Fourier coefficients

  • #1
274
1

Homework Statement


Let ##f## be a ##2\pi## periodic function. Let ##\hat{f}(n)## be the fourier coefficient of ##f## defined by
$$
\hat{f}(n)=\frac{1}{2\pi}\int_{a}^{b}f(x)e^{-inx}dx.
$$
for ##n\in\mathbb{N}##. If ##\overline{\hat{f}(n)}=\hat{f}(-n)## show that ##f## is real valued.

The Attempt at a Solution


Suppose for contradiction that ##f## is not real valued. Then
$$
\overline{\hat{f}(n)}=\frac{1}{2\pi}\overline{\int_{a}^{b}f(x)\cos nxdx}=\frac{1}{2\pi}\left[\overline{\int_{a}^{b}\Re(f(x)\cos nx)dx+i\int_{a}^{b}\Im(f(x)\cos nx)dx}\right]\\
=\frac{1}{2\pi}\left[\int_{a}^{b}\Re(f(x)\cos nx)dx-i\int_{a}^{b}\Im(f(x)\cos nx)dx\right]
$$
and
$$
\hat{f}(-n)=\frac{1}{2\pi}\left[\int_{a}^{b}\Re(f(x)\cos nx)dx+i\int_{a}^{b}\Im(f(x)\cos nx)dx\right]
$$
so ##\hat{f}(n)\not=\hat{f}(-n)## a contradiction. Can anyone verify my proof for me? Thanks!
 
Last edited:

Answers and Replies

  • #3
274
1
Woops! This was an if and only if problem and I was having trouble with the converse part. Sorry for the confusion.
 

Related Threads on Fourier coefficients

  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
8
Views
2K
  • Last Post
Replies
9
Views
703
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
10
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
11
Views
754
  • Last Post
Replies
0
Views
1K
  • Last Post
Replies
3
Views
8K
Top