Find the fourier series of (sinx)^3

[fhidiort]

I'm confused.

Trying to find the Fourier series of (sinx)^3. This is an odd function, so I try to find the Fourier sine coefficient, with integral of (sinx)^3*sinkx. However, my answer comes up with all sine terms. Of course all these terms go to zero when integrating between 0 and pi. Am I doing something wrong? Or is there some identity for sinx^3 that makes it easier to solve?

 

[AlephZero]

Of course all these terms go to zero when integrating between 0 and pi. Am I doing something wrong?

If you draw a graph of the function you are trying to integrate, you will see that you are doing something wrong, since the area under the curve isn't zero.

There are some standard identities that will help. Try expanding sin 3x = sin (2x + x), then sin 2x = sin(x+x) and cos2x = cos(x+x) and see what you get.

 

[dextercioby]

So you're telling me that

∫_{0}^{π}sin kxsin³xdx =0

necessarily?

What if k=3 or k=1 ?

Daniel.

 

[Swapnil]

I'm confused. Trying to find the Fourier series of (sinx)^3.

Actually, you don't even have to go through all that suffering. Whenever you want to find the Fourier series of a \sin^3(x) or \cos^5(x) or something, you just have to expand it out as a sum of first order sine and cosine terms--just like you would expand \sin(x)^ into \frac{1-cos(2x)}{2}.

In order to expand things like \sin^3(x) or \cos^5(x), just replace the sine and cosine term with its standard complex definition i.e.

\cos z=\frac{e^{iz}+e^{-iz}}2

\sin z=\frac{e^{iz}-e^{-iz}}{2i}

and then just simplify and convert everything back to rectangular form.

 

[HallsofIvy]

Use trig identities to write it as a "trigonometric polynomial". sin^2(x)= \frac{1}{2}(1- cos(2x)) so sin^3(x)= \frac{1}{2}(sin(x)- sin(x)cos(2x)). Since sin(a+b)= sin(a)cos(b)+ cos(a)sin(b) and sin(a-b)= sin(a)cos(b)- cos(a)sin(b), sin(a)cos(b)= (1/2)(sin(a+b)+ sin(a-b)) and, in particular, sin(x)cos(2x)= (1/2)(sin(3x)- sin(x)). That is, sin^3(x)= \frac{1}{2}sin(x)- \frac{1}{4}(sin(3x)- sin(x)= \frac{3}{4}sin(x)- \frac{1}{4}sin(3x) (modulo arithmetic errors!).