Find the fourier sine series of cosine.

[pondzo]

Homework Statement

Hi, so I am doing some past exam papers and there was this question;

Homework Equations

The Attempt at a Solution

a₀ and a_n both are equal to zero, this leaves only b_n.

Since you can only use the sine series for an odd function, and cos(t) is even, does this mean i have to find the odd half expansion of cos(t) with L=π ? This would be f(t) = -cos(t) for [-π,0] and cos(t) for [0,π]. Or do I use L= π/2 and do the integral from 0 to π for b_n?

Letting n=1 since it is the first partial sum we are talking about (can i do this?)then using the first method I get 16/(3π) and if i try the second method I get 8/(3π). But the answer is listed as 0.

Any help would be great thanks.

 

[ehild]

You should write the relevant equations.
The problem text does not define f(t) in the interval -pi<t<0. If its Fourier series contains only sine terms, that suggests that the function is odd, f(t) = -cos(t) if -pi<t<0.
What is the formula for the Fourier coefficients of an odd function?

 

[pondzo]

Hi ehild, sorry i wasn't quite sure how to typeset them.

I did write that if I perform the odd half expansion of cos(t) I get
f(t) = cos(t) for [0,π] and f(t) = -cos(t) for [-π,0]
Using this (and assuming I can let n=1 since it is this first partial sum, but I am not sure if i can do this) then $b_1=\frac{1}{\pi}[\int_{-\pi}^0-\cos{t}\sin{2t}dt + \int_{0}^{\pi}\cos{t}\sin{2t}dt] = \frac{8}{3\pi}$

Where am I going wrong?

 

[pondzo]

Oops, I found where i went wrong.
Instead of $b_1=\frac{1}{\pi}[\int_{-\pi}^0-\cos{t}\sin{2t}dt + \int_{0}^{\pi}\cos{t}\sin{2t}dt]$
It should be $b_1=\frac{1}{\pi}[\int_{-\pi}^0-\cos{t}\sin{t}dt + \int_{0}^{\pi}\cos{t}\sin{t}dt] = 0$

I would still like to know if it is correct to let n=1 in the bn integral to find b1. Or should you only sub in n=1 after you find the general form of bn?

 

[vela]

You can set n=1 before doing the integration. You might as well since all you were looking for was $b_1$.