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Prove that $\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=0$


It is very easy to prove this identity for each fixed $n$ . For example let $n = 6$; writing out all terms in a $5 \times 6$ matrix, we obtain:

$\begin{matrix} \tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13} & \tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13} & \tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13} & \tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13} & \tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13} \\ \tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13} & \tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13} & \tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13} & \tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13} & \tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13} \\ \tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13} & \tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13} & \tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13} & \tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13} & \tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13} \\ \tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13} & \tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13} & \tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13} & \tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13} & \tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13} \\ \tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13} & \tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13} & \tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13} & \tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13} & \tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13} \\ \tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13} & \tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13} & \tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13} & \tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13} & \tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13} \end{matrix}$

one can notice then, that the first column vanish the fourth one :

$\tan \dfrac {\pi } {13}\tan \dfrac {2\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {15\pi } {13}$

$\tan \dfrac {2\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {24\pi } {13}\tan \dfrac {30\pi } {13}$

$\tan \dfrac {3\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {16\pi } {13}\tan \dfrac {20\pi } {13}$

$\tan \dfrac {4\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {4\pi } {13}\tan \dfrac {5\pi } {13}$

$\tan \dfrac {5\pi } {13}\tan \dfrac {10\pi } {13}=-\tan \dfrac {8\pi } {13}\tan \dfrac {10\pi } {13}$

$\tan \dfrac {6\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {25\pi } {13}$

and the third column vanish the fifth one :

$\tan \dfrac {3\pi } {13}\tan \dfrac {4\pi } {13}=-\tan \dfrac {30\pi } {13}\tan \dfrac {36\pi } {13}$

$\tan \dfrac {6\pi } {13}\tan \dfrac {8\pi } {13}=-\tan \dfrac {5\pi } {13}\tan \dfrac {6\pi } {13}$

$\tan \dfrac {9\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {25\pi } {13}\tan \dfrac {30\pi } {13}$

$\tan \dfrac {12\pi } {13}\tan \dfrac {16\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {12\pi } {13}$

$\tan \dfrac {15\pi } {13}\tan \dfrac {20\pi } {13}=-\tan \dfrac {20\pi } {13}\tan \dfrac {24\pi } {13}$

$\tan \dfrac {18\pi } {13}\tan \dfrac {24\pi } {13}=-\tan \dfrac {15\pi } {13}\tan \dfrac {18\pi } {13}$

while the second column is self-vanishing:

$\tan \dfrac {2\pi } {13}\tan \dfrac {3\pi } {13}=-\tan \dfrac {10\pi } {13}\tan \dfrac {15\pi } {13}$

$\tan \dfrac {4\pi } {13}\tan \dfrac {6\pi } {13}=-\tan \dfrac {6\pi } {13}\tan \dfrac {9\pi } {13}$

$\tan \dfrac {8\pi } {13}\tan \dfrac {12\pi } {13}=-\tan \dfrac {12\pi } {13}\tan \dfrac {18\pi } {13}$ .

So the equality occurs. But how to generalize the proof?

  • 1
    @Micah: Your comment pointed in the right direction; see my answer.2012-07-21

4 Answers 4

25

Micah already pointed the way in a comment: The identity

$\tan a \tan b = \frac{\tan a - \tan b}{\tan(a-b)} - 1$

causes the inner sum to telescope. To make full use of this, let's extend the inner sum to $k=2n$:

$ \begin{align} \sum _{l=1}^{n}\sum _{k=1}^{2n}\tan \frac {lk\pi } {2n+1}\tan \frac {l(k+1) \pi } {2n+1} &= \sum _{l=1}^{n}\sum _{k=1}^{2n}\left(\frac{\tan \frac {l(k+1)\pi } {2n+1}-\tan \frac {lk \pi } {2n+1}}{\tan\frac {l\pi}{2n+1}}-1\right) \\ &= \sum _{l=1}^{n}\left(\frac{\tan \frac {l(2n+1)\pi } {2n+1}-\tan \frac {l\pi } {2n+1}}{\tan\frac {l\pi}{2n+1}}-2n\right) \\ &= \sum _{l=1}^{n}(0-1-2n) \\ &= -n(2n+1)\;. \end{align} $

This sum contains each of the terms we want to sum twice, with mirror symmetry, and it contains one additional term in the middle for $k=n$. Thus, for our sum to vanish, we need

$ \sum _{l=1}^{n}\tan \frac {ln\pi } {2n+1}\tan \frac {l(n+1) \pi } {2n+1}=-n(2n+1)\;. $

The arguments of the two factors add to $l\pi$, so they're negatives of each other, so we're looking for

$ -\sum _{l=1}^{n}\tan^2 \frac {ln\pi } {2n+1}\;. $

We can again extend the sum to $2n$ to double it, since the arguments form pairs that add up to $n\pi$; then, since $n$ and $2n+1$ are coprime, we can replace $ln$ by $l$ while traversing the same arguments; and then we can set the upper limit back to $n$, since the arguments still add up to $\pi$ in pairs. Thus, what we need is

