begin{align}J&=\int_{0}^{1}\frac{\arctan x}{1+x^{2}}\ln\left ( \frac{1+x^{2}}{1+x} \right )\mathrm{d}x\\
&\overset{y=\frac{1-x}{1+x}}=\frac{\pi}{4}\int_0^1\frac{1}{1+y^2}\ln\left ( \frac{1+y^{2}}{1+y} \right )\mathrm{d}y-J\\
2J&=\frac{\pi}{4}\int_0^1\frac{\ln\left ( 1+y^{2} \right )}{1+y^2}\mathrm{d}y-\frac{\pi}{4}\int_0^1\frac{\ln\left ( 1+y \right )}{1+y^2}\mathrm{d}y\\
K&=\int_0^1\frac{\ln\left ( 1+y^{2} \right )}{1+y^2}\mathrm{d}y\\&\overset{u=\frac{1}{y}}=\int_1^{\infty}\frac{\ln\left(1+u^2\right)}{1+u^2}du-2\underbrace{\int_1^{\infty}\frac{\ln u}{1+u^2}du}_{v=\frac{1}{u}}\\
2K&=\int_0^\infty\frac{\ln\left ( 1+y^{2} \right )}{1+y^2}\mathrm{d}y+2\int_0^{1}\frac{\ln u}{1+u^2}du\\
&\overset{z=\sqrt{\frac{1-\frac{u}{\sqrt{{{u}^{2}}+1}}}{1+\frac{u}{\sqrt{{{u}^{2}}+1}}}}}=-2\int_0^1 \frac{\ln\left(\frac{4z^2}{(1+z^2)^2}\right)}{1+z^2}dz-2\text{G}\\
&=2\text{G}-\pi\ln 2+4K\\
K&=\boxed{\frac{1}{2}\pi\ln 2-\text{G}}\\
L&=\int_0^1\frac{\ln\left ( 1+y \right )}{1+y^2}\mathrm{d}y\\
&\overset{u=\frac{1-y}{1+y}}=\int_0^1 \frac{\ln\left(\frac{2}{1+u}\right)}{1+u^2}du\\
2L&=\boxed{\frac{1}{4}\pi\ln 2}\\
\end{align}
Therefore,
\begin{align}\boxed{\displaystyle J=\dfrac{3}{64}\pi^2\ln 2 -\dfrac{1}{8}\pi\text{G} }\end{align}
|
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|
Multiple radicals: $(\sqrt{10}+\sqrt{11}+\sqrt{12})(\sqrt{10}+\sqrt{11}-\sqrt{12})(\sqrt{10}-\sqrt{11}+\sqrt{12})(-\sqrt{10}+\sqrt{11}+\sqrt{12})$ $(\sqrt{10}+\sqrt{11}+\sqrt{12})(\sqrt{10}+\sqrt{11}-\sqrt{12})(\sqrt{10}-\sqrt{11}+\sqrt{12})(-\sqrt{10}+\sqrt{11}+\sqrt{12})$
I don't think multiplying these out will work, and I am stuck in the beginning, without a basic concept to get started. Can anyone show me how to do these? I would appreciate more detailed responses as I have a solution to this already but since it does not explain the steps, I cannot learn from it. Thanks
Answer, if you want to check
359
|
It will be easier if you consider the expression
\begin{align}
(a+b+c)(a+b-c)(a-b+c)(-a+b+c)&=[(a+b)^2-c^2][-(a-b)^2+c^2]\\
&=(2ab+a^2+b^2-c^2)(2ab-a^2-b^2+c^2)\\
&=4a^2b^2-(a^2+b^2-c^2)^2\\
&=4a^2b^2-a^4-b^4-c^4-2a^2b^2+2a^2c^2+2b^2c^2\\
&=2(a^2b^2+b^2c^2+c^2a^2)-(a^4+b^4+c^4)
\end{align}
And then you put $a=\sqrt{10}$, $b=\sqrt{11}$ and $c=\sqrt{12}$.
|
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|
Is Sigma $\Sigma$ a mathematical way of doing a for loop? I've been a programmer for ten years, and once upon a time I was pretty good at math. Those days are long gone. I'm taking some online classes and now I find myself needing to remember the math I learned in college and it has all gone bye-bye.
Specifically, I have a question about $\Sigma$ (Sigma):
In an $\Sigma$ equation, there is an N above the $\Sigma$, and an $i = 1$ (or some number) below.
If someone could help my programmer mind - this looks like a for loop in programming. Is that essentially what is going on with everything inside the $()$ of the $\Sigma$? It simply computes all the values for each iteration from $i$ to N and then sums them all up?
|
In Mathematica I frequently switch between the Sum and the Table commands like in these examples:
Example 1:
$\sum\limits_{n=1}^{n=1} 1 = 1 = 1 $
$\sum\limits_{n=1}^{n=2} 1 = 2 = 1+1 $
$\sum\limits_{n=1}^{n=3} 1 = 3 = 1+1+1 $
$\sum\limits_{n=1}^{n=4} 1 = 4 = 1+1+1+1 $
$\sum\limits_{n=1}^{n=5} 1 = 5 = 1+1+1+1+1 $
Sum[1,{n,1,5}]
which outputs: 5
compared to:
Table[1,{n,1,5}]
which outputs: {1,1,1,1,1}
Example 2:
$\sum\limits_{n=1}^{n=1} n = 1 = 1 $
$\sum\limits_{n=1}^{n=2} n = 3 = 1+2$
$\sum\limits_{n=1}^{n=3} n = 6 = 1+2+3$
$\sum\limits_{n=1}^{n=4} n = 10 = 1+2+3+4$
$\sum\limits_{n=1}^{n=5} n = 15 = 1+2+3+4+5$
Sum[n,{n,1,5}]
which outputs: 15
compared to:
Table[n,{n,1,5}]
which outputs: {1,2,3,4,5}
Example 3:
$\sum\limits_{n=1}^{n=1} \frac{n(n+1)}{2} = 1 = 1$
$\sum\limits_{n=1}^{n=2} \frac{n(n+1)}{2} = 4 = 1+3$
$\sum\limits_{n=1}^{n=3} \frac{n(n+1)}{2} = 10 = 1+3+6$
$\sum\limits_{n=1}^{n=4} \frac{n(n+1)}{2} = 20 = 1+3+6+10$
$\sum\limits_{n=1}^{n=5} \frac{n(n+1)}{2} = 35 = 1+3+6+10+15$
Sum[n*(n+1)/2,{n,1,5}]
which outputs: 35
compared to:
Table[n*(n+1)/2,{n,1,5}]
which outputs: {1, 3, 6, 10, 15}
I don't know if it is a good answer to your question though.
|
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|
If $f(x)=\lim_{n\to\infty}[2x+4x^3+\cdots+2nx^{2n-1}]$, $0
If $f(x)=\lim_{n\to\infty}[2x+4x^3+\cdots+2nx^{2n-1}]$, $0<x<1$, then find $\int f(x)\mathrm{d}x$
$$f(x)=\lim_{n\to\infty}2x[1+2x^2+\cdots+nx^{2n}]$$
$$S=\frac{f(x)}{2x}=1+2x^2+\cdots+nx^{2n}$$
$S$ is an AGP. I used the general method for finding $S$.
$$x^2S=x^2+2x^4+\cdots+nx^{2n+2}$$
$$(1-x^2)S=1+(x^2+x^4+\cdots+x^{2n})+nx^{2n+2}$$
$$(1-x^2)S=\frac{1-x^{2n+1}}{1-x^2}+nx^{2n+2}$$
$$S=\frac{1-x^{2n+1}}{(1-x^2)^2}+\frac{nx^{2n+2}}{1-x^2}$$
$$f(x)=\lim_{n\to\infty}\left(2x\frac{1-x^{2n+1}}{(1-x^2)^2}+2x\frac{nx^{2n+2}}{1-x^2}\right)$$
I got stuck here.
|
The calculation is fine. Note that
$$\lim_{n\to\infty}x^{2n+1}=0\quad\text{and}\quad \lim_{n\to\infty} nx^{2n+2}=0.$$
Thus
$$f(x)=\frac{2x}{(1-x^2)^2}.$$
Now integrate, using the substitution $u=1-x^2$.
|
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|
Find the value of $\sqrt[4]{\alpha}-\sqrt[4]{\beta}$,where $\sqrt[4]{.}$ denotes the principal value. If $\alpha$ and $\beta$ are the roots of the equation $x^2-34x+1=0$,find the value of $\sqrt[4]{\alpha}-\sqrt[4]{\beta}$,where $\sqrt[4]{.}$ denotes the principal value.
I found out the $\alpha$ and $\beta$.
$\alpha,\beta=\frac{34\pm\sqrt{32\times 36}}{2}=17\pm12\sqrt2$ but i do not know how to find $\sqrt[4]{\alpha}-\sqrt[4]{\beta}$.
|
One may observe that
$$
(\sqrt{2}-1)^2=3-2\sqrt{2}
$$ and that
$$
(3-2\sqrt{2})^2=17-12\sqrt{2}
$$ thus
$$
(\sqrt{2}-1)^4=17-12\sqrt{2}
$$ giving
$$
\sqrt[4]{\alpha}=\sqrt[4]{17-12\sqrt{2}}=\sqrt{2}-1,
$$
similarly
$$
\sqrt[4]{\beta}=\sqrt[4]{17+12\sqrt{2}}=\sqrt{2}+1.
$$
|
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|
Inequality involving four numbers Show that, if $a$, $b$, $c$ and $d$ are four positive numbers with sum $1$, then $$\frac 3 {1-a} + \frac 3 {1-b} + \frac 3 {1-c} + \frac 3 {1-d} \ge \frac 5 {1+a} + \frac 5 {1+b} + \frac 5 {1+c} + \frac 5 {1+d}.$$ I tried to subtract the fractions, but I didn't get to any result.
|
Hint: It is enough to show that for $x\in (0,1)$,
$$f(x)=\frac3{1-x}-\frac5{1+x}+\frac{32}{15}(1-4x)=\frac{2(4x-1)^2(4x+1)}{15(1-x^2)}\ge0$$
|
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|
Fermat-Torricelli minimum distance The Fermat - Torricelli point minimizes sum of distances $S$ taken from vertices of a triangle of sides $a,b,c. $ Find $S$ in terms of $a,b,c$.
Am trying to set up problem with a Lagrange multiplier or partial derivatives for extremization but it seems tedious even with a CAS.
Although it is supposed to be known from the earliest Greek times, it is not seen (by me) in these modern times.
|
Let
*
*$a, b, c$ be the sides of $\triangle ABC$ whose angles are smaller than $120^\circ$.
*$P$ be the Fermat-Torricelli point for $\triangle ABC$.
*$\alpha = |AP|$, $\beta = |BP|$, $\gamma = |CP|$ and $S = \alpha+\beta+\gamma$.
*$\mathcal{A}$ be the area of $\triangle ABC$.
It is known that for such a triangle, $P$ is lying in its interior and
$$\angle APB = \angle BPC = \angle CPA = 120^\circ$$
Using these, we can express the sides and area of triangle as
$$\begin{cases}
a^2 = \beta^2 + \beta\gamma + \gamma^2\\
b^2 = \gamma^2 + \gamma\alpha + \alpha^2\\
c^2 = \alpha^2 + \alpha\beta + \beta^2
\end{cases}
\quad\text{and}\quad
\mathcal{A} = \frac{\sqrt{3}}{4}(\alpha\beta + \beta\gamma+\gamma\alpha)
$$
Summing the three equations from left and apply the equation from right, we find
$$a^2+b^2+c^2 = 2(\alpha^2 + \beta^2 + \gamma^2) + \frac{4}{\sqrt{3}}\mathcal{A}$$
As a result,
$$\begin{align}
S^2 &= (\alpha+\beta+\gamma)^2
= \alpha^2 + \beta^2 + \gamma^2 + \frac{8}{\sqrt{3}}\mathcal{A}
= \frac12\left(a^2 + b^2 + c^2 - \frac{4}{\sqrt{3}}\mathcal{A}\right) +
\frac{8}{\sqrt{3}}\mathcal{A}\\
&= \frac12\left(a^2 + b^2 + c^2\right) + 2\sqrt{3}\mathcal{A}\\
&= \frac12\left(a^2+b^2+c^2 + \sqrt{3(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}\,\right)
\end{align}
$$
An expression equivalent to what Quang Hoang obtained in another answer.
|
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|
The right way to cancel out the terms in the following telescoping series So how do I cancel and simplify the terms in the following telescopic series.
Been at it for hours, cant seem to figure it out.
$\sum\limits_{k = 1}^n \frac{1}{2(k+1)} -\frac{1}{k+2}+\frac{1}{2(k+3)} $
Any help would be deeply appreciated.
P.S: I need to show that its equal to the following,
$\frac{1}{12} - \frac{1}{2(n+2)} + \frac{1}{2(n+3)}$
But I cant seem to figure out the right cancellation method.
|
Here is another variant
\begin{align*}
\sum_{k=1}^n&\left(\frac{1}{2(k+1)}-\frac{1}{k+2}+\frac{1}{2(k+3)}\right)\\
&=\frac{1}{2}\sum_{k=1}^n\frac{1}{k+1}-\sum_{k=1}^n\frac{1}{k+2}+\frac{1}{2}\sum_{k=1}^n\frac{1}{k+3}\tag{1}\\
&=\frac{1}{2}\sum_{k=1}^{n}\frac{1}{k+1}-\sum_{k=2}^{n+1}\frac{1}{k+1}+\frac{1}{2}\sum_{k=3}^{n+2}\frac{1}{k+1}\tag{2}\\
&=\frac{1}{2}\left(\frac{1}{2}+\frac{1}{3}\right)-\left(\frac{1}{3}+\frac{1}{n+2}\right)+\frac{1}{2}\left(\frac{1}{n+2}+\frac{1}{n+3}\right)\tag{3}\\
&=\frac{1}{12}-\frac{1}{2(n+2)}+\frac{1}{2(n+3)}\tag{4}
\end{align*}
Comment:
*
*In (1) we split the sum
*In (2) we shift the index by $1$ resp. $2$
*In (3) we observe that the sums with index range $k=3$ up to $k=n$ cancel away
*In (4) we collect terms
|
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|
If a chord joining the points $P(a\sec\alpha,a\tan\alpha)$ and $Q(a\sec\beta,a\tan\beta)$ on the hyperbola $x^2-y^2=a^2$ is a normal to it at $P$,then If a chord joining the points $P(a\sec\alpha,a\tan\alpha)$ and $Q(a\sec\beta,a\tan\beta)$ on the hyperbola $x^2-y^2=a^2$ is a normal to it at $P$,then show that $\tan \beta=\tan\alpha(4\sec^2\alpha-1)$.
The chord joining the points $P(a\sec\alpha,a\tan\alpha)$ and $Q(a\sec\beta,a\tan\beta)$ on the hyperbola $x^2-y^2=a^2$ is given by $x\cos\frac{\alpha-\beta}{2}-y\sin\frac{\alpha+\beta}{2}=a\cos\frac{\alpha+\beta}{2}$....................$(1)$
And the normal to the hyperbola at $P(a\sec\alpha,a\tan\alpha)$ is given by
$\frac{x}{\sec\alpha}+\frac{y}{\tan\alpha}=2a$.............$(2)$
equation $(1)$ and $(2)$ are the same lines
Comparing them,gives
$\frac{\cos\frac{\alpha-\beta}{2}}{\frac{1}{\sec\alpha}}=\frac{-\sin\frac{\alpha+\beta}{2}}{\frac{1}{\tan\alpha}}=\frac{a\cos\frac{\alpha+\beta}{2}}{2a}$
$\sec\alpha=\frac{\cos\frac{\alpha+\beta}{2}}{2\cos\frac{\alpha-\beta}{2}}$ and $\tan\alpha=\frac{\cos\frac{\alpha+\beta}{2}}{-2\sin\frac{\alpha+\beta}{2}}$
I am stuck here.I do not know how to proceed further.
|
The equation of the line $PQ$ is given by
$$y-a\tan\alpha=\frac{a\tan\alpha-a\tan\beta}{a\sec\alpha-a\sec\beta}(x-a\sec\alpha),$$
i.e.
$$\frac{\sin\alpha\cos\beta-\sin\beta\cos\alpha}{\sin\alpha-\sin\beta}x+\frac{\cos\alpha-\cos\beta}{\sin\alpha-\sin\beta}y=a\tag3$$
The $(2)$ you wrote can be written as
$$\frac{\cos\alpha}{2}x+\frac{1}{2\tan\alpha}y=a\tag4$$
Comparing $(3)$ with $(4)$ gives
$$\frac{\sin\alpha\cos\beta-\sin\beta\cos\alpha}{\sin\alpha-\sin\beta}=\frac{\cos\alpha}{2},$$
i.e.
