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I want to show $$ \frac{x}{2k^2}\sum_{j=0}^{k-1}(2j+1)^3\sin \left(\frac{(2 j+1) x}{2 k}\right) = \frac{x \csc \left(\frac{x}{2 k}\right) \left(2 k \cos (x) \left(4 k^2-6 \csc ^2\left(\frac{x}{2 k}\right)+3\right)+\sin (x) \cot \left(\frac{x}{2 k}\right) \left(-12 k^2+6 \cot ^2\left(\frac{x}{2 k}\right)+5\right)\right)}{4 k^2} $$ for $x\in(0,\pi/2)$ and for all $k=1,2,3,...$. This formula is given by Mathematica. However, I have no idea how to give a rigorous proof of it. Any suggestion, idea, or comment is welcome, thanks!

LCH
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  • For real $x,;$ $\sin ((2j+1)x/2k)$ is the imaginary part of $y^{2j+1}$ where $y=\exp (x/2k).$ So you want $x/2k^2$ times the imaginary part of $\sum_{j=0}^{k-1}(2j+1)^3 y^{2j+1}$ – DanielWainfleet Mar 10 '17 at 08:19

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HINT:

Let $f(x)=\sum_{j=0}^{k-1}\cos\left(\frac{(2j+1)x}{2k}\right)$. Then, we have

$$f'''(x)=\sum_{j=0}^{k-1}\frac{(2j+1)^3}{(2k)^3}\sin\left(\frac{(2j+1)x}{2k}\right)$$

Use the result from THIS ANSWER to evaluate the sum $f(x)=\sum_{j=0}^{k-1}\cos\left(\frac{(2j+1)x}{2k}\right)$, differentiate $3$ times, and multiply by the appropriate factor.

Community
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Mark Viola
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  • Thanks! In http://math.stackexchange.com/questions/17966/how-can-we-sum-up-sin-and-cos-series-when-the-angles-are-in-arithmetic-pro/1214626#1214626, $N=n$ in the result you suggest above? – LCH Mar 10 '17 at 07:27
  • Yeah! Following your suggestion, I've completed the proof of it. The Euler formula is indeed helpful. – LCH Mar 10 '17 at 13:22
  • Pleased to hear. – Mark Viola Mar 10 '17 at 14:00