I am required to prove the following equality:


For every $x\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$.

Well, apparently, the Fourier Series of $f(x)=x$ is the RHS of the equation, which makes the equality easy to prove. The problem is that I wasn't aware of that. This question was a part of a Fourier Series exercise on the one hand, but on the other hand, I'm not sure whether am I required to know Fourier Series of some elementary functions by heart, or rather maybe there's a more sophisticated proof to the above equality.

Thank you very much!

  • $\begingroup$ math.stackexchange.com/questions/3361589/… $\endgroup$ – lab bhattacharjee Nov 13 '19 at 7:46
  • $\begingroup$ @labbhattacharjee It's more complicated here since there $n=1$, but here $n\in\mathbb Z$ $\endgroup$ – Amit Zach Nov 13 '19 at 7:51
  • $\begingroup$ I'm not sure if this is true, because at $x=\frac{\pi}{2}$, the LHS is $0$, but the RHS becomes $$\sum_{k=0}^\infty 2\frac{(-1)^k}{2k+1}=2\arctan(1) = \frac{\pi}{2}$$ $\endgroup$ – Ninad Munshi Nov 13 '19 at 8:18
  • $\begingroup$ @AmitZach, I believe you meant $n=1$ in place of $i=1$ $\endgroup$ – lab bhattacharjee Nov 13 '19 at 8:34

$$-\sum_{n=1}^\infty\dfrac{(-1)^n\sin(2rnx)}n$$ is the imaginary part of $$-\sum_{n=1}^\infty\dfrac{(-1)^ne^{i2rnx}}n=-\sum_{n=1}^\infty\dfrac{(-e^{i2rx})^n}n=\ln(1-(-e^{i2rx}))$$

Now the principal value of $-\sum_{n=1}^\infty\dfrac{(-e^{i2rx})^n}n=\ln(1+e^{i2rx})$ is $$\ln(e^{irx})+\ln(e^{irx}+e^{-irx})=\underbrace{irx}_{\text{imaginary part}}+\ln(2\cos rx)$$

Here $r=\dfrac12, r=1$

  • $\begingroup$ Thank you very much! $\endgroup$ – Amit Zach Nov 13 '19 at 9:40

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