[wazza_utd]

Hi, I have the following problem and wondered if anyone can give me some hints on where to go next.

Find the Fourier Coefficients of Sin^3(x).

Since its pretty impossible to integrate that directly, i've used the identitiy Sin (2x+x) and obtained Sin^3(x) = 3/4Sin(x) - 1/4Sin(3x).

I've said that this is an odd function as so An = 0. So using this i've got:

Integral to be (between Pi and minus Pi): (3/4Sin(x)-1/4Sin(3x))Sin(nx)

however this is where I get lost as it goes in a circle with the constant of 3/4 and 1/4 becoming larger each time using integration by parts.

Can someone point me in the right direction please?

Thanks

[Coursework.info]

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[alexpr2000]

Originally Posted by wazza_utd

Hi, I have the following problem and wondered if anyone can give me some hints on where to go next.

Find the Fourier Coefficients of Sin^3(x).

Since its pretty impossible to integrate that directly, i've used the identitiy Sin (2x+x) and obtained Sin^3(x) = 3/4Sin(x) - 1/4Sin(3x).

I've said that this is an odd function as so An = 0. So using this i've got:

Integral to be (between Pi and minus Pi): (3/4Sin(x)-1/4Sin(3x))Sin(nx)

however this is where I get lost as it goes in a circle with the constant of 3/4 and 1/4 becoming larger each time using integration by parts.

Can someone point me in the right direction please?

Thanks

sin^3x = sinx * (1-cos^2x)

let z = cosx

dz/dx = -sinx

therefore dx = dz / -sinx

so then you can just integrate (z^2 - 1) dz.

Make sure you change the limits though,

[Scipio90]

Originally Posted by wazza_utd

Since its pretty impossible to integrate that directly, i've used the identitiy Sin (2x+x) and obtained Sin^3(x) = 3/4Sin(x) - 1/4Sin(3x).

And that's the answer...

[DFranklin]

Originally Posted by Scipio90

And that's the answer...

Indeed. Although I suppose it would have been amusing to let the OP go through all the integration first...

[wazza_utd]

Originally Posted by Scipio90

And that's the answer...

But dont I then multiply by Sin(nx) to work out the fourier series? Thats where im lost

[DFranklin]

Ask yourself: Is 3/4Sin(x) - 1/4Sin(3x) a Fourier series? And does it equal sin^3 x? So...

[wazza_utd]

Originally Posted by DFranklin

Ask yourself: Is 3/4Sin(x) - 1/4Sin(3x) a Fourier series? And does it equal sin^3 x? So...

Im guessing it is
Alright so if i multiply by sin(nx) afterwards would it not just go in a circle? sorry just abit lost here

[DFranklin]

No, if you want to multply by sin nx and integrate it will all work fine. It's just completely unneccessary.

Simplest way to do the integration is to use the formula for writing sin A sin B in terms of sin((A+B)/2) and sin((A-B)/2).

[wazza_utd]

Originally Posted by DFranklin

No, if you want to multply by sin nx and integrate it will all work fine. It's just completely unneccessary.

Simplest way to do the integration is to use the formula for writing sin A sin B in terms of sin((A+B)/2) and sin((A-B)/2).

Using the trig identitiy:

1/Pi *integral between Pi and minus Pi* Sin(nt)Sin(mt) dt = 1/2pi *integral between Pi and minus Pi* Cos(n-m)t - Cos(n+m)t dt

and using Sin^3(x) = 3/4Sin(x) - 1/4Sin(3x) after integration i obtain 2 equations:

3/8Pi [2Sin(n-1)Pi/(n-1) - 2Sin(n+1)Pi/(n+1)] - Pi/8 [2Sin(n-3)Pi/(n-3) - 2Sin(n+3)Pi/(n+3)]

and from there im not sure how to obtain the fourier series?

[wazza_utd]

anyone?

[Scipio90]

The answer is what you've bolded in your first post, as we've told you...