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To find trace and determinant of matrix
Possible Duplicate:
Computing the trace and determinant of $A+B$, given eigenvalues of $A$ and an expression for $B$
Let $A$ be a $4\times 4$ matrix with real entries such that $-1,1,2,-2$ are its eigenvalues. If $B=A^{4}-5A^{2}+5I$, where $I$ denotes the $4\times 4$ identity matrix, then which of the following statements are correct?
*
*$\det (A+B)=0$
*$\det B=1$
*$\text{trace}(A-B)=0$.
*$\text{trace}(A+B)=4$.
NOTE: There may one or more options correct.
I know that trace of matrix means sum of eigenvalues of matrix and determinant means product of eigenvalues. But i dont know how to apply these things in this question?
Please help me out and explain the method.
|
The characteristic polynomial of $A$ is
$$(t+1)(t-1)(t+2)(t-2) = (t^2-1)(t^2-4) = t^4 - 5t^2 + 4.$$
Therefore, by the Cayley-Hamilton Theorem,
$$A^4 - 5A^2 + 4I = 0.$$
In particular, $B= A^4 - 5A^2 + 5I = (A^4-5A^2+4I)+I = I$.
So $B=I$, $A+B=A+I$, and $A-B=A-I$.
The eigenvalues of $A+\mu I$ are of the form $\lambda+\mu$, where $\lambda$ is an eigenvalue of $A$.
So: Since $-1$ is an eigenvalue of $A$, then $0=-1+1$ is an eigenvalue of $A+I=A+B$, so $\det(A+B)=0$.
Since $B=I$, $\det(B)=1$.
The trace of $A-B$ is the sum of the eigenvalues of $A-B$; the sum of the eigenvalues of $A-B$ is $(-1-1) + (1-1) + (2-1) + (-2-1) = -2+0+1-3 = -4$. Alternatively, it is the sum of the trace of $A$ (which is $0$, since its eigenvalues add up to $0$) and the trace of $-B$, which is $-I$, hence the trace is $-4$.
And the trace of $A+B$ is the sum of the eigenvalues of $A+B$, which is $(-1+1) + (1+1) + (2+1) + (-2+1) = 4$. Or it is $\mathrm{trace}(A)+\mathrm{trace}(B) = 0 + \mathrm{trace}(I_4) = 4$.
|
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|
Evaluating $\int\frac{x^{1/2}}{1+x^2}\,dx.$
Compute $$\int\frac{x^{1/2}}{1+x^2}\,dx.$$
All I can think of is some integration by substitution. But ran into something scary. Anyone have any tricks?
|
$$I=\int\frac{x^{\frac{1}{2}}}{1+x^2}$$
$x=t^2$, we get,
$$=\int\frac{2t^2dt}{1+t^4}$$
$$=\int\frac{t^2+1}{1+t^4}dt + \int\frac{t^2-1}{1+t^4}dt$$
upon dividing by $t^2$, we get
$$=\int\frac{1+\frac{1}{t^2}}{(t-\frac{1}{t})^2+2}dt +\int\frac{1-\frac{1}{t^2}}{(t+\frac{1}{t})^2-2}dt$$
All set now, lets integrate
$$\frac{1}{\sqrt2}\tan^{-1}\left(\frac{t-\frac{1}{t}}{\sqrt2}\right)+\frac{1}{2\sqrt2}\ln\left(\frac{t+\frac{1}{t}-\sqrt2}{t+\frac{1}{t}+\sqrt2}\right) +C$$
just replace $t$ with $x^2$ to get the final answer
This yeilds exactly the same answer as Marvis got, (except for the $\tan^{-1}$ part, which I cannot understand why?)
Exactly the same thing (the constant is also there)
Results used:
1.$\int\frac{dx}{x^2+a^2}=\frac{1}{a}\tan^{-1}(\frac{x}{a})$
2.$\int \frac{dx}{x^2-z^2}=\frac{1}{2a}\ln(\frac{x-a}{x+a})$
:)
|
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|
Calculate square root of $i \Leftrightarrow z^2=i$ Let $z = r(\cos\theta+i\sin\theta)$. In my notes there was this example to calculate the square roots of $i$. What was done was:
$z = r(\cos\theta+i\sin\theta)\\z^2 = r^2(\cos(2\theta)+i\sin(2\theta))\\=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})\\\Leftrightarrow r=1 \ \ \ \ ∧ \ \ \ \ \ 2\theta=\frac{\pi}{2}+2\pi k,\ k\in \mathbb{Z},\ \theta\in[0,2\pi)\\\Leftrightarrow r=1 \ \ \ \ ∧ \ \ \ \ \ \theta=\frac{\pi}{4}+k\pi , \ k=0,1$
Im not entirely understand what they did above, what does the 2 lines below actually tell us? $\Leftrightarrow r=1 \ \ \ \ ∧ \ \ \ \ \ 2\theta=\frac{\pi}{2}+2\pi k,\ k\in \mathbb{Z},\ \theta\in[0,2\pi)\\\Leftrightarrow r=1 \ \ \ \ ∧ \ \ \ \ \ \theta=\frac{\pi}{4}+k\pi , \ k=0,1$
I thought to calculate the square roots of iyou let $z=x+iy$ and work out $x,y$ from $(x+iy)^2=i$?
Also, how did they get that $r^2(\cos(2\theta)+i\sin(2\theta))=\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2})$?
|
$i=1(\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2}))$ and we want $z^2=i$, so $r^2(\cos(2\theta)+i\sin(2\theta))=1(\cos(\frac{\pi}{2})+i\sin(\frac{\pi}{2}))$. Then use that $r_1(\cos\theta_1+i\sin\theta_1)=r_2(\cos\theta_2+i\sin\theta_2)\implies r_1=r_2\wedge\theta_1=\theta_2+2\pi k,k\in\mathbb{Z}$.
This gives $r^2=1\wedge2\theta=\frac{\pi}{2}+2\pi k,k\in\mathbb{Z}$. Since $r$ is a nonnegative real number, $r=1$; dividing both sides of $2\theta=\frac{\pi}{2}+2\pi k$ by 2 gives $\theta=\frac{\pi}{4}+\pi k$; picking values of $k$ starting from $k=0$ and stopping when we've gone full around gives $k=0,1$ (also, we should expect exactly 2 numbers whose square is $i$).
|
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|
Prove $\gcd(a+b,a^2+b^2)$ is $1$ or $2$ if $\gcd(a,b) = 1$
Assuming that $\gcd(a,b) = 1$, prove that $\gcd(a+b,a^2+b^2) = 1$ or $2$.
I tried this problem and ended up with
$$d\mid 2a^2,\quad d\mid 2b^2$$
where $d = \gcd(a+b,a^2+b^2)$, but then I am stuck; by these two conclusions how can I conclude $d=1$ or $2$?
And also is there any other way of proving this result?
|
If $d$ divides $a+b$ then it divides $a(a+b)=a^2+ab$ so if it also divides $a^2+b^2$ then it divides $a^2+ab-(a^2+b^2)=ab-b^2=b(a-b)$. A similar calculation shows it divides $a(a-b)$. Then from $\gcd(a,b)=1$ we get $d$ divides $a-b$. But then it divides $(a+b)+(a-b)=2a$ and $(a+b)-(a-b)=2b$, so it's 1 or 2.
|
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|
$3x^3 = 24$ quadratic equation Completing the square I know by factoring
$$x^3 - 8 = 0\\
x-2 = 0$$
that one of the solutions is 2.
but the other solutions is $1 ± i \sqrt 3$.
Can someone explain to me how to get that?
|
Hints
$$3x^{3}=24\Leftrightarrow x^{3}-8=0$$
$$\frac{x^{3}-8}{x-2}=x^{2}+2x+4$$
--
Added: The second equation means that
$$x^{3}-8=\left( x-2\right) \left( x^{2}+2x+4\right) .$$
Thus we have
$$x^{3}-8=0\Leftrightarrow \left( x-2\right) \left( x^{2}+2x+4\right) =0,$$
whose solutions are the values of $x$ that make the factor $x-2$ or the
factor $x^{2}+2x+4$ equal to $0$. The equation $x-2=0$ has the solution $x=2$. The
solutions of the equation $x^{2}+2x+4=0$ may be found by the quadratic
formula or as follows:
$$\begin{eqnarray*}
x^{2}+2x+4 &=&0\Leftrightarrow \left( x+1\right) ^{2}+3=0\Leftrightarrow
\left( x+1\right) ^{2}=-3 \\
&\Leftrightarrow &x+1=\pm \sqrt{-3}=\pm \sqrt{3}i \\
&\Leftrightarrow &x=-1\pm \sqrt{3}i.
\end{eqnarray*}$$
So $$x^{3}-8=0\Leftrightarrow x=2\text{ or }x=-1\pm \sqrt{3}i.$$
|
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|
Polar equation of a circle A very long time ago in algebra/trig class we did polar equation of a circle where
$r = 2a\cos\theta + 2b\sin\theta$
Now I forgot how to derive this. So I tried using the standard form of a circle.
$$(x-a)^2 + (y - b)^2 = a^2 + b^2$$
$$(a\cos\theta - a)^2 + (b\sin\theta - b)^2 = a^2 + b^2$$
$$(a^2\cos^2 \theta + a^2 - 2a^2\cos\theta) + (b^2\sin^2 \theta + b^2 - 2b^2\sin\theta) = a^2 + b^2$$
$$a^2\cos^2 \theta + b^2\sin^2 \theta - 2a^2 \cos\theta - 2b^2 \sin\theta = 0$$
Now I am stuck, I think I was supposed to complete the square or something. Could someone finish my thought?
|
A 'backwards' answer:
The general polar equation of a circle of radius $\rho$ centered at $(r_0,\theta_0)$ is
$$r^2-2 r r_0 \cos(\theta-\theta_0) + r_0^2 = \rho^2.$$
When $r_0 = \rho$, this reduces to (ignoring the $r=0$ solution)
$$r = 2(r_0 \cos \theta_0) \cos \theta + 2(r_0 \sin \theta_0) \sin \theta,$$
choosing $(r_0,\theta_0)$ such that $a = r_0 \cos \theta_0$, $b = r_0 \sin \theta_0$ gives the required form.
|
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|
How does he get a perfect swap numerator and denominator. I'm going through a exercise, in which all the answers are given, but the tutor makes a step and I can't follow at all. A massive jump with no explanation.
Here is the question:
$\lim_{x \to 2} \frac{\frac{1}{2}-\frac{1}{x}}{x-2}$
He then simplifies:
$ \frac{x-2}{2x(x-2)}$
He said he multiplied the entire equation by 2x
How does he know 2x swaps the denominator upto the numerator
Before he gave the simplification I spent perhaps 20 minutes trying to figure something out, and failed, and then he just jumps this massive step
He simplifies because we are finding limits.
Thanks
Joseph G.
|
Note that
\begin{align*}
\frac{\frac{1}{2}-\frac{1}{x}}{x-2} &= \frac{\frac{x-2}{2x}}{x-2} \\\ &= \frac{2x \cdot \frac{x-2}{2x}}{2x \cdot (x-2)} \qquad \Bigl[\small\text{Multiplying numerator and denominator by} \ 2x \Bigr] \\\ &= \frac{(x-2)}{2x \cdot (x-2)} \\\ &= \frac{1}{2x}
\end{align*}
|
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|
Positive definite quadratic form $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ Is $\sum_{i=1}^n x_i^2 + \sum_{1\leq i < j \leq n} x_{i}x_j$ positive definite?
Approach:
The matrix of this quadratic form can be derived to be the following
$$M := \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{2} & \cdots & \frac{1}{2} \\
\frac{1}{2} & 1 & \frac{1}{2} & \cdots & \frac{1}{2} \\
\cdots & \cdots & \cdots & \cdots & \cdots \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \cdots & \frac{1}{2} \\
\frac{1}{2} & \frac{1}{2} & \frac{1}{2} & \cdots & 1 \\
\end{pmatrix}$$
It suffices to show that $\operatorname{det}M > 0$, then the claim follows.
Any hints how to show the positivity of this determinant?
|
We can write : $M= \frac 12 (I + J)$ were $I$ the unit matrix and $J$ the matrix who all entries equal $1$.
$J$ is diagonalizable as symetric real matrix with propre value $0$ , $n-1$ times because the rank of $J$ is $1$ and $n$ one time because $JX=nX$ where $X$ is the vector who has all coordinate equal $1$.
$$J=P^{-1} \text{diag}(0,..,0,n)P= P^{-1} \Delta P$$
Then $$M=\frac 12 (P^{-1} I P + P^{-1} \Delta P = \frac 12 P^{-1} \text{diag}(1,...,1,n+1)P $$
|
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|
Closed form of $\sum_{k=-\infty}^{+\infty}\frac{1}{|x-kx_0|}$ I don't know how to find an explicit form for this sum, anyone can help me?
$$\sum_{k=-\infty}^{\infty}
{1 \over \left\vert\,x - k\,x_{\atop{ \small 0}}\,\right\vert}
$$
Here are the calculations I made, but don't bring me anywhere:
(original image)
$$\begin{align}\sum_{k=-\infty}^\infty\frac{1}{|x-kx_0|}&=\frac{1}{x}+\sum_{k=-\infty}^{-1}\frac{1}{|x-kx_0|}+\sum_{k=1}^\infty\frac{1}{|x-kx_0|}\\\\ & =\frac{1}{x}+\sum_{k'=1}^{\infty}\frac{1}{|x+k'x_0|}+\sum_{k=1}^\infty\frac{1}{|x-kx_0|}\end{align}$$
If $x\neq 0$,
$$=\frac{1}{x}+\sum_{x+k'x_0>0}\frac{1}{x+k'x_0}+\sum_{x+k'x_0<0}\frac{-1}{x+k'x_0}+\sum_{x-kx_0>0}\frac{1}{x-kx_0}+\sum_{x-kx_0<0}\frac{1}{kx_0-x}$$
$$=\frac{1}{x}+\sum_{k>-\frac{x}{x_0}}\frac{1}{x+kx_0}-\sum_{k<-\frac{x}{x_0}}\frac{1}{x+kx_0}+\sum_{k<\frac{x}{x_0}}\frac{1}{x-kx_0}+\sum_{k>\frac{x}{x_0}}\frac{1}{kx_0-x}$$
$$=\frac{1}{x}+\frac{1}{x_0}\left[\sum_{k>-\frac{x}{x_0}}\frac{1}{k+\frac{x}{x_0}}-\sum_{k<-\frac{x}{x_0}}\frac{1}{k+\frac{x}{x_0}}+\sum_{k<\frac{x}{x_0}}\frac{1}{-k+\frac{x}{x_0}}+\sum_{k>\frac{x}{x_0}}\frac{1}{k-\frac{x}{x_0}}\right]$$
and my prof's version (I'm not sure he could be so quick on the absolute value)
(original image)
$$\sum_{n=-\infty}^{+\infty}\frac{1}{|x-nx_0|}=\frac{1}{x}+\sum_{n=-\infty}^{-1}\frac{1}{|x-nx_0|}+\sum_{n=1}^\infty\frac{1}{|x-nx_0|}$$
$$=\frac{1}{x}+\sum_{m=+\infty}^{+1}\underbrace{\frac{1}{x-nx_0}}_{\substack{\text{change variable }m=-n,\\\\ \Large \frac{1}{x+mx_0}}}+\sum_{n=1}^\infty\frac{1}{nx_0-x}$$
$$=\sum_{n=1}^{+\infty}\underbrace{\frac{1}{nx_0-x}-\frac{1}{x+nx_0}}_{\Large\frac{x+nx_0-nx_0+x}{n^2x_0^2-x^2}}$$
$$\frac{1}{x}+2x\sum_{n=1}^{+\infty}\frac{1}{n^2x_0^2-x^2}$$
Thanks!
|
Maybe, you could try something with this
\begin{equation}
\sum_{k=-\infty}^{\infty}\frac{1}{|x-kx_0|} = \sum_{k=1}^{\infty}\frac{1}{|x+kx_0|} + \sum_{k=0}^{\infty}\frac{1}{|x-kx_0|} =
\end{equation}
\begin{equation}
= \frac{1}{|x|}+\sum_{k=1}^{\infty}\frac{1}{|x+kx_0|} + \sum_{k=1}^{\infty}\frac{1}{|x-kx_0|} = \frac{1}{|x|}+\sum_{k=1}^{\infty}\bigg(\frac{1}{|x+kx_0|} +\frac{1}{|x-kx_0|}\bigg)
\end{equation}
|
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|
Orthogonal Trajectory of $x^2 + 2y^2 = k^2$ $x^2 + 2y^2 = k^2$
I first take the derivative like the instructions say.
$2x + 4y \frac{dy}{dx} = 0$
I am not entirely sure why a dy and dx appears but it does in the instructions so I go with it.
Now I need to solve for $y'$
$ + 4y \frac{dy}{dx} = -2x$
$ \frac{dy}{dx} = \frac{-x}{2y}$
$ \frac{dy}{dx} = \frac{-x}{2y}$
So now I need to find the inverse negative
$ \frac{dy}{dx} = \frac{2y}{x}$
And that should be my slope at each line.
And now I need to solve that.
$ \frac{1}{2}ydy = xdx$
Take the integratal and I get
$\frac{1}{4}y^2 = \frac{1}{2}x^2 + C$
$y^2 = 2x^2 + C$
$y = \sqrt{2x^2 + C}$
This is wrong and I do not know why.
|
From the line David noted you; You would have $\frac{dy}{2y}=\frac{dx}{x}$. So by integrating from both sides, you get $\frac{1}{2}$Ln$|y|$=Ln$|x|+c$:
$$\int\frac{dy}{2y}=\int\frac{dx}{x}$$ $$\frac{1}{2}\int\frac{dy}{y}=\int\frac{dx}{x}$$ $$\frac{1}{2}Ln|y|=Ln|x|+c$$ wherein $c$ is a constant. You can simplify the result as $y=cx^2$.
|
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|
Integration and Limits I suspect the following integration to be wrong. My answer is coming out to be $3/5$, but the solution says $1$.
$$\int_0^1\frac{2(x+2)}{5}\,dx=\left.\frac{(x+2)^2}{5}\;\right|_0^1=1.$$
Please help out. Thanks.
|
The integration is obtained as follows:
$$\int 2\frac{x+2}{5}dx=\frac{2}{5}\int (x+2)d(x+2)=\frac{2}{5}\int udu=\frac{2}{5}\frac{u^2}{2}=\frac 1 5 (x+2)^2$$
Since $\frac 1 5 (x+2)^2$ is a primitive of $2\frac{x+2}{5}$ we can use FTCII, and get
$$\int 2\frac{x+2}{5}dx=\frac{(\color{red}{1}+2)^2}{5}-\frac{(\color{red}{0}+2)^2}{5}= \frac 9 5- \frac 4 5 = 1$$
It seems what you did was this:
$$\int 2\frac{x+2}{5}dx=\frac{(0+2)^2}{5}-\frac{(0+1)^2}{5}= \frac 4 5- \frac 1 5 = \frac 3 5$$
|
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|
What would be the value of $\sum\limits_{n=0}^\infty \frac{1}{an^2+bn+c}$ I would like to evaluate the sum
$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c}$$
Here is my attempt:
Letting
$$f(z)=\frac{1}{az^2+bz+c}$$
The poles of $f(z)$ are located at
$$z_0 = \frac{-b+\sqrt{b^2-4ac}}{2a}$$
and
$$z_1 = \frac{-b-\sqrt{b^2-4ac}}{2a}$$
Then
$$
b_0=\operatorname*{Res}_{z=z_0}\,\pi \cot (\pi z)f(z)=
\lim_{z \to z_0} \frac{(z-z_0)\pi\cot (\pi z)}{az^2+bz+c}=
\lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b}
$$
Using L'Hopital's rule. Continuing, we have the limit is
$$
\lim_{z \to z_0} \frac{\pi\cot (\pi z)+(z_0-z)\pi^2\csc^2 (\pi z)}{2az+b}=
\frac{\pi\cot (\pi z_0)}{2az_0+b}
$$
For $z_0 \ne 0$
Similarly, we find
$$b_1=\operatorname*{Res}_{z=z_1}\,\pi \cot (\pi z)f(z)=\frac{\pi\cot (\pi z_1)}{2az_1+b}$$
Then
$$\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c} = -(b_0+b_1)=\\
-\pi\left( \frac{\cot (\pi z_0)}{2az_0+b} + \frac{\cot (\pi z_1)}{2az_1+b}\right)=
-\pi\left( \frac{\cot (\pi z_0)}{\sqrt{b^2-4ac}} + \frac{\cot (\pi z_1)}{-\sqrt{b^2-4ac}}\right)=
\frac{-\pi(\cot (\pi z_0)-\cot (\pi z_1))}{\sqrt{b^2-4ac}}=
\frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{\sqrt{b^2-4ac}}
$$
Then we have
$$\sum_{n=0}^\infty \frac{1}{an^2+bn+c} = \frac{\pi(\cot (\pi z_1)-\cot (\pi z_0))}{2\sqrt{b^2-4ac}}$$
Is this correct? I feel like I made a mistake somewhere. Could someone correct me? Is there an easier way to evaluate this sum?
|
This is almost correct, but I believe the original sum needs to range from $-\infty$ to $\infty$ instead of $0$ to $\infty$. The solution that follows considers the sum $\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c}$, and throughout I will write $\sum_{n=-\infty}^\infty f(n)$ to mean $\lim_{N\rightarrow \infty}\sum_{n=-N}^N f(n)$.
Factoring the quadratic, with your definition of $z_{0},\ z_{1}$, we have $$\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c}=\frac{1}{a}\sum_{n=-\infty}^{\infty}\frac{1}{\left(n-z_{0}\right)\left(n-z_{1}\right)}.$$ Assume that neither $z_0$ nor $z_1$ are integers, since otherwise we would have a $\frac{1}{0}$ term appearing in the sum. By applying partial fractions, remembering that $z_{0}-z_{1}=\frac{\sqrt{b^{2}-4ac}}{a}$ we get $$\frac{1}{\sqrt{b^{2}-4ac}}\sum_{n=-\infty}^{\infty}\left(\frac{1}{n-z_{0}}-\frac{1}{n-z_{1}}\right).$$ By the cotangent identity $\pi\cot\left(\pi x\right)=\sum_{n=-\infty}^{\infty}\frac{1}{n+x},$ we conclude that $$\sum_{n=-\infty}^\infty \frac{1}{an^2+bn+c}=\frac{\pi\cot\left(\pi z_{1}\right)-\pi\cot\left(\pi z_{0}\right)}{\sqrt{b^{2}-4ac}}.$$
|
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|
Maximum point of a polar function I have a curve C with polar equation
$$r^2 = a^2\cos{2\theta} $$
And I am looking to find the length $x$ when $r=max$
Judging from the equation:
$$r = \sqrt{a^2\cos{2\theta}} $$
R will be maximum at $\cos{2\theta}=1$
So the maximum value of $r$ is:
$$r = \sqrt{a^2} =a$$
However the derivative disagrees as:
$$x^2=r^2\sin^2{\theta}=a^2\cos{2\theta}\,\sin^2{\theta} \\ \frac{d}{d\theta}\left (a^2\cos{2\theta}\,\sin^2{\theta} \right )=a^2(2\sin{\theta}\cos{\theta}-8\sin^3{\theta}\cos{\theta}) \\
\sin{\theta}=\frac{1}{2} \\
\theta= \frac{\pi}{6} \\
r= \frac{a}{\sqrt{2}}$$
What am I doing wrong?
|
Differentiating $r$ wrt to $\theta$ gives $\theta \mapsto |a|\frac{\sin 2 \theta}{\cos 2 \theta}$, which is zero when $\cos 2 \theta = \pm 1$. No disagreement there!
