10
$\begingroup$

Let tangents be drawn to the curve $y=\sin x$ from the origin. Let the points of contact of these tangents with the curve be $(x_k,y_k)$ where $x_k\gt 0; k\ge 1$ such that $x_k\in (\pi k, (k+1)\pi)$ and $$a_k=\sqrt {x_k^2+y_k^2}$$ (Which is basically the distance between the corresponding point of contact and the origin i.e. the length of tangent from origin) .


I wanted to know the value of

$$\sum_{k=1}^{\infty} \frac {1}{a_k ^2}$$

Now this question has just popped out in my brain and is not copied from any assignment or any book so I don't know whether it will finally reach a conclusion or not.


I tried writing the equation of tangent to this curve from origin and then finding the points of contact but did not get a proper result which just that the $x$ coordinates of the points of contact will be the positive solutions of the equation $\tan x=x$

On searching internet for sometime about the solutions of $\tan x=x$ I got two important properties of this equation. If $(\lambda _n)_{n\in N}$ denote the roots of this equation then

$$1)\sum_n^{\infty} \lambda _n \to \infty$$ $$2)\sum_n^{\infty} \frac {1}{\lambda _n^2} =\frac {1}{10}$$

But were not of much help.

I also tried writing the points in polar coordinates to see if that could be of some help but I still failed miserably.

I could not think of any method so any other method would be openly welcomed.

Any help would be very beneficial to solve this problem.

Thanks in advance.


Edit:

On trying a bit more using some coordinate geometry I found that the locus of the points of contact is $$x^2-y^2=x^2y^2$$

Hence for sum we just need to find $$\sum_{k=1}^{\infty} \frac {\lambda _k ^2 +1}{\lambda _k ^2 (\lambda _k ^2 +2)}=\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2} -\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 (\lambda _k ^2 +2)}=\frac {1}{10} -\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 (\lambda _k ^2 +2)}=\frac {1}{10} -\sum_{k=1}^{\infty} \frac {1}{2\lambda _k ^2} +\sum_{k=1}^{\infty} \frac {1}{2(\lambda _k ^2 +2)} =\frac {1}{20}+\frac {1}{2}\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 +2} $$

Now for the second summation I did think about it to form a series but for the roots to be $\lambda _k^2 +2$ we just need to substitute $x\to \sqrt {x−2}$ in power series of $\frac {\sin x-x\cos x}{x^3}$ and then get the result but it was still a lot confusing for me.

Using $x\to\sqrt {x-2}$ in the above power series and using Wolfy I have got a series. So we need ratio of coefficient of $x$ to the constant term so is the value of second summation equal to $$\frac {5\sqrt 2\sinh(\sqrt 2)−6\cosh(\sqrt 2)}{4(2\cosh(\sqrt 2)−\sqrt 2\sinh(\sqrt 2))}?$$

Is this value correct or did I do it wrong?

I would also like to know if there is some other method to solve this problem

$\endgroup$
6
+50
$\begingroup$

The points of contact are where the tangent to $y=\sin(x)$, which has a slope of $\cos(x)$, has the same slope as the line from the origin, $\frac{\sin(x)}x$. Thus, we are looking at the points where $x_k=\tan(x_k)$.

The square of the length of the line from the origin is $x_k^2+\sin^2(x_k)=\frac{x_k^4+2x_k^2}{x_k^2+1}$. Therefore, the sum we are looking for is $$ \sum_{k=1}^\infty\frac{x_k^2+1}{x_k^4+2x_k^2}\tag1 $$


The residue of $f(z)=\frac1{\tan(z)-z}-\frac1{(z^2+2)(\tan(z)-z)}$ where $z\ne0$ and $\tan(z)=z$ is $$ \frac1{z^2}-\frac1{z^4+2z^2}=\frac{z^2+1}{z^4+2z^2}\tag2 $$

Thus, the sum of all the residues of $f(z)$ is $2$ times the sum we are seeking plus the residue of $f(z)$ at $z=0$, which is $\frac3{20}$, and the sum of the residues of $f(z)$ at $z=\pm i\sqrt2$, which is $-\frac1{2-\sqrt2\tanh(\sqrt2)}$


Note that the limit $$ \lim_{k\to\infty}\int_{\gamma_{k,\lambda}}f(z)\,\mathrm{d}z=\int_{\gamma_\lambda}f(z)\,\mathrm{d}z\tag3 $$ where $k\in\mathbb{Z}$ and the paths are $$ \scriptsize\gamma_{k,\lambda}=[k\pi+i\lambda,-k\pi+i\lambda]\cup\underbrace{[-k\pi+i\lambda,-k\pi-i\lambda]}_{\le\frac{2\lambda}{k\pi}}\cup[-k\pi-i\lambda,k\pi-i\lambda]\cup\underbrace{[k\pi-i\lambda,k\pi+i\lambda]}_{\le\frac{2\lambda}{k\pi}}\tag4 $$ and $$ \gamma_\lambda=(\infty+i\lambda,-\infty+i\lambda)\cup(-\infty-i\lambda,\infty-i\lambda)\tag5 $$ and $2\pi i$ times the sum of all the residues of $f(z)$ is $$ \lim_{\lambda\to\infty}\int_{\gamma_\lambda}f(z)\,\mathrm{d}z=-2\pi i\tag6 $$

