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Is there a simple relation for $\sum_{k=0}^{n-1}\tan^4\left({k\pi\over n}\right)$ like there is for $\sum_{k=0}^{n-1}\tan^2\left({k\pi\over n}\right)$?

Looking at Jolley, Summation of Series, formula 445: $\sum_{k=0}^{n-1}\tan^2\left(\theta+{k\pi\over n}\right)=n^2\cot^2\left({n\pi\over2}+n\theta\right)+n(n-1)$

or more simply:

$\sum_{k=0}^{n-1}\tan^2\left({k\pi\over n}\right)=n(n-1)$

Does $\sum_{k=0}^{n-1}\tan^4\left({k\pi\over n}\right)$ possibly equals some integer values?

The numerical table seems to suggest so even for higher even powers of $\tan$:

$\left( \begin{array}{ccccc} n & \sum_{k=0}^{n-1}\tan^2\left({k\pi\over n}\right)&\sum_{k=0}^{n-1}\tan^4\left({k\pi\over n}\right)&\sum_{k=0}^{n-1}\tan^6\left({k\pi\over n}\right)&\sum_{k=0}^{n-1}\tan^8\left({k\pi\over n}\right)\\ 3. & 6. & 18. & 54. & 162. \\ 5. & 20. & 180. & 1700. & 16100. \\ 7. & 42. & 742. & 14154. & 271558. \\ 9. & 72. & 2088. & 66600. & 2.14049\times 10^6 \\ 11. & 110. & 4730. & 226622. & 1.09528\times 10^7 \\ 13. & 156. & 9308. & 624780. & 4.2335\times 10^7 \\ 15. & 210. & 16590. & 1.48533\times 10^6 & 1.34314\times 10^8 \\ \end{array} \right)$

I had tried to use residue methods and I particularly like falager's method in this answer using Vieta's formula but kept getting stuck.

If one could make progress with the quartic, one could possibly find for higher even powers of tan too.

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    $\begingroup$ Well by interpolating, your table suggests the sum is: $$\dfrac{n(n-1)(n^2+n-3)}{3}$$ $\endgroup$
    – dezdichado
    Commented Nov 6, 2022 at 21:18
  • $\begingroup$ @dezdichado well spotted. Perhaps that clue can lead someone to a proof - especially the higher order powers. $\endgroup$
    – onepound
    Commented Nov 6, 2022 at 21:43
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    $\begingroup$ Also, I don't see why the Vieta's method does not work if you just split to even and odd cases and get an answer for each. Besides, the reason why all these tangent summations usually consider $\tan\dfrac{k\pi}{2n}$ where $k=0,1,\dots n-1$ is because it suffices. That is, in your sum you just need to observe that $\tan^2\alpha = \tan^2(\pi-\alpha)$ and look at only the first half of the summands. The point is that after splitting to even and odd cases, the answer given in your link must be able to work this problem out. Remember: $$\sum x_i^2 = (\sum x_i)^2-2\sum\limits_{i\neq j} x_ix_j$$ $\endgroup$
    – dezdichado
    Commented Nov 6, 2022 at 21:53
  • $\begingroup$ @dezdichado thank you for those suggestions & hint. I will give it another go. $\endgroup$
    – onepound
    Commented Nov 6, 2022 at 21:58
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    $\begingroup$ You could pull out an $n$ and then replace the summand with $\sec^2\left(\frac{k\pi}{n}\right)\left(\sec^2\left(\frac{k\pi}{n}\right)-2\right)$ using the fact that $\tan^2=\sec^2-1$ $\endgroup$
    – Kamal Saleh
    Commented Nov 6, 2022 at 22:11

1 Answer 1

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A possible approach is to use (here $n$ is still odd!) $$\prod_{k=1}^{n-1}\left(1+x^2\tan^2\frac{k\pi}{n}\right)=\left(\frac{(1+x)^n+(1-x)^n}{2}\right)^2$$ obtained by factoring the RHS over $\mathbb{C}$, or as $P_n(1+x,x-1)/P_n(1,-1)$ where $$P_n(a,b)=\prod_{k=0}^{n-1}\left(a^2+b^2-2ab\cos\frac{2k\pi}{n}\right)=(a^n-b^n)^2$$ comes from factoring $z^n-1$ over $\mathbb{C}$. Denoting $S_{2m}:=\sum_{k=1}^{n-1}\tan^{2m}(k\pi/n)$, we get \begin{align} \sum_{m=1}^\infty\frac{(-1)^m}{m}S_{2m}x^{2m}&=-\sum_{k=1}^{n-1}\log\left(1+x^2\tan^2\frac{k\pi}{n}\right) \\&=-2\log\frac{(1+x)^n+(1-x)^n}{2} \\&=-2\log\sum_{k=0}^{\lfloor n/2\rfloor}\binom{n}{2k}x^{2k} \\&=2\sum_{m=1}^\infty\frac{(-1)^m}{m}\left(\sum_{k=1}^{\lfloor n/2\rfloor}\binom{n}{2k}x^{2k}\right)^m. \end{align} Comparing the individual coefficients, we get \begin{align} S_2&=2\binom{n}{2}&&=n^2-n,\\ S_4&=2\binom{n}{2}^2-4\binom{n}{4}&&=\color{blue}{\frac{n^4-4n^2+3n}{3}},\\ S_6&=2\binom{n}{2}^3-6\binom{n}{2}\binom{n}{4}+6\binom{n}{6}&&=\frac{2n^6-10n^4+23n^2-15n}{15}, \end{align} and so on; say, $S_8=(17n^8-112n^6+308n^4-528n^2+315n)/315$.

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  • $\begingroup$ Instead of [the power series of] $\log(\dots)$, one might use Girard-Newton identities. $\endgroup$
    – metamorphy
    Commented Nov 7, 2022 at 6:00
  • $\begingroup$ Another "algorithmic recipe": let $[z^n]f(z)=a_n$ for $f(z)=\sum_{n=0}^\infty a_n z^n$; then $$S_{2m}=n\Big[(-1)^m+[z^{2m-1}]\big((z\cot z)^{2m}\tan nz\big)\Big].$$ $\endgroup$
    – metamorphy
    Commented Nov 9, 2022 at 7:01
  • $\begingroup$ thank you for the answer I think it's very interesting and comments. I thought I got the gist of it earlier in the week but I need a little help with actual calculation. How does one calculate the $\begin{align} S_4&=2\binom{n}{2}^2-4\binom{n}{4}&&=\color{blue}{\frac{n^4-4n^2+3n}{3}} \end{align}$ from $\begin{align} 2\sum_{m=1}^\infty\frac{(-1)^m}{m}\left(\sum_{k=1}^{\lfloor n/2\rfloor}\binom{n}{2k}x^{2k}\right)^m. \end{align}$ $\endgroup$
    – onepound
    Commented Nov 12, 2022 at 8:35
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    $\begingroup$ @onepound: As said, just take the coefficient of $\color{blue}{x^4}$. The LHS has $S_4/2$, the RHS is $$-2\left[\binom{n}{2}x^2+\binom{n}{4}x^4+\dots\right]+\frac22\left[\binom{n}{2}x^2+\dots\right]^2-\dots$$ (here $\dots$ stand for higher powers of $x$) and has the coefficient of $x^4$ equal to $\binom{n}{2}^2-2\binom{n}{4}$. $\endgroup$
    – metamorphy
    Commented Nov 12, 2022 at 9:55

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