From this compute $x = r \cos \theta$. Since $\theta = 0$ maximizes $r$, the corresponding $x = |a|$.
|
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|
Prove that: $\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$ Here is another interesting integral inequality :
$$\int_{0}^{1} \frac{x^{4}\log x}{x^2-1}\le \frac{1}{8}$$
According to W|A the difference between RS and LS is extremely small, namely 0.00241056. I don't know what would work here since the difference is so small.
|
Representing natural logarithm as an integral and changing the order of integration we obtain:
$$\ldots = \int_0^1 \frac{x^4}{x^2 - 1} \, dx \int_1^x \frac{dt}{t} = \int_0^1 \frac{dt}{t} \int_0^t \frac{x^4}{x^2 - 1} \, dx \\= \int_0^1 \frac{t + \frac{1}{3} t^3 - \tanh^{-1} t}{t}\, dt = \frac{10}{9} - \int_0^1 \frac{\tanh^{-1} t}{t} \, dt$$
So we now want to show that:
$$\frac{71}{72} = \frac{10}{9} - \frac{1}{8} \le \int_0^1 \frac{\tanh^{-1} t}{t} \, dt$$
Using Maclauirn series we have:
$$\frac{\tanh^{-1}}{t} = 1 + \frac{t^2}{3} + \frac{t^4}{5} + \ldots$$
Integrating first three terms yields: $\frac{259}{225} > \frac{71}{72}$.
Actually it turns out that $\frac{\tanh^{-1} t}{t} \approx 1$ is enough.
Added
Alternatively we can write:
$$\int_0^1 \frac{\tanh^{-1} t}{t} dt = \int_0^1 \frac{dt}{t} \int_0^t \frac{dx}{1-x^2} \ge \int_0^1 \frac{dt}{t} \int_0^t dx = \int_0^1 dt = 1 \ge \frac{71}{72}$$
|
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|
Integral - using Euler Substitution I've been trying to solve one simple Integral with Euler substitution several times, but can't find where I'm going wrong. The integral is (+ the answer given here, too):
$$\int\frac{1}{x\sqrt{x^2+x+1}} dx=\log(x)-\log(2\sqrt{x^2+x+1}+x+2)+\text{ constant}$$
The problem is, I cannot get this result. Below is my solution of the problem. I've checked it many times, must be something very obvious that I'm missing:
(original image)
Euler Substituion
$\displaystyle\int\frac{dx}{x\sqrt{x^2+x+1}}$
Let $\sqrt{x^2+x+1}=t-x$.
$x^2+x+1=t^2-2xt+x^2$
$x(1+2t)=t^2-1\implies x=\dfrac{t^2-1}{1+2t}$
$dx=\left(\dfrac{t^2-1}{1+2t}\right)'dt=\dfrac{2t(1+2t)-(t^2-1)2}{(1+2t)^2}=\dfrac{2t+4t^2-2t^2+2}{(1+2t)^2}=\dfrac{2(t^2+t+1)}{(1+2t)^2}$
$\sqrt{x^2+x+1}=t-x=t-\dfrac{t^2-1}{1+2t}=\dfrac{t^2+t+1}{1+2t}$
$\implies\displaystyle\int\frac{dx}{x\sqrt{x^2+x+1}}=2\int\frac{\frac{t^2+t+1}{(1+2t)^2}\;dt}{\frac{t^2-1}{1+2t}\cdot\frac{t^2+t+1}{1+2t}}=2\int \frac{1}{t^2-1}\,dt$
$\dfrac{1}{t^2-1}=\dfrac{1}{(t+1)(t-1)}=\dfrac{A}{t+1}+\dfrac{B}{t-1}\implies \begin{eqnarray}&&At-A+Bt+B=1\\&&A+B=0\implies A=-B\\ &&B-A=1\implies B=\frac{1}{2},A=-\frac{1}{2}\end{eqnarray}$
$\implies \displaystyle 2\int \frac{1}{2}\frac{1}{2t-1}\,dt-2\int\frac{1}{2}\frac{1}{t+1}\,dt=\int\frac{1}{t-1}\,dt-\int\frac{1}{t+1}\,dt=$
$=\ln|t-1|-\ln|t+1|=\ln\left|\dfrac{t-1}{t+1}\right|$
$t-x=\sqrt{x^2+x+1}\implies t=\sqrt{x^2+x+1}+x$
$\implies \ln\left|\dfrac{t-1}{t+1}\right|=\ln\left|\dfrac{\sqrt{x^2+x+1}+x-1}{\sqrt{x^2+x+1}+x+1}\right|$
I'll appreciate any help.
Thanks in advance!
|
Looks about right. In your expression, multiply top and bottom of the thing inside the log by $\sqrt{x^2+x+1}-(x-1)$.
|
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|
Error Term in Passing from Summation to Integral I encountered the following in a paper and do not understand how the error term is being bounded. In what follows, $n$ and $k$ are large integer constants.
$$ \sum_{i=0}^{n-1} \ln\left(1 - \frac{i}{k}\right) = \int_0^n
\ln\left(1 - \frac{x}{k}\right) dx \pm e(n,k) $$ where the error term
$e(n,k)$ can be bounded by $$ e(n,k) \leq \int_0^n \ln\left(1 -
\frac{x}{k}\right) - \ln\left(1 - \frac{x+1}{k}\right) dx $$
|
$$0 \le e(n,k) = \int_0^n \left(\ln \left(1 - \dfrac{\lfloor x \rfloor}{k} \right) - \ln \left(1 - \dfrac{x}{k}\right)\right) dx \le \int_0^n \left(\ln \left(1 - \dfrac{ x-1}{k} \right) - \ln \left(1 - \dfrac{x}{k}\right)\right) dx $$
So the inequality is true if $$\ln \left(1 - \dfrac{ x-1}{k} \right) - \ln \left(1 - \dfrac{x}{k}\right) \le \ln \left(1 - \dfrac{x}{k}\right) - \ln \left(1 - \dfrac{x+1}{k} \right)$$
but rearranging and taking exponential of both sides, this becomes
$$\left(1 - \dfrac{ x-1}{k}\right)\left(1 - \dfrac{x+1}{k}\right) \le \left( 1 - \dfrac{x}{k}\right)^2$$
and the left side is $\left(1 - \dfrac{x}{k}\right)^2 - \left(\dfrac{1}{k}\right)^2$.
|
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|
Want to test the series for convergence/divergence $\sum_n \frac{\sqrt{n+1} - 1}{(n+2)^3 - 1} $ Want to test the series $$\frac{\sqrt{2} - 1}{3^3 - 1} + \frac{\sqrt{3} - 1}{4^3 - 1} + \frac{\sqrt{4} - 1}{5^3 - 1} + \cdots$$ for convergence/divergence.
My attempt is
$U_n =\frac{\sqrt{n+1} - 1}{(n+2)^3 - 1} $ Now I wish to find $V_n $ which is less than (or greater than $U_n $) such that its convergence/divergence is known so that I can find the behaviour of $U_n $
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Hint: $\sqrt{n+1}-1\sim\sqrt n$ and $(n+2)^3-1\sim n^3$ (when $n\to +\infty$). Since the series have non-negative terms, we just have to deal with the convergence of $\sum\limits_{n\geq 1}\frac{\sqrt n}{n^3}$.
|
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|
Closed form solutions of $\ddot x(t)-x(t)^n=0$ Given the ODE:
$$\ddot x(t)-x(t)=0$$
the solution is:
$$x(t)=C_1\exp(-t)+C_2\exp(t)$$
If we square the $x(t)$ we have:
$$\ddot x(t)-x(t)^2=0$$
and the solution is given by:
$$x(t)=6\wp(t+C_1;0,C_2)$$
and so for $x(t)^3$ which gives:
$$x(t)=\operatorname{sn}\left(\left(\frac{1}{2}i\sqrt t +C_1\right)C_2,i\right)$$
More generally, is it possible to find closed form solutions for the equation:
$$\ddot x(t)-x(t)^n=0$$
?
Thanks.
|
$$\begin{align*}
y''-y^n&=0\\
y'y''&=y'y^n\\
\int y'y'' dx &=\int y'y^n dx\\
\frac{(y')^2}{2} &=\frac{y^{n+1}}{n+1} +c_1\\
y' &=\sqrt {\frac{2y^{n+1}}{n+1} +2c_1 }\\
\int \frac{dy}{\sqrt {\frac{2y^{n+1}}{n+1} +2c_1 }} &=\int dx =x+a\\
\frac{1}{(2c_1)^{1/2}}\int \frac{dy}{ \sqrt{1+\frac{y^{n+1}}{c_1(n+1)}}} &=x+a\\
\int \frac{dy}{ \sqrt{1+\frac{y^{n+1}}{c_1(n+1)}}} &=(2c_1)^{1/2}x+a(2c_1)^{1/2}=(2c_1)^{1/2}x+c_2
\end{align*}$$
After here you can change variable and use the binomial expansion to evaluate integral.
$u=\frac{y^{n+1}}{c_1(n+1)}$
$${(1+u)}^{\alpha}= \sum_{n=0}^\infty \binom{\alpha}{n} u^n=1+ \alpha \frac{u}{1!}+\alpha (\alpha-1) \frac{u^2}{2!}+\dots$$
I avoided doing many calculations, and I preferred to use the quick way: ask Wolfram Alpha. Here is a link to the solution.
$k=\frac{1}{c_1(n+1)}$
$\int \frac{dy}{ \sqrt{1+ky^{n+1}}} =y. _2F_1 (\frac{1}{2},\frac{1}{n+1};\frac{n+2}{n+1}; -ky^{n+1}) = y. _2F_1 (\frac{1}{2},\frac{1}{n+1};\frac{n+2}{n+1}; -\frac{y^{n+1}}{c_1(n+1)})=(2c_1)^{1/2}x+c_2 $
Solution in closed form as Hypergeometric function
$$_2F_1 \left(\frac{1}{2},\frac{1}{n+1};\frac{n+2}{n+1}; -\frac{y^{n+1}}{c_1(n+1)}\right)=\frac{1}{y}\left(\sqrt{2c_1}x+c_2\right)$$
|
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|
Squares of the form $x^2+y^2+xy$ How can I find all $(a,b,c) \in \mathbb{Z}^3$ such that $a^2+b^2+ab$, $a^2+c^2+ac$ and $b^2+c^2+bc$ are squares ?
Thanks !
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Let $c^2+a^2+ca= (c+na)^2$ where $n$ is an integer $\implies a=\frac{(2n-1)c}{1-n^2}$.
Let $a^2+b^2+ab =(a+mb)^2$ where $m$ is an integer $\implies b=\frac{(2m-1)a}{1-m^2}=\frac{(2n-1)(2m-1)}{(1-n^2)(1-m^2)}c$.
If $c|(1-n^2)(1-m^2)$,
$c=r(1-n^2)(1-m^2)$ (say, where $r$ an integer),
then, $b = r(2n-1)(2m-1)$
and $a = r(2n-1)(1-m^2)$
The above will be true if only 1st two conditions were supplied.
For all the three conditions, let $c^2+a^2+ca= r^2$ where r is an integer =>$r^2+ca$ must be perfect square and vice versa.
ca=$r^2-R^2$ for some integer R.
So, the given problem is same as finding a,b,c such that the product of any two is the difference of two squares.
Now $r^2+ca$ will be perfect square if r=$\frac{c-a}{2}$ where c-a is odd i.e., c,a are of opposite parity.
Then, $c^2+a^2+ca=(\frac{c-a}{2})^2$=>c+a=0.
Similarly, b+c=a+b=0. This provides only trivial solution (0,0,0) $ \in \mathbb{Z}^3$.
As ca=$r^2-R^2$, bc = $s^2-S^2$ and ab = $t^2-T^2$ for some integer s,S,t,T.
So, $a^2=\frac{ca.ab}{bc} =\frac{(r^2-R^2).(s^2-S^2)}{t^2-T^2} $
Now if r=$A^2+B^2$ and R= $A^2-B^2$
and if s=$C^2+D^2$ and S= $C^2-D^2$
and if t=$E^2+F^2$ and T= $E^2-F^2$ for some integers A,B,C,D,E,F.
a=±$\frac{2.A.B.2.C.D}{2E.F}$=±$\frac{2A.B.C.D}{E.F}$
b and c with be of the form ±$\frac{2.C.D.E.F}{A.B}$ and ±$\frac{2.E.F.A.B}{C.D}$
So, we need to find integers p,q,r such that p|2qr, q|2rp and r|2pq (think p=A.B, q=C.D and r=E.F).
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|
Prove that there does not exist any positive integers pair $(m, n)$ satisfying: $n(n + 1)(n + 2)(n + 3) = m{(m + 1)^2}{(m + 2)^3}{(m + 3)^4}$ How to prove that, there does not exist any positive integers pair $(m,n)$ satisfying:
$n(n + 1)(n + 2)(n + 3) = m{(m + 1)^2}{(m + 2)^3}{(m + 3)^4}$.
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This is an edited version of a partial answer that I posted sometime ago and subsequently deleted (not sure if resurrecting an answer is the correct thing to do after it has been up-voted and then deleted, perhaps someone will advise). If anyone can suggest where any of this can be improved, or point out any mistakes, I would be grateful.
Consider the equation
$$n(n+1)(n+2)(n+3) = m(m+1)^2(m+2)^3(m+3)^4$$
To avoid some trivialities later on, it is easy to check that there are no solutions with $m=1$ or $m=2$.
Using the fact that $n(n+1)(n+2)(n+3)$ is almost a square, we have
$$(n^2+3n+1)^2-1 = m(m+2)\times[(m+1)(m+2)(m+3)^2]^2$$
Putting $N = n^2+3n+1$ and $M = (m+1)(m+2)(m+3)^2$, this becomes
$$N^2-1 = m(m+2)M^2$$
so that
$$N^2-1 = [(m+1)^2-1]M^2.$$
Our approach now is to write this as
$$N^2 - [(m+1)^2-1]M^2 = 1,$$
which is Pell's equation, with $d = (m+1)^2-1$. In this case there is a particularly nice solution for the Pell equation, as the continued
fraction is very simple in this instance. For convenience we change notation slightly and
use $k = m+1$, so that we are looking at solutions of
$$x^2 - (k^2-1)y^2 = 1,$$
and bear in mind that for any solution $(x,y)$ we also
require $$y = (m+1)(m+2)(m+3)^2 = k(k+1)(k+2)^2.$$
So we investigate the properties of solutions of the Pell equation above by looking at the
standard continued fraction method.
We have $$\sqrt{k^2-1} = (k-1)+\cfrac{1}{1+\cfrac{1}{(2k-2)+\cfrac{1}{1+\cfrac{1}{(2k-2) + \ddots}}}}$$
which gives the first few solutions $(x_n,y_n)$ as $(1,0), (k,1), (2k^2-1,2k), \dots$.
Looking at the solutions for $y$, we see that they are generated by the recurrence relation
$$y_{n+2} = 2ky_{n+1} - y_n, \mbox{ with } y_0 = 0, y_1 = 1.$$
Recalling that we also need $y = k(k+1)(k+2)^2$, it is enough to prove that this last expression cannot be one
of the $y_n$ from the recurrence relation as follows.
Clearly, the values $y_n$ are strictly increasing, and we claim that
$$y_4 < k(k+1)(k+2)^2 < y_5$$
A bit of algebra gives $$y_4 = 8k^3-4k$$ so that
\begin{equation*}k(k+1)(k+2)^2-y_4 = k^4-3k^3+8k^2+8k = k^3(k-3)+8k^2+8k\end{equation*}
which is clearly positive for $k\geq 3$ and is easily checked to be positive for $k=1,2$.
For the right-hand inequality above, we have
$$y_5 = 16k^4-12k^2+1$$
and then
$$y_5 - k(k+1)(k+2)^2 = 16k^4-12k^2+1 - (k^4+5k^3+8k^2+4k)$$
$$ = 15k^4-5k^3-20k^2-4k+1$$
and by examining the graph (because I can't see an elegant way to do this bit) we see that this is
positive for $k\geq 2$, and we know that $k=1$ (corresponding to $m=2$) is not a solution of
the original equation.
This shows that $k(k+1)(k+2)^2$ cannot be one of the $y_n$, so that no solution of the original equation is possible.
I am sure that there ought to be a simpler solution, but I have been unable to find one.
|
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Prove $\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}} = \sin{A} + \cos{A}$ and $\cot{A} + \frac{\sin{A}}{1 + \cos{A}} = \csc{A}$ Can anyone help me solve the following trig equations.
$$\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}} = \sin{A} + \cos{A}$$
My work thus far
$$\frac{\frac{1}{\cos{A}}+\frac{1}{\sin{A}}}{\frac{\sin{A}}{\cos{A}}+\frac{\cos{A}}{\sin{A}}}$$
$$\frac{\frac{\sin{A} + \cos{A}}{\sin{A} * \cos{A}}}{\frac{\sin{A}}{\cos{A}}+\frac{\cos{A}}{\sin{A}}}$$
But how would I continue?
My second question is
$$\cot{A} + \frac{\sin{A}}{1 + \cos{A}} = \csc{A}$$
My work is
$$\frac{\cos{A}}{\sin{A}} + \frac{\sin{A}}{1 + \cos{A}} = \csc{A}$$
I think I know how to solve this one by using a common denominator but I am not sure.
|
Try waiting until the last minute before converting to sines and cosines.
$$\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}}$$
Recall that $\tan A\cot A = 1$ and $\tan A = \frac{\sec A}{\csc A}$
$$=\frac{\sec{A}+\csc{A}}{\tan{A} + \cot{A}}\cdot\frac{\tan A}{\tan A} $$
$$=\frac{\sec{A}\tan A+\csc{A}\tan A}{\tan^2{A} + 1} $$
$$=\frac{\sec{A}\tan A+\csc{A}\cdot\frac{\sec A}{\csc A}}{\sec^2 A} $$
$$=\frac{\sec{A}\tan A+\sec A}{\sec^2 A} $$
$$=\frac{\sec A (\tan A+1)}{\sec^2 A} $$
$$=\frac{\tan A+1}{\sec A} $$
$$=\frac{\frac{\sec A}{\csc A}}{\sec A} + \frac{1}{\sec A} $$
$$=\frac{1}{\csc A} + \frac{1}{\sec A} $$
$$=\sin A + \cos A $$
But I do like hjpotter92's answer better :)
|
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|
Minimum value of given expression What is the minimum value of the $$ \frac {x^2 + x + 1 } {x^2 - x + 1 } \ ?$$
I have solved by equating it to m and then discriminant greater than or equal to zero and got the answer, but can algebraic manipulation is possible
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let y=$\frac{x^2+x+1}{x^2-x+1}$
=>$x^2(y-1)-x(y+1)+(y-1)=0$
As x is real, the discriminant= $(y+1)^2-4(y-1)^2≥0$
=>$(y-3)(y-\frac{1}{3})≤0$
=>$\frac{1}{3}≤y≤3$
|
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Proving Quadratic Formula purplemath.com explains the quadratic formula. I don't understand the third row in the "Derive the Quadratic Formula by solving $ax^2 + bx + c = 0$." section. How does $\dfrac{b}{2a}$ become $\dfrac{b^2}{4a^2}$?
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$$ax^2+bx+c=0 - \text{divide by $a$ because $a\neq 0$ }$$ we get
$$x^2+\frac{b}{a}x+\frac{c}{a}=0$$
$$x^2+2x\frac{b}{2a}+\frac{c}{a}=0$$
$$x^2+2x\frac{b}{2a}+\frac{b^2}{4a^2}-\frac{b^2}{4a^2}+\frac{c}{a}=0$$
$$x^2+2x\frac{b}{2a}+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac{c}{a}$$
$$x^2+2x\frac{b}{2a}+\frac{b^2}{4a^2}=\frac{b^2-4ac}{4a^2}$$
$$x^2+2x\frac{b}{2a}+\left(\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$$
if in LHS we use $x=A$ and $\frac{b}{2a}=B$ then we have
$$A^2+2AB+B^2=(A+B)^2$$ or
$$\left(x+\frac{b}{2a}\right)^2=\frac{b^2-4ac}{4a^2}$$
we have two values of square roote
$$x_1+\frac{b}{2a}=+\sqrt{\frac{b^2-4ac}{4a^2}}$$and
$$x_2+\frac{b}{2a}=-\sqrt{\frac{b^2-4ac}{4a^2}}$$ or
$$x_{1,2}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
|
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|
Proof by induction of Bernoulli's inequality $ (1+x)^n \ge 1+nx$ I am working on getting the hang of proofs by induction, and I was hoping the community could give me feedback on how to format a proof of this nature:
Let $x > -1$ and $n$ be a positive integer. Prove Bernoulli's inequality:
$$ (1+x)^n \ge 1+nx$$
Proof:
Base Case: For $n=1$, $1+x = 1+x$ so the inequality holds.
Induction Assumption: Assume that for some integer $k\ge1$, $(1+x)^k \ge 1+kx$.
Inductive Step: We must show that $(1+x)^{k+1} \ge 1+(k+1)x$
Proof of Inductive Step:
$$\begin{align*}
(1+x)^k &\ge 1+kx \\
(1+x)(1+x)^k &\ge (1+x)(1+kx)\\
(1+x)^{k+1} &\ge 1 + (k+1)x + kx^2 \\
1 + (k+1)x + kx^2 &> 1+(k+1)x \quad (kx^2 >0) \\
\Rightarrow (1+x)^{k+1} &\ge 1 + (k+1)x \qquad \qquad \qquad \square
\end{align*}$$
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This looks fine to me. Just a small note on formatting of the inequalities: I would combine the third and fourth inequalities as
$$
(1+x)^{k+1} \geq 1+(k+1)x+kx^2>1+(k+1)x,
$$
so there is no need of the fifth line. Or even
$$
(1+x)^{k+1} = (1+x)(1+x)^{k} \geq (1+x)(1+kx)=1+(k+1)x+kx^2>1+(k+1)x.
$$
|
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|
How to solve $1/2 \sin(2x) + \sin(x) + 2 \cos(x) + 2 = 0$? How to solve trigonomtry function involving $\sin x \cos x$ and $\sin 2x$:
$$\frac{1}{2} \sin(2x) + \sin(x) + 2 \cos(x) + 2 = 0. $$
|
Hint:
Using the identity $\sin(2x) = 2 \sin x \cos x$ we have
$$ \sin x \cos x + \sin x + 2\cos x + 2 = 0$$
Factor
$$ (1 + \cos x) \sin x + 2(1 + \cos x) = 0 \\
(1 + \cos x)(2 + \sin x) = 0
$$
So either $1 + \cos x = 0$ or $2 + \sin x = 0.$
Solve for $x$ in each case.
|
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|
I want to find out the angle for the expression $a^3 + b^3 = c^3$. like in pythagorean theorem angle comes 90 degree for the expression $a^2 + b^2 = c^2$, however I know that no integer solution is possible.
|
There is no single angle corresponding to the relationship $a^3+b^3=c^3$.
Suppose that a triangle has sides of lengths $a,b$, and $c$ such that $a^3+b^3=c^3$. We know from the law of cosines that if $\theta$ is the angle opposite the side of length $c$, then $c^2=a^2+b^2-2ab\cos\theta$, so
$$\cos\theta=\frac{a^2+b^2-c^2}{2ab}\;.$$
Now let’s look at just a few examples. If $a=b=1$, then $c=\sqrt[3]2$, and $$\cos\theta=\frac{2-2^{2/3}}2\approx0.20630\;.$$
If $a=1$ and $b=2$, then $c=\sqrt[3]9$, and $$\cos\theta=\frac{5-9^{2/3}}4\approx0.16831\;.$$
If $a=1$ and $b=3$, then $c=\sqrt[3]{28}$, and $$\cos\theta=\frac{10-28^{2/3}}6\approx0.12985\;.$$
As you can see, these values of $\cos\theta$ are all different, so the angles themselves are also different.
|
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|
How to approximate the value of $ \arctan(x)$ for $x> 1$ using Maclaurin's series? The expansion of $f(x) = \arctan(x)$ at $x=0$ seems to have interval of convergence $[-1, 1]$
$$\arctan(x) = x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\frac{x^9}{9}-\frac{x^{11}}{11}+\mathcal{O}\left(x^{13}\right) $$
Does it mean that I cannot approximate $\arctan(2)$ using this series? Also I'm getting radius of convergence $|x| < 1$ using ratio test. How do I get $|x| \leqslant 1$?
|
If you want to approximate the value of $\arctan(2)$, then you can expand $\arctan(x)$ at a point close to $2$. For example you can have an expansion at the point $1$, but if you approximate at a point that is closer to 2 is better. Here is the Taylor series at the point $x=1$
$$ \arctan(x) = \frac{\pi}{4} +{\frac {1}{2}} \left( x-1 \right) -{\frac {1}{4}} \left( x-
1 \right) ^{2}+{\frac {1}{12}} \left( x-1 \right) ^{3}+O \left(
\left( x-1 \right) ^{4} \right)\,.$$
Another expansion at the point $ x=\frac{3}{2} $ is given by,
$$\arctan(x) = \arctan \left( {\frac {3}{2}} \right) + {\frac {4}{13}} \left( x-{
\frac {3}{2}} \right) -{\frac {24}{169}} \left( x-{\frac {3}{2}}
\right) ^{2}+{\frac {368}{6591}} \left( x-{\frac {3}{2}} \right) ^{3}
+O \left( \left( x-{\frac {3}{2}} \right) ^{4} \right)
$$
It is easier, for a hand calculations, to derive the first series. Substituting $x=1$, of course you can use the Abel's theorem to assure the convergence, you get,
$$\arctan(x) = \arctan(2) \approx \frac{\pi}{4}+\frac{1}{3} $$
As I said, if you use the second series with the same number of terms, you will have a smaller error.
|
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|
Proving $n^4 + 4 n^2 + 11$ is $16k\,$ for odd $n$
if $n$ is an odd integer, prove that $n^4 + 4 n^2 + 11$ is of the form
$16 k$.
