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I am doing some practice problems for fourier series and I don't fully understand the solution to the following problem.

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I understand part (c) and (e) but I dont understand part (b) without taking the integral (in this case, the integral is quite easy but one could be given a much more complex function to integrate and I would like to understand the relationship between the coefficients and the original function).

Originally, I had thought that $a_n = c_n + c_{-n}$ which are the coefficients corresponding to $\cos$ in the fourier series expansion. In this case, it is $c_n - c_{-n}$ and so I thought that it wouldnt affect the fact that it is real but clearly I am incorrect, so would someone mind explaining please ?

Also, is it a general condition that if the $f(x)$ is real, then $c_{n}$ is purely imaginary and so $c_n - c_{-n}$ is imaginary?

Thank you

PutsandCalls
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1 Answers1

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If $f$ is real then $f(x) = f^*(x)$ (complex conjugate). This means that if $f(x) = \sum_{n\in\mathbb{Z}} c_ne^{2\pi i n x}$ then we also have $$f(x) = f^*(x) = \sum_{n\in\mathbb{Z}} c_n^*e^{-2\pi i n x} = \sum_{n\in\mathbb{Z}} c_{-n}^*e^{2\pi i n x}$$ By uniqueness of Fourier series it follows that $c_n = c_{-n}^*$. From this it follows that $$c_n - c_{-n} = c_n - c_n^* = 2i\cdot\text{Im}[c_n]$$ which is always imaginary (or zero) so the property is a general condition for Fourier series of real functions.

Another slighty different way to answer the question is by using what you mention in the question (relating $c_n$ to the $a_n,b_n$ coefficients). A Fourier series $f(x) = \sum_{n\in\mathbb{Z}} c_ne^{2\pi i n x}$ can also be written on the form $$f(x) = a_0 + \sum_{n=1}^\infty a_n\cos(2\pi n x) + b_n\sin(2\pi nx)$$ where $a_0 = c_0$, $a_n = c_n + c_{-n}$ and $b_n = i(c_n - c_{-n})$ for $n\geq 1$. Since your (periodically continued) $f$ is odd we must have $a_n = 0$ and since it's real it follows that $b_n = i(c_n-c_{-n})$ must be real and therefore $c_n-c_{-n}$ must be purely imaginary.

Winther
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