# Exponential form of a Fourier series

Discussion in 'Math' started by amilton542, Nov 27, 2014.

1. ### amilton542 Thread Starter Active Member

Nov 13, 2010
491
64
Does anyone know why the exponential form of a Fourier series works from - infinity to +infinity?

I'm using Advanced Engineering Mathematics by Kreyszig but I find his derivations are a bit "arty farty" if that makes sense. I don't like them. He omits too much.

2. ### WBahn Moderator

Mar 31, 2012
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4,701
Your question is ambiguous.

Are you talking about -∞ to +∞ in the time domain as opposed to 0 to +∞ in the time domain for the typical Laplace transform?

If so, the reason is that the typical Laplace transform that we usually use is the one-sided Laplace transform and it blows up for t<0. There is the two-sided Laplace transform that works over all time, but it is quite a bit more troublesome to use.

3. ### studiot AAC Fanatic!

Nov 9, 2007
5,005
513

Application of the Euler formulae connecting sin and cos with a complex exponential (did you mention it was complex?) yields

$f(x) = {c_0} + \sum\limits_1^\infty {\left( {{c_n}{e^{inx}} + {k_n}{e^{ - inx}}} \right)}$

The summation runs from n=1 to n = infinity.

Put that into the first term in the sum you get all the terms from 1 to infinity.

Put it into the second and you get all the terms from minus infinity to -1,
The additional term for n=0 when the summationlimits are changed comes out of the constant before the old summation, which is then absorbed into the new summation.

If you like you can decompose it thus

$f(x) = \sum\limits_{ - \infty }^{ - 1} {\left( {{c_n}{e^{inx}}} \right)} + {c_0}{e^{0ix}} + \sum\limits_1^\infty {\left( {{c_n}{e^{inx}}} \right)}$

since both expressions have the same form these can be combined into a single expression running from minus infinity to plus infinity.

$f(x) = \sum\limits_{ - \infty }^\infty {{c_n}} {e^{inx}}$

Which is the exponential form you are asking about.

Last edited: Nov 28, 2014
4. ### WBahn Moderator

Mar 31, 2012
17,457
4,701
My apologies -- I thought you were asking about the Fourier transform, not the Fourier series. Studiot gave a good answer.

5. ### amilton542 Thread Starter Active Member

Nov 13, 2010
491
64
Yeah I get it now. Change the summation limits of the exponential with the negative exponent then it becomes positive. Another thing I was a bit skeptical about was how the latter expression "masks" the constant but I see what you've done now.

O.K thanks.