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The $a_n,b_n$ and $c_n$ are Fourier coefficients. I start by expressing $a_n$ and $b_n$ in terms of $c_n$ as follows: since for every complex number $z$, $|z|^2=z\overline{z}$ we have that

\begin{align}|a_n|^2+|b_n|^2&=a_n\overline{a_n}+b_n\overline{b_n}=(c_n+c_{-n})(\overline{c_n}+\overline{c_{-n}})+i(c_n-c_{-n})(-i)(\overline{c_n}-\overline{c_{-n}})\\ &=2c_n\overline{c_n}+2c_{-n}\overline{c_{-n}}=2\left(|c_n|^2+|c_{-n}|^2\right), \end{align}

so

$$\sum_{n\in\mathbb{Z}}|a_n|^2+|b_n|^2=2\sum_{n\in\mathbb{Z}}\left(|c_n|^2+|c_{-n}|^2\right).\tag1$$

Here I'm stuck. How do I get the $4|c_0|^2$ infront and how do I get rid of the terms $|c_{-n}|^2$ in the sum?

Parseval
  • 6,413

1 Answers1

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Assuming $a_n, b_n$ are the usual cosine and sine coefficients, they are defined only for n non-negative, with $a_0=2c_0, b_0=0$ and then the computation works - on the left n non-zero is positive, so on the right you group n and -n to get the sum over the non-zero integers, while the 0-coefficient has the right factor as above.

Conrad
  • 27,433
  • So if they are only defined for non-negative $n$, why is the sum range $\mathbb{Z}$ as indicated in the question? – Parseval Feb 02 '19 at 10:18
  • It's probably a mistake (as cos is even and sin is odd, you can immediately see that it makes sense to define the sin and cos Fourier coefficients only for natural numbers as they essentially duplicate for negative numbers) – Conrad Feb 06 '19 at 12:50