# Fourier series - complex exponential form

• Sep 9th 2010, 11:11 AM
Mollier
Fourier series - complex exponential form
Hi,

I was taking a break from numerical analysis and had a sneak peak at Fourier Series.
I want to rewrite the trigonometric form into the form with complex exponentials.
I have,

$\displaystyle f(t) = \sum^n_{k=0}a_ksin(2\pi kt) + b_kcos(2\pi kt)$,

and rewrite it using,

$\displaystyle sin(2\pi kt) = \frac{e^{2\pi ikt}-e^{-2\pi ikt}}{2i}$,

and,

$\displaystyle cos(2\pi kt) = \frac{e^{2\pi ikt}+e^{-2\pi ikt}}{2}$ to get,

$\displaystyle f(t) = \sum^n_{k=0} \frac{e^{2\pi ikt}(a_k+ib_k)-e^{-2\pi ikt}(a_k-ib_k)}{2i}$.

Not sure where to go from here. I see that there is come complex conjugate action in the numerator...
Any hints?

Thanks
• Sep 10th 2010, 07:58 AM
InvisibleMan
Your'e basically done, if a_k and b_k are real coefficients then write $\displaystyle c_k=sgn(k)(a_k+ib_k)$ and $\displaystyle c_0=b_0$ and you get a summation from -n to n.

If the coefficients aren't real, just write $\displaystyle a_k= s_k+ir_k$ and $\displaystyle b_k=d_k+it_k$.
you get that $\displaystyle a_k+ib_k= s_k+id_k-t_k+ir_k$
$\displaystyle a_k-ib_k= s_k-id_k+t_k+ir_k$

And you can do the same thing as with the real part, just replacing a_k and b_k with s_k and d_k and also with t_k and r_k.

Cheers!