$$2\sin\alpha-\cos\alpha\tan\beta=\cos\alpha\sin\alpha\sec\beta\tag5$$
Squaring the both sides and using $\sec^2\beta=1+\tan^2\beta$ gives
$$(\cos^2\alpha\sin^2\alpha-\cos^2\alpha)\tan^2\beta+4\sin\alpha\cos\alpha\tan\beta+\cos^2\alpha\sin^2\alpha-4\sin^2\alpha=0$$
and so
$$\begin{align}&\tan\beta\\&=\frac{-2\sin\alpha\cos\alpha\pm\sqrt{4\sin^2\alpha\cos^2\alpha-(\cos^2\alpha\sin^2\alpha-\cos^2\alpha)(\cos^2\alpha\sin^2\alpha-4\sin^2\alpha)}}{\cos^2\alpha\sin^2\alpha-\cos^2\alpha}\\\\&=\frac{-2\sin\alpha\cos\alpha\pm\sqrt{\cos^2\alpha\sin^2\alpha(\cos^2\alpha-2)^2}}{\cos^2\alpha\sin^2\alpha-\cos^2\alpha}\\\\&=\frac{-2\sin\alpha\cos\alpha\pm \cos\alpha\sin\alpha(\cos^2\alpha-2)}{\cos^2\alpha\sin^2\alpha-\cos^2\alpha}\\\\&=\frac{-2\sin\alpha\cos\alpha - \cos\alpha\sin\alpha(\cos^2\alpha-2)}{\cos^2\alpha\sin^2\alpha-\cos^2\alpha},\quad \frac{-2\sin\alpha\cos\alpha + \cos\alpha\sin\alpha(\cos^2\alpha-2)}{\cos^2\alpha\sin^2\alpha-\cos^2\alpha}\\\\&=\tan\alpha,\quad \tan\alpha(4\sec^2\alpha-1)\end{align}$$
Now when $\tan\beta=\tan\alpha$,
$$\begin{align}(5)&\Rightarrow \sin\alpha(\cos\beta-\cos\alpha)=0\\&\Rightarrow \sin\alpha=0\quad\text{or}\quad \cos\beta=\cos\alpha\end{align}$$
Case 1 : $\sin\alpha=0\Rightarrow \tan\alpha=\tan\beta=0\Rightarrow (\alpha,\beta)=(0,\pi),(\pi,0)\qquad (\text{$\tan\beta=\tan\alpha(4\sec^2\alpha-1)$ holds in this case})$
Case 2 : $\cos\beta=\cos\alpha\Rightarrow \sin\alpha=\sin\beta\Rightarrow \alpha=\beta\qquad (\text{this case is eliminated})$
Thus, we get that $\tan\beta=\tan\alpha(4\sec^2\alpha-1)$.
|
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|
Evaluate the line integral of a parabola How can I evaluate :
$$\int_{C} y \;dx + x^2 \; dy$$
where $C$ is the parabola define by
$$y=4x-x^2 \quad \text{from } \; (4,0) \; \text{ to } \; (1,3).$$
Do I need to parameterize the parabola?
|
$$\because y = 4x-x^2 , \therefore \frac{dy}{dx} = 4-2x , \therefore dy = (4-2x)dx $$
Hence,
$$\int_{C} y \;dx + x^2 \; dy = \int_{4}^{1} 4x-x^2 dx + \int_{4}^{1} x^2 (4-2x)dx$$
$$\implies \int_4^1 4x-x^2+4x^2-2x^3$$
$$\implies \int_4^1 4x+3x^2-2x^3$$
$$\implies (2x^2+x^3-\frac{x^4}{2})_4^1$$
$$\implies (2+1-\frac{1}{2}) - (32+64-128)$$
$$\implies 35-\frac{1}{2}$$
$$\implies \frac{69}{2}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find $\lim_{x \to 1}\frac{\sqrt{x^2+35}-6}{x-1}$ I've rationalized the numerator to
$x^2+35-36$
$$= \frac{x^2+35-36}{(x-1)\left(\sqrt{x^2+35}+6\right)} = \frac00$$
(when I substitute $x=1$)
I don't know what to do to the denominator so I can substitute $x=1$ to find the limit.
|
You already have been given good answers to the problem.
Let me show you another approach which can be of interest (even just for your curiosity at the present time).
To make life simpler, set $x=1+y$. This makes
$$A=\frac{\sqrt{x^2+35}-6}{x-1}=\frac{\sqrt{y^2+2 y+36}-6}{y}$$ Consider the first portion of the numerator $$\sqrt{y^2+2 y+36}=6 \sqrt{1+\frac y{18}+\frac {y^2}{36}}$$ Since $y\to 0$, term $y^2$ is very negligible and then $$\sqrt{1+\frac y{18}+\frac {y^2}{36}}\approx \sqrt{1+\frac y{18}}$$ May be, you already know that for small values of $z$, $\sqrt{1+z}\approx 1+\frac z2$ (you can easily check it with a pocket calculator). So $$\sqrt{1+\frac y{18}+\frac {y^2}{36}}\approx \sqrt{1+\frac y{18}}\approx1+\frac y{36}$$ All of that makes $$A\approx\frac{6(1+\frac y{36})-6} y=\frac{6+\frac y{6}-6}y=\frac 16$$ which is the limit.
|
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|
$\sqrt{x}-\sqrt{z+y}=\sqrt{y}-\sqrt{z+x}=\sqrt{z}-\sqrt{x+y}$ At a recent maths competition one of the questions was to find for which $x,y,z$ this equation holds true:
$$\sqrt{x}-\sqrt{z+y}=\sqrt{y}-\sqrt{z+x}=\sqrt{z}-\sqrt{x+y}$$
where $x,y,z \in \mathbb{R} \cup \{0\}$. So how am I supposed to approach this problem?
Also sorry for not explaining my personal progress on this problem, but there simply is none.
|
$x \ge 0; y \ge 0; z \ge 0$.
Suppose $z = 0$
Then $\sqrt{x} - \sqrt{y} = \sqrt{y} - \sqrt{x} = - \sqrt{x+y} \implies \sqrt{x} = \sqrt{y} \implies x = y; x + y = 0 \implies x = y = z = 0$.
Likewise if $x = 0$ or $y = 0$ then $x = y = z = 0$ by the same argument so either they all equal 0 or none do.
So assume no $x, y, $ or $z$ = 0:
$\sqrt{x}-\sqrt{z+y}=\sqrt{y}-\sqrt{z+x}=\sqrt{z}-\sqrt{x+y} \implies$
$(\sqrt{x}-\sqrt{z+y})^2=(\sqrt{y}-\sqrt{z+x})^2=(\sqrt{z}-\sqrt{x+y})^2 \implies$
$x+y+z -2(\sqrt{x}\sqrt{z+y})=x+y+z -2(\sqrt{y}\sqrt{z+x})=x+y+z -2(\sqrt{z}\sqrt{x+y}) \implies$
${x}({z+y})={y}({z+x})={z}({x+y}) \implies$
$xz + xy = yz + xy = xz + yz \implies$
$xz = yz; xy = xz; xy = yz $
Thus $x = y; y = z$ and $x = z$.
So any $x = y = z \ge 0$ will work.
|
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|
Contraction of an ideal
Let $f: \mathbb{Z}[X] \longrightarrow \mathbb{Z}[\sqrt{2}]$ be a ring homomorphism sending $X$ to $\sqrt{2}$.
I am asked to compute a few contractions, and I am wondering if I could get some help in understanding the reasoning. I also was told that I need to compute the kernel of $f$, which I found to be $(X^2 - 2)$.
I was also told that the kernel of $f$ is contained in every contraction. Is this true or did I possible misunderstand? If it is true, why is it so?
Here are a couple problems that I tried, and I am wondering if they are correct.
$i)$ $(0)^c = f^{-1}(0) = \ker(f) = (X^2 - 2)$
$ii)$ $(\sqrt{2})^c = f^{-1}(\sqrt{2}) = (X,X^2 - 2) = (X,2)$
(How does $(X, X^2 - 2)$ simplify to be $(X,2)$?)
$iii)$ $(2)^c = f^{-1}(2) = (X^2, X^2 - 2)$ (How does this ideal "simplify"?)
Thank you for your help!!!
|
Note that for any ring homomorphism $f \colon R \to S$, and any ideal $J \subseteq S$, we have - due to $0 \in J$ -
$$ J^c = f^{-1}[J] \supseteq f^{-1}[0] = \ker f $$
so the kernel is contained in every contraction.
In the case here, we have $\ker f = (X^2 - 2)$, as you already found. For the three problems, we have
(i) $0^c = \ker f = (X^2 - 2)$
(ii) $(\sqrt 2)^c = f^{-1}[(\sqrt 2)] = (X,X^2-2) = (X,2)$
As $f(X) = \sqrt 2$, $f(X^2 -2) = 0$, we have $(X^2 - 2, X) \subseteq (\sqrt 2)^c$.
Now let $p \in (\sqrt 2)^c$, that is $f(p) \in (\sqrt 2)$. Divide $p$ by $X^2 - 2$, giving
$$ p(X) = (X^2 - 2)q(X) + r(X) $$
where $\deg r < 2$, dividing $r$ by $X$, we get
$$ p(X) = (X^2 - 2)q(X) + q_2(X)X + r_2(X) $$
where $\deg r_2 < 1$. Under $f$ we have
$$ f(p) = 0 \cdot f(q) + f(q_2) \cdot \sqrt 2 + f(r_2) $$
hence $f(r_2) \in (\sqrt 2)$. As $\sqrt 2$ is irrational and $r_2$ is constant, we must have $r_2 = 0$. That is
$$ p(X) = (X^2 - 2)q(X) + q_2(X)X \in (X^2 - 2, X).$$
For the simplification:
Note that $X \in (X^2 - 2, X)$ and $2 = X\cdot X - (X^2 -2) \in (X^2 - 2, X)$, hence $(X,2) \subseteq (X^2-2,X)$, and on the other hand $X^2 - 2 = X \cdot X + (-1) \cdot 2 \in (X,2)$, hence $(X^2 -2, X) \subseteq (X,2)$.
(iii) $(2)^c = f^{-1}[(2)] = (X^2, X^2 - 2) = (X^2, 2)$
Along the same line as above, we have: $(X^2, X^2 - 2) \subseteq (2)^c$. Now let $p \in (2)^c$,
divide $p$ by $X^2 - 2$, giving
$$ p(X) = (X^2 - 2)q(X) + r(X) $$
where $\deg r < 2$. Under $f$ we have
$$ f(p) = f(r) $$
hence $f(r) \in (2)$. Now $r(X) = aX + b$ for some $a,b \in \mathbf Z$, as $f(r) = a\sqrt 2 + b \in (2)$, both $a$ and $b$ are even, say $a = 2a_1$, $b = 2b_1$. That is
$$ p(X) = (X^2 - 2)q(X) + 2 a_1X + 2 b_1
= (X^2 - 2)\bigl(q(X) - a_1X - b_1\bigr)
+ X^2(a_1X + b_1X) \in (X^2 , X^2- 2). $$
For the simplification we get that $X^2 \in (X^2 - 2, X^2)$ and $2 = 1\cdot X^2 - (X^2 -2) \in (X^2 - 2, X^2)$, hence $(X^2,2) \subseteq (X^2-2,X^2)$, and on the other hand $X^2 - 2 = 1 \cdot X^2 + (-1) \cdot 2 \in (X^2,2)$, hence $(X^2 -2, X) \subseteq (X^2,2)$.
|
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|
How to calculate this integral without any integration techniques?
Differentiate $f(x) = (5x+2)\ln(2x+1)$ with respect to $x$. Hence, find $\int \ln(2x+1)^3dx$.
Because of the word "Hence" I'm assuming that the question doesn't allow integration techniques such as integration by parts or substitution.
The first part is trivial. The derivative is $5\ln(2x+1) + \dfrac{2(5x+2)}{2x+1}$. Now my line of thought so far has been to somehow get this derivative to the desired result: $\ln(2x+1)^3$ and then use $f(x)$ to get the integral. However I don't see any straight way to do this.
So I decided to experiment a little. Firstly I differentiated $(3x+2)\ln(2x+1)$ and got $$3\ln(2x+1) + \dfrac{2(3x+2)}{2x+1}$$ Now at least I have the $\ln(2x+1)^3$ term but I have another complication: $\dfrac{2(3x+2)}{2x+1}$.
I decided to take a look at the answer: $3x\ln(2x+1) + \dfrac{3}{2}\ln(2x+1) - 3x + C$. I thought that if I differentiated this, I would have some idea of where I should go, and how to proceed. So that's what I did, and I got:
$$3\ln(2x+1) + \dfrac{2(3x)}{2x+1} + \dfrac{3}{2x+1} - 3$$ This is perfect because $\dfrac{2(3x)}{2x+1} + \dfrac{3}{2x+1} - 3$ cancels out perfectly. But I have no idea how to get here, starting from $f'(x)$. But I'd say that my first experiment was pretty close. Any help would be highly appreciated.
|
The derivative is
$$f'(x) = 5 \log{(2 x+1)} + \frac{10 x+4}{2 x+1} = 5 \log{(2 x+1)} + 5 - \frac1{2 x+1} $$
Integrate both sides wrt $x$:
$$f(x) = (5 x+2) \log{(2 x+1)} = 5 \int \log{(2 x+1)} dx + 5 x - \frac12 \log{(2 x+1)} +C$$
Thus
$$ 5 \int \log{(2 x+1)} dx = \left (5 x + \frac52 \right ) \log{(2 x+1)} - 5 x + C$$
or
$$\int \log{(2 x+1)^3} dx = \frac32 (2 x+1) \log{(2 x+1)}- 3 x + C'$$
|
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|
Sum of the Powers of $2$ Suppose I have a sequence consisting of the first, say, $8$ consecutive powers of $2$ also including $1$: $1,2,4,8,16,32,64,128$. Why is it that for example, $1 + 2 + 4 = 7$ is $1$ less than the next term in the series, $8$? Even if one was to try, for instance, $524,288 + 1,048,576$ ($2^{19}$ and $2^{20}$), the next term ($2^{21}$) will be $2,097,152$, which is one greater than the previous sum. I found this whilst solving the "if you are given a penny each day and it doubles" problem. Can someone please explain this to me? Thank you.
|
This is a consequence of a well-known formula, but here is an elementary proof for your case, just using distributivity and the fact that $2 - 1 = 1$:
\begin{align}
(1 + 2 + 4 + \dots + 2^n) &= (2-1)(1 + 2 + 4 + \dots + 2^n) \\
&= 2(1 + 2 + 4 + \dots + 2^n) - (1 + 2 + 4 + \dots + 2^n) \\
&= (2 + 4 + 8 + \dots + 2^{n+1}) - (1 + 2 + 4 + \dots + 2^n) \\
&= 2^{n+1} - 1
\end{align}
|
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|
Calculate the limit of: $x_n = \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$, $n \rightarrow \infty$ Is it ok to solve the following problem this way? What I have done is to solve parts of the limit first (that converges to $0$), and then solve the remaining expression? Or is this flawed reasoning?
Question
Calculate the limit of:
$$x_n = \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$$
when $n$ goes to infinity.
Answer
This can also be written as:
$$\lim_{n \to \infty} \frac{\ln(1+\sqrt{n}+\sqrt[3]{n})}{\ln(1 + \sqrt[3]{n} + \sqrt[4]{n})}$$
The denominator can be written as:
$$\ln(1 + \sqrt[3]{n} + \sqrt[4]{n}) = \ln(1 + \frac{1 + \sqrt[4]{n}}{\sqrt[3]{n}}) + \ln(\sqrt[3]{n})$$
From this we can see that:
$$\lim_{n \to \infty} \ln(1 + \frac{1 + \sqrt[4]{n}}{\sqrt[3]{n}}) \rightarrow 0$$
The numerator can be written as:
$$\ln(1 + \sqrt{n} + \sqrt[3]{n}) = \ln(1 + \frac{1 + \sqrt[3]{n}}{\sqrt{n}}) + \ln(\sqrt{n})$$
From this we can see that:
$$\lim_{n \to \infty} \ln(1 + \frac{1 + \sqrt[3]{n}}{\sqrt{n}}) \rightarrow 0$$
This means that we have the following limit:
$$\lim_{n \to \infty} \frac{\ln(\sqrt{n})}{\ln(\sqrt[3]{n})} = \lim_{n \to \infty} \frac{\ln(n^{\frac{1}{2}})}{\ln(n^{\frac{1}{3}})} = \lim_{n \to \infty} \frac{\frac{1}{2}\ln(n)}{\frac{1}{3}\ln(n)} = \lim_{n \to \infty} \frac{3 \ln(n)}{2\ln(n)} \rightarrow \frac{3}{2}$$
The limit converges towards $\frac{3}{2}.$
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Let's try the usual approach of standard limits. We have
\begin{align}
L &= \lim_{n \to \infty}\frac{\log(1 + \sqrt{n} + \sqrt[3]{n})}{\log(1 + \sqrt[3]{n} + \sqrt[4]{n})}\notag\\
&= \lim_{n \to \infty}\dfrac{\log\sqrt{n} + \log\left(1 + \dfrac{1 + \sqrt[3]{n}}{\sqrt{n}}\right)}{\log\sqrt[3]{n} + \log\left(1 + \dfrac{1 + \sqrt[4]{n}}{\sqrt[3]{n}}\right)}\tag{1}\\
&= \lim_{n \to \infty}\dfrac{\dfrac{\log n}{2} + \log\left(1 + \dfrac{1 + \sqrt[3]{n}}{\sqrt{n}}\right)}{\dfrac{\log n}{3} + \log\left(1 + \dfrac{1 + \sqrt[4]{n}}{\sqrt[3]{n}}\right)}\notag\\
&= \lim_{n \to \infty}\dfrac{\dfrac{1}{2} + \dfrac{1}{\log n}\cdot\log\left(1 + \dfrac{1 + \sqrt[3]{n}}{\sqrt{n}}\right)}{\dfrac{1}{3} + \dfrac{1}{\log n}\cdot\log\left(1 + \dfrac{1 + \sqrt[4]{n}}{\sqrt[3]{n}}\right)}\tag{2}\\
&= \dfrac{\dfrac{1}{2} + 0\cdot 0}{\dfrac{1}{3} + 0\cdot 0}\tag{3}\\
&= \frac{3}{2}\notag
\end{align}
Note that we can not go from step $(1)$ to step $(3)$ directly by just replacing the $\log$ expressions with their limits because replacing a sub-expression by its limit is justified only in two scenarios:
*
*When the sub-expression is connected to the rest of the expression in additive manner.
