Help with Fourier Sine Expansion.

[LoganS]

1. Find the Fourier sine expansion of $\phi(x)=1$
.2. Homework Equations .
In the lecture the professor worked his stuff from scratch so I was trying to do it like his example.3. The Attempt at a Solution .
I start with $\phi(x)=A_1sin(\pi x)+A_2sin(2\pi x)+\cdots+A_nsin(n\pi x),$ and then add multiply by $A_msin(m\pi x)$ term on each side and integrate from 0 to 1.
So I have $\int\phi(x)A_msin(m\pi x)=\int\left(A_1sin(\pi x)A_msin(m\pi x)+A_2sin(2\pi x)A_msin(m\pi x)+\cdots+A_nsin(n\pi x)A_msin(m\pi x)\right).$
I know that due to orthogonality you can discard the terms where m is not equal to n (but I don't really understand why so if you can explain this I would appreciate it).
Discarding those terms and using a trig relation I get, $\int\phi(x)A_msin(m\pi x)=1/2\int\left(cos((m-n)\pi x)-cos((m+n)\pi x)\right).$
I then solve the integral and try to get $A_m$ by alone, but I think I am doing something wrong cause what I get is terribly messy.
The book answer is $1=\frac{4}{\pi}\left(sin(\pi x)+\frac{1}{3}sin(3\pi x)+\frac{1}{5}sin(5\pi x) +\cdots\right).$ Any and all help is greatly appreciated.

 

[cepheid]

Welcome to PF,

You can discard those terms because ∫sin(nπx)sin(mπx)dx actually evaluates to zero for m ≠ n (with the appropriate limits of integration). Try it! That's how orthogonality of the basis functions is defined.

I don't really understand what you did. If you discarded all but the m = n terms in the sum, then you should have had

1 = ∫A_msin(mπx)A_msin(mπx)dx = ∫A_m²sin²(mπx)dx
​

That isn't the integral in your post, for some reason...

 

[LoganS]

I used the trig identity $sin(x)sin(y)=\frac{cos(x-y)-cos(x+y)}{2}.$ So my term becomes $A_m \int \left(\frac{ cos(0)-cos(2m\pi x)}{2}\right) =A_m\left(\frac{1}{2}+\frac{sin(2m\pi)}{4m\pi} \right)$ Is this not acceptable?

I managed to obtain $$A_m=\frac{4}{\pi}\left(\frac{1}{m}\right)$$ but I can't understand why the answer only has terms for odd m's. Any idea?

 

[cepheid]

I used the trig identity $sin(x)sin(y)=\frac{cos(x-y)-cos(x+y)}{2}.$ So my term becomes $A_m \int \left(\frac{ cos(0)-cos(2m\pi x)}{2}\right) =A_m\left(\frac{1}{2}+\frac{sin(2m\pi)}{4m\pi} \right)$ Is this not acceptable?

I managed to obtain $$A_m=\frac{4}{\pi}\left(\frac{1}{m}\right)$$ but I can't understand why the answer only has terms for odd m's. Any idea?

Can you post your work? After integrating, you should have obtained something of the form

some trig expression = 1

which is only true for discrete values of m.

Personally I would stick to the sine squared, since there is an easy trig identity for that.

Since cos(2a) = cos²(a) - sin²(a)

and cos²(a) = 1 - sin²(a),

it follows that:

cos(2a) = 1 - 2sin²(a)

or sin²(a) = 1/2*[1 - cos(2a)]

and this is REALLY easy to integrate.