# Determine the Fourier Series

## Homework Statement

determine the Fourier Series for f(x)=cos3x

## Homework Equations

f(x)=ao/2+(sum) an cos(nx)+ (sum) bncos(nx)

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

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

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

## The Attempt at a Solution

i worked out that ao and an are both zero, which is fine. however when i go to work out bn 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?

## Answers and Replies

have i done the integration wrong here?I feel liek i must have sence i dont really know how i would go about integrating cos3x sin(nx) OR cos3x cos(nx)

HallsofIvy
Science Advisor
Homework Helper
I don't see how you could get that "a0 and an" are 0. Since $$\displaystyle cos^3(x)sin(x)$$ 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
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
You can avoid doing any integrals by doing as HallsofIvy did for cos4 x and use trig identities to expand cos3 x.