*When the sub-expression is connected to the rest of the expression in multiplicative manner and its limit is non-zero.
We have instead done a further simplification by dividing numerator and denominator by $\log n$. By doing this we ensure that each term in numerator and denominator has a well defined limit and we can apply algebra of limits to get the final limit.
|
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|
Suppose $X, Y$ are random variables with the equal variance. Show that $X-Y$ and $X+Y$ are uncorrelated. Suppose that $X$ and $Y$ are random variables with the equal variance.
Show that $X-Y$ and $X+Y$ are uncorrelated.
I get I should use the equation $$E[XY] = E[X]E[Y]$$ For the first part I get $$E[(X-Y)(X+Y)] = E[X^2-Y^2] = E[X^2] - E[Y^2]$$ And I don't know how to follow. Someone has any ideas?
Thank you.
|
Suppose that
$$
\begin{pmatrix}
X\\
Y
\end{pmatrix}
$$
is a random vector with the covariance matrix
$$
\begin{pmatrix}
\sigma^2&\sigma_{X,Y}\\
\sigma_{X,Y}&\sigma^2
\end{pmatrix}.
$$
The covariance matrix of the random vector
$$
\begin{pmatrix}
1&1\\
1&-1
\end{pmatrix}
\begin{pmatrix}
X\\
Y
\end{pmatrix}
=
\begin{pmatrix}
X+Y\\
X-Y
\end{pmatrix}
$$
is given by
$$
\begin{pmatrix}
1&1\\
1&-1
\end{pmatrix}
\begin{pmatrix}
\sigma^2&\sigma_{X,Y}\\
\sigma_{X,Y}&\sigma^2
\end{pmatrix}
\begin{pmatrix}
1&1\\
1&-1
\end{pmatrix}
=
\begin{pmatrix}
2(\sigma^2+\sigma_{X,Y})&0\\
0&2(\sigma^2-\sigma_{X,Y})
\end{pmatrix}.
$$
This shows that $X+Y$ and $X-Y$ are uncorrelated and gives their variances.
|
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|
Is it true that $x(1-x)=y(1-y)$ if and only if $x=y$? Consider two numbers $0\leq x\leq 1$ and $0\leq y\leq 1$. Is it true that $x(1-x)=y(1-y)$ if and only if $x=y$? If not, can you give a counterexample?
|
$$x(1-x) = y(1-y) \implies -x^2+x-y(1-y)=0 \implies x^2-x+y(1-y)=0\text{.}$$
By the quadratic formula,
$$x = \dfrac{1\pm \sqrt{1-4(1)y(1-y)}}{2}=\dfrac{1 \pm \sqrt{1-4y(1-y)}}{2} = \dfrac{1\pm \sqrt{1-4y+4y^2}}{2} = \dfrac{1 \pm |2y-1|}{2}\text{,}$$
so any pair $\left(\dfrac{1 + |2y-1|}{2}, y\right)$ or $\left(\dfrac{1 - |2y-1|}{2}, y\right)$ for any $y \in \mathbb{R}$ will work. Obviously, $x = y$ isn't necessary.
|
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|
Circle inscribed in Equilateral Triangles The circle inscribed in the triangle $ABC$ touches the sides $BC$ , $CA$ , and $AB$ in the points $A_1,B_1,C_1$ respectively. Similarly the circle inscribed in the triangle $A_1B_1C_1$ touches the sides in $A_2,B_2,C_2$ respectively, and so on. If $A_nB_nC_n$ be the $n^{th}$ $\triangle$ so formed, Prove its angles are: $$\frac{\pi}{3} + (-2)^{-n}(A - \frac{\pi}{3})\ ,\ \ \frac{\pi}{3} + (-2)^{-n}(B - \frac{\pi}{3})\ ,\ \ \frac{\pi}{3} + (-2)^{-n}(C - \frac{\pi}{3})$$
Hence, prove that triangle so formed is an equilateral triangle.
|
Quang Hoang has already provided a good hint.
Let us prove that by induction on $n$.
Let $O$ be the incenter of $\triangle{ABC}$.
Then, noting that $OB\perp A_1C_1,OC\perp A_1B_1$, we have
$$\begin{align}\angle{B_1A_1C_1}&=\pi -\angle{B_1A_1C}-\angle{C_1A_1B}\\&=\pi-\left(\pi-\frac{\pi}{2}-\frac C2\right)-\left(\pi-\frac{\pi}{2}-\frac B2\right)\\&=\frac{B+C}{2}\\&=\frac{\pi-A}{2}\\&=\frac{\pi}{3}-2^{-1}\left(A-\frac{\pi}{3}\right)\end{align}$$
Similarly, we have
$$\angle{A_1B_1C_1}=\frac{\pi}{3}-2^{-1}\left(B-\frac{\pi}{3}\right),\qquad \angle{A_1C_1B_1}=\frac{\pi}{3}-2^{-1}\left(C-\frac{\pi}{3}\right)$$
Here, suppose that $$\angle{A_nB_nC_n}=\frac{\pi}{3}+(-2)^{-n}\left(B-\frac{\pi}{3}\right),\quad \angle{B_nC_nA_n}=\frac{\pi}{3}+(-2)^{-n}\left(C-\frac{\pi}{3}\right).$$
Let $O_n$ be the incenter of $\triangle{A_nB_nC_n}$.
Then, noting that $O_nB_n\perp A_{n+1}C_{n+1},O_nC_n\perp A_{n+1}B_{n+1}$, we have
$$\begin{align}&\angle{B_{n+1}A_{n+1}C_{n+1}}\\&=\pi -\angle{B_{n+1}A_{n+1}C_n}-\angle{C_{n+1}A_{n+1}B_n}\\&=\pi-\left(\frac{\pi}{2}-\frac 12\left(\frac{\pi}{3}+(-2)^{-n}\left(C-\frac{\pi}{3}\right)\right)\right)-\left(\frac{\pi}{2}-\frac 12\left(\frac{\pi}{3}+(-2)^{-n}\left(B-\frac{\pi}{3}\right)\right)\right)\\&=\frac{\pi}{3}+\frac{(-2)^{-n}}{2}(B+C)-(-2)^{-n}\cdot\frac{\pi}{3}\\&=\frac{\pi}{3}+\frac{(-2)^{-n}}{2}(\pi -A)-(-2)^{-n}\cdot\frac{\pi}{3}\\&=\frac{\pi}{3}+(-2)^{-n-1}\left(A-\frac{\pi}{3}\right)\end{align}$$
Similarly, we have
$$\small\angle{A_{n+1}B_{n+1}C_{n+1}}=\frac{\pi}{3}+(-2)^{-n-1}\left(B-\frac{\pi}{3}\right),\qquad \angle{A_{n+1}C_{n+1}B_{n+1}}=\frac{\pi}{3}+(-2)^{-n-1}\left(C-\frac{\pi}{3}\right)$$
|
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|
Derive $\tan(3x)$ in terms of $\tan(x)$ using De Moivre's theorem
Derive the following identity: $$\tan(3x)=\frac{3\tan(x)-\tan^3(x)}{1-3\tan^2(x)}$$
The way I approached the questions is that I first derived $\sin(3x)$ and $\cos(3x)$ because $\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}$. Then substituting:
$$\tan(3x)=\frac{\sin(3x)}{\cos(3x)}=\frac{-4\sin^3(x)+3\sin(x)}{4\cos^3(x)-3\cos(x)}$$
I transformed it into $$\tan(3x)=\frac{-4\sin^2(x)\tan(x)+3\tan(x)}{4\cos^2(x)-3}$$ and also to $$\tan(3x)=-\tan(x)\frac{4\sin^2(x)-3}{-4\sin^2(x)+1}$$ but I'm stuck on either of these forms. Any help?
|
Hint:
Notice that
\begin{align*}
\frac{4\sin^2 x-3}{-4\sin^2 x +1}&=\frac{1-4\cos^2 x}{4\cos^2 x-3 }\\[5pt]
&=\frac{\frac{1}{\cos^2 x}-4}{4-\frac{3}{\cos^2x}},\qquad\text{for }\cos x\neq 0\\[5pt]
&=\frac{\sec^2 x-4}{4-3\sec^2 x}\\[3pt]
&=\frac{(\color{blue}{\tan^2 x+1})-4}{4-3(\color{blue}{\tan^2 x+1})}
\end{align*}
Then, the identity follows straightforward.
|
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|
Prove: $2^k$ is the sum of two perfect squares If $k$ is a nonnegative integer, prove that $2^k$ can be represented as a sum of two perfect squares in exactly one way. (For example, the unique representation of $10$ is $3^2+1^2$; we do not count $1^2+3^2$ as different.)
I understand that $2^{2n}=0+2^{2n}$ and $2^{2n+1}=2^{2n}+2^{2n}$. But how can we prove that $2^k$ can be represented as two perfect squares in exactly one way?
|
I do not see anyone mentioning this simple aspect: if we have integers $u,v$ such that $$ u^2 + v^2 \equiv 0 \pmod 4, $$
then both $u,v$ must be even.
Which means this: take a number that is divisible by $4.$ Suppose we have
$$ x^2 + y^2 = n $$
Keep dividing $n$ by $4$ until the result, $n_0,$ is no longer divisible by $4.$ We have
$$ x^2 + y^2 = 4^k n_0, $$ where $x = 2^k x_0$ and $y = 2^k y_0.$
$$ x_0^2 + y_0^2 = n_0. $$
For you, either $n_0 = 1,$ written only as $1^2 + 0^2 = 1,$ or $n_0 = 2,$ written only as $1^2 + 1^2 = 2.$
|
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|
Integer solutions for $n$ for $|{\sqrt{n} - \sqrt{2011}}| < 1$ $$|{\sqrt{n} - \sqrt{2011}}| < 1$$
What is the number of positive integer $n$ values, which satisfy the above inequality.
My effort:
$
({\sqrt{n} - \sqrt{2011}})^2 < 1 \\n + 2011 -2\sqrt{2011n} < 1\\ n+2010<2\sqrt{2011n}\\ n^2+2 \times 2010 \times n +2010^2<4 \times 2011n \\n^2 -4024n +2010^2 < 0 $
But it seems this won't lead me for desired answer.
|
$|\sqrt n - \sqrt{2011}| <1 \iff -1 < \sqrt n - \sqrt {2011 }< 1 \iff
\sqrt{2011} - 1 <\sqrt n < \sqrt {2011} + 1 \iff (\sqrt{2011} - 1)^2 < n < (\sqrt{2011}+1)^2.$ You should be able to finish it from here.
|
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|
Is there a positive integer n such that the fraction $(9n+5)/(10n+3)$ is not in the lowest term? Is there a positive integer n such that the fraction $(9n+5)/(10n+3)$ is not in the lowest term? Please explain.
I found $n=2$ be a solution. Is it correct?
|
Your example $n=2$ is correct. We will find all possible examples.
Suppose that $d$ divides $9n+5$ and $10n+3$. Then $d$ divides
$$10(9n+5)-9(10n+3),$$
so $d$ divides $23$.
Since $23$ is prime, the only conceivable (positive) common divisors of $9n+5$ and $10n+3$ are $1$ and $23$. We will find the values of $n$ for which $23$ is a common divisor.
To make sure that $23$ divides both, all we need is to make sure that $23$ divides $9n+5$. This is because from $10(9n+5)-9(10n+3)=23$ we can conclude that if $23$ divides $9n+5$, it must divide $10n+3$.
You noticed that $23$ divides $9n+5$ when $n=2$. It follows that $23$ divides $9n+5$ if and only if $n=2+23k$ for some integer $k$. For if $23$ divides $9n+5$, then $23$ divides $9n+5-23$, so $23$ divides $9(n-2)$, and therefore $n-2$ is a multiple of $23$. Conversely, if $n-2$ is a multiple of $23$, say $n-2=23k$, then a calculation shows that $23$ divides $9n+5$.
|
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|
If $abcd=1$,where $a,b,c,d$ are positive reals,then find the minimum value of $a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd$. If $abcd=1$,where $a,b,c,d$ are positive reals,then find the minimum value of $a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd$.
Let $E=a^2+b^2+c^2+d^2+ab+ac+ad+bc+bd+cd=(a+b+c+d)^2-(ab+ac+ad+bc+bd+cd)$
I do not know how to use $abcd=1$ in this question to get the minimum value of the expression.
|
HINT
Use the AM-GM inequality, which establishes $$\frac{a_1+a_2+\dots+a_n}{n} \ge \sqrt[n]{a_1a_2\dots a_n}$$
When $a_1,a_2, \dots, a_n$ are positive reals.
For example, $a^2+b^2+c^2+d^2 \ge 4\sqrt[4]{a^2b^2c^2d^2}=4$
|
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|
How to prove $1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$ The task is to prove the following non-equality by hand:
$$1+\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) \neq 0$$
Wolframalpha shows this, but I can't prove it.
http://www.wolframalpha.com/input/?i=1%2Bcos(2pi%2F7)-4cos%5E2(2pi%2F7)-8cos%5E3(2pi%2F7)%3D0
|
Here, we will prove the equation by @AndreNicolas and @Blue in a more elementary manner.
$$ 1+4\cos \left(\frac{2\pi}{7}\right)-4\cos^2 \left(\frac{2\pi}{7}\right)-8\cos^3 \left(\frac{2\pi}{7}\right) = 0 $$
Note that since $\cos 3x=4\cos^3 x-3\cos x$, and since $\cos 2x=2\cos^2 x-1$, our equation simplifies to $$ \cos \left(\frac{\pi}{7}\right) -\cos \left(\frac{2\pi}{7}\right) + \cos \left(\frac{3\pi}{7}\right) = \frac{1}{2}$$
Let $x=\frac{\pi}{7}$.
$$\cos x - \cos 2x + \cos 3x = \frac{1}{2}$$
$$\cos x + \cos 3x + \cos 5x = \frac{1}{2}$$
$$\cos x + \cos 3x + \cos 5x + ... = \frac{{\sin 2nx}}{{2\sin x}}$$
Since $n = 3$
$$\cos x + \cos 3x + \cos 5x = \frac{{\sin 6x}}{{2\sin x}}$$
$$\frac{{\sin 6x}}{{2\sin x}} = \frac{1}{2} \Leftrightarrow \sin 6x = \sin x$$
Which is true since $\sin (\pi-x)=\sin x$.
Or,similarly if $$K=\cos x - \cos 2x + \cos 3x $$ then $$K\sin\frac{\pi}{7}=\frac{\sin\frac{2\pi}{7}}{2}+\frac{\sin\frac{4\pi}{7}-\sin\frac{2\pi}{7}}{2}+\frac{\sin\frac{6\pi}{7}-\sin\frac{4\pi}{7}}{2}=\frac{\sin\frac{6\pi}{7}}{2}\implies K=\frac{1}{2}$$
Since $ 2\sin A\cos B=\sin(A+B)+\sin(A-B)$
|
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|
Solving the following trigonometric equation: $\sin x + \cos x = \frac{1}{3} $ I have to solve the following equation:
$$\sin x + \cos x = \dfrac{1}{3} $$
I use the following substitution:
$$\sin^2 x + \cos^2 x = 1 \longrightarrow \sin x = \sqrt{1-\cos^2 x}$$
And by operating, I obtain:
$$ \sqrt{(1-\cos^2 x)} = \dfrac{1}{3}-\cos x$$
$$ 1 - \cos^2 x = \dfrac{1}{9} + \cos^2 x - \dfrac{2}{3}\cos x$$
$$ -2\cos^2 x + 2/3\cos x +\dfrac{8}{9}=0$$
$$ \boxed{\cos^2 x -\dfrac{1}{3}\cos x -\dfrac{4}{9} = 0}$$
Can I just substitute $\cos x$ by $z$ and solve as if it was a simple second degree equation and then obtain $x$ by taking the inverse cosine? I have tried to do this but I cannot get the right result. If I do this, I obtain the following results:
$$ z_1 = -0.520517 \longrightarrow x_1 = 121.4º\\
z_2= 0.8538509 \longrightarrow x_2 = 31.37º$$
I obtain $x$ from $z$ by taking the inverse cosine.
