In the following fourier series, how the red marked numbers are calculated?
Answer
The easiest way to verify this equality is to work from the right hand side. Expand it into a Fourier series, and check that the result is the series on the left. Then appeal to an appropriate theorem on the pointwise convergence of Fourier series; Dirichlet's theorem is tailor-made for this case.
But if you insist on working from left to right, then complex analysis can help. The function $\sin (2k-1)x$ is secretly the imaginary part of $\exp((2k-1) i x)=z^{2k-1}$ where $z=\exp(ix)$. Then use the Taylor series
$$2\sum_{k=1}^\infty \frac{z^{2k-1}}{2k-1} = \log\frac{1+z}{1-z}$$
but watch out: our $z$ is right on the border of the disk of convergence. Finally, use the fact that the map $w= \frac{1+z}{1-z}$ sends the unit disk onto the right half-plane, and the imaginary part of $\log$ picks up the argument of $w$. Said argument is $\pm \pi/2$ when $w$ is on the boundary of the right half-plane...
The last, but not the least: try to type formulas instead of posting images. It's not that hard.
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