And I went something like:
$$\begin{align*}
n^4 +4 n^2 +11
&= n^4 + 4 n^2 + 16 -5 \\
&= ( n^4 +4 n^2 -5) + 16 \\
&= ( n^2 +5 ) ( n^2-1) +16
\end{align*}$$
So, now we have to prove that the product of $( n^2 +5 )$ and $( n^2-1)$ is a multiple of 16.
But, how can we do this?
If anybody has any idea of how I can improve my solution, please share it here.
Edit updated to include the necessary hypothesis that $n$ is odd.
|
*
*$n=2k+1$:
$$n^4+4n^2+11\\=(n^2-1)(n^2+5)+16$$
Now, below is square of an odd integer hence it can be reperesented as (8a+1).
$$n^2$$
Therefore $$8a(8p+6) + 16\\= 64ap + 48a+16\\=16(4ap + 3a + 1)$$
This proves that when n is an odd integer it will be divisible by 16
|
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|
I know that, $S_{2n}+4S_{n}=n(2n+1)^2$. Is there a way to find $S_{2n}$ or $S_{n}$ by some mathematical process with just this one expression? $S_{2n}+4S_{n}=n(2n+1)^2$, where $S_{2n}$ is the Sum of the squares of the first $2n$ natural numbers, $S_{n}$ is the Sum of the squares of the first $n$ natural numbers.
when, $n=2$
$S_{2n}=S_{4}=1^2+2^2+3^2+4^2=30$
$S_{n}=S_{2}=1^2+2^2=5$
$S_{4}+4S_{2}=2(2*2+1)^2=50$
|
You could use induction.
Assuming that $$S_{2n}+4S_n=n(2n+1)^2$$ then add terms to both sides so that the left side increments its index:
$$
\begin{align}
&S_{2n}+4S_n+(2n+1)^2+(2n+2)^2+4(n+1)^2\\
&=n(2n+1)^2+(2n+1)^2+(2n+2)^2+4(n+1)^2\\
S_{2(n+1)}+4S_{n+1}&=n(2n+1)^2+(2n+1)^2+(2n+2)^2+4(n+1)^2\\
&=(n+1)(2n+1)^2+4(n+1)^2+4(n+1)^2\\
&=(n+1)(2n+1)^2+8(n+1)^2\\
&=(n+1)[(2n+1)^2+8(n+1)]\\
&=(n+1)[4n^2+4n+1+8n+8]\\
&=(n+1)[4n^2+12n+9]\\
&=(n+1)(2n+3)^2\\
&=(n+1)(2(n+1)+1)^2\\
\end{align}$$
The base case is established in your question.
|
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|
Seifert matrices and Arf invariant -- Cinquefoil knot I have computed the following Seifert matrix for the Cinquefoil knot:
$$ S = \begin{pmatrix} 1 & -1 & -1 & -1 \\ 0 & 1 & 0 & 0 \\
0 & 1 & 1 & 0 \\ 0 & 1& 1 & 1\end{pmatrix}$$
I also found a Seifert matrix for this knot on the internet but I still don't know how to verify the correctness of a Seifert matrix. Hence:
Question 1: How can I verify that a Seifert matrix that I computed is correct?
Now, more importantly:
I thought I could compute the Arf invariant as follows (where $K$ is my knot):
$$ A(K) = A(q) = q(x_1) q(x_2) + q(x_3) q(x_4) = 1 \cdot 1 + 1 \cdot 1 = 0$$
where $x_1 = \begin{pmatrix}1 \\ 0 \\ 0 \\ 0 \end{pmatrix}, x_2 = \begin{pmatrix} 0 \\ 1 \\ 0 \\ 0 \end{pmatrix}, x_3 = \begin{pmatrix} 0 \\ 0 \\ 1 \\ 0 \end{pmatrix} , x_4 = \begin{pmatrix} 0 \\ 0 \\ 0 \\ 1 \end{pmatrix}$ and $q(x) = x^T S x$.
But this must be wrong for the following reasons: 1. Assuming the Seifert matrix I found on the internet is correct, I get the same Arf invariant. 2. The Arf invariant of the cinquefoil knot is $1$.
3. "The Arf invariant is $0$ if a majority of classes have self-linking $0$, and is $1$ if most classes have self-linking $1$.". (Excerpt taken from here). The self-linking numbers are the diagonal entries of the Seifert matrix as far as I understand.)
Question 2: What is the correct way to compute the Arf invariant of a knot from a Seifert form / Seifert matrix?
|
What makes you think that $x_i$ form a symplectic basis for $S$? In fact they don't. Also, you don't want a symplectic basis for $S$ but rather for $I = S^T - S$.
Your Seifert matrix must be wrong since if I compute the Arf invariant using the $S$ with a symplectic basis
$$ e_1 = \begin{pmatrix} 1 \\ 0 \\ 0 \\1 \end{pmatrix}
e_2 = \begin{pmatrix} -1 \\ 1 \\ 0 \\1 \end{pmatrix}
f_1 = \begin{pmatrix} 0 \\ 0 \\ -1 \\1 \end{pmatrix}
f_2 = \begin{pmatrix} 0 \\ -1 \\ 1 \\0 \end{pmatrix}$$ I get zero when I should be getting $1$.
Using the Seifert matrix you linked,
$$ S =
\begin{pmatrix} -1 & -1 & 0 & -1 \\ 0 & -1 & 0 & 0 \\ -1 & -1 & -1 & -1 \\ 0 & -1 & 0 & -1 \end{pmatrix}$$
and a corresponding symplectic basis for
$$ I = S^T - S = \begin{pmatrix} 0 & 1 & -1 & -1 \\
-1 & 0 & -1 & -1 \\ 1 & 1 & 0 & 1 \\ -1 & 1 & -1 & 0\end{pmatrix}$$
$$ e_1 = \begin{pmatrix} -1 \\ 0 \\ 0 \\1 \end{pmatrix}
e_2 = \begin{pmatrix} 0 \\ 1 \\ 1 \\ 0 \end{pmatrix}
f_1 = \begin{pmatrix} -1 \\ 0 \\ 0 \\1 \end{pmatrix}
f_2 = \begin{pmatrix} -1 \\ 0 \\ 1 \\0 \end{pmatrix}$$
yields
$$ A(K) = e_1^T S e_1 e_2^T S e_2 + f_1^T S f_1 f_2^T S f_2 = 7 \equiv_2 1$$
as expected.
|
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|
Closed form for a sum of values of a quadratic? Today in class we were analyzing the number of half-spaces created by $n$ number of planes. For two planes there are 4 spaces, 3 there are 8, 4 there are 15, etc. our teacher challenged us to find the formula for $n$ planes. Me and my friend came up with
$$
1+\sum^n_{x=1} \left(\frac{x(x+1)}{2}+1\right)
$$
Because based on that when a line cuts a plane in half, the formula for the number of half-planes it creates is
$$
\frac{n(n+1)}{2} + 1
$$
and the difference in the number of half-spaces between each $n$ plane is equal to adding on the same number of half-planes.
+--#of planes--+--1--+--2--+--3--+--4--+--5--+--50--+
|separates into| 2 | 4 | 8 | 15 | 26 |20,876|
| _ half-spaces| | | | | | |
+--difference--+-----2-----4-----7----16----etc-----+
My teacher said that this was correct, but it would be better if it was a formula/function, where you plug in the variables rather than have to evaluate the summation. I know that the final answer is
$$
\frac{n^3+5n+6}{6}
$$
but I need to show my work, and am unsure of how to get from a sum to that formula. Am I approaching this incorrectly? How was the original formula derived?
|
The reasoning is basically right, with slight glitches in the details. The desired number is
$$2+\sum_{x=1}^{n-1} \left(\frac{x(x+1)}{2}+1\right).\tag{$1$}$$
Or else, if you want to sum from $1$ to $n$, the desired number is
$$1+\sum_{x=1}^{n} \left(\frac{x(x-1)}{2}+1\right).\tag{$2$}$$
Note that $(x+1)^3-x^3=3x^2+3x+1$. So
$$\frac{x^2+x}{2}=\frac{1}{6}\left((x+1)^3-x^3-1\right).$$
Adding $1$, we get
$$\frac{x^2+x}{2}+1=\frac{1}{6}\left((x+1)^3-x^3+5\right).$$
We will add up $(x+1)^3-x^3+4$ from $x=1$ to $x=n-1$, and divide by $6$ at the end. We have $n-1$ $5$'s, which add up to $5n-5$.
Now add up the $(x+1)^3-x^3$ from $x=1$ to $x=n-1$.
The sum is
$$(2^3-1^3)+(3^3-2^3)+(4^3-3^3)+\cdots +(n^3-(n-1)^3).$$
Note the beautiful cancellations! Almost everything disappears, and we end up with $n^3-1^3$. Add the $5n-5$. We get $n^3+5n-6$.
So our answer is $2+\dfrac{n^3+5n-6}{6}$, which simplifies to $\dfrac{n^3+5n+6}{6}$.
Remarks: $1$. If you want to work with expression $(2)$ instead of $(1)$, you may want to use the identity $x^3-(x-1)^3=3x^2-3x+1$, though the one we used works fine.
$2$. The kind of collapsing that we saw comes up surprisingly often. You may want to look into telescoping sums.
$3$. There are many many other ways to find a closed form for the sm. Here is another idea. It is known that for any quadratic $q(k)$, $\sum_{k=1}^{n-1}$ is a cubic. (And for any cubic $c(k)$, $\sum_{1}^{n-1}c(k)$ is a quartic, and so on.). So our answer must have shape $p(n)=an^3+bn^2+cn+d$. If we know the values of our function at $4$ different $n$, we get $4$ linear equations in the coefficients $a$, $b$, $c$, $d$. Solve.
|
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|
Counting Hexagons in Triangle Grid Recurrence? (This is from a long finished programming competition)
Consider a triangle grid with side N. How many hexagons can fit into it?
This diagram shows N = 4:
I need a recurrence for it:
I tried the following:
$T_1 = 0$
$T_2 = 0$
$T_3 = 1$
$T_4 = 7$
$T_n = ???$
Using inclusion/exclusion:
$T_n = 3T_{n-1} - 3T_{n-2} + T_{n-3} + 3(n-2) - 2 $
Each part is as follows:
(We consider the count of hexagons that fit in certain sub-triangles of the current triangle)
*
*$3T_{n-1}$ The three n-1 triangles in each corner
*$-3T_{n-2}$ The three n-2 triangles touching one side each. These are each double counted in (1), so we subtract one of each.
*$T_{n-3}$ The n-3 triangle in the center, this is counted 3-3=0 times from (1) and (2) so we need to add one of them.
*$3(n-2)$ The number of hexagons which take up an entire side (and don't fit in any of the above) is n-2. We count this three times for each side.
*$-2$ The maximum hexagon is counted three times in (4), so we subtract 2 copies.
However it seems I have made a mistake because for $T_7$ I get 140, whereas the expected answer is 232.
Can anyone see where I have gone wrong?
|
Starting from your idea: By inclusion-exclusion, for $n>3$
$$\tag{1}T_n = X_n +3T_{n-1}-3T_{n-2}+T_{n-3},$$
where $X_n$ is the number of "full-size" hexagons, i.e. those that touch all three sides. A full-size hexagon is determined by three positive integers $a,b,c$ (the sizes of the small triangles chopped off) with $a+b<n, b+c<n, a+c<n$.
How many such triples ($a,b,c)$ with $\max\{a+b,a+c,b+c\}<n$ are there for each $n$?
We shall assume $n\ge3$.
There are $n-1\choose 2$ possibilities to choose $x,y\ge 1$ such that $x+y<n$ (think of inserting two separators into the $n-1$ gaps between $n$ pebbles, thus gouping them into $x\ge1$, $y\ge1$ and $n-x-y\ge1$ pebbles). Therefore, there are $n-1\choose2$ triples each of the forms $(1,b,c)$ and $(a,1,c)$and $(a,b,1)$. We have to subtract the $n-2$ triples each of the forms $(a,1,1)$, $(1,b,1)$, $(1,1,c)$. Then we have to add back the single triple $(1,1,1)$. What remains are triples $(a,b,c)$ with $a,b,c\ge2$. By mapping these to $(a-1,b-1,c-1)$ we biject them with the $X_{n-3}$ triples $(x,y,z)$ with $\max\{x+y,x+z.y+z\}<n-2$.
Therefore
$$\tag{2}X_n = X_{n-2}+3{n-1\choose 2}-3(n-2)+1=X_{n-2}+3{n-2\choose 2}+1 $$
This would be the correct count of what corresponds to your (4) and (5). Without going into details, we see that $X_n$ is cubic in $n$, not linear as in your count; so that's where your error comes from.
We can eliminate $X_n$ by combinig equation $(1)$ with $n$ and $n-2$:
$$T_n-T_{n-2}=X_n-X_{n-2}+3(T_{n-1}-T_{n-3})-3(T_{n-2}-T_{n-4})+T_{n-3}-T_{n-5},$$
$$\tag{3}T_n = 3T_{n-1}-2T_{n-2}-2T_{n-3}+3T_{n-4}-T_{n-5}+3{n-2\choose 2}+1.$$
Using this recursion and $T_{-2}=\ldots=T_2=0$ as starting values, one finds
$$ T_3=1\\
T_4=7\\
T_5=29\\
T_6=90\\
T_7=232\\
T_8=524\\
T_9=1072\\
\vdots
$$
The polynomial equation $x^5=3x^4-2x^3-2x^2+3x+1$ has $+1$ as quadruple root and $-1$ as a simple root. This suggests that there is an explicit formula $T_n=a_0+a_1n+a_2n^2+a_3n^3+b(-1)^n$. However, the cubic inhomogenuous part increases the degree so that we shall find an explicit formula
$$\tag{4}T_n=a_0+a_1n+a_2n^2+a_3n^3+a_4n^4+a_5n^5+a_6n^6+b(-1)^n.$$
The coefficients can be obtained by plugging $(4)$ into $(3)$ and the inital values. It turns out that
$$ T_n = \frac{n^6}{480}-\frac{n^4}{192}-\frac{n^2}{80}+\frac1{128}-\frac{(-1)^n}{128}=\frac{4n^6 - 10n^4 - 24n^2 + 15-15\cdot(-1)^n}{1920}.$$
|
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|
Inequality. $a^7b^2+b^7c^2+c^7a^2 \leq 3 $
Let $a,b,c$ be positive real numbers such that $a^6+b^6+c^6=3$. Prove that
$$a^7b^2+b^7c^2+c^7a^2 \leq 3 .$$
|
Using AM-GM inequality, we get
$$
3 = \frac {(a^6 + b^6 + c^6)^2} 3 = \sum_{cyc} a^6 \frac {a^6 + 2b^6} 3 \geq \sum_{cyc} a^6\sqrt[3]{a^6 b^{12}} = a^8b^4 + b^8c^4 + c^8a^4
$$
Now, by means of Cauchy–Schwarz inequality we complete the proof
$$
a^3 \cdot a^4 b^2 + b^3 \cdot b^4 c^2 + c^3 \cdot c^4 a^2 \leq \sqrt{a^6 + b^6 + c^6}\sqrt{a^8 b^4 + b^8 c^4 + c^8 a^4} \leq 3
$$
|
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|
Polynomial factors Why must $x^2 + x + 1$ be a factor of $x^5+x^4+x^3+x^2+x+1$?
I know that when we divide $x^5+x^4+x^3+x^2+x+1$ by $x^3+1$ we get $x^2 + x + 1$, but is there an argument/theorem or anything that could tell that $x^2+x+1$ must divide $x^2 + x + 1$?
|
It depends on what you count as telling. It follows from an easy, basic manipulation of familiar identities, but I don’t see any way to recognize this without appealing to something at least somewhat computational. I look at the sum $x^5+x^4+x^3+x^2+x+1$ and immediately think of $x^6-1$, and from there it all falls out:
$$\begin{align*}
x^5+x^4+x^3+x^2+x+1&=\frac{x^6-1}{x-1}\\
&=\frac{(x^3)^2-1}{x-1}\\
&=\frac{(x^3-1)(x^3+1)}{x-1}\\
&=(x^3+1)\frac{x^3-1}{x-1}\\
&=(x^3+1)(x^2+x+1)\;,
\end{align*}$$
all using standard factorizations.
|
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|
Trigonometric identity and roots of a polynomial. Prove that
$$(\operatorname{cosec} A–\sin A) (\sec A–\cos A) = \frac {1}{\tan A + \cot A} $$
Also help me with this question please
If $\alpha$ and $\beta$ are zeroes of the polynomial $x^2–2x–15$ then find a quadratic polynomial whose series [roots?] are $2\alpha$ and $2\beta$
|
$\csc A -\sin A=\frac{1-\sin^2A}{\sin A}=\frac{\cos^2A}{\sin A}$
$\sec A -\cos A=\frac{1-\cos^2A}{\cos A}=\frac{\sin^2A}{\cos A}$
Multiplying we get, $\sin A \cos A$
Now $\tan A+\cot A=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}$
$=\frac{\sin^2A+\cos^2A}{\cos A\sin A}=\frac{1}{\cos A\sin A}$
So, $\alpha+\beta=\frac{2}{1}=2$ and $\alpha\beta=-15$
So, $(y-2\alpha)(y-2\beta)=0\implies y^2-2(\alpha+\beta)y+4\alpha \beta=0$
$\implies y^2-4y-60=0$
Alternatively, we need to find the equation whose roots are double to that of $x^2-2x-15=0$. If $x$ is a root of the given equation, and $y$ be a root of the required equation, then $y=2x\implies x=\frac y 2 .$
Replacing $x$ with $\frac y 2$ in $x^2-2x-15=0$ we get,
$(\frac y 2 )^2-2(\frac y 2 )-15=0\implies y^2-4y-60=0 $
|
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|
Inequality under condition $a+b+c=0$ I don't know how to prove that the following inequality holds (under condition $a+b+c=0$):
$$\frac{(2a+1)^2}{2a^2+1}+\frac{(2b+1)^2}{2b^2+1}+\frac{(2c+1)^2}{2c^2+1}\geqq 3$$
|
The following proof, as it is, only works for $a,b,c \not\in (-2,0)$.
Let $|a| \geq |b| \geq |c|$
$$
\frac{(2x+1)^2}{2x^2+1} = 1 + \frac{2x^2+4x}{2x^2+1}
$$
Then it follows that
$$
\text{left-hand side}
= 3 + \frac{2a^2+4a}{2a^2+1} + \frac{2b^2+4b}{2b^2+1} + \frac{2c^2+4c}{2c^2+1}
\geq 3 + \frac{2(a^2+b^2+c^2)+4(a+b+c)}{2a^2+1}
=
$$
$$
= 3 + \frac{2(a^2+b^2+c^2)}{2a^2+1}
\geq 3
$$
Now in the second inequality I assumed for $x\in\{a,b,c\}$ that $0 \leq 2x^2+4x = 2(x^2+2x) = 2x(x+2)$. As the parabola defined by $2x(x+2)$ only is negative between -2 and 0 exclusive we can do so savely if none of $a,b,c$ are in the interval $(-2,0)$.
As $2a^2+1\geq 2b^2+1$ and $2a^2 +1 \geq 2c^2+1$ we get by multiplying with $\frac{2b^2+1}{2a^2+1} \leq 1$ and $\frac {2c^2+1}{2a^2+1} \leq 1$ that
$$\frac{2b^2+4b}{2b^2+1} \geq \frac{2b^2+4b}{2b^2+1}\cdot\frac{2b^2+1}{2a^2+1}=\frac{2b^2+4b}{2a^2+1}$$ etc.
Reducing to a common denominator and applying $a+b+c=0$ we get the inequality.
|
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|
Relations between the roots of a cubic polynomial How do I solve the last two of these problems?
The roots of the equation $x^3+4x-1=0$ are $\alpha$, $\beta$, and $\gamma$. Use the substitution $y=\dfrac{1}{1+x}$ to show that the equation $6y^3-7y^2+3y-1=0$ has roots $\dfrac{1}{\alpha+1}$, $\dfrac{1}{\beta+1}$, and $\dfrac{1}{\gamma+1} $. $\quad \quad \quad (2)$
For the cases $n=1$ and $n=2$, find the value of
$$\dfrac{1}{(\alpha+1)^n}+\dfrac{1}{(\beta+1)^n}+\dfrac{1}{(\gamma+1)^n}. \tag{2}$$
Deduce the value of $\dfrac{1}{(\alpha+1)^3}+\dfrac{1}{(\beta+1)^3}+\dfrac{1}{(\gamma+1)^3}. \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \, \, \,(2)$
Hence show that $\dfrac{(\beta+1)(\gamma+1)}{(\alpha+1)^2}+\dfrac{(\gamma+1)(\alpha+1)}{(\beta+1)^2}+\dfrac{(\alpha+1)(\beta+1)}{(\gamma+1)^2}=\dfrac{73}{36} \quad \quad \quad \quad \quad \quad \quad (3)$
|
For the first, as Avatar described, you use that $\frac{1}{\alpha+1}$ satisfies the polynomial relation $6y^3-7y^2+3y-1=0$ to express its third power in lower powers of itself.
For the second, you rewrite it simply as $\frac{(\alpha+1)(\beta+1)(\gamma+1)}{(\alpha+1)^3}+\frac{(\alpha+1)(\beta+1)(\gamma+1)}{(\beta+1)^3}+\frac{(\alpha+1)(\beta+1)(\gamma+1)}{(\gamma+1)^3}$ which reduces the problem to calculating $(\alpha+1)(\beta+1)(\gamma+1)$, which being a symmetric rational function in the roots should be easy to relate to the coefficients of $6y^3-7y^2+3y-1$.
|
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|
The sum of the different points on $x$-axis What is the sum of the different points on $x$-axis and that intersect with the two curves :
$$x^2=x+y+4 , y^2=y-15x+36$$
|
As the comments have said, we set $y=0$. For the curve $y^2=-15x+36$, we get $x=36/15$. For the curve $x^2=x+y+4$ we get $x^2-x-4=0$.
By the Rational Roots Theorem, the only conceivable rational roots of $x^2-x-4=0$ are integers that divide $4$. None of these is in fact a root, but that is irrelevant, the point is that $36/15$ cannot be a root of $x^2-x-4=0$.
Now we still need the sum of the (real) roots of $x^2-x-4=0$. It is easy to see that there are in fact two real roots. We could compute them, using the Quadratic Formula, and then add. But in general the sum of the roots of $x^2+ax+b=0$ is $-a$. So the sum of the roots of $x^2-x-4=0$ is $1$. Add that to $36/15$.
