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The sum of the squares of the reciprocals of the positive fixed points of the tangent function is $1/10$.

I've seen this proved by means of residues, but I don't remember the details.

I've also heard it asserted that it that it can be done by means of Green's functions.

What proofs of this fact are published or otherwise known?

PS: Maybe it is of interest to note that the positive fixed points of the tangent function are also the abscissas of the extreme values of the function $x\mapsto\dfrac{\sin x} x.$

Michael Hardy
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    Have you seen [Mahajan's treatment](http://mitpress.mit.edu/books/full_pdfs/Street-Fighting_Mathematics.pdf)? (See page 113.) – J. M. ain't a mathematician Oct 23 '11 at 22:05
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    Looks like this question has been on your mind for a long time :-) http://mathforum.org/kb/message.jspa?messageID=1612788&tstart=19545 – joriki Oct 23 '11 at 22:07
  • I think it was _after_ that time that I saw it done by residues. And as I said, I've forgotten the details. – Michael Hardy Oct 24 '11 at 00:41
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    Relevant: https://groups.google.com/group/sci.math.symbolic/browse_thread/thread/c5b555272d9aca60/36a1b0fe634037cc?hl=en (there's Robert Israel 6 years ago) and http://www.ams.org/journals/mcom/1971-25-116/S0025-5718-1971-0304726-X/S0025-5718-1971-0304726-X.pdf and http://www.math.ucdavis.edu/~saito/publications/saito_rayleighfunc.pdf. – anon Oct 24 '11 at 08:13
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    The book by Mahajan is a pure marvel. Unfortunately, the link given by @J. M. is broken. See now (https://mitpress.mit.edu/sites/default/files/titles/free_download/9780262514293_Street_Fighting_Mathematics.pdf) – Jean Marie Nov 11 '16 at 18:52

6 Answers6

17

For the sake of at least having an answer, here's what's in J.M.'s link, in a nutshell anyway.

Note that $\tan(x)-x$ and $\sin(x)-x\cos(x)$ have the same zeros but the latter is holomorphic on the complex plane - the difference is that the latter has a triple root at $0$ whereas the former only has a double root, but our sum doesn't involve $x=0$ anyway. This means it affords a Weierstrass factorization. Some work can show that the coefficients of the expanded-out polynomials of the partial products in such a factorization will, in the limit, converge to those of the Taylor expansion of the function so long as we ignore the $e^{g(z)}$ and $E_n(z)$ factors. We can also show that the zeros are approximately $x_n\approx(n+\frac{1}{2})\pi$ asymptotically, and comparing this with the factorization for, say, $\sin$ tells us we don't need any $E_n$ factors. Now putting together the series expansions for sine and cosine gives

$$\left(x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\cdots\right)-x\left(1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\cdots\right)$$

$$=\frac{1}{3}x^3\left(1-\frac{1}{10}x^2+\cdots\right).$$

We ignore the $x^3$ and normalize so that $a_0=1,a_1=0,a_2=-1/10$. Now compute

$$\sum_{\tan(u)=u\ne0}u^{-2}=\left(\sum_{i}\frac{1}{\lambda_i}\right)^2-2\left(\sum_{i<j}\frac{1}{\lambda_i\lambda_j}\right)$$

$$=a_1^2-2a_2=\frac{1}{5}.$$

Note that $\tan(x)-x$ is an odd function and squaring takes out signs so by symmetry the above sum essentially double-counts every positive root. Divide by $2$ and our final answer is $1/10$.

Note that all sums-of-reciprocals of even powers of solutions to $\tan(x)=x$ can be evaluated in this way by using the Newton-Girard formulas.

anon
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    Did you define the $\lambda$s somewhere? – Michael Hardy Nov 21 '11 at 21:47
  • @Michael: Eh, not explicitly I guess. They're the nonzero roots of $\tan u=u$ of course. – anon Nov 22 '11 at 18:07
  • @anon: Recently I happened to ask the same question on MSE and was directed to your answer. I think that this method needs some sort of justification because we are not dealing with roots of polynomials. Also if we consider $\lambda$'s to be root of $\tan u = u$ then $\sum 1/\lambda_{i}$ diverges and hence can't be squared in the last step. – Paramanand Singh Jun 29 '14 at 05:57
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    @Paramanand Those are both good points. Both points are already engaged with in my answer: *"Some work can show that the coefficients of the expanded-out polynomials of the partial products in such a factorization will, in the limit, converge to those of the Taylor expansion of the function."* – anon Jul 04 '14 at 23:07
4

Here is a very different proof :

Let $0 < x_1 < x_2 < x_3 < \cdots $ be the positive roots of equation

$$\tan(x)=x \tag 0$$

These roots are such that

$$\sin(x_k)= x_k \cos(x_k) \tag 1 $$

(this relationship will be used later on).

