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This question is actually from MSE. I had to post it here due to the lack of response there even after placing a bounty. Here goes the question

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.

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 third 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 Wolfram Alpha 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

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    $\begingroup$ If $$\oint_{C(R)}\frac{f'(z)}{f(z)}g(z) dz$$ with the contour $C(R)$ a circle of radius $R$ with center the origin, tends to zero for $R\to\infty$, then the residue theorem (if applicable), yields a relation between the summation of $g(z)$ over the zeros of $f(z)$ in terms of the residues of the integrand at the poles of $g(z)$. $\endgroup$
    – Count Iblis
    Commented Sep 25, 2018 at 16:46
  • $\begingroup$ @Count Iblis: how is that related to the question? $\endgroup$
    – Qfwfq
    Commented Sep 25, 2018 at 18:22
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    $\begingroup$ Reading your title is an adventure. $\endgroup$
    – LSpice
    Commented Sep 25, 2018 at 18:26
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    $\begingroup$ @Qfwfq: The question can be posed as summing of $g(z)=\frac{1+z^2}{z^2(z^2+2)}$ over the zeros of $f(z)=\frac{\sin(z)-z\cos(z)}{z^3}$, or alike. The hope is that under an appropriate choice of $f(z)$ and $g(z)$, one can use the residue theorem to evaluate the sum. $\endgroup$
    – Max Alekseyev
    Commented Sep 25, 2018 at 18:37

2 Answers 2

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It needs to be pointed out that the series $\frac{\sin(x)-x\cos(x)}{x^3}$ comes from the MSE answer, and as I understand the claim there this series equals $\frac{1}{3}\prod_{k\geq 1} (1-\frac{x^2}{\lambda_k^2})$. I have not verified this claim, but if it's true, then your approach does make sense. UPD. Precise argument can be seen in this answer.

I've also verified your analytical expression numerically, and it looks fine. Namely, the analytical expression gives $\sum_k a_k^{-2} = 0.0973745978\ldots$, while numerically we have $\sum_{k=1}^{10^5} \frac {\lambda _k ^2 +1}{\lambda _k ^2 (\lambda _k ^2 +2)} = 0.09737358469\ldots$ with the difference $\approx 10^{-6}$.

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One perhaps more direct way could be to use your expression $\frac {1}{20}+\frac {1}{2}\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2 +2}$, expand it ($\frac {1}{20}+\frac {1}{2}\sum_{k=1}^{\infty} \frac {1}{\lambda _k ^2} (1-\frac{2}{\lambda_k^2} + \cdots)$) and calculate this sum by using the sums-of-reciprocals of even powers of $\lambda_k$ which can be calculated in a simliar way as the sum of reciprocal squares of $\lambda_k$ (see the last sentence of the accepted answer of https://math.stackexchange.com/questions/75206/sum-of-the-squares-of-the-reciprocals-of-the-fixed-points-of-the-tangent-functio).

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  • $\begingroup$ Thanks for the pointer on the factorization argument! Although I doubt that computing sum of reciprocals of even powers of $\lambda_k$ based on Newton-Girard formulas is that simple, at least it's not simpler than using substitution $x\mapsto \sqrt{x-2}$. $\endgroup$
    – Max Alekseyev
    Commented Sep 25, 2018 at 22:45

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