The correct result should be around 329º which corresponds to 4.165 rad. My question is if what I am doing is wrong because I have tried multiple times and I obtain the same result (or in the worst case, I have done the same mistake multiple times).
|
$$\cos(x)+\sin(x)=\frac{1}{3}\Longleftrightarrow$$
Use:
$$\cos(x)+\sin(x)=\sqrt{2}\left[\frac{\cos(x)}{\sqrt{2}}+\frac{\sin(x)}{\sqrt{2}}\right]=$$
$$\sqrt{2}\left(\sin\left(\frac{\pi}{4}\right)\cos(x)+\cos\left(\frac{\pi}{4}\right)\sin(x)\right)=\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)$$
$$\sqrt{2}\sin\left(\frac{\pi}{4}+x\right)=\frac{1}{3}\Longleftrightarrow$$
$$\sin\left(\frac{\pi}{4}+x\right)=\frac{1}{3\sqrt{2}}$$
Now, when we take the inverse sine of both sides, we got two options, with $n_1\space\wedge\space n_2\in\mathbb{Z}$:
*
*$$\frac{\pi}{4}+x=\pi-\arcsin\left(\frac{1}{3\sqrt{2}}\right)+2\pi n_1\Longleftrightarrow x=\frac{3\pi}{4}-\arcsin\left(\frac{1}{3\sqrt{2}}\right)+2\pi n_1$$
*$$\frac{\pi}{4}+x=\arcsin\left(\frac{1}{3\sqrt{2}}\right)+2\pi n_1\Longleftrightarrow x=\arcsin\left(\frac{1}{3\sqrt{2}}\right)+2\pi n_2-\frac{\pi}{4}$$
|
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|
integrate $\int \frac{\tan^4x}{4}\cos^3x$
$$\int \frac{\tan^4x}{4}\cos^3x$$
$$\int \frac{\tan^4x}{4}\cos^3x=\frac{1}{4}\int \frac{\sin^4x}{\cos^4x}\cos^3x=\frac{1}{4}\int\frac{\sin^4x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot\sin^2x}{\cos x}=\frac{1}{4}\int\frac{\sin^2x\cdot(1-\cos^2x)}{\cos x}=\frac{1}{4}\int \frac{\sin^2x}{\cos x}-\frac{\cos^2x}{\cos x}=\frac{1}{4}\int \frac{1-\cos^2x}{\cos x}-\cos x=\frac{1}{4} \int \frac{1}{\cos x}-2\cos x=\frac{1}{4}(\ln(\tan x+\sec x)+2\sin x$$
Is it correct?
|
Omitting the $\frac{1}{4}$, we have that
$\displaystyle\tan^4 x\cos^3 x=\frac{\sin^4 x}{\cos x}=\frac{(1-\cos^2 x)^2}{\cos x}=\frac{1-2\cos^2 x+\cos^4 x}{\cos x}=\sec x-2\cos x+\cos^3 x$
$\hspace{.85 in}\displaystyle=\sec x-2 \cos x+\cos x(1-\sin^2x)=\sec x-\cos x-\sin^2 x\cos x$,
so $\displaystyle\int\tan^4x\cos^3x \;dx=\ln|\sec x+\tan x|-\sin x-\frac{1}{3}\sin^3x+C$
|
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|
How to integrate $\int \frac{ev+f}{av^2 + bv +c} dv$? I want to integrate
\begin{align}
\int \frac{ev+f}{av^2 + bv +c} dv
\end{align}
can you give me some hints or detail procedure for this integral?
From mathematca, i have
\begin{align}
\frac{-\frac{2 (b e-2 a f) \text{ArcTan}\left[\frac{b+2 a v}{\sqrt{-b^2+4 a c}}\right]}{\sqrt{-b^2+4 a c}}+e \text{Log}[c+v (b+a v)]}{2 a}
\end{align}
which seems uncomfortable for me..
|
$$
\int \frac{ev+f}{av^2 + bv +c} dv = \frac{e}{2a}\int \frac{d(av^2 + bv + c)}{av^2 + bv +c} + \left(f-\frac{be}{2a}\right)\int \frac{1}{av^2 + bv +c} dv
$$
$$
= \frac{e}{2a} \ln(av^2+bv+c) + \left(f-\frac{be}{2a}\right)\int \frac{1}{(\sqrt{a}v+\frac{b}{2\sqrt{a}})^2 - \frac{b^2}{4a} +c} dv
$$
$$
\int \frac{1}{(\sqrt{a}v+\frac{b}{2\sqrt{a}})^2 - \frac{b^2}{4a} +c} dv = \frac{1}{\sqrt{a}}\int \frac{d(\sqrt{a}v+\frac{b}{2\sqrt{a}})}{(\sqrt{a}v+\frac{b}{2\sqrt{a}})^2 - \frac{b^2}{4a} +c}
$$
$$
c \geq \frac{b^2}{4a} \rightarrow k^2 = c - \frac{b^2}{4a}
$$
$$
\frac{1}{\sqrt{a}}\int \frac{d(\sqrt{a}v+\frac{b}{2\sqrt{a}})}{(\sqrt{a}v+\frac{b}{2\sqrt{a}})^2 - \frac{b^2}{4a} +c} = \frac{1}{\sqrt{a}} \int \frac{du}{u^2+k^2} = \frac{1}{k\sqrt{a}}\arctan\left(\frac{u}{k}\right) = \\ \frac{1}{k\sqrt{a}}\arctan\left(\frac{\sqrt{a}v+\frac{b}{2\sqrt{a}}}{k}\right) = \\
\frac{1}{\sqrt{c - \frac{b^2}{4a}}\sqrt{a}}\arctan\left(\frac{\sqrt{a}v+\frac{b}{2\sqrt{a}}}{\sqrt{c - \frac{b^2}{4a}}}\right)
$$
Then:
$$
\int \frac{ev+f}{av^2 + bv +c} dv = \frac{e}{2a} \ln(av^2+bv+c) + \left(f-\frac{be}{2a}\right)[\frac{1}{\sqrt{c - \frac{b^2}{4a}}\sqrt{a}}\arctan\left(\frac{\sqrt{a}v+\frac{b}{2\sqrt{a}}}{\sqrt{c - \frac{b^2}{4a}}}\right)] + C
$$
which is what you have.
$$
c \leq \frac{b^2}{4a} \rightarrow k^2 = \frac{b^2}{4a} - c
$$
$$
\frac{1}{\sqrt{a}}\int \frac{d(\sqrt{a}v+\frac{b}{2\sqrt{a}})}{(\sqrt{a}v+\frac{b}{2\sqrt{a}})^2 - \frac{b^2}{4a} +c} = \frac{1}{\sqrt{a}} \int \frac{du}{u^2-k^2} = \frac{1}{2k\sqrt{a}} \int {\frac{1}{u-k}-\frac{1}{u+k}} du =
$$
$$
\frac{1}{2k\sqrt{a}}\ln (\frac{u-k}{u+k}) = \frac{1}{2\sqrt{\frac{b^2}{4a} - c}\sqrt{a}}\ln (\frac{\sqrt{a}v+\frac{b}{2\sqrt{a}}-\sqrt{\frac{b^2}{4a} - c}}{\sqrt{a}v+\frac{b}{2\sqrt{a}}+\sqrt{\frac{b^2}{4a} - c}})
$$
Then:
$$
\int \frac{ev+f}{av^2 + bv +c} dv = \frac{e}{2a} \ln(av^2+bv+c) + \left(f-\frac{be}{2a}\right)[\frac{1}{2\sqrt{\frac{b^2}{4a} - c}\sqrt{a}}\ln (\frac{\sqrt{a}v+\frac{b}{2\sqrt{a}}-\sqrt{\frac{b^2}{4a} - c}}{\sqrt{a}v+\frac{b}{2\sqrt{a}}+\sqrt{\frac{b^2}{4a} - c}})] + C
$$
|
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|
In $\triangle ABC$, if $\cos A\cos B\cos C=\frac{1}{3}$, then $\tan A\tan B+\tan B \tan C+\tan C\tan A =\text{???}$
In $\triangle ABC$, if
$$\cos A \cos B \cos C=\frac{1}{3}$$
then can we find value of
$$\tan A\tan B+\tan B \tan C+\tan C\tan A\ ?$$
Please give some hint. I am not sure if $\tan A \tan B+\tan B \tan C+\tan C \tan A$ will be constant under given condition.
|
The answer to the question is no. The maximum value of $\cos A \cos B \cos C$, where $A$, $B$, and $C$ are the angles of a triangle in the plane, is $\frac{1}{8}$, so there is no plane triangle for which $\cos A \cos B \cos C=\frac{1}{3}$.
The product $\cos A \cos B \cos C$ equals $\frac{1}{8}$ for an equilateral triangle, and the fact that this is a maximum follows from the fact that for any acute* triangle $\triangle ABC$, the product is greater for the “more equilateral” triangle with angles $\frac{A+B}{2}$, $\frac{A+B}{2}$, and $C$, because $$\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A+B}{2}\right)\cos C-\cos A\cos B\cos C= \frac{1-\cos(A+B)}{2}\cdot\cos C>0.$$
*We can assume the triangle is acute (and $\cos C>0$), because otherwise the product $\cos A \cos B \cos C \le 0$ (only one angle in a given triangle can be non-acute, so only one of the cosines can be non-positive) and $\triangle ABC$ can’t possibly be one for which $\cos A \cos B \cos C$ is a maximum.
This might be an interesting question for triangles on a surface of negative curvature, however.
|
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|
Use differentiation to find a power series Use differentiation to find a power series of
$f(x) = \frac{1}{(8+x)^2}$
$ f'(x) = \frac{-2}{(8+x)^3} $
how do I find the power series of this? I can not go next step.
|
$$
g(x) = \frac{1}{(8+x)} = \sum_{n=0}^{\infty} \frac{(-1)^nx^n}{8^{n+1}}
$$
Taking the derivative in both sides: (n=0 is constant)
$$
g'(x) = \frac{-1}{(8+x)^2} = \sum_{n=1}^{\infty} \frac{n(-1)^nx^{n-1}}{8^{n+1}}
$$
$$
-f(x) = \frac{-1}{(8+x)^2} = \sum_{n=0}^{\infty} \frac{(n+1)(-1)^{n+1}x^{n}}{8^{n+2}}
$$
$$
f(x) = \sum_{n=0}^{\infty} \frac{(n+1)(-1)^{n}x^{n}}{8^{n+2}}
$$
|
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|
Last Digit of $x^0 + x^1 + x^2 + \cdots + x^{p-1} + x^p$
Given $x$ and $p$. Find the last digit of $x^0 + x^1 + x^2 + \cdots + x^{p-1} + x^p$
I need a general formula. I can find that the sum is equal to
$\dfrac{x^{p+1}-1}{x-1}$
But how to find the last digit.
P.S: $x\leq 999999$ and $p \leq 10^{15}$
|
Assuming that $x>0$.
Let $d$ denote the last digit of the series.
You can split the answer into $10$ different cases.
The following cases are rather simple:
*
*$x\equiv0\pmod{10} \implies d=1$
*$x\equiv1\pmod{10} \implies d=(1+p)\bmod{10}$
*$x\equiv4\pmod{10} \implies d=1+4(p\bmod{2})$
*$x\equiv5\pmod{10} \implies d=1+5(p\bmod{2})$
*$x\equiv9\pmod{10} \implies d=1-1(p\bmod{2})$
The following cases are a little more complicated:
*
*$x\equiv2\pmod{10} \implies d=\frac{-4(p\bmod{4})^3+15(p\bmod{4})^2-5(p\bmod{4})+3}{3}$
*$x\equiv3\pmod{10} \implies d=\frac{+1(p\bmod{4})^3-9(p\bmod{4})^2+17(p\bmod{4})+3}{3}$
*$x\equiv6\pmod{10} \implies d=\frac{-5(p\bmod{5})^4+40(p\bmod{5})^3-100(p\bmod{5})^2+83(p\bmod{5})+3}{3}$
*$x\equiv7\pmod{10} \implies d=\frac{+1(p\bmod{4})^3-15(p\bmod{4})^2+35(p\bmod{4})+3}{3}$
*$x\equiv8\pmod{10} \implies d=\frac{+11(p\bmod{4})^3-54(p\bmod{4})^2+67(p\bmod{4})+3}{3}$
|
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|
If the sum to 4 terms of a geometric progression is 15 and the sum to infinity is 16 find the possible values of the common ratio. I can't find a way to get an answer for this. I have tried using the formula for the sum to infinity and dividing it by the sum to 4 terms but i can't get it to work.
|
Let the sequence have initial term $a_1$ and common ratio $r$. Then $a_k = a_1r^{k - 1}$. The sum of the first n terms of the geometric series is
$$\sum_{k = 1}^{n} a_1r^{k - 1} = a_1(1 + r + r^2 + \cdots + r^{n - 1}) = a_1 \frac{1 - r^n}{1 - r}$$
provided that $r \neq 1$. If $r = 1$, then the series would not converge unless $a_1 = 0$, which cannot be the case here since the sum of the series is not equal to zero. Since the sum of the first four terms is $15$, we have
$$a_1 \frac{1 - r^4}{1 - r} = 15 \tag{1}$$
If the series converges, then its limit is
$$\sum_{k = 1}^{\infty} a_1r^{k - 1} = \frac{a_1}{1 - r}$$
Since the series has sum $16$, we have
$$\frac{a_1}{1 - r} = 16 \tag{2}$$
Dividing equation 1 by equation 2 yields
$$1 - r^4 = \frac{15}{16}$$
Solving for $r$ yields
\begin{align*}
1 - \frac{15}{16} & = r^4\\
\frac{1}{16} & = r^4\\
\pm \frac{1}{2} & = r
\end{align*}
Check:
Substituting $r = 1/2$ into equation 1 yields
\begin{align*}
a_1 \cdot \frac{1 - \frac{1}{16}}{1 - \frac{1}{2}} & = 15\\
a_1 \cdot \frac{\frac{15}{16}}{\frac{1}{2}} & = 15\\
a_1 \cdot \frac{15}{8} & = 15\\
a_1 & = 8
\end{align*}
Substituting $a_1 = 8$ and $r = 1/2$ into equation 2 yields
$$\frac{a_1}{1 - r} = \frac{8}{\frac{1}{2}} = 16$$
If $r = -1/2$, then
\begin{align*}
a_1 \cdot \frac{1 - \frac{1}{16}}{1 + \frac{1}{2}} & = 15\\
a_1 \cdot \frac{\frac{15}{16}}{\frac{3}{2}} & = 15\\
a_1 \cdot \frac{15}{16} \cdot \frac{2}{3} & = 15\\
a_1 \cdot \frac{5}{8} & = 15\\
a_1 & = 24
\end{align*}
Substituting $a_1 = 24$ and $r = -1/2$ into equation 2 yields
$$\frac{24}{1 + \frac{1}{2}} = \frac{24}{\frac{3}{2}} = 24 \cdot \frac{2}{3} = 16$$
Thus, both $r = 1/2$ and $r = -1/2$ satisfy the given conditions.
|
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|
Using squeeze thorem find $ \lim_{n \to \infty}{\frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n}}$ It is already solved here at Math.stackexchange, but we haven't learned Stirling's approximation (at our school), so can it be solved using only squeeze theorem?
$$\lim_{n \to \infty}{\frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n}}$$
My attempt,
Let $y_n = \frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n}$, we see that $ \frac{1 \cdot 3 \cdot 5 \cdot ...\cdot (2n-1)}{2\cdot 4 \cdot 6\cdot ...\cdot 2n} > \frac{1 \cdot 2 \cdot 2 \cdot ... \cdot 2}{2\cdot 4 \cdot 6 \cdot ...\cdot 2n} = \frac{1}{1 \cdot 2 \cdots .. \cdot (n-1)\cdot2n} = x_n$
$$\lim_{n \to \infty}{x_n} = 0 $$
Now I just need to find a sequnce $z_n>y_n$, so, $\lim_{n \to \infty} z_n = 0$.
|
Let
$$
b_n=\frac{1\cdot 3\cdots (2n-1)}{2\cdot 4\cdots2n}.
$$
We shall show inductively that
$$
\frac{2}{\sqrt{2n+1}}>b_n.
$$
For $n=1$ it clearly holds.
Assume that
$$
\frac{2}{\sqrt{2k+1}}>\frac{1\cdot 3\cdots (2k-1)}{2\cdot 4\cdots2k}.
$$
Then
$$
\frac{2}{\sqrt{2k+1}}\cdot\frac{2k+1}{2k+2}>\frac{1\cdot 3\cdots (2k-1)\cdot(2k+1)}{2\cdot 4\cdots(2k)\cdot(2k+2)}.
$$
But
$$
\sqrt{\frac{2k+3}{2k+1}}=\sqrt{1+\frac{2}{2k+1}}<1+\frac{1}{2k+1}=\frac{2k+2}{2k+1}
$$
and hence
$$
\frac{2}{\sqrt{2k+3}}>\frac{2}{\sqrt{2k+1}}\cdot\frac{2k+1}{2k+2}>b_{k+1}.
$$
Another way to do it is by showing that
$$
a_n=\frac{2\cdots 4\cdots 2n}{1\cdots 3\cdots (2n-1)}\to\infty.
$$
Now, using the fact that,
$$
x\in[0,1]\Longrightarrow \log(1+x)=\int_0^x\frac{dt}{1+t}\ge \frac{1}{2}\int_0^x\frac{dt}{1+t}=\frac{x}{2},
$$
we obtain
$$
\log a_n=\sum_{k=1}^n\log \left(\frac{2k}{2k-1}\right)=\sum_{k=1}^n\log \left(1+\frac{1}{2k-1}\right)\ge \frac{1}{2}\sum_{k=1}^n \frac{1}{2k-1}.
$$
It only remains to show that
$$
1+\frac{1}{3}+\frac{1}{5}+\cdots+\frac{1}{2n-1}\to\infty.
$$
|
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|
Compositeness test for repunits Is this proof acceptable ?
Definition
Let $R_p=\frac{10^p-1}{9} $ with $p$ prime be a repunit number .
Theorem
If $R_p$ is prime then $7^{\frac{R_p-1}{2}} \equiv -1 \pmod {R_p}$
Proof
Let $R_p$ be a prime , then by Euler's criterion :
$7^{\frac{R_p-1}{2}} \equiv \left(\frac{7}{R_p}\right) \pmod {R_p}$ , where $\left(\frac{7}{R_p}\right)$ denotes Legendre symbol .
If $R_p$ is prime then $R_p \equiv 1,2,4 \pmod 7$ and therefore : $\left(\frac{R_p}{7}\right)=1$
Since $R_p \equiv 3 \pmod 4$ according to the law of quadratic reciprocity it follows that : $\left(\frac{7}{R_p}\right)=-1$ .