We have to show that :

$$\sum_{k=1}^\infty \frac{1}{x_k^2}=\frac{1}{10} \tag 2$$

The framework is Hilbert space $L^2[0,1]$ of square integrable functions on $[0,1]$ with bilinear form $\langle f|g\rangle:=\int_0^1fg$.

Let us consider its closed subspace defined by

$$E=\{f\mid f(0)=0 \ \ \& \ \ f(1)=f'(1) \ \ \& f \in C^2\}, \tag 3 $$

itself a Hilbert space.

Operator $\Delta : f \mapsto f''$ is a self-adjoint operator on $E$, due to the fact that a double integration by parts gives :

$$\int f g'' = \int f'' g \ \ \iff \ \ \langle f|\Delta g \rangle=\langle \Delta f| g \rangle \tag 4 $$

The kernel of $\Delta$ is one-dimensional : it is the set of linear functions $f(x)=ax$. Let $\varphi_0(x)=x$ be the basis of this kernel.

Functions defined by $\varphi_k(x)=\sin(x_kx)$ (with $x_k$ as defined in (2)), due to property (1), belong to space $E$ and moreover are such that :

$$\Delta \varphi_k = x_k^2 \varphi_k \tag 5 $$

i.e., they are eigenvectors of operator $\Delta$ with associated eigenvalues $x_k^2$.

Their norms are found to be

$$\|\varphi_0\|^2=\frac{1}{3} \ \ \text{and} \ \ \|\varphi_k\|^2=\frac{1}{2}\sin^2(x_k)\tag{6}$$

Being eigenvectors of a self-adjoint operator, functions $\varphi_k$ are mutually orthogonal. We admit that they constitute an Hilbert basis of $E$, allowing us to write, for any $f \in E$ :

$$f=\sum_{k=0}^\infty \frac{1}{\|\varphi_k\|^2} \langle f \mid \varphi_k \rangle\tag{7}$$

Now consider the particular function $f(x)=x^3-2x^2 \in E$.

We have the following chain of equalities :

  • in the case [$k>0$] :

$$\langle f\mid \varphi_k \rangle=\langle f\mid \frac{1}{x_k^2}\Delta\varphi_k \rangle=\frac{1}{x_k^2}\langle \Delta f\mid\varphi_k \rangle=\frac{1}{x_k^2}(-6\langle \varphi_0 \mid \varphi_k \rangle+ \langle 4\mid\varphi_k \rangle) = \frac{4}{x_k^2}(1-\cos(x_k))\tag{8}$$

  • in the case [$k=0$] :

$$\langle f\mid \varphi_0 \rangle=\int_0^1 x(x^3-2x^2) \, dx=-\tfrac{3}{10}\tag{9}$$

Now, using (7)

$$f(x):=x^3-2x^2=-\frac{9}{10}x+\sum_{k=1}^\infty \frac{8}{x_k^2(1+\cos(x_k))}\sin(x_kx)\tag{10}$$

Let us terminate by computing, thanks to (10), the expression $f'(0)+f'(1)$ in two ways (with the LHS and the RHS), from which we deduce (2).

Jean Marie
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3

I answered this similar question a while ago, so I thought I would answer this question in the same manner, with more detail.


Residue at $\boldsymbol{z=0}$

Compute the series near $x=0$ for $\tan(x)$ as the ratio of the series for $\sin(x)$ and $\cos(x)$: $$ \begin{align} \tan(x) &=\frac{x-\frac16x^3+\frac1{120}x^5+O\!\left(x^7\right)}{1-\frac12x^2+\frac1{24}x^4+O\!\left(x^6\right)}\\ &=\left(x-\frac16x^3+\frac1{120}x^5+O\!\left(x^7\right)\right)\left(1+\frac12x^2+\frac5{24}x^4+O\!\left(x^6\right)\right)\\ &=x+\frac13x^3+\frac2{15}x^5+O\!\left(x^7\right)\tag1 \end{align} $$ compute the series near $x=0$ for $\frac1{\tan(x)-x}$ using $(1)$: $$ \begin{align} \frac1{\tan(x)-x} &=\frac1{\frac13x^3+\frac2{15}x^5+O\!\left(x^7\right)}\\ &=\frac3{x^3}\left(1-\frac25x^2+O\!\left(x^4\right)\right)\\ &=\frac3{x^3}-\frac6{5x}+O(x)\tag2 \end{align} $$ Therefore, $$\newcommand{\Res}{\operatorname*{Res}} \Res_{z=0}\left(\frac1{\tan(z)-z}\right)=-\frac65\tag3 $$