Remark: Consider the polynomial equation $a_0x^n+a_1x^{n-1}+\cdots+a_n=0$, where $a_0\ne 0$. Then the sum of all the roots of the equation in the complex numbers is $-a_1/a_0$. In this formula, a root of multiplicity greater than $1$ counts as many times as it occurs. For example, $x^2-6x+9=0$ as a "double root" at $x=3$. The formula predicts, correctly, that the sum of the roots is $-(-6)$.
|
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|
Solving $ \sin x + \sqrt 3 \cos x = 1 $ - is my solution correct? I have an equation that I'm trying to solve:
$$ \sin x + \sqrt 3 \cos x = 1 $$
After pondering for a while and trying different things out, this chain of steps is what I ended up with:
$$ \sin x + \sqrt 3 \cos x = 1 $$
$$ \sin x = 1 - \sqrt 3 \cos x $$
$$ \left(\sin x \right)^2 = \left(1- \sqrt 3 \cos x\right)^2 $$
$$ \sin^2 x = 1 - 2 \sqrt 3 \cos x + 3 \cos^2 x $$
$$ 2 \sqrt 3 \cos x - 3 \cos^2 x = 1 - \sin^2 x $$
$$ 2 \sqrt 3 \cos x - 3 \cos^2 x = \cos^2 x $$
$$ 2 \sqrt 3 \cos x = \cos^2 x + 3 \cos^2 x $$
$$ 4 \cos^2 x = 2 \sqrt 3 \cos x $$
$$ \frac{4 \cos^2 x}{\cos x} = 2 \sqrt 3 $$
$$ 4 \cos x = 2 \sqrt 3 $$
$$ \cos x = \frac{2 \sqrt 3}{4} $$
$$ \cos x = \frac{\sqrt 3}{2} $$
The fraction $ \frac{\sqrt 3}{2} $ can be rewritten as $ \cos \left(\pm \frac{\pi}{6}\right) $, so my solutions are:
$$ \cos x = \cos \left(\frac{\pi}{6}\right) \quad \text{or} \quad \cos x = \cos \left(-\frac{\pi}{6}\right) $$
$$ x = \frac{\pi}{6} + 2\pi n \quad \text{or} \quad x = -\frac{\pi}{6} + 2\pi n $$
Since I earlier on exponentiated both sides I have to check my solutions:
$$ x = \frac{\pi}{6} + 2\pi \Rightarrow \text{LHS} = \sin \left(\frac{\pi}{6} + 2\pi\right) + \sqrt 3 \cos \left(\frac{\pi}{6} + 2\pi\right) = 2 \not = \text{RHS} $$
$$ x = -\frac{\pi}{6} + 2\pi \Rightarrow \text{LHS} = \sin \left(-\frac{\pi}{6} + 2\pi\right) + \sqrt 3 \cos \left(-\frac{\pi}{6} + 2\pi\right) = 1 = \text{RHS} $$
Leaving $ x = -\frac{\pi}{6} + 2\pi n $ as the answer since its positive counterpart was not equal to $ 1 $.
$$ \text{Answer:} \: x = -\frac{\pi}{6} + 2\pi n $$
Have I done anything wrong or does this look good? I haven't really done this before so I feel uncertain not just about the solution, but also my steps and notation...
|
You can collapse the left-hand side into a single sine function:
$$\sin(x)+\sqrt3\cos(x) = 2\sin(x+\pi/3)$$
Then, dividing by two, all that remains is to solve the following:
$$\sin(x+\pi/3) = \frac{1}{2}$$
Wikipedia has an article on useful trigonometric identities, including linear combinations of sin and cos.
|
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|
Showing that $\frac{\sqrt[n]{n!}}{n}$ $\rightarrow \frac{1}{e}$ Show:$$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}= \frac{1}{e}$$
So I can expand the numerator by geometric mean. Letting $C_{n}=\left(\ln(a_{1})+...+\ln(a_{n})\right)/n$. Let the numerator be called $a_{n}$ and the denominator be $b_{n}$ Is there a way to use this statement so that I could force the original sequence into the form of $1/\left(1+\frac{1}{n}\right)^n$
|
Here is a rather direct calculation of the limit using squeezing. It needs
*
*$\ln k < \int_k^{k+1}\ln x \; dx < \ln (k+1)$
*$ \int \ln x \; dx = x(\ln x - 1) \color{grey}{+C} $
*$\lim_{n \rightarrow \infty}\frac{\ln (n+1)}{n} = 0$
Set
$$
x_n = \ln \frac{\sqrt[n]{n!}}{n} = \frac{1}{n}\sum_{k=1}^n \ln k - \ln n$$
So, to show is $\color{blue}{x_n \stackrel{n\rightarrow \infty}{\longrightarrow} -1}$. We have
$$
\int_1^n \ln x \; dx < \sum_{k=1}^{n-1} \ln (k+1) =\sum_{k=1}^n \ln k < \int_1^{n+1}\ln x \; dx
$$
We now squeeze:
$$
\color{blue}{L_n} := \frac{1}{n}\int_1^n \ln x \; dx - \ln n \color{blue}{<x_n < } \frac{1}{n} \int_1^{n+1}\ln x \; dx - \ln n =: \color{blue}{R_n}
$$
\begin{align*}
\color{blue}{L_n} & = \frac{1}{n}\left( n(\ln n - 1) +1 \right) - \ln n \\
& = -1 +\frac{1}{n} \color{blue}{\stackrel{n \rightarrow \infty}{\longrightarrow} -1} \\
& \\
\color{blue}{R_n} & = \frac{1}{n}\left( (n+1)(\ln (n+1) - 1) +1 \right) - \ln n \\
& = \left( 1 + \frac{1}{n} \right) \ln (n+1) - \left( 1 + \frac{1}{n} \right) + \frac{1}{n} - \ln n \\
& = -1 + \ln \left( 1+\frac{1}{n} \right) + \frac{\ln (n+1)}{n} \color{blue}{\stackrel{n \rightarrow \infty}{\longrightarrow} -1}
\end{align*}
|
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|
Proof that $\sqrt[3]{3}$ is irrational Proof that $\sqrt[3]{3}$ (Just need a check)
Let $\sqrt[3]{3}= \frac{a}{b}$ both a,b are integers of course. $\Rightarrow 3=\frac{a^{3}}{b^{3}}$ $\Rightarrow$ $3b^{3}=a^{3}$ $\Rightarrow$ 3b=a $\Rightarrow$ $3b^{3}=27b^{3}$ and this is a contradiction because the cubing function is a 1-1 function. Does this work?
|
No, $3b^3=a^3$ does not imply that $a=3b$. In fact if $a=3b$, then $a^3=(3b)^3=3^3b^3=27b^3$. What $3b^3=a^3$ does imply is that $3\mid a^3$, which then further implies that $3\mid a$. Then you have $a=3m$ for some integer $m$, and $3b^3=a^3=(3m)^3=27m^3$. Divide through by $3$ to get $b^3=9m^3$. Now $3\mid b^3$; what does this tell you about $b$? If you had begun by assuming (as you certainly may) that $\frac{a}b$ was in lowest terms, how would this be a contradiction?
|
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|
Check my workings: Prove the limit $\lim\limits_{x\to -2} (3x^2+4x-2)=2 $ using the $\epsilon,\delta$ definition. Prove the limit $\lim\limits_{x\to -2} (3x^2+4x-2)=2 $ using the $\epsilon,\delta$ definition.
Precalculations
My goal is to show that for all $\epsilon >0$, there exist a $\delta > 0$, such that
$$0<|x+2|<\delta\ \ \text{implies}\ |3x^2+4x-2-2|<\epsilon$$
$|3x^2+4x-2-2|=|3(x+2)^2-8x-16|$
$=|3(x+2)^2-4(x+2)|$
$\leq3|x+2|^2+4|x+2|$ by triangle inequality
$<3\delta^2+4\delta$
Hence, it is sufficient to show that $3\delta^2+4\delta=\epsilon$
Proof
For all $\epsilon>0$, choose $\delta=\min\left(\sqrt{\dfrac{\epsilon}{6}},\dfrac{\epsilon}{8}\right)$
$$\begin{align*}0<|x+2|<\delta\ \ \to\ \ &|3x^2+4x-2-2|<3\delta^2+4\delta\\&<3\left(\sqrt{\frac{\epsilon}{6}}\right)^2+4\delta\\&=\frac{\epsilon}{2}+4\delta\\&<\frac{\epsilon}{2}+4\frac{\epsilon}{8}\\&=\frac{\epsilon}{2}+\frac{\epsilon}{2}\\&=\epsilon\end{align*}$$
Therefore proven? Hehe. Not sure this will work or not.
My doubts lies in the steps.
Hence, it is sufficient to show that $3\delta^2+4\delta=\epsilon$
choose $\delta=\min\left(\sqrt{\dfrac{\epsilon}{6}},\dfrac{\epsilon}{8}\right)$
And hey, I am looking out for other possible ways to do this question too.
|
Here's how I would do it:
For any $\epsilon>0$, choose $\delta=\min\left(1,\dfrac{\epsilon}{11}\right)$. Then:
$$\begin{align*}
0<|x+2|<\delta\ \ \implies\ \ |(3x^2+4x-2)-2| &= |3x^2+4x-4| \\
&= |3x-2||x+2| \\
&= 3\left|x-\frac{2}{3}\right||x+2| \\
&= 3\left|x+2-\frac{8}{3}\right||x+2| \\
&\le 3\left(|x+2|+\left|\frac{-8}{3}\right|\right)|x+2| \text{ }\text{ }\text{ by triangle ineq.}\\
&< 3\left(1 + \frac{8}{3}\right)|x+2| \text{ }\text{ }\text{ since }|x+2|<\delta\le1\\
&=11|x+2| \\
&<11\left(\frac{\epsilon}{11}\right) \text{ }\text{ }\text{ since }|x+2|<\delta\le\frac{\epsilon}{11}\\
&=\epsilon\end{align*}$$
|
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|
How to integrate $1/(16+x^2)^2$ using trig substitution My answer: $1/128 (\tan^{-1}(x/4) + 4x/(16+x^2)^4)$
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The main question is how to write $16-x^2$ as a square. Note that $\cosh^2 \theta-\sinh^2 \theta = 1$ hence $1-\tanh^2\theta = \text{sech}^2 \theta$ thus $16-16\tanh^2\theta = 16\text{sech}^2 \theta$. This suggests a $x=4\tanh \theta$ substitution. Observe $dx = 4\text{sech}^2(\theta) d\theta$. Putting this all together,
$$ \int \frac{dx}{(16-x^2)^2} = \int \frac{4\text{sech}^2(\theta) d\theta}{(16)^2\text{sech}^4 \theta} = \frac{1}{64}\int \cosh^2 \theta d\theta $$
Now, $\cosh^2 \theta = \frac{1}{2}(\cosh 2 \theta+1)$
$$ \frac{1}{64}\int \cosh^2 \theta d\theta = \frac{1}{128}\int (\cosh 2 \theta +1)d\theta = \frac{1}{256}(\sinh 2 \theta + 2\theta)+c $$
Finally, $\theta = \tanh^{-1}(x/4)$ thus,
$$ \int \frac{dx}{(16-x^2)^2} = \frac{1}{256}\biggl(\sinh \bigl[2 \tanh^{-1}(x/4)\bigr] + 2\tanh^{-1}(x/4)\biggr)+c $$
Of course you can write the first term as an algebraic function by studying the identities for hyperbolic functions and the given hyperbolic subsitution.
( I like the sine or cosine substitution for this problem, but I include this answer for breadth)
|
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|
Finding the derivative of $ h(x) = \dfrac {2\sqrt{x}}{x^2+2}$ $ h(x) = \dfrac {2\sqrt{x}}{x^2+2}$ Find the derivative
How do I tackle this? My answer is totally different from the correction model, but I have tried for half an hour to show my answer in lateX but I don't know how to, it's too complicated, so, can someone please give a step to step of the solution? the correction model's solution is $ \dfrac {2-3x^2}{\sqrt{x} (x^2+2)^2} $
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Alternatively, whenever you have a quotient, you can turn it into a product. This doesn't always make things easier, but sometimes it does. You be the judge in this case. For example
$$h(x) = \frac{2\sqrt{x}}{x^2+2} = 2\sqrt{x} (x^2 + 2)^{-1}$$
So, now we can do the product rule to find the derivative
$$\begin{align*}
h'(x) &= 2\sqrt{x} \frac{d}{dx}(x^2 + 2)^{-1} + (x^2 + 2)^{-1} \frac{d}{dx} (2\sqrt{x}) \\
&= 2\sqrt{x} (-1)(x^2 + 2)^{-2}(2x) + (x^2 + 2)^{-1} x^{-1/2} \\
&= \frac{-4x^{3/2} + x^{3/2} + 2x^{-1/2}}{(x^2+2)^2} \\
&= \frac{2 - 3x^2}{\sqrt{x}(x^2+2)^2}
\end{align*}
$$
|
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Find all real numbers $t$ such that the quadratic form $f$ is positive definite. Where $$f(x_1,x_2,x_3)=2x_1^2+x_2^2+3x_3^2+2tx_1x_2+2x_1x_3$$.
This is a problem in my Matrix Analysis homework. Below is my effort.
Let $x=(x_1,x_2,x_3)^T$, then we have $$f=x^*Sx$$, in which $$S=\left(\begin{matrix}2&t&1\\t&1&0\\1&0&3\end{matrix}\right)$$. $f$ is positive definite is equivalent to $S$ is positive definite which is equivalent to all the eigenvalues of $S$ is positive.
The characteristic polynomial of $S$ is: $$\begin{align}|\lambda I-S|&=-\lambda^3+6\lambda^2+(3t^2-10)\lambda+(-3t^2+5)\\&=(-3+3\lambda)t^2+(-\lambda^3+6\lambda^2-10\lambda+5)\end{align}$$.
Now the only problem left is that how do I find all the possible real values of $t$ that makes this polynomial of $\lambda$ only has positive roots?
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We can solve it writing $x^tSx$ as a sum of squares. We have
\begin{align}
x^tSx&=(x_2+tx_1)^2-t^2x_1^2+2x_1^2+3x_3^2+2x_1x_3\\
&=(x_2+tx_1)^2+3\left(x_3^2+\frac 23x_1x_3\right)+(2-t^2)x_1^2\\
&=(x_2+tx_1)^2+3\left(x_3+\frac{x_1}3\right)^2-3\frac{x_1^2}9+(2-t^2)x_1^2\\
&=(x_2+tx_1)^2+3\left(x_3+\frac{x_1}3\right)^2+\left(\frac 53-t^2\right)x_1^2.
\end{align}
|
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|
Sum of a floor function Given that $\sum_{k=1}^n \lfloor \sqrt{k} \rfloor = Nn-\frac{N(N-1)(2N+5)}{6}$ where $N=\lfloor \sqrt{n} \rfloor$, for which values of N is the sum $\sum_{k=1}^n \lfloor \sqrt{k} \rfloor$ divisible by N?
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The question amounts to asking for what values of $N$ the expression $$Nn-\frac{N(N-1)(2N+5)}{6}$$ is divisible by $N$. Since $Nn$ is certainly divisible by $N$, this is the same as asking when $$\frac{N(N-1)(2N+5)}{6}=N\cdot\frac{(N-1)(2N+5)}6$$ is divisible by $N$. Clearly this is the case precisely when $(N-1)(2N+5)$ is divisible by $6$.
A number is divisible by $6$ if and only if it’s divisible by both $2$ and $3$. $2N+5$ is always odd, so it’s never divisible by $2$. Thus, $(N-1)(2N+5)$ is divisible by $2$ if and only if $N-1$ is even, i.e., if and only if $N$ is odd. Thus, $N$ must be congruent to $1,3$, or $5\bmod 6$. If $N\equiv 3\pmod 6$, then $$(N-1)(2N+5)\equiv 2\cdot5\equiv 4\pmod 6\;,$$ so $(N-1)(2N+5)$ is not divisible by $6$. If $N\equiv 1\pmod 6$, $$(N-1)(2N+5)\equiv 0\pmod 6\;,$$ so $(N-1)(2N+5)$ is certainly divisible by $6$. And if $N\equiv 5\pmod 6$, then $$(N-1)(2N+5)\equiv 4\cdot3\equiv 0\pmod 6\;,$$ and again $(N-1)(2N+5)$ is divisible by $6$.
Thus, $(N-1)(2N+5)$ is divisible by $6$ if and only if $N\equiv 1\pmod 6$ or $N\equiv 5\pmod 6$; this condition can be expressed a little more compactly by saying that $N\equiv\pm1\pmod6$. Equivalently, $(N-1)(2N+5)$ is divisible by $6$ if and only if $N$ is divisible by neither $2$ nor $3$: if either $2$ or $3$ divides $N$, then $(N-1)(2N+5)$ is not divisible by $6$.
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|
Inverse function of a polynomial What is the inverse function of $f(x) = x^5 + 2x^3 + x - 1?$ I have no idea how to find the inverse of a polynomial, so I would greatly appreciate it if someone could show me the steps to solving this problem. Thank you in advance!
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This is an experimental way of working out the inverse.
We can treat the polynomial like an expansion
\begin{equation}
f(x) = -1 + x + 0x^2 + 2x^3 + 0x^4 + x^5 + 0x^6 + 0x^7 + \cdots
\end{equation}
then we can perform a Series Reversion on this to give the inverse series (as an infinite expansion) \begin{equation}
f^{-1}(x) = (1+x) -2(1+x)^3 +11(1+x)^5-80(1+x)^7+665(1+x)^9-\cdots
\end{equation}
at a glance this doesn't seem to helpful, but if we search the coefficients in the OEIS we seem to get a hit! It would appear (conjecture) that the general coefficient is then \begin{equation}
a(n) = \binom{5n+1}{n}\frac{(-1)^n}{2n+1}
\end{equation}
and we could write that \begin{equation}
f^{-1}(x) = \sum_{n=0}^\infty \binom{5n+1}{n}\frac{(-1)^n}{2n+1}(x+1)^{2n+1}
\end{equation}
if we evaluate that in Mathematica, it gives \begin{equation}
f^{-1}(x)=(1+x)\;_4F_3\left(\frac{2}{5},\frac{3}{5},\frac{4}{5},\frac{6}{5}\bigg|\frac{3}{4},\frac{5}{4},\frac{6}{4}\bigg|-\frac{5^5}{4^4}(1+x)^2 \right)
\end{equation}
a generalised hypergeometric function. If we plot the composition of these two i.e $f(f^{-1}(x))$ or $f^{-1}(f(x))$ the plot seems to indicate \begin{equation}
f(f^{-1}(x))=f^{-1}(f(x))=x
\end{equation}
of course this is not a proof, and evaluating the inverse function may become numerically unstable if $x$ becomes too large.
|
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|
about a series $\sum_{n \geq1}{\frac{2n^2}{3^n}} $ How is this series ?
$$\sum_{n \geq1}{\frac{2n^2}{3^n}} ?$$
How I made:
$$a_{n}=\frac{2n^2}{3^n}$$ and then $$\frac{a_{n+1}}{a_n} \leq 1$$ so the series is convergence .
Is ok ?
thanks :)
|
Since $2n^2 < 4\cdot 2^n$, the series is dominated by a geometric series, so it is convergent. Moreover, we have:
$$ \forall x:|x|<3,\quad \sum_{n\geq 1}\frac{x^n}{3^n} = \frac{x}{3-x}, $$
$$ \forall x:|x|<3,\quad \sum_{n\geq 1}\frac{n x^n}{3^n} = x\cdot\frac{d}{dx}\left(\frac{x}{3-x}\right) = \frac{3x}{(x-3)^2}$$
$$ \forall x:|x|<3,\quad \sum_{n\geq 1}\frac{n^2 x^n}{3^n} = x\cdot\frac{d}{dx}\left(\frac{3x}{(x-3)^2}\right) = \frac{3x(3+x)}{(3-x)^3}$$
so:
$$\sum_{n\geq 1}\frac{2n^2}{3^n}=2\cdot\frac{3\cdot 4}{8} = 3.$$
|
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|
Determine $\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$, without L'Hospital or Taylor How can I prove that $$\lim_{x \to 0}{\frac{x-\sin{x}}{x^3}}=\frac{1}{6}$$
without using L'Hospital or Taylor series?
thanks :)
|
Let $L = \lim_{x \to 0} \dfrac{x - \sin(x)}{x^3}$. We then have
\begin{align}
L & = \underbrace{\lim_{y \to 0} \dfrac{3y - \sin(3y)}{27y^3} = \lim_{y \to 0} \dfrac{3y - 3\sin(y) + 4 \sin^3(y)}{27y^3}}_{\sin(3y) = 3 \sin(y) - 4 \sin^3(y)}\\
& = \lim_{y \to 0} \dfrac{3y - 3\sin(y)}{27 y^3} + \dfrac4{27} \lim_{y \to 0} \dfrac{\sin^3(y)}{y^3} = \dfrac{3}{27} L + \dfrac4{27}
\end{align}
This gives us $24L = 4 \implies L = \dfrac16$
|
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|
primitive residue classes modulo 32 $\mathbb Z_{32}^*$ is the primitive residue classes modulo 32. How is it possible to show that $\mathbb Z_{32}^*$ is generated by 5 and -1, without showing it for every element of $\mathbb Z_{32}^*$=$\{1,3,5,7,9,11,...,31\}$
|
The number $32$ is small, so it is easy to show by computing that $5$ has order $8$. Note that we only need to compute $5^{2^k}$ modulo $32$. For $\varphi(32)=16$, so the order of $5$ must be a power of $2$.
The following is severe overkill. We show that for any $n\ge 3$, $\mathbb{Z}^\ast_{2^n}$ is generated by $5$ and $-1$.
Lemma: Let $n\ge 3$. Then $5^{2^{n-3}}\not\equiv 1 \pmod{2^n}$, while $5^{2^{n-2}}\equiv 1\pmod{2^n}$.
Proof of Lemma: We show by induction on $n$ that if $n \ge 3$, then $5^{2^{n-3}}\equiv 1+2^{n-1}\pmod{2^n}$. This easily yields both parts of our assertion.
The result is easy to check for $n=3$. Suppose now that
$$5^{2^{k-3}}\equiv 1+2^{k-1}\pmod{2^{k}}.\tag{$1$}$$ We show that $5^{2^{k-2}}\equiv 1+2^{k}\pmod{2^{k+1}}$.
Square both sides of $(1)$, and simplify modulo $2^{k+1}$. We get
$$5^{2^{k-2}}=(1+2^{k-1} +n2^k)^2\equiv 1+2^{k}\pmod{2^{k+1}},$$
since $2^{2k-2}$ is divisible by $2^{k+1}$ if $k\ge 3$. This ends the proof of the lemma.
Now it is easy to finish. The order of $5$ modulo $2^n$ divides $\varphi(2^n)=2^{n-1}$, so the order of $5$ is a power of $2$. It follows from the results above that the order of $5$ is exactly $2^{n-2}$.
No power of $5$ is congruent to $-1$ modulo $2^n$. It follows that if $n \ge 3$, then $5$ and $-1$ together generate all of $\mathbb{Z}^\ast_{2^n}$.
|
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|
Mixing things- ratios Suppose I take two things, A and B. C is made from a $1/4$ ratio of $A$ to $B$, while D is made from a $4/3$ ratio. If I want to know what ratio of $C$ to $D$ will give a $5/6$ ratio of $A$ to $B$, do I just solve the system
$A+4B=C, 4A+3B=D, 5A+6B=z$
to get
$z= \frac{9C+14D}{13} \rightarrow \frac{5A}{6B} = \frac{9C}{14D}?$ If not what am I doing wrong? What is the right method?
|
Each unit of $C$ contains $\frac15$ unit of $A$ and $\frac45$ unit of $B$. Each unit of $D$ contains $\frac47$ unit of $A$ and $\frac37$ unit of $B$. Thus, $c$ units of $C$ and $d$ units of $D$ contain $\frac{c}5+\frac{4d}7=\frac{7c+20d}{35}$ units of $A$ and $\frac{4c}5+\frac{3d}7=\frac{28c+15d}{35}$ units of $B$. (Sanity check: these two quantities do in fact sum to $c+d$.)
You want these two quantities to be in the proportion $5:6$, so you want
$$\frac{7c+20d}{35}=\frac56\cdot\frac{28c+15d}{35}\;,$$
or $6(7c+20d)=5(28c+15d)$. Simplifying yields $45d=98c$, or $\dfrac{c}d=\dfrac{45}{98}$: the correct proportion of $C$ to $D$ is $45:98$.