Hence , $7^{\frac{R_p-1}{2}} \equiv -1 \pmod {R_p}$
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Looks all good but as Crostul I feel that it is good to add why $R_p \equiv 1,2,4 \pmod 7$. It is easy to prove that $R_p \equiv 1,2 \pmod 7$ for odd $p$:
As $16 \cdot 63 = 1008$, we have $10^3 \equiv -8 \pmod {63}$ and thus $10^6 \equiv 1 \pmod {63}$. This implies $$10^{1+6n} \equiv 10 \pmod {63}$$ and $$10^{5+6n} \equiv 10^5 \pmod {63} \equiv -800 \pmod {63} \equiv 19 \pmod {63}$$ and thus (any odd $p$ is $\equiv 1,5 \pmod 6$):
either $R_p \equiv 1 \pmod 7$ or $R_p \equiv 2 \pmod 7$.
And of course, when $p=2$, then $R_p = 11 \equiv 4 \pmod 7$
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proof limit of $\frac{1}{(x^{2}-1)}$ = -1 using $\epsilon - \delta$ as $x_{0} \to 0$ Question: proof limit of $\frac{1}{(x^{2}-1)}$ = -1 using $\epsilon - \delta$ as $x_{0} \to 0$
Hi... I am stumped on an intro analysis problem.
so it is stated Show $lim \frac{1}{x^{2}-1}$ = -1 as $x_{0} \to 0$.
here is my work, and I can't quite simplify the $|f(x) - L | < \epsilon$ to get an expression for $\delta$.
|$\frac{1}{x^{2}-1} + 1| \leq |\frac{1}{(x+1)(x-1)} + 1 | \leq | \frac{1}{x-1} - \frac{1}{x+1} + 2| \leq \epsilon $ not sure if I should use partial fractions here.
I am not sure the neighborhood to select for $x_0$ , so I'm guessing $|x| < 1$
this choice yields $1 < \frac{1}{x+1} < \frac{1}{x} < \frac{1}{x-1} $
so the expression of $\epsilon$ can be written:
$|\frac{1}{x-1} - \frac{1}{x+1} + 2 | < |\frac{2}{x-1} + 2 | $ this can be simplified to
$| x - 0 | < \frac{\epsilon}{2}(x-1)$ where given our choice of $x_{0}$ (x-1) $< 2$
so I chose $\delta = min( 1, \epsilon)$.
but I am not sure this work is any good, and am having trouble writing the proof.
sincerely thank you!
Here is the Proof
let $\delta = min\{\frac{1}{2},\frac{5\epsilon}{2} \}$
let $\delta = \frac{1}{2} \implies x^{2} < \frac{|x|}{2} <\frac{1}{4}$
where $|x| < \frac{1}{2}$
then $\frac{-5}{4} < x^{2} - 1 < \frac{-3}{4} \implies \frac{5}{4} < |x^{2}-1| < \frac{3}{4} \implies \frac{1}{x^{2}-1}<\frac{4}{5}$
then $\frac{x^{2}}{x^{2}-1} < \frac{4|x|}{2*5} < \frac{4*1}{4*5} = \frac{1}{5} < \epsilon$
where if $\delta = \frac{1}{2} \implies \frac{1}{2} < \frac{5\epsilon}{2} \implies \frac{1}{5}<\epsilon$
for $\delta = \frac{5\epsilon}{2} $
let $|x - 0| < \frac{5\epsilon}{2} $ and for $|\frac{x}{2}|<\frac{5 \epsilon}{4}$ then
$x^{2}<|\frac{x}{2}| \implies |\frac{x^{2}}{x^{2}-1}| \leq \frac{2|x|}{5}$
so $x^{2} < \frac{|x|}{2} < \frac{5\epsilon}{4}$
then
$|\frac{x^{2}}{x^{2}-1}| < \frac{4|x|}{5} <\frac{4}{5}*\frac{5\epsilon}{4} = \epsilon$
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Note that $$|f(x)-L|=\left|\frac1{x^2-1}-1 \right|=\left|\frac{x^2}{(x-1)(x+1)} \right|=\frac{x^2}{|x-1||x+1|}, $$ so if $|x|<\frac12$, then $|x-1||x+1|>2|x|$ and $|f(x)-L|<\frac12|x|$. So choosing $\delta=\min\left\{\varepsilon,\frac12\right\}$ we see that $|x|<\delta$ implies $$|f(x)-L|<\frac12|x|<\varepsilon. $$
|
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|
Function parameters and cartesian curves Given
$$x = \cos t + \cos 2t \,,\; y = \sin t + \sin 2t ,$$
find the tangent line for the parameter at point $(-1, 1),$ and draw a graph of the curve.
To find the point you could simply do $\dfrac{dx}{dt}$ and $\dfrac{dy}{dt}$ and find $\dfrac{dy}{dx}$ by taking the quotient, so you get $$ \dfrac{dy}{dx} =\dfrac{\cos t+2 \cos 2t}{-\sin t-2 \sin 2t}$$
How exactly would you define parameter $t$ in this case as a value to find the equation of the tangent? And when trying to draw the graph, you would need the Cartesian curve, which can be fetched by eliminating the parameter by solving for $t$. Is there a trigonometric identity that would make it possible to solve for $t$ using $X$ and $Y$ equations?
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Another way by finding implicit equation
\begin{align*}
x^{2} &= \cos^{2} t+2\cos t \cos 2t+ \cos^{2} 2t \\
y^{2} &= \sin^{2} t+2\sin t \sin 2t+ \sin^{2} 2t \\
x^{2}+y^{2} &= 2+2(\cos t \cos 2t+\sin t \sin 2t) \\
&= 2+2\cos t \\
\cos t &= \frac{x^{2}+y^{2}}{2}-1 \\
x &= \cos t+2\cos^{2} t-1 \\
&=(\cos t+1)(2\cos t-1) \\
&= \frac{(x^{2}+y^{2})}{2} (x^{2}+y^{2}-3) \\
0 &= (x^{2}+y^{2})(x^{2}+y^{2}-3)-2x \\
0 &= x^{4}+2x^{2}y^{2}+y^{4}-3x^{2}-3y^{2}-2x \\
\end{align*}
In principle, the tangent can be found by implicit differentiation and plug in the point $(-1,1)$.
The rest of work is left as an exercise, but showing the graph instead:
P.S.: The curve is actually a shifted limaçon of Pascal,
$r=2\cos \theta+1$.
|
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Change in Interval of convergence if center of convergence changes So I have to find a power series that is centered at $-2^{1/2}$
If I choose to use the power series expansion for $e^x$ which converges for all $x$, and change $x$ to $x + 2^{1/2}$ does the interval of convergence remain the same?
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If the radius of convergence of a power series is positive and finite, it is the distance from the center of the series expansion to the nearest singularity or branchpoint. If the center changes, the distance to the nearest singularity or branchpoint may change and so the radius of convergence will also change.
Considering for example the geometric series expansion at $x=0$, we obtain a power series with radius of convergence equal to $1$, since we have a singularity, a pole of order $1$ at $x=1$.
\begin{align*}
\frac{1}{1-x}=\sum_{n=0}^\infty x^n\qquad\qquad|x|<1
\end{align*}
If we expand the geometric series at $x=\frac{1}{2}$ we obtain a power series with radius of convergence equal to $\frac{1}{2}$, since then the distance to the nearest singularity $1$ from the center $x=\frac{1}{2}$ is equal to $\frac{1}{2}$.
\begin{align*}
\frac{1}{1-x}=\frac{2}{1-2\left(x-\frac{1}{2}\right)}=\sum_{n=0}^\infty2^{n+1}\left(x-\frac{1}{2}\right)^n\qquad\qquad\left|x-\frac{1}{2}\right|<\frac{1}{2}
\end{align*}
The situation is different, when we consider an entire function which is convergent in the whole complex plane. Since the radius of convergence is $\infty$ in this case, it will not change if we change the center. The exponential function is an entire function and so
\begin{align*}
e^x&=\sum_{n=0}^\infty\frac{x^n}{n!}\qquad\qquad\qquad\qquad x\in \mathbb{C}\\
e^x=e^{x_0}e^{x-x_0}&=e^{x_0}\sum_{n=0}^\infty \frac{\left(x-x_0\right)^n}{n!}\qquad\qquad x\in\mathbb{C}
\end{align*}
|
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Find remainder when $1^{5} + 2^{5} \cdots +100^{5}$ divided by 4 I'm studding D.M Burton & want to solve: Find remainder when $1^{5} + 2^{5} \cdots +100^{5}$ divided by $4$. . Please help me by giving your solution to it. I'm new comer to number theory so please don't use theorems above Theory of Congruence.
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You can do it even without knowledge of congruences, we have that $1^5 + 2^5 + ... + n^5= \dfrac {(n(n+1))^2(2n^2+2n-1)}{12}$, now set $n=100$ to get $1^5 + 2^5 + ... + 100^5 = \dfrac {(100 \cdot 101)^2 (2 \cdot 100^2 + 2 \cdot 100 - 1)}{12}=4 \cdot \dfrac {4 \cdot 25^2 \cdot 101^2 \cdot 3 \cdot 6733}{12}$ so your number is divisible by $4$.
|
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|
solve $\sin 2x + \sin x = 0 $ using addition formula
$\sin 2x + \sin x = 0 $
Using the addition formula, I know that
$\sin 2x = 2\sin x \cos x$
=> $2\sin x \cos x + \sin x = 0$
=> $\sin x(2\cos x + 1) = 0$
=> $\sin x = 0$ and $\cos x = -\frac{1}2 $
I know that $\sin x = 0$ in first and second quadrant so $x = 0$ and $x = 180$
What I do not know is what to do with $\cos x = -\frac{1}2$ and which quadrants this applies to.
The book I got the question from gives the following answer which does not make sense to me:
0, 120, 180, 240, 360
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$$\sin(2x)+\sin(x)=0 $$
First, you have to remember that:
$$\sin(a)+\sin(b)=2\sin(\frac{a+b}{2})\cos(\frac{a-b}{2})$$
Aplying this in the equation:
$$2\sin(\frac{3x}{2})\cos(\frac{x}{2})=0$$
$$\sin(\frac{3x}{2})\cos(\frac{x}{2})=0$$
Finally, you have two equations:
$$\sin(\frac{3x}{2})=0$$
$$\cos(\frac{x}{2})=0$$
For the first, the solution are $x=\frac{2n\pi}{3} / n=0,1,2 $
And for the second are $x=n\pi / n=1,3 $
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How many strings of length 8 The question - how many strings of length 8, from 4 letter alphabet, using each letter twice. There is to be exactly one pair of same letters next to each other (example of valid string: AABCDBCD).
I tried to check how this develops by drawing a tree, but it seems that this gets really unwieldy really fast. What counting technique can be used here to make it easier?
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If there were no restrictions, we could arrange the letters AABBCCDD in
$$\binom{8}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} = \frac{8!}{2!6!} \cdot \frac{6!}{4!2!} \cdot \frac{4!}{2!2!} \cdot \frac{2!}{0!2!} = \frac{8!}{2!2!2!2!}$$
distinguishable ways since we can place the two A's in two of the eight available positions, the two B's in two of the six remaining positions, the two C's in two of the four remaining positions, and the two D's in the last two open positions.
We choose which of the four letters is doubled, placing the pair in a box so that we have seven objects - the box containing the double letter, and the other six letters. We choose which of the seven positions will be occupied by that box. The remaining letters can be arranged in $\binom{6}{2}\binom{4}{2}\binom{2}{2}$ ways. Hence, the number of ways of arranging the letters AABBCCDD so that two adjacent letters are the same is
$$\binom{4}{1}\binom{7}{1}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$
However, we have counted arrangements in which more than one pair of identical letters are adjacent. We must exclude these from the total.
Suppose two letters are doubled. We choose two of the four letters to be doubled, placing each double letter in a distinct box. This gives us six objects, the two boxes and four remaining letters. We choose which two of the six positions will be occupied by the boxes, arrange them in these positions in $2!$ ways, and then arrange the remaining four letters in $\binom{4}{2}\binom{2}{2}$ ways. This gives
$$\binom{4}{2}\binom{6}{2} \cdot 2! \cdot \binom{4}{2}\binom{2}{2}$$
arrangements with two double letters.
Suppose three letters are doubled. We choose three of the four letters to be doubled, placing each double letter in a distinct box. This gives us five objects, the three boxes and the two single letters. We choose three of the five positions for the letters, arrange the boxes in these positions in $3!$ ways, then arrange the remaining two letters in $\binom{2}{2}$ ways. This gives
$$\binom{4}{3}\binom{5}{3} \cdot 3! \cdot \binom{2}{2}$$
arrangements with three double letters.
Suppose all four letters are doubled. We place each double letter in a distinct box. The four boxes can be arranged in $4!$ orders.
By the Inclusion-Exclusion Principle, the number of arrangements of AABBCCDD in which there is exactly one pair of adjacent letters that are the same is
$$\binom{4}{1}\binom{7}{1}\binom{6}{2}\binom{4}{2}\binom{2}{2} - \binom{4}{2}\binom{6}{2} \cdot 2! \cdot \binom{4}{2}\binom{2}{2} + \binom{4}{3}\binom{5}{3} \cdot 3! \cdot \binom{2}{2} - 4!$$
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Alternative "Fibonacci" sequences and ratio convergence So the well known Fibonacci sequence is
$$
F=\{1,1,2,3,5,8,13,21,\ldots\}
$$
where $f_1=f_2=1$ and $f_k=f_{k-1}+f_{k-2}$ for $k>2$. The ratio of $f_k:f_{k-1}$ approaches the Golden Ratio the further you go:
$$\lim_{k \rightarrow \infty} \frac{f_k}{f_{k-1}} =\phi \approx 1.618$$
Let's define a class of similar sequences $F_n$ where each $f_k$ is the sum of the previous $n$ numbers, $f_k=f_{k-1} + f_{k-2} + \dots + f_{k-n}$ so that the traditional Fibonacci sequence would be $F_2$ but we can talk about alternatives such as
$$F_3 = \{1,1,1,3,5,9,17,\dots \}$$
where we initialized the values $f_1$ through $f_3$ to be $1$ and we can show that in this case
$$
\lim_{k \rightarrow \infty} \frac{f_k}{f_{k-1}} \approx 1.839286755
$$
The following table gives some convergences for various values of $n$:
$$
\begin{matrix}
F_n & \text{Converges to} \\ \hline
F_2 & \phi \\
F_3 & 1.839286755 \\
F_4 & 1.927561975 \\
F_5 & 1.965948237 \\
F_{6} & 1.983582843 \\
F_{10} & 1.999018626
\end{matrix}
$$
Just by inspection, it seems that the convergence values are converging toward $2$ as $n \rightarrow \infty$.
So my primary question is:
What is the proof that the convergence converges to 2 (assuming it does).
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From the standard theory of linear recurrence,s $F_n$ is the positive real root of the equation $z^n = z^{n-1} + z^{n-2} + \cdots + z + 1$. Multiplying both sides by $1-z$ and rearranging, you get that $F_n$ is the positive real root of $f_n(z) = z^{n+1} - 2z^n + 1 = 0$ that is not $z = 1$. (By Descartes' rule of signs (https://en.wikipedia.org/wiki/Descartes%27_rule_of_signs), there are either 0 or 2 positive real roots of this polynomial; we know there is at least 1 , so there must be 2.)
Now, we have
$$ f_n(2)= 2^{n+1} - 2 \cdot 2^{n} + 1 = 1 $$
and
$$f_n(2 - 1/2^{n-1}) = (2 - 1/2^{n-1})^{n+1} - 2 \cdot (2-1/2^{n-1})^n + 1. $$
We aim to show that $f_n(2-1/2^{n-1}) < 0$; then $f_n$ must have a root between $2 - 1/2^{n-1}$ and $2$.
Factoring out powers of 2, we get
$$f_n(2 - 1/2^{n-1}) = 2^{n+1} (1-1/2^n)^{n+1} - 2^{n+1} (1-1/2^n)^n + 1.$$
Factoring out $2^{n+1} (1-1/2^n)^n$ from the first two terms gives
$$f_n(2 - 1/2^{n-1}) = 2^{n+1} (1-1/2^n)^n (-1/2^n) + 1$$
or, finally,
$$f_n(2 - 1/2^{n-1}) = 1-2(1-1/2^n)^n. $$
So the result that $f_n$ has a root in the interval $[2-1/2^{n-1}, 2]$, for $n$ sufficiently large, follows from the fact that $(1-1/2^n)^n > 1/2$ for $n$ sufficiently large. Since $\lim_{n \to \infty} (1-1/2^n)^n = 1$ (use for example L'Hopital's rule) this follows.
Thus $F_n \in [2 - 1/2^{n-1}, 2]$ and the desired result follows, for example, from the squeeze theorem.
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Calculate the sum of the infinite series $ 1 + \frac{1+2}{2!} + \frac{1+2+3}{3!} .... $ Calculate the sum of the infinite series $ 1 + \frac{1+2}{2!} + \frac{1+2+3}{3!} .... $
My attempt : I recognised that this series can be decomposed into the taylor expansion of $ e $ around 0.
So I thought of writing the series as
$ 1 + \frac{1}{2!} + \frac{1}{3!} ...$ + $ 2[ \frac{1}{2!} + \frac{1}{3!}...]$ $ +$ $ 3[ \frac{1}{3!} + \frac{1}{4!} ...] + ...$
However I got stuck here and couldn't proceed further.
Any hints on how to proceed further , or a better method to solve the question would be appreciated.