Residue at Other Singularities

At any point $z_k\ne0$ where $\tan(z_k)=z_k$, $$ \begin{align} \Res_{z=z_k}\frac1{\tan(z)-z} &=\lim_{z\to z_k}\frac{z-z_k}{\tan(z)-z}\\ &=\lim_{z\to z_k}\frac1{\sec^2(z)-1}\\ &=\frac1{\tan^2(z_k)}\\ &=\frac1{z_k^2}\tag4 \end{align} $$ Enumerate the positive roots of $\tan(z)=z$ as $\{z_k\}_{k=1}^\infty$. Note that $\{-z_k\}_{k=1}^\infty$ are the negative roots of $\tan(z)=z$.


Behavior Near $\boldsymbol{\infty}$

Since $\tan(n\pi+iy)=i\tanh(y)$ for $n\in\mathbb{Z}$, we get that $$\newcommand{\Re}{\operatorname{Re}}\newcommand{\Im}{\operatorname{Im}} \begin{align} |\tan(n\pi+iy)| &=|\tanh(y)|\\ \le1\tag5 \end{align} $$ and $$ \begin{align} |\tan(x+iy)| &=\left|\frac{e^{ix-y}-e^{-ix+y}}{e^{ix-y}+e^{-ix+y}}\right|\\ &\le\left|\frac{\cosh(y)}{\sinh(y)}\right|\\ &=|\coth(y)|\tag6 \end{align} $$ Thus, on the path $\max(|\Re(z)|,|\Im(z)|)=n\pi$, $n\in\mathbb{Z}$, we have $$ |\tan(z)|\le\coth(\pi)\tag7 $$ Thus, on that path $$ \begin{align} \frac1{\tan(z)-z} &=\frac1{O(1)-z}\\ &=-\frac1z+O\!\left(\frac1{z^2}\right)\tag8 \end{align} $$ Thus, the integral counter-clockwise around that path is $-2\pi i$.


The Sum

Putting $(3)$, $(4)$, and $(8)$ together, we get $$ -2\pi i=2\pi i\left(-\frac65+2\sum_{k=1}^\infty\frac1{z_k^2}\right)\tag9 $$ which gives $$ \sum_{k=1}^\infty\frac1{z_k^2}=\frac1{10}\tag{10} $$

robjohn
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2

I just saw a link here, so here's a solution of mine that dates back to 2008. Original version in this AoPS thread, which also includes the Weierstrass factorization solution. The solution here is modified from the original, most importantly by using a different contour so I don't have to worry about a principal value at a 3rd order pole.

Consider the function $$f(z)=\frac{\sin z}{z(\sin z-z\cos z)}=\frac{1}{z-z^2\cot z}$$ This function has simple poles at $z=x_n$ for nonzero roots $x_n=\tan x_n$ with residue $\frac{\sin x_n}{x_n}\cdot \frac1{x_n\sin x_n}=\frac1{x_n^2}$. Note that this residue is the same at the positive root $x_n$ and the negative root $-x_n$. It also has a third-order pole at $z=0$, where $f$ has Laurent series expansion \begin{align*}f(z) &= \frac1{z-z^2\frac{1-\frac12z^2+\frac1{24}z^4+\cdots}{z-\frac16z^3+\frac1{120}z^5+\cdots}}=\frac{z-\frac16z^3+\frac1{120}z^5+\cdots}{z^2-\frac16z^4+\frac1{120}z^6-z^2+\frac12z^4-\frac1{24}z^6+\cdots}\\ &= \frac{z-\frac16z^3+\frac1{120}z^5+\cdots}{\frac13z^4-\frac1{30}z^6+\cdots}=3z^{-3}\left(1-\frac16z^2+\cdots\right)\left(1+\frac1{10}z^2+\cdots\right)\\ f(z) &= 3z^{-3}-\frac15z^{-1}+\cdots\end{align*} for a residue of $-\frac15$.

