Finding the Fourier Series of a Step Function

[stunner5000pt]

Find the Fourier Series in terms of \phi_{n} = \sin{nx} of the step function

f(x) = 0 for 0 \leq x \leq \frac{1}{2} \pi
= 1 for \frac{1}{2} \pi < x \leq \pi

Solution
Fourier Series is \Sigma c_{n} \phi_{n}
and for hte interval for x between 0 and 1/2 pi
c_{n} = \frac{\int_{0}^{\frac{1}{2}\pi} 0 dx}{\int_{0}^{\frac{1}{2}\pi} (\sin{nx})^2 dx
because rho is assumed to be 1
the numerator is a constant and assumed to be 1
the denominator
\int_{0}^{\frac{1}{2}\pi} (\sin{nx})^2 dx = \int_{0}^{\frac{1}{2} \pi} (\frac{e^{ix}-e^{-ix}}{2i})^2 = \frac{-1}{4} [\frac{e^{2ixn}}{2in}- \frac{e^{-2inx}}{2in} -2x]_{0}^{\frac{1}{2} \pi} = \frac{\sin{n \pi} - \pi}{-4} = \frac{\pi}{4}


P.s. Not done typing this yet but can you tell me if I am going in the right direction?

 

[fourier jr]

i don't think it's that complicated. just use integration stuff you learned in 1st-year calculus to deal with that sine integral. that's all you need to know. i think there's some identity that will change that sin^2 into cos something. i can't remember it off the top of my head but i think that how i learned to reduce an even power of a trig function.

& what is rho? i don';t think that appears in anything you've written there.

 

[stunner5000pt]

rho is supposed to be the weight

what is the integral of (Sin nx)^2??
can anyone help!

 

[fourier jr]

aha I found the identity in my 2nd edition of edwards/penney's calculus text, in the section on trig integrals:
sin^{2}n\theta = \frac{1 - cos(2n\theta)}{2}