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Using the well-known formula for $1+2+\dots\;$your sum is
$$\sum_{n=1}^\infty \frac{n(n+1)}{2}\frac{1}{n!}=
\sum_{n=1}^\infty \frac{(n+1)}{2(n-1)!}=
\sum_{n=0}^\infty \frac{(n+2)}{2n!}=
\sum_{n=0}^\infty \left(\frac{n}{2n!} + \frac{2}{2n!}\right)$$
$$=\frac{1}{2}\sum_{n=0}^\infty \frac{n}{n!} + \sum_{n=0}^\infty \frac{2}{2n!}$$
The first term of the first sum is zero, so we omit this term, start the summation at $n=1$ and divide out $n$ in the fraction
$$=\frac{1}{2}\sum_{n=1}^\infty \frac{1}{(n-1)!} + \sum_{n=0}^\infty \frac{1}{n!}$$
$$=\frac{1}{2}\sum_{n=0}^\infty \frac{1}{n!} + \sum_{n=0}^\infty \frac{1}{n!}$$
$$=\frac{1}{2}e + e = \frac{3}{2}e$$
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|
Let $x \ge 0$. Determine a condtion on $|x-4|$ that'll assure $|\sqrt{x} - 2| < 10^{-2}$ I'm trying to understand the logic of this proof.
Let $x \ge 0$. Determine a condition on $|x-4|$ that'll assure $|\sqrt{x} - 2| < 10^{-2}$
Proof
$|\sqrt{x} - 2| = \frac{|(\sqrt{x} - 2)(\sqrt{x} + 2)|}{\sqrt{x} + 2} = \frac{x-4}{\sqrt{x} + 2} \le \frac{|x-4|}{2}$
For $x$ satisfying $|x-4| = 2 \cdot 10^{-2}$
we have $|\sqrt{2} -2| \le \frac{2 \cdot 10^{-2}}{2}=10^{-2}$
Does the $|\sqrt{2} -2|$ just come from the 2 in $|x-4| = 2 \cdot 10^{-2}$? Is there more too it? Or is there a better, more complete proof?
Thank you, in advance.
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What you need to do is find bounds for $\frac{1}{\sqrt x + 2}$
Suppose $|x - 4| < \delta$. Then we can argue that
\begin{align}
|x - 4| < \delta
&\implies4-\delta < x < 4 + \delta \\
&\implies \sqrt{4-\delta} < \sqrt x < \sqrt{4 + \delta}\\
&\implies \sqrt{4-\delta} + 2 < \sqrt x + 2 < \sqrt{4 + \delta} + 2\\
\text{If we also require that }\; &0 < \delta < 3 \;
\text{then we can continue with}\\
&\implies 3 < \sqrt x + 2 \\
&\implies \dfrac{1}{\sqrt x + 2}< \dfrac 13 \\
&\implies |\sqrt x - 2| < \dfrac{|x - 4|}{\sqrt x + 2} \\
&\implies |\sqrt x - 2| < \dfrac{\delta}{3} \\
\end{align}
So choose $\delta = \min\{3, \epsilon\}$ and you are done.
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How do I show that $\sum_{cyc} \frac {a^6}{b^2 + c^2} \ge \frac {abc(a + b + c)}2?$ Let $a, b, c$ be positive real numbers, show that
$$\frac {a^6}{b^2 + c^2} + \frac {b^6}{c^2 + a^2} + \frac {c^6}{a^2 + b^2} \ge \frac {abc(a + b + c)}2.$$
I think this is likely to turn out to be proved by Hölder, but I can't see how. Any hints will be appreciated.
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Note that both sides are homogeneous of degree 4: ie, if we replace $a,b,c$ by $ta,tb,tc$ where $t>0$, both sides get multiplied by $t^4$ which has no effect on the inequality.
If we rescale by $t=1/(a+b+c)$, the effect is the same as requiring that $a+b+c=1$. Often this is formulated as the statement that we can choose $a+b+c=1$ without loss of generality. Alternatively, we could have chosen $a^2+b^2+c^2=1$ without loss of generality, which would be convenient for reformulating the LHS. Unfortunately, we cannot get both these assumptions at the same time. However, we can use then in a step-wise manner to break the inequality into three steps:
$$
\sum_{\text{cycle}(a,b,c)}\frac{a^6}{b^2+c^2}
\ge\frac{(a^2+b^2+c^2)^2}{6}
\ge\frac{(a+b+c)^4}{2\cdot3^3}
\ge\frac{abc(a+b+c)}{2}
$$
The way I got to this was the realisation that the RHS expression could be bounded from above more easily under the assumption $a+b+c=1$. Basically, if $a+b+c=1$, we have $abc\le1/3^3$, and so $RHS\le1/(2\cdot3^3)$. However, this inequality now only applies to $a+b+c=1$, so the return to the general case of any $a,b,c>0$, multiply by $(a+b+c)^4$, which is just $1$ when $a+b+c=1$ to make it
$$
\frac{abc(a+b+c)}{2}\le\frac{1}{2\cdot3^3}=\frac{(a+b+c)^4}{2\cdot3^3}
\quad\text{when}\quad a+b+c=1.
$$
However, the outermost terms, $abc(a+b+c)/2\le(a+b+c)^4/(2\cdot3^3)$ is now homogeneous of degree 4 in $a,b,c$ and so holds for all $a,b,c>0$.
The LHS can be made easier to address if we assume $a^2+b^2+c^2=1$. Then, $a^6/(b^2+c^2)=a^6/(1-a^2)$. On this we can apply Jensen's inequality, but in terms of $u=a^2$, $v=b^2$, $w=c^2$ with $u+v+w=1$. Again, we work under the restriction $u+v+w=1$, and then multiply in the appropriate power of $u+v+w$ at the end to get homogeneous expressions of the same degree.
After the two simplifying bounds, we get the inequality in the middle of the topmost chain of inequalities, which shouldn't be too hard.
|
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|
In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is In a $\triangle ABC,a^2+b^2+c^2=ac+ab\sqrt3$,then the triangle is
$(A)$equilateral
$(B)$isosceles
$(C)$right angled
$(D)$none of these
The given condition is $a^2+b^2+c^2=ac+ab\sqrt3$.
Using sine rule,
$a=2R\sin A,b=2R\sin B,c=2R\sin C$,we get
$\sin^2A+\sin^2B+\sin^2C=\sin A\sin C+\sin A\sin B\sqrt3$
I am stuck here.
|
Considering cosine law:
\begin{align*}
\sum_{abc} (b^2+c^2-a^2) &= \sum_{abc} 2bc\cos A \\
a^2+b^2+c^2&= 2bc\cos A+2ca\cos B+2ab\cos C
\end{align*}
One possibility is $\cos A=0$, $2\cos B=1$ and $2\cos C=\sqrt{3}$.
Hence $A=90^{\circ}$, $B=60^{\circ}$ and $C=30^{\circ}$.
Proof of Uniqueness of the above solution:
Rearrange the equality as a quadratic in $b$:
$$b^2-(a\sqrt{3})b+(a^2-ac+c^2) =0 $$
The discriminant:
\begin{align*}
\Delta &= (-a\sqrt{3})^2-4(a^2-ac+c^2) \\
&=-a^2+4ac-4c^2 \\
&=-(a-2c)^{2} \\
&\leq 0
\end{align*}
To admit a real solution for $b$, $\Delta$ has to be zero. That is
$$a=2c$$
Now, $$3c^2-2bc\sqrt{3}+b^2=0 \implies b=c\sqrt{3}$$
Therefore, $$\fbox{$a^2=b^2+c^2$}$$
|
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|
Find the limit a matrix raised to $n$ when $n$ goes to infinity Let $ A $ be a $ 3\times3 $ matrix such that
$$A
\left( \begin{array}{ccc}
1 \\
2 \\
1 \end{array} \right)=\left( \begin{array}{ccc}
1 \\
2 \\
1 \end{array} \right),~~~A
\left( \begin{array}{ccc}
2 \\
2 \\
0 \end{array} \right)=\left( \begin{array}{ccc}
1 \\
1 \\
0 \end{array} \right),~~A
\left( \begin{array}{ccc}
3 \\
0 \\
6 \end{array} \right)=\left( \begin{array}{ccc}
-1 \\
0 \\
2 \end{array} \right)
$$
Find $$ \lim_{n\to\infty}A^n \left( \begin{array}{ccc}
6 \\
7 \\
0 \end{array} \right)$$
So, do I first find $ A $ by letting A =$ \left( \begin{array}{ccc}
a&b&c \\
d&e&f \\
g&h&i \end{array} \right) $ and using the given information to solve the corresponding linear equations and then solve the actual problem of finding the limit? Is there a more efficient way of doing this? Also, I am not quite sure how to find the limit so any hints would be greatly appreciated. Thanks!
|
Use diagonalization. $A=PDP^{-1}$, where $P=\pmatrix{1&2&3\cr 2&2&0\cr 1&0&6\cr}$ and $D=\pmatrix{1&0&0\cr 0&1/2&0\cr 0&0&-1/3\cr}$. Then
$$\lim_{n\to\infty} A^n \pmatrix{6\cr 7\cr 0\cr} = \lim_{n\to\infty} (PD^n P^{-1}) \cdot \pmatrix{6\cr 7\cr 0\cr} = P \lim_{n\to\infty} D^n \cdot P^{-1} \cdot \pmatrix{6\cr 7\cr 0\cr} = P \cdot \pmatrix{1&0&0\cr 0&0&0\cr 0&0&0\cr} \cdot P^{-1} \cdot \pmatrix{6\cr 7\cr 0\cr}.$$
|
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|
Limit of tan function This is a question from an old tutorial for a basic mathematical analysis module.
Show that $$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = e^2$$
My tutor has already gone through this in class but I am still confused. Is there anything wrong with the following reasoning?
Since $\frac{\pi}{4}+\frac{1}{n} \to \frac{\pi}{4}$ as $n \to \infty$,
it seems to me that $\tan(\frac{\pi}{4}+\frac{1}{n}) \to \tan(\frac{\pi}{4}) = {1}$,
and hence $\tan^n(\frac{\pi}{4}+\frac{1}{n}) \to 1^n = 1$.
Additionally, my tutor has given a hint, to use Squeeze theorem along with the definition $e = \lim\limits_{n \to \infty }(1 + {1\over n})^n$, but I can't see how these are to be used.
Edit: Here's another attempt I've made.
After using addition formula for tangent, we get
$$\frac{1 + \tan{\frac{1}{n}}}{1- \tan{\frac{1}{n}}} = 1 - \frac{2\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}} = 1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}} = [(1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}})^\frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}}]^\frac{2\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}}$$
so given that $\lim\limits_{n \to \infty}{\frac{\tan{\frac{1}{n}}}{\frac{1}{n}}}=1$, we have
$$\lim\limits_{n\to \infty}\tan^n(\frac{\pi}{4}+\frac{1}{n}) = \lim\limits_{n \to \infty}[(1 - \frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}})^\frac{1}{\frac{1-\tan{\frac{1}{n}}}{2\tan{\frac{1}{n}}}}]^\frac{2n\tan{\frac{1}{n}}}{1-\tan{\frac{1}{n}}} = e^\frac{2*1}{1-0} = e^2$$
I'm really hoping that this method works as well! So sorry for the ugly formatting, I couldn't figure out some parts.
|
If the only standard limit available is $$\lim_{n \to \infty}\left(1 + \frac{1}{n}\right)^{n} = e$$ then we are out of luck here. It is better to assume the following standard limits $$\lim_{x \to 0}\frac{\tan x}{x} = 1 = \lim_{x \to 0}\frac{\log(1 + x)}{x}$$ and then we can easily evaluate the desired limit. We have
\begin{align}
\log L &= \log\left\{\lim_{n \to \infty}\tan^{n}\left(\frac{\pi}{4} + \frac{1}{n}\right)\right\}\notag\\
&= \log\left\{\lim_{n \to \infty}\left(\frac{1 + \tan(1/n)}{1 - \tan(1/n)}\right)^{n}\right\}\notag\\
&= \log\left\{\lim_{n \to \infty}\left(1 + \frac{2\tan(1/n)}{1 - \tan(1/n)}\right)^{n}\right\}\notag\\
&= \lim_{n \to \infty}\log\left(1 + \frac{2\tan(1/n)}{1 - \tan(1/n)}\right)^{n}\text{ (via continuity of log)}\notag\\
&= \lim_{n \to \infty}n\log\left(1 + \frac{2\tan(1/n)}{1 - \tan(1/n)}\right)\notag\\
&= \lim_{n \to \infty}n\cdot\dfrac{2\tan(1/n)}{1 - \tan(1/n)}\cdot\dfrac{\log\left(1 + \dfrac{2\tan(1/n)}{1 - \tan(1/n)}\right)}{\dfrac{2\tan(1/n)}{1 - \tan(1/n)}}\notag\\
&= 2\lim_{n \to \infty}\dfrac{\tan(1/n)}{1/n}\cdot\dfrac{1}{1 - \tan(1/n)}\cdot\lim_{x \to 0}\dfrac{\log\left(1 + x\right)}{x}\notag\\
&= 2\cdot 1\cdot 1\cdot 1\notag\\
&= 2\notag
\end{align}
Hence $L = e^{2}$.
|
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|
Is the following solution correct? Question: $ \sqrt{x^2 + 1} + \frac{8}{\sqrt{x^2 + 1}} = \sqrt{x^2 + 9}$
My solution: $(x^2 + 1) + 8 = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) = \sqrt{x^2 + 9} \sqrt{x^2 + 1}$
$=> (x^2 + 9) - \sqrt{x^2 + 9} \sqrt{x^2 + 1} = 0$
$=> \sqrt{x^2 + 9} (\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$
So, either $\sqrt{x^2 + 9} = 0$ or $(\sqrt{x^2 + 9} - \sqrt{x^2 + 1}) = 0$
From the first expression, I get $x = \pm 3 i$ and from the second expression, I get nothing.
Now, notice how in the 2nd step, I could've divided both the sides by $\sqrt{x^2 + 9}$, but I didn't because I learned here that we must never do that and that we should always factor: Why one should never divide by an expression that contains a variable.
So, my question is: is the solution above correct? Would it have been any harm had I divided both the sides by $\sqrt{x^2 + 9}$?
|
It would indeed , because then you wont have $x=\pm 3i$ as the root because that would make the denominator zero , which is the greatest offense one can do in algebra!! or more broadly in Mathematics (:P)
|
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|
Find the limit $\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$ Find the limit if it exists. (this exercise is taken from Calculus - The Classic Edition by Swokowski Chapter 10, section 1, no. 9, p.498)
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$$
since the limit is $0/0$ therefore, we use L'Hopital's rule, that is,
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x} = \lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 }$$
since $\lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 } = \frac {1 - 1}{ 1- 1} = \frac00$. Thus, we use the L'Hopital's rule again. that is,
$$\lim_{x\rightarrow 0}\frac{\cos x - 1}{\sec^2 x - 1 } = \lim_{x\rightarrow 0}\frac{-\sin x}{2 \sec^2 x \sec x \tan x}$$
since $\lim_{x\rightarrow 0} \frac{-\sin x}{2 \sec^2 x \sec x \tan x}= \frac{0}{2(1)(0)} = \frac00$. Thus, it always goes to zero by zero. But the answer for this question is $\frac{-1}2$. How is that?!! there must be something missing that either I forget or misunderstand.
|
You can make it a little simpler by rewriting
$$\frac{\sin(x)-x}{\tan(x)-x}=\cos(x)\frac{\sin(x)-x}{\sin(x)-x\cos(x)}.$$
The first factor tends to $1$ and can be ignored. Then, applying L'Hospital three times,
$$\frac{\sin(x)-x}{\sin(x)-x\cos(x)}\to\frac{\cos(x)-1}{\cos(x)-\cos(x)+x\sin(x)}\to\frac{-\sin(x)}{\sin(x)+x\cos(x)}\to\frac{-\cos(x)}{\cos(x)+\cos(x)-x\sin(x)}\to-\frac12.$$
Alternatively,
$$\frac{\cos(x)-1}{x\sin(x)}=\frac{-2\sin^2\left(\dfrac x2\right)}{4\left(\dfrac{x}2\right)^2}\frac x{\sin(x)}\to-\frac24\cdot1^2\cdot1.$$
|
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|
Solve for a in Exponential equation Is it possible to solve for $a$ in the following equation: $a^\alpha=b^\alpha-a$? Currently, I have resorted to using Excel to approximate $a$ (I am given values for $b$ and $\alpha$), but am wondering if it is possible to pinpoint $a$ exactly.
|
Although it can not be solved generally, it can be solved if $\alpha \in \{ -1,0, \frac 12,1,2 \}$
Let's consider these in turn:
*
*$\alpha = -1$
$a^{-1} =b^{-1}-a$
$b =a-a^2b$
$a^2b-a+b=0$
$a= \displaystyle \frac {1 \pm \sqrt{(-1)^2-4(b)(b)}}{2b}$
$a= \displaystyle \frac {1 \pm \sqrt{1-4b^2}}{2b}$
*$\alpha = 0$
$a^0 =b^0-a$
$1 =1-a$
$a=0$
*$\alpha = \frac 12$
$a^{\frac 12} =b^{\frac 12}-a$
Let $x=a^{\frac 12} \Rightarrow x^2=a$
$x =b^{\frac 12}-x^2$
$x^2+x-b^{\frac 12}=0$
$x= \displaystyle \frac {-1 \pm \sqrt{1+4b^{\frac 12}}}{2}$
$a= \displaystyle \left(\frac {-1 \pm \sqrt{1+4b^{\frac 12}}}{2}\right)^2$
*$\alpha = 1$
$a^{1} =b^{1}-a$
$a =b-a$
$2a=b$
$a= \displaystyle \frac b2$
*$\alpha = 2$
$a^{2} =b^{2}+a$
$a^2-a-b^2=0$
$a= \displaystyle \frac {1 \pm \sqrt{(-1)^2-4(-b^2)}}{2}$
$a= \displaystyle \frac {1 \pm \sqrt{1+4b^2}}{2}$
|
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|
Given that $\cos A + \cos B + \cos C = 0$ and $\sin A + \sin B + \sin C = 0$. If $\cos A + \cos B + \cos C = 0$ and $\sin A + \sin B + \sin C = 0$.
