How to Determine the Fourier Series for f(x)=cos3x

[lycraa]

Homework Statement

determine the Fourier Series for f(x)=cos³x

Homework Equations

f(x)=a_o/2+(sum) a_n cos(nx)+ (sum) b_ncos(nx)

a_o= (integral) f(x)dx (from -\pi to\pi)

a_n= (integral) f(x)cos(nx)dx (from -\pi to\pi)

b_n= (integral) f(x)sin(nx)dx (from -\pi to\pi)

The Attempt at a Solution

i worked out that a_o and a_n are both zero, which is fine. however when i go to work out b_n i get answers that are divided by (n-1) and (n-3) which means that when i try and find b₁ and b₃ I'm dividing by zero. i don't know what to do now! can this function be made into a Fourier Series?

 

[lycraa]

have i done the integration wrong here?I feel liek i must have sense i don't really know how i would go about integrating cos³x sin(nx) OR cos³x cos(nx)

 

[HallsofIvy]

I don't see how you could get that "a₀ and a_n" are 0. Since [math]cos^3(x)sin(x)[/math] is an odd function, it immediately follows that the integral from -\pi to \pi is 0- that is that b_n= 0 for all n.

And, it is easy to show that a_0= 0 but
a_1= \int_{-\pi}^{\pi}cos^4(x)dx

Since cos^2(x)= (1/2)(1+ cos(2x)), cos^4(x)= (1/4)(1+ cos(2x))^2=1/4+ (1/2)cos(2x)+ (1/4)cos^2(2x))
= 1/4+ (1/2)cos(2x)+ (1/4)(1+ cos(4x))=(3/4)+ (1/2)cos(2x)+ (1/4)cos(4x)

Now, since the integral of cosine is sine and sine is 0 at any integer multiple of \pi, those two cosine integrals will be 0 but the integral of 3/4, from -\pi to \pi will be 3\pi/2, not 0.

By the way, I think you are missing the normalzing factor of 1/(2\pi) in front of the integrals. a_1= (3\pi/2)(1/2\pi)= 3/4, no 0.

For higher order terms, to integrate3 cos^3(x)cos(nx), with n> 1, you will need to reduce either cos^3(x) or cos(nx).

 

[vela]

You can avoid doing any integrals by doing as HallsofIvy did for cos⁴ x and use trig identities to expand cos³ x.