Now, let $C(N,M)$ be the rectangular contour with corners at $N\pi+iM$, $-N\pi+iM$, $-N\pi-iM$, and $N\pi-iM$, for some large $M$ and large integer $N$. Define $I(N,M)=\int_{C(N,M)} f(z)\,dz$.

On the vertical segments, note that $\cot(\pm N\pi + iy)=\frac{\cos \pm N\pi\cosh y}{i\cos \pm N\pi\sinh y}=\frac1{i\tanh y}$. This is greater than $1$ in absolute value, so $|z^2\cot z-z|>|z|^2-|z|$ on those segments. Estimating $|z|\ge N\pi$, we get an integral of at most $\frac{2M}{N\pi(N\pi-1)}$ in absolute value on each vertical segment.

On the horizontal segments, $\cot(x+iy)$ tends to $\mp i$ uniformly as $y\to\pm\infty$. If $M$ is large enough that we're within $\epsilon$, this gives $|f(z)|\le \frac{1}{(1-\epsilon)^2|z|^2-|z|}$. Estimating $|z|\ge M$, this lead to an integral of at most $\frac{2N\pi}{M(M(1-\epsilon)^2-1)}$ in absolute value on each horizontal segment.

As long as neither $M$ nor $N$ gets too far ahead of the other - say, $M=\pi N$ - these both tend to zero as $M$ and $N$ go to $\infty$ together. We have shown that $\lim_{N\to\infty} I(N,\pi N)=0$.

Now, we calculate that integral with residues. Inside the contour $C(N,\pi N)$, there are $2N-1$ poles: the pole at zero, the first $N-1$ positive roots $x_1,x_2,\dots,x_{N-1}$ of $\tan x=x$, and their negatives $-x_1,-x_2,\dots,-x_{N-1}$. Therefore $$I(N,\pi N) = 2\pi i\left(-\frac15 + \sum_{k=1}^{N-1}\frac1{x_k^2}+\sum_{k=1}^{N-1}\frac1{(-x_k)^2}\right)$$ $$0 = 2\pi i\left(-\frac15 + 2\sum_{k=1}^{\infty}\frac1{x_k^2}\right)$$ $$\sum_{k=1}^{\infty}\frac1{x_k^2} = \frac1{10}$$

Connected : https://www.math.ucdavis.edu/~saito/publications/saito_rayleighfunc.pdf

Jean Marie
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jmerry
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This is not an answer of mine It is an image of a solution with a higher degree of generalization (with references to the original question) that I have found on AOS [Explanation : As AOS site is no longer active, I fear the link I give here will be disactivated] (pseudo of the author : PolyaPal). This thread is as well mentionned by @jmerry.

Independant remark : function $f$ can be written using Bessel function $J_{3/2}$ (see this).

enter image description here

Jean Marie
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The answer I got was not $\frac{1}{10}$ so it is likely that I have made a mistake somewhere.

$$x=\tan x$$

$$x^{2}\cos^{2}{x} =\sin^{2}{x}$$

$$x^{2}(1-\frac{x^{2}}{2!}+\frac{x^{4}}{4!}+\cdots)^{2}=(x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}+\cdots)^{2}$$

$$x^{2}-x^{4}+\frac{x^{6}}{4}+\frac{x^{6}}{12}-\frac{x^{8}}{24}+\cdots=x^{2}-\frac{x^{4}}{3}+\frac{x^{6}}{36}+\frac{x^{6}}{60}+\cdots$$

$t=x^{2}$

$$t^{2}(\frac{2}{3}-\frac{13t}{45}+\cdots)=0$$

In a polynomial of the form $a_0+a_1x+\cdots+a_nx^{n}$ with roots $x_1,\ldots,x_n$, $\frac{a_0}{a_n}=x_{1}x_{2}\cdots$ and $\frac{a_2}{a_n}=-x_3x_4x_5-\cdots-x_1x_4-\cdots-x_1x_2x_5-\cdots-x_2x_4x_5-\cdots-\cdots$

If the root with $t=0$ is excluded, the sum of the reciprocals $=\frac{\frac{13}{45}}{\frac{2}{3}}=\frac{13}{30}$. $\tan{-x}=-\tan{x}$ therefore the reciprocal sums of positive and negative sums are equal, so for positive roots the sum is $\frac{13}{60}$.

Michael Hardy
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Angela Pretorius
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