The value of $ \sin^3A+\sin^3B+\sin^3C$
What I can see here is that as $\sin A + \sin B + \sin C = 0$ hence $ \sin^3A+\sin^3B+\sin^3C=3\sin A \sin B\sin C$ but I am not able to achieve a constant value. Please give some hint.
|
Hint: If the centroid of a triangle coincides with it circumcenter, the triangle is equilateral.
Answer: $\sin^3(A)+\sin^3(B)+\sin^3(C)=3\sin^3(A)-\frac{9}{4}\sin(A)=-\frac{3}{4}\sin(3A)$.
|
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|
Surface of sphere above/below ellipse I am struggling with the following problem: Find the surface area of $x^2+y^2+z^2=a^2$ enclosed by the cylinder $\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1$ $(a>b>0)$. The solution of the problem is supposed to be $4 \pi a^2 - 8a^2 \arcsin\left(\frac{\sqrt{a^2-b^2}}{a}\right)$.
I tried polar coordinates, I tried $x= a \, r \, cos(\theta)$, $y= b \, r \, sin(\theta)$ (with the Jacobian of this transformation equal to $abr$, and the limits of integration $0≤\theta ≤ \frac{\pi}{2}$ and $0≤ r ≤1$).
Time and time again, I get integranda which are impossible to integrate with respect to $\theta$ (and which also definitely won't give me an $\arcsin$ as a result).
Any help would be greatly appreciated.
|
It turns out we don't really need any calculus to solve this problem.
For completeness, let us first solve this problem the calculus way.
Method 1 - using spherical polar coordinates.
By symmetry, we only need to compute the area on the upper hemisphere and then multiply the result by $2$.
Let $c^2 = a^2-b^2$ and parametrize the sphere by spherical polar coordinate
$$(x,y,z) = (a\sin\theta\cos\phi,a\sin\theta\sin\phi,a\cos\theta)$$
In the upper hemisphere, the portion of unit sphere within the ellipsoidal cylinder $$\left\{ (x,y,z) : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \right\}$$
is given by
$$\frac{(a\sin\theta\cos\phi)^2}{a^2} + \frac{(a\sin\theta\sin\phi)^2}{b^2} \le 1
\iff \sin\theta \le \frac{1}{\sqrt{\cos^2\phi + \frac{a^2}{b^2}\sin^2\phi}}
= \frac{1}{\sqrt{1 + \frac{c^2}{b^2}\sin^2\phi}}
$$
This means as $\phi$ varies from $0$ to $2\pi$, $\theta$ can take values from $0$ to $\theta_0$ (dependent of $\phi$) where
$\theta_0 \in [ 0, \frac{\pi}{2} ]$ is determined by
$$\sin\theta_0 = \frac{1}{\sqrt{1+\frac{c^2}{b^2}\sin^2\phi}}
\quad\iff\quad
\cos\theta_0 =
\frac{\frac{c}{b}|\sin\phi|}{\sqrt{1+\frac{c^2}{b^2}\sin^2\phi}}
= \frac{\frac{c}{a}|\sin\phi|}{\sqrt{1-\frac{c^2}{a^2}\cos^2\phi}}
$$
The area on the upper hemisphere we seek is then given by the integral
$$\begin{align}
\verb/Area/_{up}
&= a^2\int_0^{2\pi}\int_0^{\theta_0} \sin\theta d\theta d\phi
= a^2\int_0^{2\pi}(1-\cos\theta_0) d\phi\\
&= a^2\left[ 2\pi - \int_0^{2\pi}\frac{\frac{c}{a}|\sin\phi|}{\sqrt{1-\frac{c^2}{a^2}\cos^2\phi}} d\phi \right]
= a^2\left[ 2\pi - 4\int_0^{\pi/2}
\frac{\frac{c}{a}\sin\phi}{\sqrt{1-\frac{c^2}{a^2}\cos^2\phi}} d\phi \right]
\end{align}
$$
Change variable to $t = \frac{c}{a}\cos\phi$, we get
$$\verb/Area/_{up}
= a^2\left[ 2\pi - 4\int_0^{c/a} \frac{dt}{\sqrt{1-t^2}} \right]
= a^2\left[ 2\pi - 4\sin^{-1}\left(\frac{c}{a}\right)\right]
$$
As a result, the area we seek is
$$\verb/Area/ = 2\verb/Area/_{up} = 4\pi a^2 - 8a^2\sin^{-1}\left(\frac{\sqrt{a^2-b^2}}{a}\right)$$
Method 2 - the geometric way.
As mentioned in beginning of this answer, we don't need any calculus to derive
the result.
For any point $(x,y,z)$ on the intersection of the sphere and the ellipsoidal cylinder, we have
$$\frac{x^2}{a^2} + \frac{y^2}{a^2} + \frac{z^2}{a^2} = 1 = \frac{x^2}{a^2} + \frac{y^2}{b^2}
\iff \frac{z^2}{a^2} = \frac{y^2}{b^2} - \frac{y^2}{a^2} = \frac{y^2c^2}{a^2b^2}
\iff z = \pm \frac{c}{b} y
$$
The RHS is the equation for a pair of planes.
What this means is the portion of upper hemisphere within the ellipsoidal cylinder is the portion of sphere above the two planes $z \ge \pm \frac{c}{b} y$.
Since this two planes are making an angle $\tan^{-1}\left(\frac{c}{b}\right)$ with $xy$-plane, we find:
$$\begin{align}
\verb/Area/ = 2\verb/Area/_{up}
&= 4a^2\left[\pi - 2\tan^{-1}\left(\frac{c}{b}\right)\right]
= a^2\left[4\pi - 8\sin^{-1}\left(\frac{\frac{c}{b}}{\sqrt{1+\frac{c^2}{b^2}}}\right)\right]\\
&= a^2\left[4\pi - 8\sin^{-1}\left(\frac{c}{a}\right)\right]
\end{align}
$$
The same result we obtained by Method 1 above.
|
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|
How to find continuities with square root? I don't understand how to find $$\frac{4-x^2}{3-\sqrt{x^2+5}}$$
The book says to multiply the equation by $\frac{3 + \sqrt {x^2+5}}{3 + \sqrt {x^2+5}}$. I don't understand where that comes from. It says the multiplication simplifies to "$3 + \sqrt {x^2+5}$" - I don't see how that's possible. Is the book wrong?
|
Just use the binomial identity
$$(a-b)(a+b)=a^2-b^2,$$
which for $a=3$ and $b=\sqrt{x^2+5}$ yields
$$\frac{(4-x^2)(3+\sqrt{x^2+5})}{(3-\sqrt{x^2+5})(3+\sqrt{x^2+5})}=\frac{(4-x^2)(3+\sqrt{x^2+5})}{9-(x^2+5)}=\frac{(4-x^2)(3+\sqrt{x^2+5})}{(4-x^2)}$$
which is your desired result $(3+\sqrt{x^2+5}).$
|
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|
Expanding a complex function in Taylor series Expand the function
$$ f(z) = \frac {2(z + 2)} {z^2 − 4z + 3} $$
in a Taylor series about the point $ z = 2 $ and find the circle C inside of which the series converges. Find a Laurent series that converges in the region outside of C.
I tried writing the denominator as $ (z-3)(z-1) $ to find the singularities $ z = 1,3 $ both simple. There exists a circle around $ z = 2 $ such that $f(z) $ is analytic so we can write it as a Taylor series. I got stuck trying to make the function look like a known Taylor series like $e^z$ or $\frac {1} {1-z}$
|
Split them into partial fraction and upon solving you will get,
$ \frac {2(z+2)}{(z-1)(z-3)} = \frac {3} {(z-(-1))} + \frac {5} {z-3} $
taylor's series expansion is
$ \frac {1}{1-x} = \frac{1}{1-a} + \frac {x-a}{(1-a)^2} + \frac {(x-a)^2}{(1-a)^3} + ...$
so for $\frac {3}{(z-(-1)} $ , value of a=-1 i.e., $ \mathcal\ z ^\left(-1 \right) $ $\left[ \frac {1}{z-a} \right] $
the taylor's series for
$\frac {3}{z-(-1)}$ = $ 3 \{ \frac{1}{2} + \frac{(x+1)}{4} +\frac {(x+1)^2}{8} + ...\}$
similaryly for $\frac {5}{(z-3)} $ , value of a=3 i.e., $ \mathcal\ z ^\left(-1 \right) $ $\left[ \frac {1}{z-a} \right] $
the taylor's series for
$\frac {5}{z-3}$ = $ 5 \{ \frac{-1}{2} + \frac{(x-3)}{4} +\frac {(x-3)^2}{8} + ...\}$
|
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|
Find the sum using The question is as follows:
Find the sum:
$1\cdot2 + 2\cdot3 + ... + (n-1)n$
What I have tried so far:
We can write $(n-1)n$ as $\frac{(n+1)!}{(n-1)!}$ which we can also write as $2\cdot\dbinom{n+1}{2}$
I believe it is possible to use the binomial theorem here, setting $a = b = 1$ in $(a+b)^n$. I am not sure how to proceed however.
|
Hint: Rewrite $k(k+1)$ as $\frac{1}{3}((k+1)^3-k^3)-\frac{1}{3}$. Observe the mass cancellation (telescoping) when we add up.
Another way: As remarked in the post, we want
$$2\left(\binom{2}{2}+\binom{3}{2}+\binom{4}{2}+\cdots+\binom{n}{2}\right).$$
We show that the sum of the binomial coefficients is $\binom{n+1}{3}$. That will show that our sum is $2\cdot\binom{n+1}{3}$, which simplifies to $\frac{(n+1)(n)(n-1)}{3}$.
There are $n+1$ different doughnuts in a row. We want to choose $3$ of them for a healthy breakfast. There are $\binom{n+1}{3}$ ways to do the choosing.
Let us count another way. Maybe the leftmost doughnut we choose is the first doughnut. Then there are $\binom{n}{2}$ ways to choose the other $2$.
Maybe the leftmost doughnut we choose is the second one. There are then $\binom{n-1}{2}$ ways to choose the other $2$. Continue. Maybe the leftmost doughnut we choose is the third from the right end. Then there are $\binom{2}{2}$ ways to choose the other $2$. We have shown that
$$\binom{n+1}{3}=\binom{n}{2}+\binom{n-1}{2}+\cdots +\binom{2}{2}.$$
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Find all function that satisfy $(f(x) + xy) \cdot f(x - 3y) + (f(y) + xy) \cdot f(3x - y) = (f(x + y))^2$ Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that for all real numbers $x$ and $y$,$$(f(x) + xy) \cdot f(x - 3y) + (f(y) + xy) \cdot f(3x - y) = (f(x + y))^2.$$
|
Letting $x=y=0$, we have $ f(0)^2+f(0)^2=f(0)^2$, hence $$\tag1f(0)=0.$$
If $a=x-3y$ and $b=3x-y$ then $x+y=\frac{b-a}2$, whence
$$\tag2f(a)=f(b)=0\implies f\left(\frac{b-a}2\right)=0$$
and in combination with $(1)$
$$ \tag3f(x)=0\implies f\left(\pm\tfrac12x\right)=0.$$
Assume $f(a)=0$ (with $a\ne 0$). Then for $x=\frac12a$, $y=-\frac12a$, we have $f(x)=f(y)=0$ by $(3)$ and so $2xyf(2a)=f(0)^2$, i.e.,
$$\tag4f(a)=0\implies f(2a)=0. $$
From $(1)$, $(2)$, and $(4)$ we see that $A:=\{\,x\in\Bbb R\mid f(x)=0\,\}$ is a subgroup of $\Bbb R$, which is closed under division by $2$ because of $(3)$.
Let $B=(\Bbb R\setminus A)\cup \{0\}$.
As $A$ is a group and $(3)$, we have that $$\tag{$\star$} a\in A, b\in B\implies -b\in B,2b\in B,\frac12b\in B,b\pm a\in B.$$
For $y=-x$, we have
$$(f(x)-x^2)f(4x)+(f(-x)-x^2)f(4x)=f(0)^2,$$ so (using $f(4x)=0\iff f(\pm x)=0$)
$$ \tag5x\in B\implies f(x)+f(-x)=2x^2.$$
Letting $x=y$, we have $$(f(x)+x^2)(f(-2x)+f(2x))=f(2x)^2,$$and likewise ($x\leftrightarrow-x$)
$$(f(-x)+x^2)(f(-2x)+f(2x))=f(-2x)^2,$$ hence by subtracting
$$(f(x)-f(-x))(f(-2x)+f(2x))=f(2x)^2-f(-2x)^2. $$
If $x\notin A$, we can divide off $f(-2x)+f(2x)=8x^2$ to obtain
$$\tag6 f(x)-f(-x)=f(2x)-f(-2x)$$
(which trivially also holds for $x\in A$). Adding $8x^2$ and using $(5)$
$$\tag7 x\in B\implies f(x)+3x^2=f(2x)$$
Revisiting the case $y=x\in B$ now gives
$$\begin{align}x\in B\implies\quad (f(x)+x^2)(f(2x)+f(-2x))&=f(2x)^2\\
(f(x)+x^2)8x^2&=(f(x)+3x)^2\\
8x^2f(x)+8x^4&= f(x)^2+6x^2f(x)+9x^4\\
(f(x)-x^2)^2&=0\end{align} $$
i.e.,
$$x\in B\implies f(x)=x^2.$$
Assume $x,y\in B$ and $x+y\in A$. Then by $(\star)$, $$3x-y=2\cdot2\cdot x-(x+y)\in B$$ and likewise $x-3y\in B$, so that $$f(x+y)^2=(x^2+xy)(3x-y)^2+(y^2+xy)(x-3y)^2=(x+y)^4,$$ i.e., also $x+y\in B$. We conclude that $B$ is also a group.
But the union of two subgroups is the whole group $\Bbb R$ only if one already equals $\Bbb R$. Therefore $A=\Bbb R$ or $B=\Bbb R$, i.e., the only two solutions $f_1,f_2$ of the functional equation are $f_1(x)=0$ and $f_2(x)=x^2$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$ - I keep getting imaginary numbers $$\sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 = 38$$
My attempt
$\sqrt{x+938^2} + \sqrt{x + 140^2} = 1116$
$(\sqrt{x+938^2} + \sqrt{x + 140^2})^2 = (1116)^2$
$x+938^2 + 2*\sqrt{x+938^2}*\sqrt{x + 140^2} + x + 140^2 = 1116^2$
$2x + 2*\sqrt{x+938^2}*\sqrt{x + 140^2} = 1116^2 - 938^2 - 140^2$
$x + \sqrt{x^2 + 2(938^2 + 140^2)x+(938*140)^2} = 1116^2 - 938^2 - 140^2$
At this point trying to solve for x inside the sqrt in the quadratic gives me an imaginary number. How is it possible to solve this?
|
Repeated squaring is not necessary. Let $y > -140$ be such that $x = y^2 + 280y$, so that $$\sqrt{x+140^2} - 140 = \sqrt{(y + 140)^2} - 140 = y.$$ Then $$\begin{align*} 38 &= \sqrt{x+938^2} - 938 + \sqrt{x + 140^2} - 140 \\ &= \sqrt{y^2 + 280y + 938^2} - 938 + y, \end{align*}$$ or equivalently, $$976-y = \sqrt{y^2 + 280y + 938^2}.$$ Now we square only once, giving $$y^2 - 2(976)y + 976^2 = y^2 + 280y + 938^2.$$ The quadratic terms cancel, and we get $$2(140+976)y = 976^2 - 938^2 = (976-938)(976+938) = 38(1914),$$ or $$y = \frac{6061}{186}.$$ Then substitute back to find the solution for $x$.
|
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|
Prove that $\sqrt[3]{2} +\sqrt{5}$ is algebraic over $\mathbb Q$ Prove that $\sqrt[3]{2} +\sqrt{5}$ is algebraic over $\mathbb{Q}$ (by finding a nonzero polynomial $p(x)$ with coefficients in $\mathbb{Q}$ which has $\sqrt[3] 2+\sqrt 5$ as a root).
I first tried letting $a=\sqrt[3]{2} +\sqrt{5}$ and then square both sides. But I keep on going into a loop by continuing to square it over and over again. Then I tried $a^3=(\sqrt[3]{2} +\sqrt{5})^3$. Just can't seem to get rid of the radicals.
|
$x = \sqrt[3]{2} + \sqrt{5}$
$x - \sqrt{5} = \sqrt[3]{2}$
$x^3 - 3\sqrt{5}x^2 + 15x - 5\sqrt{5} = 2$
$x^3 - 15x - 2 = \sqrt{5}(3x^2 + 5)$
$(x^3 - 15x - 2)^2 = 5(3x^2 + 5)^2$
$x^6 -75x^4 - 4x^3 - 400x^2 + 60x - 121 = 0$
Now, ignoring accuracy and assuming arithmetical errors are both inevitable and irrelevant, there is another correctly figured out 6th degree polynomial with $ \sqrt[3]{2} + \sqrt{5}$ as a solution.
