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Find the value of

$\dfrac{\sum_\limits{k=0}^{6}\csc^2\left(x+\dfrac{k\pi}{7}\right)}{7\csc^2(7x)}$

when $x=\dfrac{\pi}{8}$.

The Hint given is: $n\cot nx=\sum_\limits{k=0}^{n-1}\cot\left(x+\dfrac{k\pi}{n}\right)$

I dont know how it comes nor how to use it

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  • $\begingroup$ Another hint : $\csc^2 x = 1 + \cot^2 x$. $\endgroup$ – A---B Feb 21 '17 at 14:02
  • $\begingroup$ is this question from FIITJEE AITS? $\endgroup$ – Navin Feb 21 '17 at 14:24
  • $\begingroup$ indeed fiitjee aits $\endgroup$ – Maverick Feb 21 '17 at 14:25
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with the help of hint $\displaystyle \sum^{n-1}_{k=0}\cot\left(x+\frac{k\pi}{n}\right) = n\cot(nx)$

differentiate with respect to $x$

$\displaystyle \displaystyle -\sum^{n-1}_{k=0}\csc^2\left(x+\frac{k\pi}{n}\right) = -n^2\csc^2(nx)$

$\displaystyle \sum^{n-1}_{k=0}\csc^2\left(x+\frac{k\pi}{n}\right) = n^2\csc^2(nx)$

put $n=7$ and $\displaystyle x = \frac{\pi}{8}$

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  • $\begingroup$ Very nice. Can we derive $n\cot nx=\sum_\limits{k=0}^{n-1}\cot\left(x+\dfrac{k\pi}{n}\right)$ $\endgroup$ – Maverick Feb 21 '17 at 14:07
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    $\begingroup$ Using $\displaystyle \prod^{n-1}_{k=0}\sin \left(x+\frac{k\pi}{n}\right) = \frac{\sin (nx)}{2^{n-1}}$ . Taking $\ln$ and Differentiate with respect to $x$ $\endgroup$ – DXT Feb 21 '17 at 14:11
  • $\begingroup$ math.stackexchange.com/questions/2024464/… $\endgroup$ – DXT Feb 21 '17 at 14:11
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    $\begingroup$ @PankajSinha, See math.stackexchange.com/questions/1607134/… $\endgroup$ – lab bhattacharjee Feb 21 '17 at 17:53
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For future reference with this problem being tagged complex-numbers we show how to evaluate the sum using residues. Suppose we are interested in

$$S(n) = \sum_{k=0}^{n-1} \csc^2\left(x+\frac{k\pi}{n}\right) = \sum_{k=0}^{n-1} \frac{2}{1-\cos\left(2x+\frac{2k\pi}{n}\right)}.$$

where we take $x$ to be a real number.

With $$f(z) = \frac{4}{2-\exp(2ix)z-1/\exp(2ix)/z} \frac{nz^{n-1}}{z^n-1}$$

or alternatively

$$f(z) = \frac{4}{2-\exp(2ix)z-1/\exp(2ix)/z} \frac{1}{z} \frac{n}{z^n-1} \\ = \frac{4}{2z-\exp(2ix)z^2-1/\exp(2ix)} \frac{n}{z^n-1} \\ = -\frac{4\exp(-2ix)}{z^2 - 2z\exp(-2ix) + \exp(-4ix)} \frac{n}{z^n-1} \\ = -\frac{4\exp(-2ix)}{(z-\exp(-2ix))^2} \frac{n}{z^n-1}$$

we get for the sum with $\zeta_k = \exp(2\pi i k/n)$

$$\sum_{k=0}^{n-1} \mathrm{Res}_{z=\zeta_k} f(z)$$

which means we can evaluate the sum using the negative of the residues at $z=\exp(-2ix)$ and at infinity. Note however that with $R$ the radius of a circle going to infinity we get that $f(z)$ is $\theta(1/R^{n+2})$ and $2\pi R \times 1/R^{n+2} = 2\pi \times 1/R^{n+1}$ vanishes so the residue at infinity is zero. That leaves for the other residue

$$\left. \left(-\frac{4n\exp(-2ix)}{z^n-1}\right)'\right|_{z=\exp(-2ix)} = \left.\frac{4n\exp(-2ix)}{(z^n-1)^2} \times n z^{n-1} \right|_{z=\exp(-2ix)} \\ = \frac{4n^2\exp(-2inx)}{(\exp(-2inx)-1)^2} = -\frac{(2i)^2 n^2}{(\exp(-inx)-\exp(inx))^2}.$$

We obtain

$$S(n) -\frac{(2i)^2 n^2}{(\exp(inx)-\exp(-inx))^2} = 0$$

or

$$\bbox[5px,border:2px solid #00A000]{ S(n) = n^2 \csc^2(nx).}$$

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