Consider the following Fourier series

$ f\left(t\right)=\sum_{k=1}^{N}c_{k}e^{i\frac{2\pi}{T}kt} $

Can we prove that the function which its Fourier series is $ f $ is a complex function due to the fact that its Fourier series only contains positive indices?

Also, are there other cool properties of a function that we can tell just from looking at its Fourier series? I know that for real functions we can tell if the function is odd/even by looking at the Fourier series in the form of the $ \sin + \cos $ orthogonal system. Are there any other properties?

Thanks in advance.


1 Answer 1


If $f(t)$ is real valued then $$\sum_{k=1}^{N}c_{k}e^{i\frac{2\pi}{T}kt} =\sum_{k=1}^{N}c_{k}e^{-i\frac{2\pi}{T}kt} $$ and orthogonality of the functiosn $e^{i\frac{2\pi}{T}kt}$, $k \in \mathbb Z$ on $[0,T]$ show that each $c_k$ must be $0$.

  • $\begingroup$ How does the orthogonality shows it? $\endgroup$
    – FreeZe
    Jul 12, 2021 at 16:24
  • $\begingroup$ Take inner product on both sides with $e^{i\frac {2pi} T mt}$ for any particuar $m$ (with $|m| \leq N$) . All terms except one will vanish and you get $c_m=0$. @FreeZe $\endgroup$ Jul 12, 2021 at 23:14
  • $\begingroup$ It is a basic fact that orthogonal functions are linearly independent. @FreeZe $\endgroup$ Jul 12, 2021 at 23:27

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