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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
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

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



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



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

$$x=\tan x$$

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





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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