$ \sum _{l=1}^{n}\tan^2 \frac {l\pi } {2n+1}=n(2n+1)\;. $

How to find this sum is shown at Prove that $\sum\limits_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$. Adapting the argument in the accepted answer there for our odd denominator, we obtain

$ \left(\cos\frac{k\pi}{2n+1}+\mathrm i\sin\frac{k\pi}{2n+1}\right)^{2n+1}=(-1)^k\;, $

taking the imaginary part,

$ \sum_{r=0}^n\binom{2n+1}{2r+1}\left(\cos\frac{k\pi}{2n+1}\right)^{2n-2r}\left(\mathrm i\sin\frac{k\pi}{2n+1}\right)^{2r+1}=0\;, $

dividing by $\left(\cos\frac{k\pi}{2n+1}\right)^{2n+1}$,

$ \sum_{r=0}^n\binom{2n+1}{2r+1}\left(\mathrm i\tan\frac{k\pi}{2n+1}\right)^{2r+1}=0\;, $

and dividing by $\tan\frac{k\pi}{2n+1}$ and letting $x=\tan^2\frac{k\pi}{2n+1}$,

$ \sum_{r=0}^n\binom{2n+1}{2r+1}(-x)^r=0\;. $

Then Vieta's formula shows that the sum of the roots of this equation is

$\frac{\binom{2n+1}2}{\binom{2n+1}0}=n(2n+1)\;,$

as required.

[Update:]

This answer suggests an alternative, more elementary way to calculate the sum of the squares of the tangents: On the grid $\frac{l\pi}{2n+1}$, the tangent decomposes into $n$ mutually orthogonal sines, each of whose dot product with itself is $2n+1$, so the dot product of the tangent with itself is $n(2n+1)$.

  • 0
    @robjohn: I didn't actually compute the sum of the squared sines in that answer, but that's easy to do using geometric series; no roots of polynomials or contour integration required :-)2012-07-22
23

Starting with $ \tan(x-y) = \frac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}\tag{1} $ we get $ \tan(x)\tan(y)=\frac{\tan(x)-\tan(y)}{\tan(x-y)}-1\tag{2} $ Thus, $ \tan\left(\frac{l(k+1)\pi}{2n+1}\right)\tan\left(\frac{lk\pi}{2n+1}\right) =\frac{\tan\left(\frac{l(k+1)\pi}{2n+1}\right)-\tan\left(\frac{lk\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}-1\tag{3} $ Therefore, because of the telescoping sum, $ \begin{align} \sum_{k=1}^{n-1}\tan\left(\frac{l(k+1)\pi}{2n+1}\right)\tan\left(\frac{lk\pi}{2n+1}\right) &=\frac{\tan\left(\frac{ln\pi}{2n+1}\right)-\tan\left(\frac{l\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}-(n-1)\\ &=\frac{\tan\left(\frac{ln\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}-n\tag{4} \end{align} $ Note that $ \frac{\tan\left(\frac{(2n+1-l)n\pi}{2n+1}\right)}{\tan\left(\frac{(2n+1-l)\pi}{2n+1}\right)} =\frac{\tan\left(\frac{ln\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)}\tag{5} $ so that by replacing the odd $l$s with even $2n+1-l$s and using $\tan(2x)=\frac{2\tan(x)}{1-\tan^2(x)}$, we get $ \begin{align} \sum_{l=1}^n\frac{\tan\left(\frac{ln\pi}{2n+1}\right)}{\tan\left(\frac{l\pi}{2n+1}\right)} &=\sum_{l=1}^n\frac{\tan\left(\frac{2ln\pi}{2n+1}\right)}{\tan\left(\frac{2l\pi}{2n+1}\right)}\\ &=-\sum_{l=1}^n\frac{\tan\left(\frac{l\pi}{2n+1}\right)}{\tan\left(\frac{2l\pi}{2n+1}\right)}\\ &=\frac12\sum_{l=1}^n\left(\tan^2\left(\frac{l\pi}{2n+1}\right)-1\right)\tag{6} \end{align} $


Using contour integration, we will compute $ \sum_{l=1}^n\tan^2\left(\frac{\pi l}{2n+1}\right)=n(2n+1)\tag{7} $ Note that $ \frac{(2n+1)/z}{z^{2n+1}-1}\tag{8} $ has simple poles. It has residue $1$ at $z=e^{\large\frac{2\pi li}{2n+1}}$ for each $l$ and residue $-(2n+1)$ at $z=0$.