The calculations below resulted from a misreading of the problem; they answer a different question from the one actually asked, but I’ll leave them up, since someone may at some point find them useful.
From $1$ unit of $A$ and $4$ units of $B$ you get $5$ units of $C$. From $4$ units of $A$ and $3$ units of $B$ you get $7$ units of $D$. If you scale that second combination down by a factor of $\frac67$, you find that $\frac67\cdot4=\frac{24}7$ units of $A$ and $\frac67\cdot3=\frac{18}7$ units of $B$ will give you $\frac67\cdot7=6$ units of $D$. These data are summarized in first four rows of the table below.
$$\begin{array}{cccc|l}
A&B&C&D\\ \hline
1&4&5&-&*\\
4&3&-&7\\
24/7&18/7&-&6&*\\ \hline
31/7&46/7&5&6&**
\end{array}$$
Now add the starred rows to get the bottom row: $\frac{31}7$ units of $A$ and $\frac{46}7$ units of $B$ give you $5$ units of $C$ and $6$ of $D$. That’s $11$ units altogether, so for a single unit of a $5:6$ $CD$ mixture divide the quantities of $A$ and $B$ by $11$: you need $\frac{31}{11\cdot7}=\frac{31}{77}\approx0.4026$ units of $A$ and $\frac{46}{11\cdot7}=\frac{46}{77}\approx0.5974$ units of $B$.
|
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|
Computing the indefinite integral $\int x^n \sin x\,dx$ $\newcommand{\term}[3]{
\sum_{k=0}^{\lfloor #1/2 \rfloor} (-1)^{#2} x^{#3} \frac{n!}{(#3)!}
}$
I am trying to prove that for $n \in\mathbb N$,
$$
\int x^n \sin x \, dx
= \cos x \term{n}{k+1}{n-2k}
+ \sin x \term{(n-1)}{k}{n-2k-1}
$$
I started with differentiation, and this is what I got:
If we define $f(x)$ as
$$
f(x)
= \cos x \term{n}{k+1}{n-2k}
+ \sin x \term{n-1}{k}{n-2k-1}
$$
then we have
$$
\begin{align*}
f’(x)
&= \cos x \term{n}{k+1}{n-2k-1}
- \sin x \term{n}{k+1}{n-2k} \\
&\qquad
+ \sin x \term{(n-1)}{k}{n-2k-2}
+ \cos x \term{(n-1)}{k}{n-2k-1} \\[8pt]
&= \cos x \left[
\term{n}{k+1}{n-2k-1}
+ \term{(n-1)}{k}{n-2k-1}
\right] \\
&\qquad
+ \sin x \left[
\term{(n-1)}{k}{n-2k-2}
- \term{n}{k+1}{n-2k}
\right]
\end{align*}
$$
I don't know how to go on, because of the different limits of the sum with $\lfloor{n/2}\rfloor$ and $\lfloor{(n-1)/2}\rfloor$.
|
Well.. I think I have a different way
$$ \int_a^b e^{izt} dz = \frac{e^{izt} }{it} |_{z=a}^{z=b}$$
Differentiate both sides with $t$ n times (*) and apply leibniz product rule(**) and some rearranging::
$$ \int_a^b z^n e^{izt} dz = \frac{1}{i^{n+1}} \sum_{k=0}^n \frac{1}{t^{k+1}} \binom{n}{k} (-1)^k k! (iz)^{n-k} e^{izt} |_{z=a}^{z=b}$$
Simplfying and putting t=1:
$$ \int_a^b z^n e^{iz} dz= \sum_{k=0}^n \frac{n!}{(n-k)!} (i)^{k-1} z^{n-k} e^{iz} |_{z=a}^{z=b}$$
Let's see if this formula really works, sub in $a=0$ and $b= \frac{\pi}{2}$ and $n=2$:
$$ \int_0^{\frac{\pi}{2}} z^2 e^{iz} dz= \sum_{k=0}^2 \frac{2!}{(2-k)!} (i)^{k-1} z^{2-k} e^{iz} |_{z=0}^{z= \frac{\pi}{2}}= 2!e^{iz} \left[ \frac{-iz^2}{2!} + \frac{z}{1!} + \frac{i}{0!}\right]_{0}^{\frac{\pi}{2} }=e^{iz} \left[ 2i + 2z -iz^2 \right]_0^{\frac{\pi}{2} }= i(2i + \pi - i\frac{\pi^2}{4}) - (2i)=( \frac{\pi^2}{4} -2)+i(\pi -2) $$
Hence, by taking imaginary on both sides we get:
$$ \int_0^{\frac{\pi}{2} } z^2 sin(z) dz = \pi -2$$
And similar results for cosine :)
P.s: I nominate the above result to be called the trigonometric gamma function (incase this wasn't discovered before)
Oh btw I this also gives the results for $x^n sin(ax)$ :P
*: Feynman's trick
**: Leibniz product rule
|
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|
The roots of $x^3+4x-1=0$ are $a$, $b$, $c$. Find $(a+1)^{-3}+(b+1)^{-3}+(c+1)^{-3}$ This is a question in A level Further Pure mathematics pastpaper Nov 2010.
The roots of the equation $x^3+4x-1=0$ are $a$, $b$ and $c$.
i) Use the substitution $y=1/(1+x)$ to show that the equation $6y^3-7y^2+3y-1=0$ has roots $1/(a+1)$, $1/(b+1)$ and $1/(c+1)$.
ii) For the cases $n=1$ and $n=2$, find the value of $$1/(a+1)^n+1/(b+1)^n+1/(c+1)^n$$
iii) Deduce the value of $$1/(a+1)^3+1/(b+1)^3+1/(c+1)^3$$
I know how to obtain the answer of i) and ii) by substituting $x=(1-y)/y$ and $\sqrt{y}=1/(1+x)$.
The equation for $n=2$ is $36y^3-13y^2-5y-1=0$
To find the value of the case $n=3$, I used to use a formula such as $S^3=\left(\sum a\right )^3-3\sum a \sum ab+3abc$
But there's another formula given by the Examiner Report for Teachers:
$$6S^3=7S^2-3S+3$$
It seems to be a nice method to solve this problem but it baffles me for a long time and I still can't find out how to get this...
|
Let the equation
$$6y^3 - 7y^2 + 3y - 1 = 0\tag{*}$$
have roots $\alpha$, $\beta$ and $\gamma$.
Using Vieta's formulas:
$$\sum \alpha = \alpha + \beta + \gamma = - \frac{-7}{6} = \frac{7}{6}\tag{1}$$
$$\sum \alpha\beta = \alpha\beta + \beta\gamma + \gamma\alpha = \frac{3}{6} = \frac{1}{2}$$
Using the identity
$\sum \alpha^2 = \left ( \sum\alpha \right )^2 - 2\sum\alpha\beta$ gives
$$\sum \alpha^2 = \frac{49}{36} - 2\times\frac{1}{2} = \frac{13}{36}\tag{2}$$
Rearranging $(*)$
$$6y^3 = 7y^2 - 3y + 1$$
so that
$$6\left (\sum\alpha^3\right ) = 7\left(\sum\alpha^2\right) - 3\left (\sum\alpha\right ) + 1 + 1 + 1.$$
(after adding $6\alpha^3 = 7\alpha^2 - 3\alpha + 1$, $6\beta^3 = 7\beta^2 - 3\beta + 1$ and $6\gamma^3 = 7\gamma^2 - 3\gamma + 1$)
Applying $(1)$ and $(2)$
$$\sum\alpha^3 = \frac{1}{6}\left ( 7\times\frac{13}{36} - 3\times\frac{7}{6} + 3 \right ) = \frac{73}{216}$$
|
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|
How to prove $\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$ With $a + b + c = 3$ and $a, b, c>0$ prove these inequality:
1)$$\frac{a}{a+bc}+\frac{b}{b+ac}+\frac{c}{c+ab}\geq \frac{3}{2}$$
2)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+abc\geq 4$$
3)$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b}+\frac{9}{4}abc\geq \frac{21}{4}$$
|
We need to prove that:
$$\sum_{cyc}\frac{a}{a+bc}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\left(\frac{a}{a+bc}-\frac{1}{2}\right)\geq0$$ or
$$\sum_{cyc}\frac{a-bc}{a+bc}\geq0$$ or
$$\sum_{cyc}\frac{a(a+b+c)-3bc}{a+bc}\geq0$$ or
$$\sum_{cyc}\frac{(a-b)(a+3c)-(c-a)(a+3b)}{a+bc}\geq0$$ or
$$\sum_{cyc}(a-b)\left(\frac{a+3c}{a+bc}-\frac{b+3c}{b+ac}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2c(a+b+3c-3)}{(a+bc)(b+ac)}\geq0$$ or
$$\sum_{cyc}\frac{(a-b)^2c^2}{(a+bc)(b+ac)}\geq0.$$
Done!
|
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|
probability problem and There are three caskets of treasure. The first casket contains 3 gold coins, the second casket contains 2 gold coins and 2 bronze coins, and the third casket contains 2 gold coins and 1 silver coin. You choose one casket at random and draw a coin from it. The probability that the coin you drew is gold has the value $\frac ab$, where $a$ and $b$ are coprime positive integers. What is the value of $a+b$?
|
If we draw from the first, our probability of gold is $\dfrac{3}{3}$. If we draw from the second, our probability of gold is $\dfrac{2}{4}$. And if we draw from the third, our probability of gold is $\dfrac{2}{3}$. So the required probability is
$$\frac{1}{3}\cdot\frac{3}{3}+\frac{1}{3}\cdot \frac{2}{4}+\frac{1}{3}\cdot\frac{2}{3}.$$
This simplifies to $\dfrac{13}{18}$. The sum of numerator and denominator is $31$.
|
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|
What is the integer part of $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{3}} +\cdots + \frac{1}{\sqrt{(2n+1)^2}}$ I tried to solve the following problem.
What is the integer part of
$$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{3}} + \cdots + \frac{1}{\sqrt{(2n+1)^2}}=\sum_{k=0}^{2(n^2+n)} \frac 1{\sqrt{2k+1}} ?$$
I tried using some inequalities( by grouping 1,3,5,7 / 9,11,13,...,25/ ),
but I failed.
How can I compute the integer part of this sum?
|
A more general sum can be bounded as follows:
$$
\int_0^{n+1} \frac{1}{\sqrt{2 x + 1}} dx \leq \sum_{k=0}^n \frac{1}{\sqrt{2 k + 1}} \leq 1 + \int_0^n \frac{1}{\sqrt{2 x + 1}}dx
$$
or, computing the integrals
$$
\sqrt{2 n + 3} -1 \leq \sum_{k=0}^n\frac{1}{\sqrt{2 k + 1}} \leq \sqrt{2 n + 1}.
$$
The second inequality is strict if $n > 0$. If $2n + 1$ is a square, say $m^2$ then
$$
m^2 < 2n+3 = m^2 + 2 < (m+1)^2
$$
and so the integral part of $\sqrt{2n+3}$ is $m$. From this it follows that the integral part of the sum is $m - 1$ if $n > 0$ and $1$ if $n=0$.
|
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|
solving the equation $x^4-5x^3+11x^2-13x+6=0$ by given condition 1.(a) solve the equation $x^4-5x^3+11x^2-13x+6=0$ , given that two of its roots $p$ & $q$ are connected by the relation $3p+2q=7$
(b) solve the equation $x^4-5x^3+11x^2-13x+6=0$ which has two roots whose difference is $1$
did I need to solve these problems by taking roots $x_1,x_2,x_3,x_4$ and using the relation between roots and coefficients and the given facts. it will be then a very lengthy process.is there any alternative short process
|
$$x^4-5x^3+11x^2-13x+6=0$$
$$x^4-x^3-4x^3+4x^2+7x^2-7x-6x+6=0$$
$$(x-1)x^3-(x-1)4x^2+(x-1)7x-(x-1)6=0$$
$$(x-1)(x^3-4x^2+7x-6)=0$$
$$(x-1)(x^3-2x^2-2x^2+4x+3x-6)=0$$
$$(x-1)\left( (x-2)x^2-(x-2)2x+(x-2)3 \right)=0$$
$$(x-1)(x-2)(x^2+2x+3)=0$$
|
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|
Inverse Laplace Transform Assistance How do I compute the following transform?
$$\frac {s-1}{2s^2+s+6}$$
I've gotten this far:
$$\frac {1}{2}\cdot \frac {s-1}{\left(s+\frac{1}{4}\right)^2 + \frac{47}{16}}$$
|
$$ F(s)=\frac {s-1}{2s^2+s+6}=\frac {s-1}{2(s^2+\frac{s}{2} +3)} $$
$$ ax^2+bx=a\left[ (x+\frac{b}{2a})^2-(\frac{b}{2a})^2 \right] $$
So:
$$ 2(s^2+\frac{s}{2} +3)=2 \left[ (s+\frac{1}{4})^2-(\frac{1}{4})^2 +3 \right]=2 \left[ (s+\frac{1}{4})^2+(\frac{47}{15}) \right] $$
therefor:
$$ F(s)=(\frac{1}{2}) \frac{s-1 {\color{red} { +\frac{5}{4} -\frac{5}{4}} } }{(s+\frac{1}{4})^2+(\frac{47}{15})} = (\frac{1}{2}) \left[ \frac{s+\frac{1}{4}}{(s+\frac{1}{4})^2+(\frac{47}{15})}-\frac{\frac{5}{4}}{(s+\frac{1}{4})^2+(\frac{47}{15})} \right] $$
We know:
$$ \mathcal{L}\left[ e^{-at}cos(bt) \right]=\frac{s+a}{(s+a)^2+b^2} $$
$$ \mathcal{L}\left[ e^{-at}sin(bt) \right]=\frac{b}{(s+a)^2+b^2} $$
therefor:
$$ f(t)=(\frac{1}{2})e^{\frac{-t}{4}} \left[ cos(\frac{\sqrt{47}}{4}t) - \frac{5}{\sqrt{47}} sin(\frac{\sqrt{47}}{4}t)\right] $$
|
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|
Evaluate the sum? problem.. I need to evaluate the sum: can someone help?
The series is as follows:
$$
\frac 14 - \frac {1}{2(4)^2} + \frac{1}{3(4)^3} - \cdots + \frac{(-1)^{(x+1)}}{x(4)^x}
$$
|
We know, $$\sum_{0\le s\le n-1}y^s=\frac{y^n-1}{y-1}$$
Integrating either sides wrt $y$, we get $$\sum_{1\le r\le n}\frac{y^r}r=\int \left(\frac{y^n-1}{y-1}\right) dy$$
If $\sum_{1\le r\le n}\frac{y^r}r=S(y,n),$
$\frac 14 - \frac {1}{2(4)^2} + \frac{1}{3(4)^3} - \cdots + \frac{(-1)^{(x+1)}}{x(4)^x}=-S(-\frac14, x+1)$
The formula for the general summation is explained here.
|
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|
Why are these two expressions different in this induction problem? Prove with $n \ge 1$:
$$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot4}+\cdots+\frac{n+2}{n(n+1)2^n} = 1 - \frac{1}{(n+1)2^n}$$
First, I prove it for $n=1$:
$$\left(\frac{1+2}{1(1+1)2^1} = 1-\frac{1}{(1+1)2^1}\right) \implies \left(\frac{3}{4} = 1- \frac{1}{4}\right) \implies \left(\frac{3}{4} = \frac{3}{4}\right)$$
Which is true.
So I will now assume this:
$$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} = 1 - \frac{1}{(n+1)2^n}$$
And I want to prove it for $n+1$, i.e:
$$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}} = 1 - \frac{1}{((n+1)+1)2^{n+1}}$$
This is how I tried to prove it:
$$\frac{3}{1\cdot2\cdot2} + \frac{4}{2\cdot3\cdot2}+\cdots+\frac{n+2}{n(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}} =$$
$$ 1 - \frac{1}{(n+1)2^n} + \frac{n+3}{(n+1)(n+2)2^{n+1}}$$
Before continuing, I usually like to grab a calculator, give a value to $n$ and evaluate my current expression with the expression I want to reach. If both values are equal, it means I'm doing okay.
So I took $n=5$ and evaluated the expression I want to reach:
$$1 - \frac{1}{(5+2)\cdot2^5+1} = \frac{447}{448}$$
Then, still with $n = 5$ I evaluated my current expression:
$$1 - \frac{1}{(5+1)\cdot2^5}+\frac{5+3}{(5+1)(5+1)\cdot2^{5+1}} = \frac{575}{576}$$
So I got $\frac{447}{448}$ for the expression I want to reach and $\frac{575}{576}$ for what I got so far. Something went wrong.
My problem with this is that I haven't done any calculations yet. All my steps so far were rather mechanical - things I always do with mathematical induction.
Maybe I simply evaluated them wrongly. But I can't see it - I've tested it many times already.
Why are both expressions different? They should be the same.
|
There is a flaw in the logic in this posting:
begin quote:
First, I prove it for $n=1$:
$$\left(\frac{1+2}{1(1+1)2^1} = 1-\frac{1}{(1+1)2^1}\right) \implies \left(\frac{3}{4} = 1- \frac{1}{4}\right) \implies \left(\frac{3}{4} = \frac{3}{4}\right)$$
Which is true.
end of quote
This is not valid reasoning. You're saying "Let's prove $A$, as follows: If $A$ then $3/4 = 3/4$, which is true." Very well then, I shall prove that Socrates was John F. Kennedy: If Socrates was John F. Kennedy, then Socrates died in 1963, and hence, Socrates was mortal. Which is true." To say "If B then A. And A is true. Therefore B." is a standard erroneous form of reasoning. Writers on logic since the time of Aristotle have identified this as a fallacy. A correct way of reasoning is this:
$$
\frac{1+2}{1(1+1)2^1} = \frac34 = 1 - \frac14 = 1-\frac{1}{(1+1)2^1}.
$$
Therefore the proposition in case $n=1$.
And then you repeat the same fallacy in your inductive step.
You should write "$=$" between things ALREADY KNOWN TO BE EQUAL. Then when you write
$$
A = B = C = \cdots = Z,
$$
then that proves that $A=Z$. You should not write
$$
A = Z
$$
therefore
$$
B=Y
$$
therefore
$$
C=X
$$
etc. etc. etc. etc.
Therefore
$$
3=3
$$
which is true. Therefore $A=Z$.
|
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|
Integral by partial fractions
$$ \int \frac{5x}{\left(x-5\right)^2}\,\mathrm{d}x$$ find the value of the constant when the antiderivative passes threw (6,0)
factor out the 5, and use partial fraction
$$ 5 \left[\int \frac{A}{x-5} + \frac{B}{\left(x-5\right)^2}\, \mathrm{d}x \right] $$
Solve for $A$ and $B$.
$A\left(x-5\right) + B = x$ Then $B-5A$ has to be zero and $A$ has to be 1.
Resulting in
$$ 5 \left[\int \frac{1}{x-5} + \frac{5}{\left(x-5\right)^2}\, \mathrm{d}x \right]$$
$$ \Rightarrow 5 \left[ \ln \vert x - 5 \vert -\frac{5}{x-5}\right] + C$$
However, this approach doesn't give the answer in the book.
Book's Answer
$$ \frac{5}{x-5} \left(\left(x-5\right) \ln \vert x - 5 \vert - x \right) + C $$
The value should be 30, according to the book.
|
Probably they missed include a constant in book's answer.
If we include a constant $k$, the book's answer will change to:
$$\frac{5}{x-5}((x-5)\ln|x-5|-x)+k$$
But with some algebra we get
$$\frac{5}{x-5}((x-5)\ln|x-5|-x)+k=$$ $$=5(\frac{(x-5)}{x-5}\ln|x-5|-\frac{x}{x-5})+k= 5(\ln|x-5|-\frac{x}{x-5}+1)-5+k=$$ $$=5(\ln|x-5|-\frac{x}{x-5}+\frac{x-5}{x-5})-5+k=5(\ln|x-5|-\frac{5}{x-5})-5+k=$$
$$=5(\ln|x-5|-\frac{5}{x-5})+C$$
Which is your answer, where C is a new constant, such that $C=k-5$.
|
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|
Converting $x = \sin \frac{t}{2}, y = \cos \frac{t}{2}$ to Cartesian form How can we transform these parametric equations to Cartesian form?
$$x = \sin \frac{t}{2}, \quad y = \cos \frac{t}{2}, \quad -\pi \leq t \leq \pi.$$
|
$$x = \sin \frac{t}{2}, \quad y = \cos \frac{t}{2}, \quad -\pi \leq t \leq \pi.$$
$$x^2+y^2=(\sin \frac{t}{2})^2+(\cos \frac{t}{2})^2=1$$ so
$$x^2+y^2=1$$ is equation of some circle
|
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|
Computation of a certain integral I would like to compute the following integral. This is for a complex analysis course but I managed to around some other integrals using real analysis methodologies. Hopefully one might be able to do for this one too.
$$\int_{0}^{2\pi} \frac{1}{a-\cos(x)}dx, \text{ with } a > 1.$$
Any suggestion will be greatly appreciated.
|
Let $\tan(x/2) = t$. We then get that $$\sec^2(x/2) dx = 2dt \implies dx = \dfrac{2dt}{1+t^2}$$
Also, $\cos(x) = \dfrac{1-\tan^2(x/2)}{1+\tan^2(x/2)} = \dfrac{1-t^2}{1+t^2}$.
Hence,
\begin{align}
\dfrac{dx}{a- \cos(x)} & = \dfrac{2dt}{1+t^2} \dfrac1{a - \dfrac{1-t^2}{1+t^2}}\\
& = \dfrac{2dt}{a(1+t^2) - (1-t^2)}\\
& = \dfrac{2dt}{(a+1) t^2 + (a-1)}
\end{align}
Hence, $$I = \int \dfrac{2dt}{(a+1) t^2 + (a-1)} = \dfrac2{a+1} \int \dfrac{dt}{t^2 + \dfrac{a-1}{a+1}} = \dfrac2{\sqrt{a^2-1}} \arctan \left(t\sqrt{\dfrac{a+1}{a-1}}\right) + c$$
Writing it in terms of $x$, we get that
$$I = \dfrac2{\sqrt{a^2-1}} \arctan \left(\tan(x/2)\sqrt{\dfrac{a+1}{a-1}}\right) + c$$
|
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|
Inequality. $(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a)$ Let $a,b,c$ be three real positive(strictly) numbers. Prove that:
$$(a^2+bc)(b^2+ca)(c^2+ab) \geq abc(a+b)(b+c)(c+a).$$
I tried :
$$abc\left(a+\frac{bc}{a}\right)\left(b+\frac{ca}{b}\right)\left(c+\frac{ab}{c}\right)\geq abc(a+b)(b+c)(c+a) $$
and now I want to try to prove that for example $$a+\frac{bc}{a} \geq a+b$$
but I don't know if is is a good idea.
Thanks:)
|
We have$:$ $$ \left( {a}^{2}+bc \right) \left( ac+{b}^{2} \right) \left( ab+{c}^{
2} \right) -abc \left( a+b \right) \left( b+c \right) \left( c+a
\right) $$
$$=ac \left( {b}^{2}+{c}^{2} \right) \left( a-b \right) ^{2}+ab \left( {a}^{2}+{c}^
{2} \right) \left( b-c \right) ^{2} +bc \left( {
a}^{2}+{b}^{2} \right) \left( a-c \right) ^{2}\geqq 0$$
Done.
|
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|
Does multiplying polynomials ever decrease the number of terms? Let $p$ and $q$ be polynomials (maybe in several variables, over a field), and suppose they have $m$ and $n$ non-zero terms respectively. We can assume $m\leq n$. Can it ever happen that the product $p\cdot q$ has fewer than $m$ non-zero terms?
