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Is there any identity for $\sum_{k=1}^{n}\tan\left(\theta+\frac{k\pi}{\color{red} {2n+1}}\right)$ or $\sum_{k=1}^{n}\tan\left(\frac{k\pi}{\color{red} {2n+1}}\right)$ ?

I thought maybe wrongly that given:

$\sum_{k=1}^{n}\cot\left(\frac{k\pi}{\color{red} {2n+1}}\right)=\sum_{k=1}^{n}-\tan\left(\frac{\pi}{ {2}}+\frac{k\pi}{\color{red} {2n+1}}\right)$

and since $n \cot(nx)$ is the logarithmic derivative of $\sin(nx)$ and $\cot\left(x+\frac{\pi k}{n}\right)$ is the logarithmic derivative of $\sin\left(x+\frac{\pi k}{n}\right)$, I tried manipulating the identity:

$2^n \prod_{k=1}^n \sin\left(\frac{k\pi}{2n+1}\right)=\sqrt{2n+1}$

but I kept getting stuck.

The variation $\sum_{k=0}^{n-1}\tan\left(\theta+\frac{k\pi}{n}\right)=−n\cot\left(\frac{n\pi}{2}+n\theta\right)$ is seen to work quite nicely.

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    $\begingroup$ Complex analysis is very useful for those things, caracterizing $\tan(z)$ as the only $\pi$-periodic meromorphic function with simple poles of residue $1$ at $\pi/2+\pi \Bbb{Z}$, analytic and bounded elsewhere and $\lim_{\Im(z)\to +\infty} \tan(z)=i$. $\endgroup$
    – reuns
    Feb 13, 2022 at 15:37
  • $\begingroup$ @reuns thank you I've updated the tag to complex-analysis and will try to give some thought on that idea you mention. $\endgroup$
    – onepound
    Feb 13, 2022 at 15:55
  • $\begingroup$ The reason why you have an identity for sine is that sine function can be easily rewrite as a simple sum of power function. $\endgroup$
    – High GPA
    Feb 13, 2022 at 18:14
  • $\begingroup$ Mathematica says for general a and b $$\sum _{k=1}^n \tan (a k+b)=i n+\frac{2 i \psi _{e^{2 i a b}}\left(1-\frac{i \pi }{\ln \left(e^{2 i a b}\right)}\right)}{\ln \left(e^{2 i a b}\right)}-\frac{2 i \psi _{e^{2 i a b}}\left(1+n-\frac{i \pi }{\ln \left(e^{2 i a b}\right)}\right)}{\ln \left(e^{2 i a b}\right)}$$ where: $\psi _{e^{2 i a b}}\left(1+n-\frac{i \pi }{\ln \left(e^{2 i a b}\right)}\right)$ is QPolyGamma function. $\endgroup$ Feb 15, 2022 at 17:13
  • $\begingroup$ @High GPA on the balance of things this does seem to be true. $\endgroup$
    – onepound
    Feb 21, 2022 at 18:07

1 Answer 1

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Disclaimer: I thought this would turn out to be a routine, olympiad type exercise on manipulating roots of unity and so I started writing an answer but then I got stuck. Leaving it here for now if someone else/myself manage to fix it and will remove it before the bounty period ends.

Let $\theta = \dfrac{\pi}{2n+1}.$ Then,

$$i\tan\theta = \dfrac{e^{i\theta} - e^{-i\theta}}{e^{i\theta} + e^{-i\theta}} = 1 - \dfrac{1}{1+e^{2i\theta}},$$ so it suffices to find them sum: $$\sum_{k=1}^n\dfrac{1}{1+e^{2ik\theta}}.$$ If we let $z_k = e^{2ik\theta}$, then the plan of attack is we find what unique, monic, degree $n$ polynomial have $z_1,z_2,\dots z_n$ as roots. If we could find a simple expression $f(z),$ then $f(z-1)$ has roots $1+z_k$'s and from there we can write a simple Vieta ratio to find the desired sum.

Note that $z_k = e^{2\pi i\tfrac{k}{2n+1}} = \rho^{k},$ where is a primitive root of unity of order $2n+1.$ This means that:

$$\dfrac{z^{2n+1}-1}{z-1} = \prod_{k=1}^n(z-\rho^k)\prod_{k=n+1}^{2n}(z-\rho^k)=f(z) \prod_{k=1}^n(z-\rho^{k+n}).$$ Some basic manipulation tells us: $$\prod_{k=1}^n(z-\rho^{k+n}) = \prod_{k=1}^n(z-\rho^{2n+1-k})=\prod_{k=1}^n(z-\rho^{-k}) = \rho^{-n(n+1)/2}\prod_{k=1}^n\rho^{k}(z-\rho^{-k}) = $$ $$=\rho^{-n(n+1)/2}\cdot (-1)^n\prod_{k=1}^n(1-\rho^kz) = \rho^{-n(n+1)/2}\cdot (-z)^n\prod_{k=1}^n\left(\frac 1z-\rho^k\right) = $$ $$ \rho^{-n(n+1)/2}\cdot (-z)^n f\left(\frac 1z\right),$$ so we obtain: $$z^nf(z)f\left(\frac{1}{z}\right) = \dfrac{z^{2n+1}-1}{z-1}\cdot\rho^{n(n+1)/2}\cdot (-1)^n.$$

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  • $\begingroup$ Thank you for your effort sketching this out it helps me understand the problem much clearer than before and the difficulties. I have since yesterday found that $\sum_{l=1}^n\tan^2\left(\frac{\pi l}{2n+1}\right)=n(2n+1)$ works as shown here math.stackexchange.com/questions/173447/… and here $\endgroup$
    – onepound
    Feb 14, 2022 at 8:57
  • $\begingroup$ math.stackexchange.com/questions/2339/… but as robjohn and yourself has explained it is not so simple in this case. My attempts were in manipulating the sine product into sum tangent but I could only make progress with the sine squared. $\endgroup$
    – onepound
    Feb 14, 2022 at 8:57
  • $\begingroup$ @onepound yes I quickly rediscovered that after writing this. At this point, my feeling is it does not admit a simple formula. I also considered the Chebyshev polynomial of second kind $$2^{2n}U_{2n}(x) = \prod _{k=0}^{2n}\left(x-\cos\dfrac{k\pi}{2n+1}\right),$$ which you can manipulate further to get $\tan\dfrac{k\pi}{2n+1}$ on the right hand side, but that only yielded similar equation as the one above I wrote. $\endgroup$
    – dezdichado
    Feb 14, 2022 at 18:05
  • $\begingroup$ How many olympiad problems are about calculating sums using complex numbers? $\endgroup$ Feb 19, 2022 at 13:33
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    $\begingroup$ @AitorIribarLopez I definitely learned Chebyshev and Cyclotomic polynomials as part of IMO training. I guess most of the problems won't be just about summing things as that would be too simple at IMO level. But cyclotomic polynomials solve a lot of number theory problems. There was a problem about summing some inverse cosine's in my TST one time which I failed big time because I had not seen Chebyshev in 9th grade. $\endgroup$
    – dezdichado
    Feb 19, 2022 at 19:52

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