Furthermore, at $z=e^{i\theta}$, $ -\left(\dfrac{z-1}{z+1}\right)^2=\tan^2(\theta/2)\tag{9} $ Because the total residue of $ f(z)=\left(\frac{z-1}{z+1}\right)^2\frac{(2n+1)/z}{z^{2n+1}-1}\tag{10} $ is $0$, we get that the sum of its residues at $z=0$ and $z=-1$ equals $ \sum_{l=1}^{2n}\tan^2\left(\frac{\pi l}{2n+1}\right)=2\sum_{l=1}^n\tan^2\left(\frac{\pi l}{2n+1}\right)\tag{11} $ First, $ \mathrm{Res}_{z=0}f(z)=-(2n+1)\tag{12} $ Next, $ \begin{align} \mathrm{Res}_{z=-1}f(z) &=\mathrm{Res}_{z=0}f(z-1)\\ &=\mathrm{Res}_{z=0}(2n+1)\left(\frac{z-2}{z}\right)^2\frac1{1-z}\frac1{1+(1-z)^{2n+1}}\\ &=\mathrm{Res}_{z=0}(2n+1)\left(1-\frac4z+\frac4{z^2}\right)(1+z+\dots)\frac12\left(1+\frac{2n+1}{2}z+\dots\right)\\ &=\mathrm{Res}_{z=0}\frac{2n+1}{2}\left(1-\frac4z+\frac4{z^2}\right)\left(1+\frac{2n+3}{2}z+\dots\right)\\ &=\mathrm{Res}_{z=0}\frac{4n+2}{z^2}+\frac{(2n+1)^2}{z}+\dots\\ &=(2n+1)^2\tag{13} \end{align} $ Combining $(11)$, $(12)$, and $(13)$, yields $(7)$.


Combining $(4)$, $(6)$, and $(7)$ yields $ \sum_{k=1}^{n-1}\tan\left(\frac{l(k+1)\pi}{2n+1}\right)\tan\left(\frac{lk\pi}{2n+1}\right)=0\tag{14} $

  • 6
    @joriki: Just as $\pi\cot(\pi z)$ has residue $1$ at all integers, $\dfrac{n/z}{z^n-1}$ has residue $1$ at all $n^{\text{th}}$ roots of unity (and residue $-n$ at $0$). They are nice tools for computing sums on $\mathbb{Z}$ and $\mathbb{Z}/n\mathbb{Z}$.2012-07-22
7

This isn't an answer, just some numerical results that don't fit in the comments.

If you first perform the sum over $l$, the result is always an integer multiple of $2n+1$. I don't see a pattern in the multipliers, but I thought I'd post them in case someone else does. The numbers in the table are the multipliers $m$ in

$ \sum _{l=1}^{n}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=m(2n+1)\;, $

with $n$ increasing downward and $k$ to the right.

$ \begin{array}{r|rr} &1&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16&17\\\hline 2&0\\ 3&-1&1\\ 4&0&-1&1\\ 5&-1&-1&1&1\\ 6&0&0&-2&0&2\\ 7&-1&0&-2&1&0&2\\ 8&0&-2&2&-2&0&0&2\\ 9&-1&1&-1&-3&1&1&-1&3\\ 10&0&-1&-1&2&-3&0&-1&1&3\\ 11&-1&-1&-1&1&-3&1&1&-1&1&3\\ 12&0&0&0&-2&2&-4&0&0&0&0&4\\ 13&-1&0&0&-1&0&-4&3&-1&1&-1&0&4\\ 14&0&-2&0&-2&2&2&-4&0&0&-2&2&0&4\\ 15&-1&1&-3&3&-3&1&-5&3&-1&1&-1&-1&1&5\\ 16&0&-1&1&-2&-1&1&2&-5&-1&1&0&-1&0&1&5\\ 17&-1&-1&1&-1&-3&2&0&-5&3&-1&1&1&-2&0&1&5\\ 18&0&0&-2&0&2&-4&0&4&-6&2&-2&0&-2&2&0&0&6\\ \end{array} $

-4

To prove that,

$\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=0$

Consider the following,

$\tan ( \frac {l(k+1)\pi } {2n+1} - \frac {l(k)\pi } {2n+1} ) = \frac { \tan \frac {l(k+1)\pi } {2n+1} - \tan \frac {l(k)\pi } {2n+1} } {1 + \tan \frac {l(k+1))\pi } {2n+1}\tan \frac {l(k) \pi } {2n+1}}$

i.e. $\tan ( \frac {l\pi } {2n+1} ) = \frac { \tan \frac {l(k+1)\pi } {2n+1} - \tan \frac {l(k)\pi } {2n+1} } {1 + \tan \frac {l(k+1))\pi } {2n+1}\tan \frac {l(k) \pi } {2n+1}}$

i.e. $\ ( 1 + \tan \frac {l(k+1))\pi } {2n+1}\tan \frac {l(k) \pi } {2n+1}) = \frac { \tan \frac {l(k+1)\pi } {2n+1} - \tan \frac {l(k)\pi } {2n+1} } {\tan \frac {l\pi } {2n+1}}$

i.e. $\tan \frac {l(k+1))\pi } {2n+1}\tan \frac {l(k) \pi } {2n+1} = (\frac { \tan \frac {l(k+1)\pi } {2n+1} - \tan \frac {l(k)\pi } {2n+1} } {\tan \frac {l\pi } {2n+1}} - 1)$

Hold l = 1 and take $\sum _{k=1}^{n-1}$

Next hold l = 2 and take $\sum _{k=1}^{n-1}$

.......

Continue this till l exhausts for l = n

Now sum all the terms you get and we see that the sum is indeed equal to 0

i.e. $\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \frac {lk\pi } {2n+1}\tan \frac {l( k+1) \pi } {2n+1}=0$

  • 0
    There are like terms which cancel out simplifying the process of summation.2012-07-21