I ask this because I vaguely recall seeing a positive answer in book somewhere (probably about computation or algorithms since the polynomials were unwieldy). If anyone knows what book this is from it would be much appreciated.
|
Here's an elementary example. Start with the well-known identity $x^n - 1 = (x-1) (x^{n-1} + x^{n-2} + \ldots + x + 1)$. If $n$ is odd, we can factor $x^n+1$ in a similar way by flipping the signs: $x^n + 1 = (x+1) (x^{n-1} - x^{n-2} + \ldots - x + 1)$. Now mix and match the two:
$$\begin{align*}
x^{2n} - 1 &= (x^n - 1) (x^n + 1) \\
&= (x-1) (x^{n-1} + x^{n-2} + \ldots + x + 1) (x+1) (x^{n-1} - x^{n-2} + \ldots - x + 1) \\
&= (x+1) (x^{n-1} + x^{n-2} + \ldots + x + 1) (x-1) (x^{n-1} - x^{n-2} + \ldots - x + 1) \\
&= (x^n + 2x^{n-1} + 2x^{n-2} + \ldots + 2x + 1) (x^n - 2x^{n-1} + 2x^{n-2} - \ldots + 2x - 1)
\end{align*}$$
I don't see an obvious generalization to even values of $n$.
|
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|
Subsitution in Integrals Okay, so I'm working on definite integrals and I calculated the indefinte integral of $\frac{x}{\sqrt{x+1}}$ to be $\frac{2}{3}u^{3/2}-2\sqrt{u}$ where $u=x+1$. The definite integral is on the interval $[0,5]$ so I used my $u$-subsitution equation on the interval $[1,6]$. I keep getting $7.9981$ but I know that's incorrect. What am I doing wrong?
|
Same answer, but with the useful trick of "adding $0$":
\begin{align}
\int \frac{x}{\sqrt{x+1}}dx = \int \frac{x+1 -1}{\sqrt{x+1}}&= \int \left(\frac{x+1}{\sqrt{x+1}} - \frac{1}{\sqrt{x+1}} \right)dx
\\
&= \int\left( (x+1)^{\frac{1}{2}} - (x+1)^{- \frac{1}{2}}\right)dx
\\
&= \frac{2}{3} (x+1)^{\frac{3}{2}} - 2 (x+1)^{\frac{1}{2}}.
\end{align}
|
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|
Number of ordered sets of integers How many ordered sets of integers $(x,y,z)$ satisfying $$x,y,z \in [-10,10]$$ are solutions to the following system of equations:
$$x^2y^2+y^2z^2=5xyz$$
$$y^2z^2+z^2x^2=17xyz$$
$$z^2x^2+x^2y^2=20xyz$$
By a quite cumbersome method, which is too long for me to type down here, I arrived at the answer 5: the ordered sets being $(0,0,0), (8,2,4), (-8,-2,4), (-8,2,-4)$ and $(8,-2,-4)$. But this does not seem to be the right answer. What I basically did was divide the system of equations with $(xyz)$, then by a process of elimination of the variables, found the possible values of x, y and z. However, it seems I've somehow missed out other solutions as well. Can anyone help me out with this?
Thanks!
|
Let $(x,y,z)$ be a solution. Then the three numbers
$$
u_1=x^2y^2+y^2z^2-5xyz, \ u_2=y^2z^2+z^2x^2-17xyz, \ u_3=z^2x^2+x^2y^2=20xyz
$$
are all zero.
Now
$$
\begin{array}{lcl}
(1) \ -u_1+u_2+u_3 &=& 2x^2z^2-32xyz=32xz\bigg( \frac{xz}{16}-y\bigg) \\
(2) \ u_1-u_2+u_3 &=& 2x^2y^2-8xyz=8xy\bigg( \frac{xy}{4}-z\bigg) \\
(3) \ u_1+u_2-u_3 &=& 2y^2z^2-2xyz=2yz\bigg( yz-x\bigg)
\end{array}
$$
If $x,y$ and $z$ are all nonzero, we see that $x=yz, y=\frac{xz}{16}$ and $z=\frac{xy}{4}$. Multiplying, we deduce $xyz=\frac{(xyz)^2}{64}$, so $xyz=64$. Using
$x^2=xyz, y^2=\frac{xyz}{16}$ and $z^2=\frac{xyz}{4}$, we deduce $x^2=64,y^2=4,z^2=16$.
This produces four solutions : $(8\varepsilon_1,2\varepsilon_2,4\varepsilon_1\varepsilon_2)$ where each of $\varepsilon_1$ and $\varepsilon_2$ can be $+1$ or $-1$.
If $z=0$, then by (2) we have $2(xy)^2=0$ and hence $x=0$ or $y=0$. Similarly when $x=0$ or $y=0$ : this will produce the solutions found by Johan.
|
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|
If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint: $a^2 -ab +b^2 = (a+b)^2 -3ab.$
I know we can say that there exists an $x,y$ such that $ax + by = 1$. So in this case,
$(a+b)x + ((a+b)^2 -3ab)y =1.$
I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$ Any suggestions?
|
Suppose that the positive integer $d$ divides both $a+b$ and $a^2-ab+b^2$. You showed that $d$ divides $3ab$.
Note that $d$ is relatively prime to $ab$. Otherwise, there is a prime $p$ that divides both $d$ and $ab$. Since $p$ divides $ab$, it follows that $p$ divides one of $a$ or $b$, say $a$. But since $p$ divides $d$, it follows that $p$ divides $a+b$. Since $p$ divides $a$, it follows that $p$ divides $b$. This contradicts the fact that $a$ and $b$ are relatively prime.
Since $d$ divides $3ab$, and $d$ is relatively prime to $ab$, we conclude that $d$ divides $3$. So if $d$ is positive, then $d=1$ or $d=3$.
Note that both $1$ and $3$ are possible. For let $a=1$ and $b=1$. Then $\gcd(a+b,a^2-ab+b^2)=1$.
Now let $a=2$ and $b=1$. Then $a^2-ab+b^2=3$, so in this case the gcd is $3$.
|
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|
How to compute $7^{7^{7^{100}}} \bmod 100$? How to compute $7^{7^{7^{100}}} \bmod 100$? Is $$7^{7^{7^{100}}} \equiv7^{7^{\left(7^{100} \bmod 100\right)}} \bmod 100?$$
Thank you very much.
|
$7^4\equiv1\pmod{100}$
Let $$y=7^{7^{7^{100}}}$$
So, $7y=7^{(7^{7^{100}}+1)}\equiv1\pmod{100}$ as $(7^{7^{100}}+1)$ is divisible by $(7+1)=8$
Let us use the property of the successive convergents of a continued fraction.
Now, $\frac{100}7=14+\frac27=14+\frac1{\frac72}=14+\frac1{3+\frac12}$
So, the previous convergent of $\frac{100}7$ is $14+\frac13=\frac{43}3$
So, $100\cdot3-7\cdot43=-1\implies y\equiv43\pmod {100}$
|
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|
Can someone show me why this factorization is true? $$x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \dots + xy^{n-2} + y^{n-1})$$
Can someone perhaps even use long division to show me how this factorization works? I honestly don't see anyway to "memorize this". I like to see some basic intuition behind this
|
It’s easier to verify that it multiplies back together correctly:
$$x\left(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}\right)=x^n+\color{red}{x^{n-1}y+\ldots+x^2y^{n-2}+xy^{n-1}}\;,\tag{1}$$
and
$$y\left(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}\right)=\color{red}{x^{n-1}y+x^{n-2}y^2+\ldots+xy^{n-1}}+y^n\;.\tag{2}$$
The two red blocks are identical, so when you subtract $(2)$ from $(1)$, all that remains is $x^n-y^n$.
Note that the identity has a very simple form in summation notation:
$$x^n-y^n=(x-y)\sum_{k=0}^{n-1}x^{n-1-k}y^k\;.$$
In each term the exponents of $x$ and $y$ sum to $n-1$, and there is a term for every possible pair of non-negative exponents with sum $n-1$. This observation may make it easier to remember.
That said, it is possible to do the long division. Here’s how it starts, enough to show the pattern:
$$\begin{array}{c|ll}
&x^{n-1}&+&x^{n-2}y&+&x^{n-3}y^2&+&\dots&+&y^{n-1}\\ \hline
x-y&x^n&&&&&&\dots&&&-&y^n\\
&x^n&-&x^{n-1}y\\ \hline
&&&x^{n-1}y&&&&&&&-&y^n\\
&&&x^{n-1}y&-&x^{n-2}y^2\\ \hline
&&&&&x^{n-2}y^2&&&&&-&y^n\\
&&&&&x^{n-2}y^2&-&x^{n-3}y^3\\ \hline
&&&&&&&x^{n-3}y^3\\
&&&&&&\vdots\\ \hline
&&&&&&&&&xy^{n-1}&-&y^n\\
&&&&&&&&&xy^{n-1}&-&y^n\\ \hline
\end{array}$$
|
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|
Find the rotation axis and angle of a matrix $$A=\frac{1}{9}
\begin{pmatrix}
-7 & 4 & 4\\
4 & -1 & 8\\
4 & 8 & -1
\end{pmatrix}$$
How do I prove that A is a rotation ? How do I find the rotation axis and the rotation angle ?
|
You have $A^T A = I$. Hence $A$ is a rotation. Since $\det A = 1$, it is proper.
By inspection, $A \begin{bmatrix} 1 \\ 2 \\ 2\end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\2\end{bmatrix}$, which gives the axis of rotation.
Inspection also shows that $\begin{bmatrix} 2 \\ -1 \\ 0 \end{bmatrix}$ and $\begin{bmatrix} 2 \\ 4 \\ -5 \end{bmatrix}$ are orthogonal eigenvectors corresponding to the (repeated) eigenvalue $-1$. Hence we see that the rotation angle is $\pi$.
Explicitly, if we let $R = \begin{bmatrix} 1 & 2 & 2 \\ 2 & -1 & 4 \\ 2 & 0 & -5 \end{bmatrix}$, then $R^{-1} = \frac{1}{405} \begin{bmatrix} 45 & 90 & 90 \\ 162 & -81 & 0 \\ 18 & 36 & -45\end{bmatrix}$, and $R^{-1} A R = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1 \end{bmatrix}$, from which we see that the rotation angle is $\pi$.
|
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|
$2^{10 - x} \cdot 2^{10 - x} = 4^{10-x}$ $$2^{10 - x} \cdot 2^{10 - x} = 4^{10-x}$$
Is that correct?
I would've done
$$
2^{10 - x} \cdot 2^{10 - x}\;\; = \;\;
(2)^{10 - x + 10 - x} \; = \; (2)^{2 \cdot (10 - x)} \;=\; 4^{10 - x}\tag{1}
$$
Is that allowed?
If so, can I say that
$$
\frac{4^x}{2^y} = 2^{x - y} \tag{2}
$$
|
$$\text{Yes:}\;\;\;\large 2^{10 - x} \cdot 2^{10 - x} =\; 4^{10-x}\tag{1.a}$$
$$\underbrace{\large 2^{10 - x} \cdot 2^{10 - x}
= (2)^{10 - x + 10 - x} = (2)^{2 \cdot (10 - x)} = 4^{10 - x}}\tag{1.b}
$$
$$\text{Yes, that is that is allowed.}$$
$$\underbrace{
\large \frac{4^x}{2^y} = 2^{x - y}}\;\;\; ?\tag{2 ?}$$
$$\text{NOT correct. See below:}$$
$$(2)\quad \large \frac{4^x}{2^y} \;=\; 4^x \cdot 2^{-y} \;= \;(2^2)^x \cdot 2^{-y} \;=\; 2^{(2x)} \cdot 2^{(-y)} \;=\;2^{(2x-y)}$$
|
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|
Convergence of parameterized series $\sum_{n=2}^{+\infty} \left(( \sqrt{n+1} - \sqrt{n})^p \cdot \ln\left( \frac{n-1}{n+1}\right) \right)$ $$\sum_{n=2}^{+\infty} \left(( \sqrt{n+1} - \sqrt{n})^p \cdot \ln\left( \frac{n-1}{n+1}\right) \right)$$
I guess that more useful form is:
$$\sum_{n=2}^{+\infty} \left(( \sqrt{n+1} + \sqrt{n})^{-p} \cdot \left( \ln(n-1) - \ln(n+1) \right) \right)$$
The question is, for what $ p \in \mathbb{R}$ does it converge?
I've made some experimental random checks, and there seems that there's something happening for -1 and -2, I think it converges for $p \in (-2; -1) \cup (-1; +\infty)$, but then again it's just some random test results.
How could I find for which $p$ it works, and which method of examining convergence should I use?
|
HINT: Rewrite the series as
$$ \begin{eqnarray}
\mathcal{S} &=& \sum_{n=2}^\infty \left( \sqrt{n+1}-\sqrt{n} \right)^p \cdot \ln \left(\frac{n-1}{n+1} \right) \\ &=& \sum_{n=2}^\infty \left( \frac{\sqrt{n+1}^2-\sqrt{n}^2}{\sqrt{n+1}+\sqrt{n}} \right)^p \cdot \ln \left(1-\frac{2}{n+1} \right)\\ &=& \sum_{n=2}^\infty \frac{ \ln \left(1-\frac{2}{n+1} \right)}{\left(\sqrt{n+1} + \sqrt{n} \right)^p}
\end{eqnarray} $$
|
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|
How to calculate $I=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y}dy$? How do I integrate this guy? I've been stuck on this for hours..
$$I=\frac{1}{2}\int_{0}^{\frac{\pi }{2}}\frac{\ln(\sin y)\ln(\cos y)}{\sin y\cos y}dy$$
|
Let $\sin(y) = t$. Recall that $\cos^2(y) = 1-t^2$. Then $\cos(y) dy = dt$. Hence,
$$I = \dfrac14 \int_0^1 \dfrac{\ln(\sin(y)) \ln(\cos^2(y)) \cos(y) dy}{\sin(y) \cos^2(y)}$$
$$4I = \int_0^1 \dfrac{\ln(t) \ln(1-t^2) dt}{t (1-t^2)}$$
$$\dfrac{\ln(t) \ln(1-t^2)}{t (1-t^2)} = -\dfrac{\ln(t)}{t(1-t^2)} \sum_{k=1}^{\infty} \dfrac{t^{2k}}k = - \sum_{k=1}^{\infty}\dfrac{\ln(t)}{(1-t^2)} \dfrac{t^{2k-1}}k$$
$$\dfrac{\ln(t)}{(1-t^2)} \dfrac{t^{2k-1}}k = \dfrac1k \sum_{l=0}^{\infty}t^{2k+2\ell-1} \ln(t)$$
Hence,
$$\dfrac{\ln(t) \ln(1-t^2)}{t (1-t^2)} = -\sum_{k=1}^{\infty} \dfrac1k\sum_{l=0}^{\infty}t^{2k+2\ell-1} \ln(t)$$
Recall that $$\int_0^1 t^m \log(t) dt = -\dfrac1{(m+1)^2}$$
Hence, $$4I = \sum_{k=1}^{\infty} \sum_{\ell=0}^{\infty} \dfrac1k \dfrac1{(2k+2 \ell)^2} = \dfrac14 \sum_{k=1}^{\infty} \sum_{\ell=0}^{\infty} \dfrac1k \dfrac1{(k+\ell)^2} = \dfrac14 2 \zeta(3)$$
Hence, $$I = \dfrac{\zeta(3)}8$$
EDIT
Consider the sum
$$S = \sum_{k=1}^{\infty} \sum_{l=0}^{\infty} \dfrac1k\dfrac1{(k+l)^2}$$
We have that
$$S = \underbrace{\sum_{k=1}^{\infty} \sum_{l=k}^{\infty} \dfrac1k\dfrac1{l^2} = \sum_{l=1}^{\infty} \sum_{k=1}^l \dfrac1k\dfrac1{l^2}}_{\text{Change order of summation}} = \sum_{l=1}^{\infty} \dfrac{H_l}{l^2}$$
Also, note that
$$S = \sum_{k=1}^{\infty} \dfrac1k\dfrac1{k^2} + \sum_{k=1}^{\infty} \sum_{l=1}^{\infty} \dfrac1k\dfrac1{(k+l)^2} = \zeta(3) + \sum_{k=1}^{\infty} \sum_{l=1}^{\infty} \dfrac1k\dfrac1{(k+l)^2}$$
Now let us evaluate the second sum in the above line.(To evaluate this, we follow the technique @sos440 has in his answer.)
\begin{align*}
\sum_{k=1}^{\infty} \sum_{l=1}^{\infty} \frac{1}{k} \frac{1}{(k+l)^2}
&= \frac{1}{2}\sum_{k,l=1}^{\infty} \left\{ \frac{1}{k} \frac{1}{(k+l)^2} + \frac{1}{l} \frac{1}{(k+l)^2} \right\} \\
&= \frac{1}{2}\sum_{k,l=1}^{\infty} \frac{1}{kl(k+l)}
= \frac{1}{2}\sum_{k=1}^{\infty} \frac{1}{k^2}\sum_{l=1}^{\infty} \left\{ \frac{1}{l} - \frac{1}{k+l}\right\} \\
&= \frac{1}{2}\sum_{k=1}^{\infty} \frac{H_{k}}{k^2}
\end{align*}
Hence, we have $$S = \sum_{n=1}^{\infty} \frac{H_{n}}{n^2} = \zeta(3) + \frac{1}{2}\sum_{k=1}^{\infty} \frac{H_{k}}{k^2}$$
This gives us $$S = 2 \zeta(3)$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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|
What is the value of $N$? $N$ is the smallest positive integer such that for any integer $n > N$, the quantity $n^3 – 7n^2 + 11n – 5$ is positive. What is the value of $N$?
Note: $N$ is a single digit number.
|
$\text{We have}\quad f(n)=n^3-7n^2+11n-5\tag{1}.$
$\textrm{We want to solve for}\;\; n\;\textrm{ when}\;\;f(n) = n^3 - 7n^2 + 11n - 5 = 0\tag{2}$
Note that replacing $n$ with $1$ in equation $(2)$, and then summing, gives us:
$$1(1^3) - 7(1^2 + 11(1) - 5 = 1 - 7 + 11 - 5 = 0,$$ and so $n = 1$ is a solution to $(2)$, and so $(n - 1)$ is at least one factor of the polynomial $(1)$.
Dividing $\;f(n) = n^3-7n^2+11n-5\;$ by $\;(n - 1)$ gives us $f(n)/(n-1) = (n^2 - 6n + 5).$
We are then able to factor $(1)$:
$$f(n)=(n-1)(n^2-6n+5)=(n-1)[(n-1)(n-5)]=(n-1)^2(n-5).\tag{3}$$
Now we look at the roots of $f(n)$; that is, when is it true that $f(n) = (n-1)^2(n-5) = 0\;$?.
- We already know that when $n = 1$, $f(n) = 0$.
- It is also the case that when $n = 5$, $f(n) = 0$.
So what must the value of $n$ be if we must have $f(n) > 0$?.
*
*This is true when $n < 1$, but then, since $n$ is an integer, $n = 0$ is the greatest integer less than $1$, and we know that $n > N$, meaning $N \le -1$. But we need a positive $N$.
*When $1 < n < 5$, $f(n) < 0.$ No good.
*When $n > 5$, $f(n) > 0$. So we must have $n = 6$. This is the smallest positive integer for which $f(n)>0$.
Key Note: Now, knowing $n$ is NOT ENOUGH! This $n$ is NOT the answer to your question.
*
*We need to use $n$ to determine the integer $N$. We want the smallest positive integer $N$ such that for any $n > N$, $f(n)$ is positive.
*When $N = 5$, we have that for all $n > 5$, $f(n) > 0$: this is your answer to this question.
|
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|
A relationship between matrices, bernoulli polynomials, and binomial coefficients We define the following polynomials, for $n≥0$:
$$P_n(x)=(x+1)^{n+1}-x^{n+1}=\sum_{k=0}^{n}{\binom{n+1}{k}x^k}$$
For $n=0,1,2,3$ this gives us,
$$P_0(x)=1\enspace P_1(x)=2x+1\enspace P_2(x)=3x^2+3x+1\enspace P_3(x)=4x^3+6x^2+4x+1$$
We then define the set $P_{(3)}=\{P_0,P_1,P_2,P_3\}$. It can be easily shown that this set is a basis over the vector space of polynomials of degree $3$ and lower. We take $3$ for the sake of brevity.
Taking the coefficients of these polynomials and turning them into column vectors, we can construct the matrix (coefficients from the lowest term to the highest term)
$$\large{M_{P_{(3)}}}=\begin{pmatrix}
1 & 1 & 1 & 1 \\
0 & 2 & 3 & 4 \\
0 & 0 & 3 & 6 \\
0 & 0 & 0 & 4
\end{pmatrix}$$
We'll call this matrix the pascal in the context of this post, and the above polynomials as pascal polynomials.
The inverse of this matrix is the matrix,
$$M_{P_{(3)}}^{-1}=\begin{pmatrix}
1 & -\frac{1}{2} & \frac{1}{6} & 0 \\
0 & \frac{1}{2} & -\frac{1}{2} & \frac{1}{4} \\
0 & 0 & \frac{1}{3} & -\frac{1}{2} \\
0 & 0 & 0 & \frac{1}{4}
\end{pmatrix}$$
We'll factor this matrix into two matrices as follows:
$$M_{P_{(3)}}^{-1}=\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & \frac{1}{2} & 0 & 0 \\
0 & 0 & \frac{1}{3} & 0 \\
0 & 0 & 0 & \frac{1}{4}
\end{pmatrix}×\begin{pmatrix}
1 & -\frac{1}{2} & \frac{1}{6} & 0 \\
0 & 1 & -1 & \frac{1}{2} \\
0 & 0 & 1 & -\frac{3}{2} \\
0 & 0 & 0 & 1
\end{pmatrix}$$
We can see the Bernoulli numbers in the first row of the matrix. Every column is a coefficient vector of a Bernoulli polynomial
The following are extended versions of these matrices:
*
*Pascal Matrix
*Inverse Pascal Matrix
*Last, Bernoulli Matrix
Are there accepted names for these matrices and polynomials? What is the meaning of these relationships?
In particular, is there some treatment of using these matrices as change of basis transformations between representations of polynomials? E.g. from a linear combination of pascal polynomials to a linear combination of monomial terms.
|
Regarding the interpretation of the matrices:
Let
$$X =
\left(\begin{array}{c}
1 \\
x \\
\vdots \\
x^n
\end{array}\right)
\qquad \textrm{and}\qquad
A =
\left(\begin{array}{c}
a_0 \\
a_1 \\
\vdots \\
a_n
\end{array}\right)$$
so
$$X^T A = \sum_{k=0}^n a_k x^k.$$
(I've written $X^T A$ instead of the preferred $A^T X$ to agree with your notation and avoid things like $(M^{-1})^T$.)
Then
$$\begin{eqnarray*}
X^T A &=& X^T M M^{-1} A \\
&=& {X'}^T A'
\end{eqnarray*}$$
where
$X' = M^T X$
is the new basis and $A' = M^{-1} A$ are the coefficients in the new basis.
This interpretation holds for any invertible transformation $M$.
The formalism can be generalized in a straightforward way to transformations between any two bases of polynomials.
For your transformation we find the coefficients of
$\sum_{k=0}^3 a_k x^k$ in the "Pascal basis" are
$$M_{P_{(3)}}^{-1}A = \left(
\begin{array}{c}
a_0 - \frac{1}{2} a_1 + \frac{1}{6} a_2 \\
\frac{1}{2}a_1 - \frac{1}{2} a_2 + \frac{1}{4} a_3 \\
\frac{1}{3} a_2 - \frac{1}{2} a_3 \\
\frac{1}{4} a_3
\end{array}
\right).$$
Thus,
$$\begin{eqnarray*}
\sum_{k=0}^3 a_k x^k &=&
\left(a_0 - \frac{1}{2} a_1 + \frac{1}{6} a_2\right) P_0(x) \\
&&+ \left(\frac{1}{3} a_2 - \frac{1}{2} a_3\right) P_1(x) \\
&&+ \left(\frac{1}{3} a_2 - \frac{1}{2} a_3\right) P_2(x) \\
&&+ \frac{1}{4} a_3 P_3(x).
\end{eqnarray*}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
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|
How to recognize, where function has no derivative? For example this function: $f(x) = (x + 1)|x + 3| + 2$ it has no derivative in $x = -3$, but how can I discover it?
|
Recall that the $\vert y \vert$ is not differentiable at $y=0$. Hence, in your case, $f(x) = (x+1) \vert x + 3\vert + 2$ i.e.
$$f(x) = \begin{cases} (x+1)(x+3) + 2 & x \geq -3\\ -(x+1)(x+3) + 2 & x \leq -3 \end{cases}$$
Hence, the only point where derivative need not exist is at $x+3 = 0$ i.e. $x=-3$, since away from $x=-3$, the function is a nice quadratic function.
