Why the series is divergent, but the equation holds? $$\sum\limits_{i=1}^{\infty }{\sin kx}=\frac{1}{2}\cot \left( \frac{x}{2} \right)$$
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3This question is incorrect. What kind of equality is this? Approximately everywhere, in distributions, pointwise? – Norbert Feb 02 '13 at 10:41
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2The LHS is defined on $2\pi\mathbb{Z}$ while the RHS isn't! – HorizonsMaths Feb 02 '13 at 12:37
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Because you assumed that the series from which you derived it could be differentiated, when it could not.
Ron Gordon
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1@vonbrand: The Fourier series is not necessarily convergent at individual points; rather, it converges with respect to the $L^2(-\pi,\pi)$ metric. So even though a Fourier series representing a function does converge, the FS representing the derivative of that function is not guaranteed to converge. So there are additional conditions on the function, i.e., the piecewise continuity of the original function. See, for example, http://tutorial.math.lamar.edu/Classes/DE/ConvergenceFourierSeries.aspx – Ron Gordon Feb 02 '13 at 17:15
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1that I (mostly) knew, but I'm at a loss at the series you claim OP differentiated. – vonbrand Feb 02 '13 at 17:21
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http://math.stackexchange.com/questions/292468/fourier-series-of-log-sine-and-log-cos/292477#292477 – Ron Gordon Feb 02 '13 at 17:25