2

I have a question that gives me a periodic function $f(x)$ and asks me to find the complex Fourier series (which I think I have done correctly) and then asks me to obtain from that the regular Fourier series. I assume by 'regular' it just means real. I have no idea how to turn this complex series into a real series. I worked out what the real series should be by using Euler's formulae but that isn't obtaining it using the complex series (like the question asks).

The initial $f(x)$ given is: $$f(x) = \begin{cases} 0 & -\pi < x \leq 0\\ x & 0 < x < \pi \end{cases}$$

The complex Fourier series I found is:

$$f(x) = \frac{\pi}{4} + \sum_{n = -\infty}^{\infty}\frac{1}{2\pi n^2}\left(1-\frac{n\pi}{i} - e^{inx}\right)$$

The 'regular' series I found is:

$$f(x) = \frac{\pi}{4} +\sum_{n=1}^{\infty}\frac{1}{\pi n^2}((-1)^{n}-1)\cos(nx)-\frac{1}{n}(-1)^n\sin(nx)$$

How do I get from the complex Fourier series to the regular series?

(Sorry I couldn't get nice formatting done above)

sion
  • 157

2 Answers2

3

Just use: $$e^{ikx} = \cos(kx) + i\sin(kx)$$ Generally, if the fourier transform is real, you'll have terms like $e^{ikx}+e^{-ikx}$ that cancel out the imaginary part. In your case, divide the infinite series into three parts: $$\sum_{k=-\infty}^{\infty} =\sum_{k=-\infty}^{-1} + \sum_{k=1}^{\infty} + \sum_{k=0}^{0}$$

Notice how you now get terms like $e^{ikx}+e^{-ikx}=2\cos(kx)$, or $e^{ikx}-e^{-ikx}=2i\sin(kx)$.

Pratyush Sarkar
  • 3,734
Nathaniel Bubis
  • 33,018
  • Splitting the sums up like you suggested really helped to see what was could be done. I ended up with $$f(x) = \frac{\pi}{4} +\sum_{n=1}^{\infty}\frac{1-cos(nx)}{\pi n^2}$$ Im not sure if that is the same as the regular expression I found before (in the original post) but its much neater so I'll just go with it. – sion Jun 03 '13 at 21:55
  • Are there any websites that can calculate the series in these expansion forms when given a piecewise formula? I can find plenty of sites that do it term by term but then I have to try and calculate the formula for the expanding sum from those terms... – sion Jun 03 '13 at 21:57
0

The correct complex Fourier series is given below.

$$\ f(x)= \sum_{n=-∞}^∞ \frac{1}{2πn^2}(-1^n-1−\frac{nπ}{i}(-1)^n) *e^{inx}$$

To get real Fourier series from complex, try to expand +n and negative n and add them the imaginary part will vanish to give the correct real FS

rakeshKM
  • 1