At $x=-3$, we have
\begin{align}
\lim_{h \to 0^+} \dfrac{f(-3+h) - f(-3)}h & = \lim_{h \to 0^+} \dfrac{((-3+h+1)(-3+h+3)+2) - ((-3+1)(-3+3)+2)}{h}\\
& = \lim_{h \to 0^+} \dfrac{h(h-2) +2 - 2}h = \lim_{h \to 0^+} \dfrac{h(h-2)}h = -2
\end{align}
\begin{align}
\lim_{h \to 0^-} \dfrac{f(-3+h) - f(-3)}h & = \lim_{h \to 0^-} \dfrac{(-(-3+h+1)(-3+h+3)+2) - (-(-3+1)(-3+3)+2)}{h}\\
& = \lim_{h \to 0^-} \dfrac{h(2-h)+2 - 2}h = \lim_{h \to 0^-} \dfrac{h(2-h)}h = 2
\end{align}
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Smallest possible value on Fibonacci Function Suppose $f$ is a polynomial with integer coefficients, such that for all non-negative integers $n$ the $n$-th Fibonacci number $u_n$ divides $f(u_{n+1})$. Find the smallest possible positive value of $f(4)$.
|
This is a really nice problem! The answer is $255$. First note that (by taking integer linear combinations of polynomials with the given property) the problem is equivalent to "find the GCD of all possible nonzero values of $f(4)$". Now observe that $g(x)=x^4-1$ is one such polynomial (left as an exercise! see hint below), and $g(4)=255$. So the answer is some divisor of $255 = 3 \times 5 \times 17$.
To conclude, we need to show that $f(4)$ must always be divisible by $3$, by $5$, and by $17$.
But $21 \mid f(34)$, so $f(34)$ is divisible by $3$; and $f(34) \equiv f(4) \pmod{3}$ (because $34-4$ is divisible by $3$), so $f(4)$ is also divisible by $3$.
Similarly $34 \mid f(55)$, so $f(55)$ is divisible by $17$; and $f(55) \equiv f(4) \pmod{17}$ (because $55-4$ is divisible by $17$), so $f(4)$ is also divisible by $17$.
Finally $55 \mid f(89)$, so $f(89)$ is divisible by $5$; and $f(89) \equiv f(4) \pmod{5}$ (because $89-4$ is divisible by $5$), so $f(4)$ is also divisible by $5$. Done.
|
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"timestamp": "2023-03-29T00:00:00",
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|
for a $3 \times 3$ matrix A ,value of $ A^{50} $ is I f
$$A= \begin{pmatrix}1& 0 & 0 \\
1 & 0 & 1\\
0 & 1 & 0 \end{pmatrix}$$
then $ A^{50} $ is
*
*$$ \begin{pmatrix}1& 0 & 0 \\
50 & 1 & 0\\
50 & 0 & 1 \end{pmatrix}$$
*$$\begin{pmatrix}1& 0 & 0 \\
48 & 1 & 0\\
48 & 0 & 1 \end{pmatrix}$$
*$$\begin{pmatrix}1& 0 & 0 \\
25 & 1 & 0\\
25 & 0 & 1 \end{pmatrix}$$
*$$\begin{pmatrix}1& 0 & 0 \\
24 & 1 & 0\\
24 & 0 & 1\end{pmatrix}$$
I am stuck on this problem. Can anyone help me please...............
|
Simplify your problem as follows. The Jordan Normal form of your matrix $A$ is
$$J = \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1& 1 \\ 0& 0& 1 \end{pmatrix},$$
where the matrix $P$ is such that $P^{-1}AP = J$ is given by
$$P = \begin{pmatrix} 0 & 0& 2 \\ -1 & 1 & 1 \\ 1 & 1 &0 \end{pmatrix}.$$
Then $A^{50} = PJ^{50}P^{-1}$. Now what is $J^{50}$?
Edit: Here's how you compute the Jordan Normal form of $A$ and the matrix $P$. First note that it has eigenvalues -1 and 1(with multiplicity 2). The Jordan normal form of $A$ is not diagonal because $A$ is not diagonalisable. It has eigenvectors $(0,1,-1)$ and $(0,1,1)$.
Now by my answer here we already know what the Jordan Normal Form of $A$ looks like, i.e. we know the matrix $J$. How do we obtain the matrix $P$? We now write $J$ like this:
$$J = \left(\begin{array}{c|cc} -1 & 0 & 0 \\ \hline 0 & 1& 1 \\ 0& 0& 1 \end{array}\right).$$
I have put in the grid to emphasize the fact that the first column of the matrix $P$ has to be the eigenvector $(0,-1,1)$ associated to the eigenvalue $-1$. The second column of $P$ looking at the second column of $J$ must now be the eigenvector associated to the eigenvalue $1$. What about the third column of $P$? This is of course given by a basis element for the kernel of $(A - I)^2$. We find that
$$(A - I)^2 = \begin{pmatrix} 0 & 0 & 0 \\ -1 & 2 & -2 \\ 1 & -2 & 2 \end{pmatrix} $$
which has a basis for its kernel the vector $(2,1,0)$ and so this completes the computation of $P$.
|
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|
correct combination of differentiation rules I am trying to calculate the derivative of a rather complex function for my homework. I think I have found the solution, it just seems too bulky for my taste. See the bottom for specific questions I have regarding my solution.
$ f(x)=\frac{\overbrace{\sin x}^\text{u(x)}\cdot \overbrace{e^x+x^3}^\text{v(x)}}{\underbrace{x^3+2x+2}_\text{w(x)}} $
$ u'(x)=\cos x $, $ v'(x)=e^x+3x^2 $, $ w'(x)=3x^2+4x $
$ \begin{align*}
u'v'(x)&=(\sin x\cdot e^x+3x^2)+(\cos x\cdot x^3)\\
&= e^x\cdot\sin x+3x\cdot\sin x+x^3\cdot\cos x
\end{align*} $
$ \begin{align*}
f'(x)&= \frac{u'v'(x)\cdot w(x)-w'(x)\cdot uv(x)}{w(x)^2}\\
&= \frac{(e^x\cdot\sin x+3x\cdot\sin x+x^3\cdot\cos x)\cdot (x^3+2x^2+2)-(3x^2+4x)(\sin x\cdot e^x+x^3)}{(x^3+2x^2+2)^2}
\end{align*} $
The obvious question: Is this derivative correct?
Especially:
*
*Is it really possible to calculate the derivative by splitting the function into part-functions and calculating them together following the differentiation rules, the way I did it?
*Is it possible to reduce the summands by applying some rule I am not aware of? E.g. reducing $(e^x\cdot\sin x+3x\cdot\sin x+x^3\cdot\cos x)$ to something with just one $\sin$ or something?
|
Your solution is pretty close, but remember that you're not multiplying $\sin x$ by $(e^x + x^3)$ in your numerator, as you do when you try to multiply $u'(x)$ by $v'(x).$ Instead, try differentiating your numerator in the following form: $e^x\sin x + x^3.$ I believe the rest will follow from the quotient rule.
$n(x) = e^x\sin x$
$d(x) = x^3 + 2x + 2$
$n'(x) = e^x \cos x + e^x\sin x$
$d'(x) = 3x^2 + 2$
$\frac{d}{dx} \frac{n(x)}{d(x)} = \frac{n'(x)d(x) - d'(x)n(x)}{d(x)^2}$
$ = \frac{(e^x\cos x + e^x\sin x)(x^3 + 2x + 2) - (3x^2+2)(e^x\sin x)}{(x^3 + 2x + 2)^2}$
which is messy but can be simplified.
|
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|
$(\tan^2(18^\circ))(\tan^2(54^\circ))$ is a rational number Assuming $$\cos(36^\circ)=\frac{1}{4}+\frac{1}{4}\sqrt{5}$$ How to prove that $$\tan^2(18^\circ)\tan^2(54^\circ)$$ is a rational number? Thanks!
|
$$\tan18^\circ\tan54^\circ=\frac{2\sin18^\circ\sin54^\circ}{2\cos18^\circ\cos54^\circ}$$
$$=\frac{\cos36^\circ-\cos72^\circ}{\cos36^\circ+\cos72^\circ}$$ (applying $\cos(A\pm B)$ formulae)
$$=\frac{\frac{\sqrt5+1}4-\frac{\sqrt5+1}4}{\frac{\sqrt5+1}4+\frac{\sqrt5+1}4}$$ as $$\cos 72^\circ=2\cos^236^\circ-1=\frac{\sqrt5-1}4$$
$$\implies \tan18^\circ\tan54^\circ=\sqrt5$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/275151",
"timestamp": "2023-03-29T00:00:00",
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|
Proving $f(x+y)=f(x)f(y)$ for exponential function $f$ I'm sorry if the title was a bit misleading but I don't know how else to phrase it.
I'm having trouble understanding some lecture notes and I couldn't find extra information by Googling.
Anyways, the exponential function is represented as
$$\lim_{n}\left(1+\frac{x}{n}\right)^n$$
Then, this inequality follows:
$$\left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right)=\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)\ge 1+\frac{x+y}{n}$$
Ok, following so far.
Then there's this thing:
$$\begin{align*}
\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n-\left(1+\frac{x+y}{n}\right)^n&=\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^n-\left(1+\frac{x+y}{n}\right)^n\\
&=\frac{xy}{n^2}\left[\sum\limits_{k=0}^{n-1}\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k\right]
\end{align*}$$
That last one amost broke me. After figuring out it's just the formula for
$$a^n-b^n$$
I finally get to the part I just don't understand.
It says "Because of the last inequality follows:"
$$\begin{align*}
\frac{xy}{n}\left(1+\frac{x+y}{n}\right)^{n-1}&\le\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n-\left(1+\frac{x+y}{n}\right)^n\\
&\le\frac{xy}{n}\left(1+\frac{x}{n}\right)^{n-1}\left(1+\frac{y}{n}\right)^{n-1}
\end{align*}$$
I'm supposed to study for an exam on Monday and I still have a ton of stuff to go through but I've been staring at the page for the last 2 hours. I don't know if this is supposed to be one of the harder proofs or maybe I'm just missing something (in my brain).
Help is much appreciated.
P.S Sorry if my Latex is bad, this is the second time I'm using it.
|
We have
$$\begin{align*}
&\frac{xy}{n^2}\left[\sum\limits_{k=0}^{n-1}\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k\right]\\
&\qquad=\frac{xy}n\cdot\frac1n\sum_{k=0}^{n-1}\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k\;.
\end{align*}$$
Your first inequality indicates that you’re assuming that $x,y\ge 0$, so
$$\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k\ge\left(1+\frac{x+y}n\right)^{n-1}$$
for $k=0,\dots,n-1$. There are $n$ terms, so
$$\frac1n\sum_{k=0}^{n-1}\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k\ge\left(1+\frac{x+y}n\right)^{n-1}\;,$$
and hence
$$\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n-\left(1+\frac{x+y}{n}\right)^n\ge\frac{xy}n\left(1+\frac{x+y}n\right)^{n-1}\;.$$
The rest comes from your first inequality.
|
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|
$\displaystyle\sum_{n=2}^{\infty}\frac{2}{(n^3-n)3^n} = -\frac{1}{2}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot 3^n}$ Please help me, to prove that
$$
\sum_{n=2}^{\infty}\frac{2}{(n^3-n)3^n} =
-\frac{1}{2}+\frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot 3^n}.
$$
|
\begin{align*}
\sum_{n=2}^\infty \frac{2}{3^n( n^3 - n) } &= \sum_{n=2}^\infty \frac{1}{3^n}\left( \frac{-2}{n} + \frac{1}{n+1} + \frac{1}{n-1} \right) \\
&= \sum_{n=1}^\infty \frac{1}{3^{n+1}n} - \sum_{n=1}^\infty \frac{2}{3^{n+1} (n+1)}+\sum_{n=1}^\infty \frac{1}{(n+2)3^{n+1}} \\
&= \frac{1}{3}\sum_{n=1}^\infty\frac{1}{n 3^n} - 2 \left ( \sum_{n=1}^\infty \frac{1}{n 3^n} - \frac{1}{3} \right ) + 3 \left( \sum_{n=1}^\infty \frac{1}{n 3^n} - \frac{1}{3} -\frac{1}{3^2 \cdot 2}\right ) \\
&= \sum_{n=1}^{\infty}\frac{1}{n 3^n}\left( \frac{1}{3} - 2 + 3 \right ) + \left( \frac{2}{3} - 1 - \frac{1}{6} \right ) \\
&= \frac{4}{3}\sum_{n=1}^{\infty}\frac{1}{n\cdot 3^n} -\frac{1}{2}
\end{align*}
|
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|
Cardano's Formulas help I am working on solving this cubic: $x^3 +x^2 - 2 = 0$ using Cardano's explicit formulas:
$$
A = \sqrt[3]{{-27\over 2}q + {3 \over 2} \sqrt{-3D}} \qquad B = \sqrt[3]{{-27\over 2}q - {3 \over 2} \sqrt{-3D}}
$$
where for a negative discriminant, I will get a real root $\alpha = {A+ B \over 3}$ for $p,q,D$ defined as follows:
Start with an equation of the form $x^3 + ax^2 + bx + c = 0$, transform using $y = x-a/3$ to get an equation of the form $y^3 + py + q = 0$, and the discriminant of the cubic: $D = -4p^3 - 3q^2$.
My Work
I transformed the above cubic to $y^3 - {1\over 3}y - {52\over 27}$ using the substitution $y = x-1/3$. I then got a discriminant of $D = -{100}$. However, when I plug this value of $D$ into my formulas for $A$ and $B$, I get $A = \sqrt[3]{26 + 15\sqrt 3}, B = \sqrt[3]{26 - 15\sqrt 3}$. And I know that one of the roots is given by $\alpha = {A+ B \over 3}$
An additional hint is given:
$(2 + \sqrt 3)^3 = 26 + 15\sqrt 3$
So my formula for $\alpha = {A+ B \over 3} = {4\over 3}$. But I am supposed to check that this equation has the real root
${1\over 3} (\sqrt[3]{26 + 15\sqrt 3} + \sqrt[3]{26 - 15\sqrt 3} \mathbf{ - 1})$
That would make sense because then it would give me a real root of $1$ which can be checked directly to be a root of this cubic. I suppose my question is: where does the $1$ come from?
|
It should be $y=x+\frac{1}{3}$. Then after we have found a solution $y_0$ of the cubic in $y$, the corresponding $x_0$ is $y_0-\frac{1}{3}$. That's where the $-1$ comes from.
|
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|
Prove the following identity I am having some trouble proving following identity without use of induction, with which it is trivial.
$$\sum_{n=1}^{m}\frac{1}{n(n+1)(n+2)}=\frac{1}{4}-\frac{1}{2(m+1)(m+2)}$$
I did expand the expression:
$$\sum_{n=1}^{m}\left( \frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)} \right)$$
I have no idea how to proceed further.
|
Hint: by partial fraction
$$\frac{1}{n(n+1)(n+2)}=\frac{A}{n}+\frac{B}{n+1}+\frac{C}{n+2}\Longrightarrow$$
$$1=A(n+1)(n+2)+Bn(n+2)+Cn(n+1)$$
Choosing smartly values for $\,n\,$ above , we get
$$1=2A\Longrightarrow A=\frac{1}{2}\;\;,\;\;1=-B\Longrightarrow B=-1\;\;,\;\;1=2C\Longrightarrow C=\frac{1}{2}$$
so
$$\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2}\right)$$
So summing up
$$\sum_{n=1}^m\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\sum_{n=1}^m\left(\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2}\right)=$$
$$=\frac{1}{2}\left[\sum_{n=1}^m\left(\frac{1}{n}-\frac{1}{n+1}\right)-\left(\frac{1}{n+1}-\frac{1}{n+2}\right)\right]$$
Now just check that there's a lot of cancellation up there...
|
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|
Let $a_{i} \in\mathbb{R}$ ($i=1,2,\dots,n$), and $f(x)=\sum_{i=0}^{n}a_{i}x^i$ such that if $|x|\leqslant 1$, then $|f(x)|\leqslant 1$. Prove that: Let $a_{i} \in\mathbb{R}$ ($i=1,2,\dots,n$), and $f(x)=\sum_{i=0}^{n}a_{i}x^i$ such that if $|x|\leqslant 1$, then $|f(x)|\leqslant 1$. Prove that:
*
*$|a_{n}|+|a_{n-1} | \leqslant 2^{n-1}$.
*$|a_{n}|+|a_{n-1}|+\cdots+|a_{1}|+|a_{0}|\leqslant\dfrac{(1+\sqrt{2})^n+(1-\sqrt{2})^n}{2}$.
|
Part (a)
By replacing $f(x)$ with $\pm f(\pm x)$ if necessary, we may assume that $a_n, a_{n-1}$ are both non-negative. Let $T_n(x)$ be the Chebyshev polynomial of the first kind, of degree $n$, i.e. it satisfies
$$T_n(\cos \theta) = \cos (n\theta)$$
Notice that $T_n(x)$ is a polynomial of degree $n$ with leading coefficient $2^{n-1}$, and that $T_n(x) = \pm 1$ for $x = \cos \frac{0\pi}{n}, \cos \frac{\pi}{n}, \cdots, \cos \frac{n\pi}{n}$.
We first show that $a_n \leq 2^{n-1}$. Suppose the contrary, i.e. $a_n > 2^{n-1}$. Then consider
$$g(x) = \frac{a_n}{2^{n-1}}T_n(x) - f(x)$$
would be positive at $\cos \frac{0\pi}{n}$, negative at $\cos\frac{\pi}{n}$, positive at $\cos\frac{2\pi}{n}$ and so forth. (Here we used $|f(x)| \leq 1$ whenever $|x| \leq 1$) Intermediate value theorem says that $g(x)$ would have at least $n$ distinct roots, but $g(x)$ has degree less than $n$. Thus $g(x) \equiv 0$, i.e. $f(x) = \frac{a_n}{2^{n-1}}T_n(x)$. But then
$$f(1) = \frac{a_n}{2^{n-1}} T_n(\cos 0) = \frac{a_n}{2^{n-1}} > 1$$
Contradicting the assumption of $f$.
We then show that $a_n + a_{n-1} \leq 2^{n-1}$. Suppose the contrary, i.e.
$$a_{n-1} > 2^{n-1} - a_n \hspace{5mm} (*)$$
Consider
$$g(x) = (1+\epsilon) T_n(x) - f(x)$$
where $\epsilon > 0$ is sufficiently small and will be picked later.
The leading terms of $g(x)$ are
$$g(x) = ((1+\epsilon)2^{n-1} - a_n)x^n - a_{n-1} x^{n-1} + \cdots$$
This implies that the sum of roots of $g(x)$ are $\frac{a_{n-1}}{(1+\epsilon)2^{n-1} - a_n}$. By picking a small enough $\epsilon$ and (*), the sum of roots of $g(x)$ is at least 1.
The condition on $f(x)$ (i.e. $|f(x)| \leq 1$ whenever $|x| \leq 1$) would again imply that $g(x)$ is positive at $\cos \frac{0\pi}{n}$, negative at $\cos\frac{\pi}{n}$, positive at $\cos\frac{2\pi}{n}$ and so forth. Intermediate value theorem says that $g(x)$ would have at least $n$ roots, one lying between $\cos \frac{k\pi}{n}$ and $\cos \frac{(k-1)\pi}{n}$ for $k = 1, \cdots, n$. But $g(x)$ has degree $n$, so these are exactly the roots of $g(x)$. But then the sum of roots is strictly smaller than
$$\cos \frac{0 \pi}{n} + \cos \frac{\pi}{n} \cdots + \cos \frac{(n-1) \pi}{n} = 1$$
Contradiction. This shows part (a).
|
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|
Probability of number of events where $p_{n+1} = \frac{1}{5}p_n$ For $ n \ge 0, p_{n+1} = \frac{1}{5}p_n $. Where $p_n$ is the probability of an envent occuring n times in a period.
What is the probability that more than one event occurs?
|
$1= \sum\limits_{n=0}^{\infty}(\frac{1}{5})^{n}p_0 $
$(\frac{1}{5})(1) = (\frac{1}{5})\sum\limits_{n=0}^{\infty}(\frac{1}{5})^{n}p_0 $
$1 - (\frac{1}{5})(1) = p_0 (\frac{1}{5})^{0}$
$ \frac{4}{5} = p_0$
using the formula in the question,
$p_1 = \frac{1}{5}(\frac{4}{5})$
$p_1 = \frac{4}{25} $
$P(n\ge 2) = 1 - P(n=0) - P(n=1)$
$P(n\ge 2) = 1 - \frac{4}{5} - \frac{4}{25}$
$P(n\ge 2) = 0.04$
|
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|
Find $\lim\limits_{n \to \infty}\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)$
Find the limit $$\lim_{n \to \infty}\left[\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\cdots\left(1-\frac{1}{n^2}\right)\right]$$
I take log and get $$\lim_{n \to \infty}\sum_{k=2}^{n} \log\left(1-\frac{1}{k^2}\right)$$
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The $r$th term is $\frac{r^2-1}{r^2}=\frac{(r-1)}r\frac{(r+1)}r$
So, the product of $n$ terms is $$\frac{3.1}{2^2}\frac{4.2}{3^2}\frac{5.3}{4^2}\cdots \frac{(n-1)(n-3)}{(n-2)^2}\frac{(n-2)n}{(n-1)^2}\frac{(n-1)(n+1)}{n^2}$$
$$=\frac12\frac32\frac23\frac43\cdots\frac{n-2}{n-1}\frac n{n-1}\frac{n-1}n\frac{n+1}n=\frac12\frac{(n+1)}n$$
as the 1st half of any term is cancelled by the last half of the previous term except for the 1st & the last term.
|
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|
Equations over permutations Let $\sigma=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 1 & 4 & 2 \end{pmatrix} \in S_4$ and $\theta=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 2 & 1 & 4 & 3 \end{pmatrix} \in S_4$.
Solve the following equations $(x \in S_4)$:
a) $x \sigma = \sigma x$;
b) $x^2 = \sigma$;
c) $x^2 = \theta$.
I'm writing here my work. I want to know if something goes wrong in my proof.
We denote the number of inversions in $σ$ with $N(\sigma)$ and the sign of permutation $\sigma$ with $\text{sgn}(\sigma)$.
b) We have
$\text{sgn}(\sigma)=(-1)^{N(\sigma)}=(-1)^3=-1$, and $\text{sgn}(x^2)=\text{sgn}(x) \cdot \text{sgn}(x)=(-1)^{2 \cdot N(x)}=1$.
Relation $\text{sgn}(\sigma) \ne \text{sgn}(x^2)$ imply that the equation $x^2 = \sigma$ has no solutions in $S_4$.
Ideas for the rest of the exercise, please?
|
c) Let $x=\begin{pmatrix} 1 & 2 & 3 & 4 \\ a & b & c & d \end{pmatrix} \in S_4$.
Case 1. If $a=1$, then $x(1)=1$, which implies that $x(x(1))=x(1)=1$. On the other hand, $\theta(1)=2$.
We obtained that $x^2(1) \ne \theta(1)$. Therefore, in this case, the equation $x^2=\theta$ has no solutions.
Case 2. If $a=2$, then $x(1)=2$, which implies that $x(x(1))=x(2)=b$. On the other hand, $\theta(2)=1$. Hence, $b=1$.
Now, $x(2)=1$, which implies that $x(x(2))=x(1)=a$. On the other hand, $\theta(2)=1$. Hence $a=1$, contradiction with $a=2$.
Case 3. If $a=3$, then $x(1)=3$, which implies that $x(x(1))=x(3)=c$. On the other hand, $\theta(1)=2$. Hence $c=2$.
Now, $x(3)=2$, which implies that $x(x(3))=x(2)=b$. On the other hand, $\theta(3)=4$. Hence $b=4$.
Now, $x(2)=4$, which implies that $x(x(2))=x(4)=d$. On the other hand, $\theta(4)=3$. Hence $d=3$.
In this case we obtain the solution $x=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix}$.
