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I am considering the sum $$ A_m = \sum_{j=0}^m \sin^4\left(\frac{j}{m}\cdot\frac{\pi}{2}\right). $$ I think that for $m>1$ it holds $$ A_m = \frac{3m+4}{8}, $$ but I can't really get to it.

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  • $\begingroup$ So $A_1$ would be negative ? $\endgroup$ Apr 28, 2016 at 9:38
  • $\begingroup$ Write $\sin{x} = \frac{e^{ix}-e^{-ix}}{2i}$ and try to use the binomial theorem and the sum of a geometric sequence. $\endgroup$
    – KoA
    Apr 28, 2016 at 9:58
  • $\begingroup$ @Claude: Fixed the wrong minus sign $\endgroup$ Apr 28, 2016 at 10:05

1 Answer 1

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Start by linearizing $\sin^{4}(x)$ :

$$ \forall x \in \mathbb{R}, \, \sin^{4}(x) = \frac{1}{8}\Big( \cos(4x) - 4\cos(2x) + 3 \Big). $$

This leads to :

$$ \sum_{j=1}^{m} \sin^{4}\Big( \frac{j\pi}{2m} \Big) = \frac{1}{8}\Bigg( \sum_{j=1}^{m} \cos\Big( \frac{2j\pi}{m} \Big) - 4 \sum_{j=1}^{m} \cos\Big( \frac{j\pi}{m} \Big) + 3m \Bigg). $$

The sums $\displaystyle \sum_{j=1}^{m} \cos\Big( \frac{2j\pi}{m} \Big)$ and $\displaystyle \sum_{j=1}^{m} \cos\Big( \frac{j\pi}{m} \Big)$ can be computed using complex numbers. For $z \in \mathbb{C}$, $\Re(z)$ denotes the real part of $z$. Indeed, note that :

$$ \sum_{j=1}^{m} \cos \Big( \frac{2j\pi}{m} \Big) = \Re\Bigg( \sum_{j=1}^{m} e^{2ij\pi / m} \Bigg). $$

and, since $\exp(2i\pi /m) \neq 1$ for all $m$, we have :

$$ \sum_{j=1}^{m} e^{2ij\pi / m} = e^{2i\pi / m} \frac{1 - \big( e^{2i\pi / m} \big)^{m} }{1 - e^{2i\pi / m} } = 0. $$

Similarily : $\displaystyle \sum_{j=1}^{m} \cos\Big( \frac{j\pi}{m} \Big) = -1$.

As a consequence :

$$ \sum_{j=1}^{m} \sin^{4}\Big( \frac{j\pi}{2m} \Big) = \frac{1}{8}(3m+4). $$

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