Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
4  
+1 Kudos for writing down all that! I liked in particular your optimism, where you write "One can notice then that..."...I must confess I noticed nothing. –  DonAntonio Jul 21 '12 at 1:42
4  
By using the identity $\tan a \tan b = \frac{\tan a - \tan b}{\tan(a-b)} - 1$, you can telescope the inner sum, showing that your quantity is equal to $-n^2+\sum_{l=1}^n \frac{\tan [nl\pi/(2n+1)]}{\tan [l\pi/(2n+1)]}$. I'm not sure where to go from there, but this sure looks like a simplification... –  Micah Jul 21 '12 at 4:49
    
Fixed typo noted by @anon. –  joriki Jul 21 '12 at 7:17
2  
From your example, one might think that this is a number-theoretic effect that can be explained just in terms of residues, but that seems to be the case only if $2n+1$ is prime; if $2n+1$ is composite, there are terms that don't cancel any other term directly. The simplest such case is $n=4$ with $2n+1=9$, where the four terms for $k=1$ have no cancelling partners, but the sum of their contributions nevertheless vanishes, and this seems to always be the case. Thus the answer will have to involve more specific properties of the tangent than just its symmetries. –  joriki Jul 21 '12 at 8:33
1  
@Micah: Your comment pointed in the right direction; see my answer. –  joriki Jul 21 '12 at 16:57

4 Answers 4

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 $\sum_{k=1}^{n-1}\tan^{2}\frac{k \pi}{2n} = \frac{(n-1)(2n-1)}{3}$. 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 $n(2n+1)$.

share|improve this answer
3  
+1. Extending the sum up to $k=2n$ is brilliant. –  Did Jul 21 '12 at 17:02
    
@did: I actually first started from Micah's result and got the sum of squared tangents through further trigonometric transformations before I saw this more direct way. –  joriki Jul 21 '12 at 17:16
    
Your proof is correct. It is approximatively what I've done: having the value of the sum $\sum _{l=1}^{n}\tan ^{2}\dfrac {l\pi } {2n+1}=n\left( 2n+1\right)$, one can infer the value of $\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \dfrac {lk\pi } {2n+1}\tan \dfrac {l\left( k+1\right) \pi } {2n+1}$. The real difficulty occur when we want to prove $\sum _{l=1}^{n}\tan ^{2}\dfrac {l\pi } {2n+1}=n\left( 2n+1\right)$ WITHOUT USING COMPLEX NUMBERS THEORY –  Maria Mikolayevskaya Jul 21 '12 at 19:09
    
and finding an appropriate polynominal, what I tried to do as an exercice...In this case the only difficult thing to manipulate is proving in a direct way, that $\sum _{l=1}^{n}\sum _{k=1}^{n-1}\tan \dfrac {lk\pi } {2n+1}\tan \dfrac {l\left( k+1\right) \pi } {2n+1}=0$...But nobody, as me too, know how to do it... –  Maria Mikolayevskaya Jul 21 '12 at 19:09
1  
I was just about to post a similar proof, when I saw that you had posted this (darn!). So I had to work up another proof using contour integration. Very nice! (+1) –  robjohn Jul 22 '12 at 12:18

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 $$ \sum_{k=1}^{n-1}\tan\left(\frac{l(k+1)\pi}{2n+1}\right)\tan\left(\frac{lk\pi}{2n+1}\right)=0\tag{14} $$

share|improve this answer
3  
How on Earth did you come up with this? :-) –  joriki Jul 22 '12 at 12:27
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 $\mathbb{Z}/n\mathbb{Z}$. –  robjohn Jul 22 '12 at 13:10

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.

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

share|improve this answer

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

share|improve this answer
    
You need to enclose the $\TeX$ code in dollar signs to get it formatted. Single dollar signs for inline formulas, double dollar signs for displayed equations. –  joriki Jul 21 '12 at 11:08
    
$\displaystyle \tan\frac{l\pi}{2n+1}\ne0$. (Also you need parentheses around the difference argument of the tangent.) –  joriki Jul 21 '12 at 11:14
    
@Maria: You need double dollar signs to get displayed equations. Using inline formulas for expressions with double fractions is very hard on the eyes. –  joriki Jul 21 '12 at 11:39
    
You can use \left[ and \right] instead of the plain brackets to get the right size of bracket appropriate to the expression contained. All this formatting notwithstanding, as I wrote above, $\displaystyle\tan\frac{l\pi}{2n+1}$ isn't zero, so I don't think any of this works. –  joriki Jul 21 '12 at 11:42
    
There are like terms which cancel out simplifying the process of summation. –  Rajesh K Singh Jul 21 '12 at 12:29

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.