# Sum of the squares of the reciprocals of the fixed points of the tangent function

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

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 '14 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 '14 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