# PDE: How to use Fourier Series to express a real function?

1. Mar 11, 2015

### iScience

actually have two questions:

here we have a fourier series..
$$f(t) = \sum c_k e^{2\pi ikt}$$ (c is complex)
if we're trying to express a real function via fourier series, and we do it the following way..

Impose condition: $$\overline{c_k} = c_{-k}$$

$$f(t) = \sum\limits_{k= -n}^n c_k e^{2\pi ikt}$$
then, although the constants would become real, wouldn't the exponents go to zero? why not?

could someone give me more of an intuitive understanding for the "orthogonality" of sine functions? i know the integral definition, but how exactly does this integral dictate what can & can't be expressed in terms of what (ie could you relate the integral definition to the classical sense of the word 'orthogonal')

Last edited: Mar 11, 2015
2. Mar 11, 2015

### Strum

Maybe you should try and explain why the exponent would go to zero. Or why the constants become real for that matter. I dont understand you last paragraph.

3. Mar 11, 2015

### iScience

(i accidentally left out the k)

actually.. you'd essentially be having this

$$c(u+iv)e^{2\pi i k t} +c(u-iv)e^{-2\pi i k t}$$ (since the condition above would imply k is part of v)

so i don't even know how you'd add the two to get a real anymore..

4. Mar 11, 2015

### Strum

Try and write out the exponentials as cos and sin. Also complex number plus its conjugate is real.

5. Mar 11, 2015

### iScience

$$c(u+iv)(cosg+ising)+ c(u-iv)(-cosg-ising) = (cosg+ising)(c(u+iv)-c(u-iv))$$

yes, but instead of a plus i get a minus, what'd i do wrong?

6. Mar 11, 2015

### Strum

exp(-ix) = cos(x) - isin(x)

7. Mar 11, 2015

### iScience

oops, okay and after that, since all the sin terms go to zero they're irrelevant. thanks