1
$\begingroup$

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

enter image description here

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

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.