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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?

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Have you seen Mahajan's treatment? (See page 113.) –  J. M. Oct 23 '11 at 22:05
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Looks like this question has been on your mind for a long time :-) mathforum.org/kb/… –  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 0:41
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2 Answers 2

up vote 12 down vote accepted

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.

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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 at 5:57
    
@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 4 at 23:07

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 $\frac{13}{60}$.

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The question is "sums of squares of reciprocals..." –  anon Oct 27 '11 at 7:21

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