# Hexus.Help.Maths-Coursework :p (Fourier Series)

• 26-11-2005, 07:15 PM
Twigman
Hexus.Help.Maths-Coursework :p (Fourier Series)
Anyhting that involves Fourier is gay. Everyone knows this. But...

Quote:

Compute the Fourier series representation of

f(x) = exp(ax), -pi <= x < pi.
Got the a0 term:

(sinh(a*pi))/(a*pi)

For an and bn i just keep going roun in circles for the integration by parts and splitting it up with different limits, evena nd odd functions etc isnt helping :(

Any ideas?
• 26-11-2005, 11:17 PM
dave87
If I understood what that meant, I'd help. But I dont, so i suggest teh following:

• 27-11-2005, 05:26 PM
Twigman
Cheer for the useful replies :p
Still as clueless as i was before tbh.
• 29-11-2005, 09:18 PM
Twigman
Ok..well its to do with the exponential trig identities.
(im an engineer so j is imaginary...learn to cope with it :p)

cos(nx) = { exp(jnx) + exp(-jnx) } / 2
and equivalent for sin,
and
exp(pi*j) = -1
(and exp(-pi*j) = 1)

but still i just get a mass of crap and i dont really know how to simplify it...

MATHS IS OFFICIALLY TEH GAY!!!!1111
• 29-11-2005, 09:32 PM
herulach
all that says is cos(pi*i) = 0, which im not sure is true anyway.
what you actually mean is the integral -pi to pi is 0, whcih is not the same
• 29-11-2005, 09:37 PM
herulach
writing it out, its not that difficult an integration, use the exponential identites, multiply by your initially function (rembering that exp(ax)*exp(bx) = exp((a+b)x) and it really isnt that difficult to integrate.
• 29-11-2005, 10:47 PM
Twigman
Quote:

Originally Posted by herulach
all that says is cos(pi*i) = 0, which im not sure is true anyway.

Where did you get that from? stupid maths person :p ;)
Quote:

what you actually mean is the integral -pi to pi is 0, whcih is not the same
...?
What?
integral of what? integral of sin is 0 yeah...as is cos...

its not the integration part that screws me over...its once you have done that, put in the limits...simplifying it seems to be beyond me :(