My question is about Dirichlet kernel identity. Why is the following true?
$$\sum_{k=-n}^{n}e^{ikx}=1+ 2\sum_{k=1}^{n}\cos(kx)$$
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Hint: $e^{ikx}+e^{i(-k)x}=2cos(kx)$, for all $1\le k\le n$.
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$\begingroup$ Can I just check, if you have the representation $2\sum^{n}_{k=0}\cos(kx)$ then if you evaluate the first term ($k=0$) would that not give the an extra $2$ rather than $1$ in the identity? $\endgroup$ – user230715 Sep 4 '15 at 13:55
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$\begingroup$ @GeorgeSimpson, Yes we have an extra 2 in that case. $\endgroup$ – Woria Sep 9 '15 at 19:06
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$\begingroup$ Sorry to be really stupid, but with that in mind, how come there is a $+1$ in the trigonometric Dirichlet representation rather than a $+2$? $\endgroup$ – user230715 Sep 10 '15 at 8:52
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$\begingroup$ @GeorgeSimpson, If $k=0$ then $k=-k$ and there is one $+1$ not two. $\endgroup$ – Woria Sep 10 '15 at 14:37
