# Infinite sum of reciprocals of squares of lengths of tangents from origin to the curve $y=\sin x$

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.

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

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}

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.

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.

• 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. – Rohan Shinde Sep 18 '18 at 16:59
• Simplified the answer to original question, which also answers the modified questions. – user10354138 Sep 18 '18 at 17:22
• So I must try finding the answer to the third Sum right? – Rohan Shinde Sep 18 '18 at 17:30
• I suspect using the infinite product representation and nasty binomial expansion could work, but I haven't tried it. – user10354138 Sep 18 '18 at 17:34
• 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 $$\frac {5\sqrt 2 \sinh (\sqrt 2)- 6\cosh (\sqrt 2)}{4(2\cosh (\sqrt 2)-\sqrt 2 \sinh (\sqrt 2))}$$? – Rohan Shinde Sep 20 '18 at 15:34