$(6)$ means the sum of the residues of $f(z)$ over all singularities is $-1$. This is $2$ times the sum we are looking for plus $\frac3{20}-\frac1{2-\sqrt2\tanh(\sqrt2)}$


Therefore, $$ \bbox[5px,border:2px solid #C0A000]{ \begin{align} \sum_{k=1}^\infty\frac{x_k^2+1}{x_k^4+2x_k^2} &=-\frac{23}{40}+\frac1{4-2\sqrt2\tanh\left(\sqrt2\right)}\\ &=0.097374597898595746715 \end{align} }\tag5 $$


Numerical Check

Note that each of the roots is a little less than an odd multiple of $\frac\pi2$:

$x_1=4.4934094579090641753\approx\frac{3\pi}2$
$x_2=7.7252518369377071642\approx\frac{5\pi}2$
$x_3=10.904121659428899827\approx\frac{7\pi}2$
$x_4=14.066193912831473480\approx\frac{9\pi}2$

Thus, we can under-approximate the sum using $$ \begin{align} \sum_{k=1}^\infty\frac{x_k^2+1}{x_k^4+2x_k^2} &\approx\sum_{k=1}^\infty\frac{(2k+1)^2\pi^2/4+1}{(2k+1)^4\pi^4/16+(2k+1)^2\pi^2/2}\\ &=0.092481600740508343614 \end{align} $$

$\endgroup$
1
$\begingroup$

Answer to original question

Simple bound $\pi k\leq a_k \leq \sqrt{\pi^2(k+\frac12)^2+1}$ shows that $\dfrac{a_k}{a_{k+1}}\to 1$. So both sums diverge.

Answer to modified question

Again, (1) diverges. (2) also diverges since squaring the ratio doesn't change $\to 1$. (3) converges since you have $\pi k<\lambda_k=x_k<a_k$ giving $\dfrac{1}{a_k^2}\leq\dfrac{1}{\lambda_k^2}\leq\dfrac{1}{k^2}$. This is of course a very loose bound.

Finding $\displaystyle\sum_{k=1}^\infty\frac{1}{\lambda_k^2+2}$ for use in $\sum a_k^{-2}$

Recall one way of finding $\displaystyle\sum_{k=1}^\infty\lambda_k^{-2}=\frac{1}{10}$ is write down the series expansion of $$ \sin x-x\cos x $$ and set that to zero, reading off the lowest terms $$ x^3\left(\frac{1}{3}-\frac{x^2}{30}+\frac{x^4}{840}+\dots\right)=0 $$ and after cancelling $x^3$ factor in front, you read off $\dfrac{1/30}{1/3}$ in a reciprocal Viete's formula way (except to make it rigourous you need to do it properly with infinite products, but that's another story).

So now we want to do this with $\lambda_k^2+2$. You want to construct a series whose roots are $\lambda_k^2+2$ The simplest heuristic way is to use the full series expansion above (ignoring the $x^3$) and try to express it as a power series in $x^2+2$, and read off the sum of reciprocals of roots.

$\endgroup$
  • $\begingroup$ Oh so sorry I didn't reread my question. I had to make a few edits so did it now. You can check the sequences now. $\endgroup$ – Rohan Shinde Sep 18 '18 at 16:59
  • $\begingroup$ Simplified the answer to original question, which also answers the modified questions. $\endgroup$ – user10354138 Sep 18 '18 at 17:22
  • $\begingroup$ So I must try finding the answer to the third Sum right? $\endgroup$ – Rohan Shinde Sep 18 '18 at 17:30
  • 1
    $\begingroup$ I suspect using the infinite product representation and nasty binomial expansion could work, but I haven't tried it. $\endgroup$ – user10354138 Sep 18 '18 at 17:34
  • 1
    $\begingroup$ Using $x\to \sqrt {x-2}$ using Wolfy I have got a series. So we need ratio of coefficient of $x$ to the constant term so is the value of second sum( where I need sum of reciprocals of $\lambda _k ^2 +2$) equal to $$\frac {5\sqrt 2 \sinh (\sqrt 2)- 6\cosh (\sqrt 2)}{4(2\cosh (\sqrt 2)-\sqrt 2 \sinh (\sqrt 2))}$$? $\endgroup$ – Rohan Shinde Sep 20 '18 at 15:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.