Unless, I didn't make an arithmetical error calculating this. I might not have. There's a first time for everything.
====
So obviously I made an arithmetical error. I knew I would. But it doesn't matter as the solving for the polynomial doesn't affect that there will be such a polynomial.
$x = \sqrt[3]{2} + \sqrt{5} \iff$
$x + k_1\sqrt{5} = \sqrt[3]{2} \iff$
$x^3 + k_2\sqrt{5}x^2 + k_3*5x + k_4*5*\sqrt{5} = 2 \iff$
$x^3 + k_5x + k_6 = \sqrt{5}(k_7x^2 + k_8) \iff $
$(x^3 + k_5x + k_6)^2= 5(k_7x^2 + k_8)^2 \iff $
$x^6 + k_9x^5 + k_{10}x^4 + k_{11}x^3 + k_{12}x^2 + k_{13}x + k_{14} = 0$
Where $k_i$ are integers and easily calculatable by someone who can count to 20 with his shoes on.
|
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|
Find all integer values of $x$ such that $x^2 + 13x + 3$ is a perfect integer square.
Question: Find all integer values of $x$ such that $x^2 + 13x + 3$ is a perfect integer square.
What I have attempted;
For $x^2 + 13x + 3$ to be a perfect integer square let it equal $k^2$ where $k \in \mathbb{Z} $
Hence $$x^2 + 13x + 3 = k^2$$
$$ \Leftrightarrow x^2 + 13x + (3-k^2)=0$$
$$ \Leftrightarrow x={-b\pm\sqrt{b^2-4ac} \over 2a}
$$
$$ \Leftrightarrow x={-13\pm\sqrt{13^2-4(1)(3-k^2)} \over 2(1)}$$
$$ \Leftrightarrow x={-13\pm\sqrt{169-12+4k^2} \over 2}$$
$$ \Leftrightarrow x={-13\pm\sqrt{157+4k^2} \over 2}$$
We also want $157+4k^2$ to be a perfect square hence
$$ 157 + 4k^2 = n^2 $$
$$ n^2 -4k^2 = 157 $$
$$ (n-2k)(n+2k) = 157 $$
What should I do now?
I have looked at this; Find all integers $x$ such that $x^2+3x+24$ is a perfect square.
but I am confused what I should do to continue..
|
Observe that $x^2+13x+3=(x+\frac{13}{2})^2-\frac{157}{4}$
As you have assumed, let $(x+\frac{13}{2})^2-\frac{157}{4}=k^2$
Hence, by simplifying, we get that $(2x+13)^2-157=4k^2$
Further simplifying by factorisation yields $(2x+13-2k)(2x+13+2k)=157=157\cdot 1$
So, there are only two possibilities:
*
*$(2x+13-2k)=157$ and $(2x+13+2k)=1$
*$(2x+13-2k)=1$ and $(2x+13+2k)=157$
The first one yields $k=-39$ and the second one yields $k=39$.
Can you find $x$ now??
|
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|
Derivation of Spherical Law of Cosines I am trying to get a derivation of the spherical law of cosines. The Wikipedia page [https://en.wikipedia.org/wiki/Spherical_law_of_cosines ] contains a proof that I don't understand because there are not enough intermediate steps shown.
The Wikipedia page says that for unit vectors $\mathbf{u},\mathbf{v},\mathbf{w}$ :
$$
\begin{align}
\cos(a) = \mathbf{u}\cdot\mathbf{v}
\end{align}
$$
\begin{align}
\cos(b) = \mathbf{u}\cdot\mathbf{w}
\end{align}
$$
\begin{align}
\cos(c) = \mathbf{v}\cdot\mathbf{w}
\end{align}
$$
The unit vector $\mathbf{t}_a$ is defined as the unit vector perpendicular to $\mathbf{u}$ in the u-v plane, whose direction is given by the component of v perpendicular to u. Wikipedia explains that
$$
\begin{align}
\mathbf{t}_a =
{
{\mathbf{v}-\mathbf{u}(\mathbf{u \cdot v})}
\over
{
\left\lVert
{\mathbf{v}-\mathbf{u}(\mathbf{u \cdot v})}
\right\rVert
}}
=
{{{\mathbf{v}-\mathbf{u}\cos(a)}}
\over
{\sin(a)}}
\end{align}
$$
Similarly,
$$
\begin{align}
{\mathbf{t}_b}
=
{{{\mathbf{w}-\mathbf{u}\cos(b)}}
\over
{\sin(b)}}
\end{align}
$$
Without any further justification, their proof ends by the claim:
$$
\begin{align}
{\mathbf{t}_a} \cdot {\mathbf{t}_b}
=
{{\cos(c)-\cos(a)\cos(b)}
\over
{\sin(a)\sin(b)}}
\end{align}
$$
It is the final step, which yeilds the formula for $\mathbf{t}_a\cdot\mathbf{t}_b$, that I do not comprehend. Can somebody please explain to me the justification for this expression?
|
$$\mathbf{t}_a \cdot \mathbf{t}_b = \frac{\mathbf{v} - \mathbf{u} \cos(a)}{\sin(a)} \cdot \frac{\mathbf{w} - \mathbf{u} \cos(b)}{\sin(b)} \\ = \frac{\mathbf{v} \cdot \mathbf{w} - \mathbf{v} \cdot \mathbf{u} \cos(b) - \mathbf{u} \cdot \mathbf{w} \cos(a) + \mathbf{u} \cdot \mathbf{u}\cos(a)\cos(b)}{\sin(a)\sin(b)} \\ = \frac{\cos(c) - \cos(a)\cos(b) - \cos(a)\cos(b) + \cos(a)\cos(b)}{\sin(a)\sin(b)} = \frac{\cos(c) - \cos(a)\cos(b)}{\sin(a)\sin(b)}.$$
Note that $\mathbf{u} \cdot \mathbf{u} = 1$ since it's a unit vector.
|
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|
Diophantine $4x^3-y^2=3$ I am interested in how to tackle this Diophantine equation:
$$4x^3-y^2=3$$
The solutions I have found so far are $(1,1)$ and $(7,37)$. Are there any more?
I have looked up various material on cubic Diophantines but most of what I’ve found is on equations where the coefficients of $x^3$ and $y^2$ are the same. In this particular problem, if both coefficients were equal to $1$, it would just be a nice Mordell’s equation. But the coefficient of the cubic variable is not $1$ – which is why it’s so frustrating. Still, would I be right in saying that if the solutions were to lie on an elliptic curve, there would only be finitely many of them? What if they don’t lie on an elliptic curve? Will the number of solutions still be finite?
|
Let $x$ and $y$ be integers such that $4x^3-y^2=3$. Reducing mod $2$ shows that $y$ is odd, so $y=2z+1$ for some integer $z$. Then in the Eisenstein integers $\Bbb{Z}[\omega]$, where $\omega^2+\omega+1=0$, we have
\begin{eqnarray*}
x^3&=&\frac{y^2+3}{4}
=\left(\frac{y+\sqrt{-3}}{2}\right)\left(\frac{y-\sqrt{-3}}{2}\right)\\
&=&\left(\frac{y+1+2\omega}{2}\right)\left(\frac{y-1-2\omega}{2}\right)\\
&=&(z+1+\omega)(z-\omega).
\end{eqnarray*}
The greatest common divisor of the two factors on the right hand side also divides their sum and difference, which are $y$ and $1+2\omega$, respectively. Note that $1+2\omega$ is prime in $\Bbb{Z}[\omega]$ with $\mathcal{N}(1+2\omega)=3$, and that $y$ is coprime to $3$ as otherwise
$$4x^3=y^2+3\equiv3\pmod{9},$$
which is clearly impossible. This shows that the two factors are coprime, and hence that both are perfect cubes in $\Bbb{Z}[\omega]$, up to units. That is to say, there exist $a,b\in\Bbb{Z}$ such that
$$z-\omega=\omega^k(a+b\omega)^3,$$
for some $k\in\{0,1,2\}$. Expanding the right hand side then yields
\begin{eqnarray*}
z-\omega&=&a^3-3ab^2+b^3+(3a^2b-3ab^2)\omega,\\
z-\omega&=&(3ab^2-3a^2b)+(a^3-3a^2b+b^3)\omega,\\
z-\omega&=&-(a^3-3a^2b+b^3)-(a^3-3ab^2+b^3)\omega,
\end{eqnarray*}
for $k=0,1,2$ respectively. In particular, comparing the coefficients of $\omega$, we get the equations
\begin{eqnarray*}
-1&=&3a^2b-3ab^2=3ab(a-b),\\
-1&=&a^3-3a^2b+b^2,\\
-1&=&-(a^3-3ab^2+b^3).
\end{eqnarray*}
The first clearly has no integral solutions, and solutions to the second and third equations are in bijection by $(a,b)\ \leftrightarrow\ (-b,-a)$. So it suffices to find all integers $a$ and $b$ such that
$$a^3-3a^2b+b^3=-1.$$
This is a Thue equation, for which there exist effective methods to find all integral solutions. PARI/GP tells me that they are
$$(-3,-1),\quad(-1,0),\quad(0,-1),\quad(1,-2),\quad(1,1),\quad(2,3).$$
Then the integral solutions to the third equation are
$$(1,3),\quad(0,1),\quad(1,0),\quad(2,-1),\quad(-1,-1),\quad(-3,-2).$$
The corresponding values of $z$ for the second equation are
$$0,\quad0,\quad0,\quad18,\quad0,\quad18,$$
corresponding to $y=1,37$. For the third equation they are
$$-1,\quad-1,\quad-1,\quad-19,\quad-1,\quad-19,$$
unsurprisingly corresponding to $y=-1,-37$. This shows that you have found all solutions.
|
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|
How to prove that $(x^2+y)(y^2+z)(z^2+x)+2xyz \leqslant 10$, where $x,y,z \in \mathbb{R}^+$ and $x^2+y^2+z^2+xyz=4$ $x,y,z \in \mathbb{R}^+$ and $x^2+y^2+z^2+xyz=4$, prove
$$(x^2+y)(y^2+z)(z^2+x)+2xyz \leqslant 10$$
I try several trig substitutions but feel hopeless with the cyclic term here. The condition $x^2+y^2+z^2+xyz=4$ made it too difficult to homogenize the inequality. I don't even know how to do brutal force either.
|
$\color{brown}{\textbf{Trigonometrical substitution.}}$
From the given conditions should
\begin{cases}
x,y,z \in [0,2]\\[4pt]
(2z+xy)^2 = (4-x^2)(4-y^2).\tag1
\end{cases}
Taking in account $(1),$ can be applied substitution
$$x=2\sin a,\quad y=2\sin b\quad \Rightarrow \quad z = 2\cos(a+b) = 2\sin c,\tag2$$
where
$$a\ge 0,\quad b\ge 0,\quad c\ge 0,\quad a+b+c=\dfrac\pi2.\tag3$$
Then the given inequality takes the form of
$$4(2\sin^2a+\sin b)(2\sin^2b+\sin c)(2\sin^2c+\sin a) + 8\sin a\sin b\sin c \le 5.\tag4$$
$\color{brown}{\textbf{The proof.}}$
Taking in account $(3),$ one can get
\begin{align}
&2\sin^2 a + \sin b = \sin^2 a + 1 - \cos^2 a + \cos (c+a)\\[4pt]
&= 1 + \cos a (\cos c - \cos a) + \sin a (\sin a -\sin c)
= 1 - A \sin (a+\varphi),
\end{align}
where
\begin{align}
&A = \sqrt {(\cos c -\cos a)^2 + (\sin a - \sin c)^2}
= \sqrt {2 - 2\cos (a-c)} = 2 \sin \dfrac{|a-c|}2,\\[4pt]
&\tan \varphi = \dfrac{\cos c -\cos a}{\sin a - \sin c}
= \dfrac{2\sin \frac{a-c}2 \sin \frac{a+c}2}{2\sin \frac{a-c}2 \cos \frac{a+c}2}
= \tan \frac{a+c}2.
\end{align}
Therefore,
$$2\sin^2 a + \sin b = 1 - 2 \sin \frac {|a-c|}2 \sin \frac{3a+c}2.$$
Using the symmetry of task by $a,b,c$, should
$$2\sin^2 a + \sin b \le 1,\quad 2\sin^2 b + \sin c \le 1,\quad
2\sin^2 c + \sin a \le 1.\tag5$$
On the other hand, is known the identity
\begin{align}
\sin(a+b+c) = \cos a \cos b \sin c + \cos a \sin b \cos c + \sin a \cos b \cos c - \sin a \sin b \sin c.
\end{align}
Taking in account $(3),$ one can get
\begin{align}
&\sin a \sin b \sin c = \cos a \cos b \sin c + \cos a \sin b \cos c
+ \sin a \cos b \cos c - \sin(a+b+c)\\[4pt]
&= \dfrac12(\cos a \sin(b+c) + \cos c \sin(a+b) + \cos b \sin(c+a)) - 1\\[4pt]
&= \dfrac12(\cos^2 a + \cos^2 c + \cos^2 b) - 1
= \dfrac14(\cos 2a + \cos 2b + \cos 2c - 1)\\[4pt]
&= \dfrac18 (\cos(a+b)\cos(a-b) + \cos(b+c)\cos(b-c) + \cos(c+a)\cos(c-a) - 2)\\[4pt]
&= \dfrac18 (\sin c\cos(a-b) + \sin a\cos(b-c) + \sin b\cos(c-a) - 2),
\end{align}
$$\sin a \sin b \sin c \le \dfrac18.\tag6$$
Since from $(5),(6)$ should $(4),$ then the given inequality is proved.
|
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|
Evaluation of $\int_{0}^{1}\frac{x^4(1-x^2)^5}{(1+x^2)^{10}}dx$
Evaluation of $\displaystyle \int_{0}^{1}\frac{x^4(1-x^2)^5}{(1+x^2)^{10}}dx$
$\bf{My\; Try::}$ Let $$I = \int_{0}^{1}\frac{x^4(1-x^2)^5}{(1+x^2)^{10}}dx$$
Now Put $x=\tan \theta\;,$ Then $dx = \sec^2 \theta d\theta$ and changing limits, We get
$$I = \int_{0}^{\frac{\pi}{4}}\frac{\tan^4 \theta\cdot \left(1-\tan^2 \theta\right)^5}{(1+\tan^2\theta)^{10}}\cdot \sec^2 \theta d\theta = \int_{0}^{\frac{\pi}{4}}\sin^4 \theta\cdot \cos^4 \theta \left(\cos 2 \theta\right)^5 d\theta$$
So $$I = \frac{1}{16}\int_{0}^{\frac{\pi}{4}}(\sin 2\theta)^4\cdot (\cos 2 \theta)^4\cdot \cos 2 \theta d\theta$$
Now Put $\sin 2 \theta = t\;,$ Then $2\cos 2 \theta d\theta = dt$ and changing limits, We get
$$I = \frac{1}{64}\int_{0}^{1}t^4(1-t^2)^2 dt$$
Now after expanding , We can integrate it,
My question is can we solve it any other way, If yes then plz explain here
Thanks
|
There was a small error in the development in the OP. The integral $I$ should be $$I=\frac{1}{32}\int_0^1 t^4(1-t^2)^2\,dt$$
We can evaluate the integral of interest in terms of the Beta function, $$B(x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\,dt$$and its relationship to the Gamma function, $$B (x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$$
Note that
$$\begin{align}
I&=\frac{1}{32}\int_0^1 t^4(1-t^2)^2\,dt\\\\
&=\frac{1}{64}\int_0^1 t^{3/2}(1-t)^2 \,dt\,\,\dots\text{enforcing the substitution}\,\,t\to \sqrt{t}\\\\
&=\frac{1}{64}B(5/2,3)\\\\
&=\frac{1}{64}\frac{\Gamma(5/2)\Gamma(3)}{\Gamma(11/2)}\\\\
&=\frac{1}{64}\frac{\left(3\frac{\sqrt{\pi}}{4}\right)\,(2!)}{945\frac{\sqrt{\pi}}{32}}\\\\
&=\frac{1}{1260}
\end{align}$$
|
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|
Prove: $\cos^2 x (\sec x - 1)(\sec x + 1) = (1 - \cos x)(1 + \cos x)$
Prove the trigonometric identity
$$\cos^2 x (\sec x - 1)(\sec x + 1) = (1 - \cos x)(1 + \cos x)$$
I've searched high and low on the net and cannot find identities where there is $+$ or $- 1$'s in the equation. Any help is appreciated.
Edit after reviewing comments.
$\cos^2 {x} (\sec {x} -1)(\sec {x} +1) = (1 - \cos {x})(1 + \cos {x})\quad$
$\cos^2 {x} (\sec^2 {x}- 1) = 1 - \cos^2 {x}\quad$
$\cos^2 {x}·tan^2 {x} = \sin^2 {x} \quad$
$\cos^2 {x} ·\frac{\sin^2 {x}}{\cos^2 {x}} = \sin^2 {x} \quad$
$\frac{\sin^2 {x}}{\cos^2 {x} }=\frac{\sin^2 {x}}{\cos^2 {x} }\quad$
How's that? I tried the $\sec {x }= \frac1{\cos {x}}\quad$ method. However I was unsuccessful.
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hint: $(\sec x -1)(\sec x + 1) = \sec^2 x - 1 = \tan^2 x = \dfrac{\sin^2 x}{\cos^2 x}$, and $(1-\cos x)(1+ \cos x) = 1 - \cos^2 x = \sin^2 x$. I hope you can pull it off...
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"url": "https://math.stackexchange.com/questions/1777739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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