Case 4. If $a=4$, then $x(1)=4$, which implies that $x(x(1))=x(4)=d$. On the other hand, $\theta(1)=2$. Hence $d=2$.
Now, $x(4)=2$, which implies that $x(x(4))=x(2)=b$. On the other hand, $\theta(4)=3$. Hence $b=3$.
Now, $x(2)=3$, which implies that $x(x(2))=x(3)=c$. On the other hand, $\theta(2)=1$. Hence $c=1$.
In this case we obtain the solution $x=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 3 & 1 & 2 \end{pmatrix}$.
Conclusion: The equation $x^2=\theta$ has two solutions, $x=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 3 & 4 & 2 & 1 \end{pmatrix}$, and $x=\begin{pmatrix} 1 & 2 & 3 & 4 \\ 4 & 3 & 1 & 2 \end{pmatrix}$.
|
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How to show $AB^{-1}A=A$
Let $$A^{n \times n}=\begin{pmatrix} a & b &b & \dots & b \\ b & a &b & \dots & b \\ b & b & a & \dots & b \\ \vdots & \vdots & \vdots & & \vdots \\ b & b & b & \dots &a\end{pmatrix}$$
where $a \neq b$ and $a + (n - 1)b = 0$.
Suppose $B=A+\frac{11'}{n}$ , where $1=(1,1,\dots,1)'$ is an $n \times 1$ vector. Show that $AB^{-1}A=A$
Trial: Here $B^{-1}=\frac{1}{\alpha-\beta}I_n-\frac{\beta~11'}{(\alpha-\beta)(\alpha+(n-1)\beta)}$ Where $(\alpha-\beta)=(a+\frac{1}{n}-b-\frac{1}{n})=(a-b)$ and $\alpha+(n-1)\beta=1$ So, $B^{-1}=\frac{1}{a-b}[I_n-(b+\frac{1}{n})11']$. Then pre and post multipling $A$ I can't reach to the desire result . Please help.
|
Note that
$$A = (a-b)I + b e e^T$$
where $I$ is the identity matrix, $e = \begin{bmatrix} 1&1&1& \cdots &1 \end{bmatrix}^T$. We have
$$B = A+ \dfrac{ee^T}n = (a-b)I + \left(b + \dfrac1n\right)ee^T$$
Now make use of Sherman-Morrison Woodbury formula to compute $AB^{-1}A$ and conclude what you want.
I have added the details below. Hoever you mouse over the gray area below for more details.
Using Sherman-Morrison Woodbury formula, we get that$$B^{-1} = \dfrac{I}{a-b} - \dfrac{ee^T}{(a-b)^2 + n(a-b)}$$ Hence, $$AB^{-1} = I + \dfrac{b}{a-b} ee^T - \dfrac{(a-b)ee^T}{(a-b)^2+n(a-b)} - \dfrac{bn ee^T}{(a-b)^2+n(a-b)}\\ = I + \left(\dfrac{b}{a-b} - \dfrac{a}{(a-b)^2+n(a-b)}\right)ee^T\\ = I + \dfrac{ab-b^2+nb-a}{(a-b)^2+n(a-b)}ee^T$$$$AB^{-1}A = \left(I + \dfrac{ab-b^2+nb-a}{(a-b)^2+n(a-b)}ee^T \right) \left( (a-b)I + b e e^T\right)\\=(a-b)I + \left(\dfrac{ab-b^2+nb-a}{(a-b)^2+n(a-b)}(a-b) + b + \color{blue}{\underbrace{\dfrac{nb(ab-b^2+nb-a)}{(a-b)^2+n(a-b)}}_{\text{This term makes use of the fact that $e^Te = n$}}}\right) ee^T\\=(a-b)I + \left(\dfrac{ab-b^2+nb-a}{(a-b)^2+n(a-b)} \color{red}{\underbrace{(a-b + nb)}_{\text{This term is zero}}} + b\right) ee^T\\= (a-b)I + bee^T = A$$
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|
Show $\sum_{i=1}^N {\frac{i}{p^i}}\le \frac{p}{(p-1)^2}$ Let $p$ be a prime number. How can I show that, for any positive integer $N$,
$$\sum_{i=1}^N {\frac{i}{p^i}}\le \frac{p}{(p-1)^2}?$$
I can see that
$$\sum_{i=1}^N {\frac{1}{p^i}}\lt \sum_{i=1}^\infty {\frac{1}{p^i}} = \frac{1}{p-1}$$
by the infinite sum of a geometric series, but not sure if this is useful.
Thank you.
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Note that if $|x| <1$ $$\begin{align}\sum_{i=0}^{\infty}i\cdot x^i &=x+2x^2+3x^3+4x^4+\cdots \\ &=(x+x^2+x^3+x^4+\cdots)+(x^2+x^3+x^4+\cdots)+\cdots \\ &=\frac{x}{1-x}+\frac{x^2}{1-x}+\cdots \\&=\frac{1}{1-x}(x+x^2+x^3+\cdots)\\&=\frac{1}{1-x}\frac{x}{1-x}\\&=\frac{x^2}{1-x}\end{align}$$
Here $x=\frac{1}{p-1}<1$ So $\sum_{i=1}^N {\frac{i}{p^i}} \leq\sum_{i=1}^{\infty} {\frac{i}{p^i}}=\frac{p}{(p-1)^2}$
|
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Find five positive integers whose reciprocals sum to $1$ Find a positive integer solution $(x,y,z,a,b)$ for which
$$\frac{1}{x}+ \frac{1}{y} + \frac{1}{z} + \frac{1}{a} + \frac{1}{b} = 1\;.$$
Is your answer the only solution? If so, show why.
I was surprised that a teacher would assign this kind of problem to a 5th grade child. (I'm a college student tutor) This girl goes to a private school in a wealthy neighborhood.
Please avoid the trivial $x=y=z=a=b=5$. Try looking for a solution where $ x \neq y \neq z \neq a \neq b$ or if not, look for one where one variable equals to another, but explain your reasoning. The girl was covering "unit fractions" in her class.
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I am posting another solution since one of my students found this method.
(He is 16 years old but as you see he is gifted)
Start with $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}$. No write $\frac{1}{6}=\frac{3}{18}=\frac{1}{18}+\frac{2}{18}=\frac{1}{9}+\frac{1}{18}$ and you get $\frac{1}{2}+\frac{1}{3}+\frac{1}{9}+\frac{1}{18}=1$. Proceed like this: Write the last fraction (which will have an even denominator) of the sum as $\frac{1}{2n}=\frac{1}{6n}+\frac{2}{6n}=\frac{1}{3n}+\frac{1}{6n}$ and so on!
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|
Inequality. $\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \geq \dfrac{3}{2}. $ Let $x,y,z$ be real positive numbers such that $x^2+y^2+z^2=3$. Prove that :
$$\frac{x^3}{y^2+z^2}+\frac{y^3}{z^2+x^2}+\frac{z^3}{x^2+y^2} \geq \dfrac{3}{2}. $$
I try to write this expression as:
$$\frac{x^4}{x(y^2+z^2)}+\frac{y^4}{y(z^2+x^2)}+\frac{z^4}{z(x^2+y^2)}$$ and then I try to apply Cauchy-Buniakowsky but still nothing.
I need a proof/idea without derivatives.
thanks for your help.
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We need to prove that
$$\sum_{cyc}\frac{x^3}{3-x^2}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\left(\frac{x^3}{3-x^2}-\frac{1}{2}\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)(2x^2+3x+3)}{3-x^2}\geq0$$ or
$$\sum_{cyc}\left(\frac{(x-1)(2x^2+3x+3)}{3-x^2}-2(x^2-1)\right)\geq0$$ or
$$\sum_{cyc}\frac{(x-1)^2(2x^2+6x+3)}{3-x^2}\geq0.$$
Done!
|
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Compute the limit $\displaystyle \lim_{x\rightarrow 0}\frac{n!.x^n-\sin (x).\sin (2x).\sin (3x).......\sin (nx)}{x^{n+2}}\;\;,$ How can i calculate the Given limit
$\displaystyle \lim_{x\rightarrow 0}\frac{n!x^n-\sin (x)\sin (2x)\sin (3x)\dots\sin (nx)}{x^{n+2}}\;\;,$ where $n\in\mathbb{N}$
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We know for small $y,$ $$\sin y=y-\frac{y^3}{3!}+\frac{y^5}{5!}-\cdots=y\left(1-\frac{y^2}{3!}+\frac{y^4}{5!}-\cdots\right)$$
So, $$\prod_{1\le r\le n}\sin rx=\prod_{1\le r\le n}rx\left(1-\frac{(rx)^2}{3!}+\frac{(rx)^4}{5!}+\cdots\right)=n!x^n\prod_{1\le r\le n}\left(1-\frac{(rx)^2}{3!}+\frac{(rx)^4}{5!}+\cdots\right)$$
$$=n!x^n\prod_{1\le r\le n}\left( 1-\frac1{3!} r^2x^2 +O(x^4) \right)$$
So, $$n!x^n-\prod_{1\le r\le n}\sin rx=n!x^n \left(\frac{x^2}{3!}(1^2+2^2+\cdots+n^2) +O(x^4)\right)$$
So, $$\lim_{x\to0}\frac{n!x^n-\prod_{1\le r\le n}\sin rx}{x^{n+2}}=n!\frac{1^2+2^2+\cdots+n^2}{3!}=n!\frac{n(n+1)(2n+1)}{36}$$
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Integrate: $\int x(\arctan x)^{2}dx$ I'm not sure how to start I think we have to use integration by parts
$$\int x(\arctan x)^{2}dx$$
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Using integration by parts, we get
$$\begin{align} \int dx \: x (\arctan{x} )^2 &= \frac{1}{2} x^2 (\arctan{x} )^2 - \int dx \: \frac{x^2}{1+x^2} \arctan{x} \\ &= \frac{1}{2} x^2 (\arctan{x} )^2 - \int dx \: \arctan{x} + \int dx \: \frac{\arctan{x}}{1+x^2} \\ &= \frac{1}{2} x^2 (\arctan{x} )^2 - x \arctan{x} + \int dx \: \frac{x}{1+x^2} + \frac{1}{2} (\arctan{x})^2 \\ &= \frac{1}{2} [(1+x^2) (\arctan{x})^2 + \log{(1+x^2)}] - x \arctan{x} + C \end{align}$$
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How can I evaluate $\lim_{x\to1}\frac{\sqrt{5-x}-2}{\sqrt{2-x}-1}$ without invoking l'Hôpital's rule? In the math clinic I work at, somebody in a Calculus 1 class asked for help with this limit problem. They have not covered basic differentiation techniques yet, let alone l'Hôpital's rule.
$$\lim_{x\to1}\frac{\sqrt{5-x}-2}{\sqrt{2-x}-1}$$
We have tried various algebraic techniques, such as multiplying the top and bottom of the fraction by the conjugate of the denominator, but haven't had any success at getting rid of the indeterminate form.
How can this limit be solved, using only techniques that would be available to a beginning Calculus 1 student?
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$$\begin{align}
& \hphantom{=} \lim_{x\to1}\frac{\sqrt{5-x}-2}{\sqrt{2-x}-1} \\
& = \lim_{x\to 1} \frac{\sqrt{5-x}-2}{\sqrt{2-x}-1}\frac{\sqrt{2-x}+1}{\sqrt{2-x}+1} \\
& = \lim_{x\to 1} \frac{(\sqrt{5-x}-2)(\sqrt{2-x}+1)}{1-x} \\
& = \lim_{x\to 1} \frac{(\sqrt{5-x}-2)(\sqrt{2-x}+1)}{1-x}\frac{\sqrt{5-x}+2}{\sqrt{5-x}+2} \\
& = \lim_{x\to 1} \frac{(1-x)(\sqrt{2-x}+1)}{1-x}\frac{1}{\sqrt{5-x}+2} \\
& = \lim_{x\to 1} \frac{\sqrt{2-x}+1}{\sqrt{5-x}+2} \\
& = \frac{1}{2}
\end{align}$$
When in doubt, multiply by the "conjugate".
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/291818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
}
|
Missing and parasite roots in the trigonometric equation. I have this equation:
$$\boxed{\cos(2x) - \cos(8x) + \cos(6x) = 1}$$
RIGHT
And its right solution from the textbook is:
$$
\begin{align}
\cos(2x)+\cos(6x)&=1+\cos(8x)\\\\
2\cos(4x)*\cos(2x)&=2\cos^2(4x)\\\\
\cos(4x)*(\cos(2x)-\cos(4x))&=0\\\\
\cos(4x)*2\sin(3x)*\sin(x)=&0
\end{align}
$$
*
*$\cos(4x) = 0 \implies 4x = \frac{\pi}{2} + \pi k \implies \boxed{x=\frac{\pi}{8} + \frac{\pi k}{4}}$
*$\sin(3x) = 0 \implies 3x = \pi k \implies \boxed{x=\frac{\pi k}{3}}$
*$\sin(x) = 0 \implies \boxed{x=\pi k}$
WRONG
And this is the way how I have been tried to solve it (with different approach) and here I need a help to determine where I'm wrong:
$$
\begin{align}
\cos(2x) - \cos(8x) + \cos(6x) &= 1\\
\cos(2x) + \cos(6x) &= 1 + \cos(8x)
\end{align}
$$
Then using this formula:
$$\bbox[Beige]{\boxed{\cos\alpha + \cos\beta = 2\cos(\frac{\alpha + \beta}{2}) \cos(\frac{\alpha - \beta}{2})}} \tag{1}$$
$$2\cos(4x)\cos(2x) = 1 + \cos(8x)$$
And then similarly:
$$\bbox[Beige]{\boxed{\cos2\alpha=2\cos^2\alpha-1}} \tag{2}$$
$$2\cos(4x)\cos(2x) = 1 + 2\cos^2(4x) - 1$$
After that I subtracted $1$ from $1$ in the right part and divided both right and left parts by $2$:
$$\cos(4x)\cos(2x) = \cos^2(4x)$$
Then once again I divided both parts. This time by $\cos(4x)$ and with the aim not to lose a root I checked whether $\cos(4x)$ can be equal to zero.
$$\cos(4x) = 0 \implies 4x = \frac{\pi}{2} + \pi k \implies \boxed{x=\frac{\pi}{8} + \frac{\pi k}{4}}$$
so here I found the first root. And using formula (2):
$$\cos(2x) = \cos(4x) \implies \cos(2x) = 2\cos^2(2x) - 1$$
then I simply introduced new variable and solved quadratic equation:
$$u = \cos(2x)$$
$$
2u^2 - u -1 = 0\\
u_1=1, \, u_2=-\frac{1}{2}
$$
So finally:
*
*$\cos(2x) = 1 \implies 2x = \pi k \implies \boxed{x=\frac{\pi k}{2}}$ (wrong)
*$\cos(2x) = -\frac{1}{2} \implies 2x = \pm \frac{2 \pi}{3} + 2\pi k \implies \boxed{x=\frac{\pi}{3} + \pi k}$ (wrong)
But none of the two last roots fits to the original equation. And I lost another 2 roots. Where is my mistake? Any suggestions will be greatly appreciated. Thanks for attention.
|
If $\cos 2x=1\implies 2x=2n_1\pi,x=n_1\pi$
If $\cos2x=-\frac12=\cos \frac{2\pi}3\implies 2x=2n_2\pi\pm \frac{2\pi}3= \frac{2\pi}3(3n_2\pm1)\implies x=\frac\pi3(3n\pm1)$
If we compare these $x=n_1\pi,\frac\pi3(3n_2\pm1)$ with $x=k_1\pi,k_2\frac\pi3,$
the apparent mismatch is when $3\mid k_2$
But, in that case the result $k_2\frac\pi3 \in x=n_1\pi$ where $n_1,n_2,k_1,k_2$ are integers.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/292559",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
}
|
Evaluate $P[X^2> Y^2]$
Let $X$ and $Y$ have the joint probability density function
$$f(x,y)= \begin{cases}cxye^{-(x^2+2y^2)}, & x>0,y>0 \\ 0, &\text{otherwise} \end{cases}$$
Evaluate $c$ and $P[X^2> Y^2]$.
Trial: We know that
\begin{align}
c\int_{0}^{\infty}\int_{0}^{\infty}xye^{-(x^2+2y^2)}\,dx\,dy&=1\\ \implies c(\int_{0}^{\infty}xe^{-x^2}\,dx)(\int_{0}^{\infty}ye^{-2y^2})\,dy&=1 \\ \implies c &=8
\end{align}
Now $$P[X^2>Y^2]= \iint_{x^2>y^2} 8xye^{-(x^2+2y^2)}\,dx\,dy$$
Then I try to solve it by polar transformation. But I can not find limits of $r$ and $\theta$.Mainly I want to evaluate $P[X^2>Y^2]$. Please help.
|
The transformation you want is $x=r \cos{\theta}$, $y= \frac{r}{\sqrt{2}} \sin{\theta}$, and $dx \, dy = \frac{r}{\sqrt{2}} dr \, d \theta$. As far as the integration limits go, $x^2 > y^2$ implies $0 < \tan{\theta} < \sqrt{2}$, while $r \in [0,\infty)$.
The integral then becomes
$$\frac{8}{2} \underbrace{\int_0^{\infty} dr \: r^3 e^{-r^2}}_{1/2} \underbrace{\int_0^{\arctan{\sqrt{2}}} d \theta \: \sin{\theta} \cos{\theta}}_{(1/2) \sin^2{\arctan{\sqrt{2}}}=(1/2)(2/3) = 1/3} = \frac{2}{3} $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/295209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Closed formula for linear binomial identity I have the following identity:
\begin{equation}
m^4 = Z{m\choose 4}+Y{m\choose 3}+X{m\choose 2}+W{m\choose 1}
\end{equation}
I solved for the values and learned of the interpretation of W, X, Y, and Z in my last post:
Combinatorial reasoning for linear binomial identity
Now, I am interested in using the above to find a closed form solution for:
\begin{equation}
\sum\limits_{k=1}^nk^4 \\
\end{equation}
I do not see how to get a closed form solution to the above sum. What steps can I take? I especially would like to use the work from my last post.
Thanks!
|
If $r>1$:
$\sum_{k=1}^{n}\left(\begin{array}{c}
k\\
r
\end{array}\right)=\sum_{k=1}^{n}\left(\begin{array}{c}
k-1\\
r
\end{array}\right)+\sum_{k=1}^{n}\left(\begin{array}{c}
k-1\\
r-1
\end{array}\right)=\\
=\sum_{k=0}^{n-1}\left(\begin{array}{c}
k\\
r
\end{array}\right)+\sum_{k=0}^{n-1}\left(\begin{array}{c}
k\\
r-1
\end{array}\right)=\\
=\sum_{k=1}^{n}\left(\begin{array}{c}
k\\
r
\end{array}\right)-\left(\begin{array}{c}
n\\
r
\end{array}\right)+\left(\begin{array}{c}
0\\
r
\end{array}\right)+\sum_{k=1}^{n}\left(\begin{array}{c}
k\\
r-1
\end{array}\right)-\left(\begin{array}{c}
n\\
r-1
\end{array}\right)+\left(\begin{array}{c}
0\\
r-1
\end{array}\right)=\\
=\sum_{k=1}^{n}\left(\begin{array}{c}
k\\
r
\end{array}\right)-\left(\begin{array}{c}
n\\
r
\end{array}\right)+\sum_{k=1}^{n}\left(\begin{array}{c}
k\\
r-1
\end{array}\right)-\left(\begin{array}{c}
n\\
r-1
\end{array}\right)=\\
=\sum_{k=1}^{n}\left(\begin{array}{c}
k\\
r
\end{array}\right)+\sum_{k=1}^{n}\left(\begin{array}{c}
k\\
r-1
\end{array}\right)-\left(\begin{array}{c}
n+1\\
r
\end{array}\right)
$
$\sum_{k=1}^{n}\left(\begin{array}{c}
k\\
r
\end{array}\right)=\sum_{k=1}^{n}\left(\begin{array}{c}
k\\
r
\end{array}\right)+\sum_{k=1}^{n}\left(\begin{array}{c}
k\\
r-1
\end{array}\right)-\left(\begin{array}{c}
n+1\\
r
\end{array}\right)
$
$\sum_{k=1}^{n}\left(\begin{array}{c}
k\\
r-1
\end{array}\right)=\left(\begin{array}{c}
n+1\\
r
\end{array}\right)
$
Or $\sum_{k=1}^{n}\left(\begin{array}{c}
k\\
r
\end{array}\right)=\left(\begin{array}{c}
n+1\\
r+1
\end{array}\right)
$ (for $r>0$)
So, if $\begin{equation}
m^4 = Z{m\choose 4}+Y{m\choose 3}+X{m\choose 2}+W{m\choose 1}
\end{equation}$ then $\sum_{k=1}^{n}k^{4}=X\left(\begin{array}{c}
n+1\\
5
\end{array}\right)+Y\left(\begin{array}{c}
n+1\\
4
\end{array}\right)+Z\left(\begin{array}{c}
n+1\\
3
\end{array}\right)+W\left(\begin{array}{c}
n+1\\
2
\end{array}\right)
$
I remember it from book http://en.wikipedia.org/wiki/Concrete_Mathematics, where Knuth and Co did the same thing what you want to do now.
Also I suggest related book $A=B$ (about summing of hypergeometric sums) which can be legally downloaded here - http://www.math.upenn.edu/~wilf/Downld.html
So, if I got it right, then in general case
$$\sum_{k=1}^{n}k^{m}=\sum_{k=1}^{m}k!\left\{ \begin{array}{c}
m\\
k
\end{array}\right\} \left(\begin{array}{c}
n+1\\
k+1
\end{array}\right)
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/299758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
}
|
Evaluating $\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$ Inspired by Ramanujan's problem and solution of $\sqrt{1 + 2\sqrt{1 + 3\sqrt{1 + \ldots}}}$, I decided to attempt evaluating the infinite radical
$$
\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
$$
Taking a cue from Ramanujan's solution method, I defined a function $f(x)$ such that
$$
f(x) = \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}}
$$
We can see that
$$\begin{align}
f(0) &= \sqrt{2^0 + \sqrt{2^1 + \sqrt{2^2 + \sqrt{2^3 + \ldots}}}} \\
&= \sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}
\end{align}$$
And we begin solving by
$$\begin{align}
f(x) &= \sqrt{2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}}} \\
f(x)^2 &= 2^x + \sqrt{2^{x+1} + \sqrt{2^{x+2} + \sqrt{2^{x+3} + \ldots}}} \\
&= 2^x + f(x + 1) \\
f(x + 1) &= f(x)^2 - 2^x
\end{align}$$
At this point I find myself stuck, as I have little experience with recurrence relations.
How would this recurrence relation be solved? Would the method extend easily to
$$\begin{align}
f_n(x) &= \sqrt{n^x + \sqrt{n^{x+1} + \sqrt{n^{x+2} + \sqrt{n^{x+3} + \ldots}}}} \\
f_n(x)^2 &= n^x + f_n(x + 1)~\text ?
\end{align}$$
|
Just a comment on such similar cases. From this short answer,
$$x+2^n=\sqrt{4^{n}+x\sqrt{4^{\left(n+1\right)}+x\sqrt{4^{\left(n+2\right)}+x\sqrt{...}}}}\tag{1}$$
Let $N=\sqrt{1 + \sqrt{2 + \sqrt{4 + \sqrt{8 + \ldots}}}}$. Then
$$\begin{aligned}
N &< \sqrt{1+\sqrt{2+\sqrt{4^{1}+\sqrt{4^{2}+\sqrt{4^{3}+...}}}}} \\
N &< \sqrt{1+\sqrt{5}} \\
&< 1.79890743995\\
\end{aligned}$$
For comparison, $N ≃ 1.78316580926410..$. Also from Eq. $1$, we can get;
$$\boxed{1+\sqrt{2}=\sqrt{2^{1}+\sqrt{2^{3}+\sqrt{2^{5}+\sqrt{2^{7}...}}}}}$$
$$\begin{aligned}
N &< \sqrt{1+\sqrt{2^{1}+\sqrt{2^{3}+\sqrt{2^{5}+\sqrt{2^{7}...}}}}} \\
&< \sqrt{2+\sqrt{2}} \\
\end{aligned}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/300299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 6,
"answer_id": 2
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.