Homework Help: Fourier Integral

1. Nov 7, 2005

Cyrus

Can someone please show me how to solve these three integrals, or proivde a link that does. I cant think of a way to solve it right now and I dont have any free time to do so. Thank you!

$$\int cos(nx)cos(mx)dx$$

$$\int sin(nx)sin(mx)dx$$

$$\int sin(nx)cos(mx) dx$$

The thing that makes it tricky is the different coefficients of m and n in each trig term, so you cant make them cos^2(x) or sin^2(x) or use trig identities. (at least I cant see how). Anywho, thanks once more.

Cyrus.

Last edited: Nov 7, 2005
2. Nov 7, 2005

HallsofIvy

Trig identities.

sin(a+ b)= sin(a)cos(b)+ cos(a)sin(b)
sin(a- b)= sin(a)cos(-b)+ cos(a)sin(-b) but cos(-b)= cos(b) sin(-b)= -sin(b)
= sin(a)cos(b) - cos(a)sin(b)

To integrate sin(nx)cos(mx) let a= nx, b= mx to write
sin(nx)cos(mx)= (1/2)sin((m+n)x)+ (1/2)sin((m-n)x).

cos(a+ b)= cos(a)cos(b)- sin(a)sin(b)
cos(a- b)= cos(a)cos(b)+ sin(a)sin(b)

To integrate cos(nx)cos(mx), let a= nx, b= mx to write
cos(nx)cos(mx)= (1/2)cos((m+n)x)+ (1/2)cos((m-n)x)

Subtracting, cos(a+b)- cos(a-b)= 2sin(a)sin(b)
To integrate sin(nx)sin(mx), let a= nx, b= mx to write
sin(nx)sin(mx)= (1/2)cos((m+n)x)- (1/2)cos((m-n)x)

3. Nov 7, 2005

Cyrus

Im not following how this is true, pluging in a value for b and none of them are comming out equal?

Ahhh, I figure your just missing a comma inbetween there. You meant cos(-b) = cos(b) and sin(-b)= - sin(b).

Thats a VERY clever trick to solve it. I would never have spotted it! Thanks!

Last edited: Nov 7, 2005
4. Nov 10, 2005

Cyrus

A follow up, can you show me how the series is reduced, its been a while since i touched a series, a long while.

$$cos(n \pi) = (-1) ^n$$

My book says $$cos(x) = (-1)^n x^{2n} / (2n!)$$

what happened to the fractional part?

The n is different for each equation, poor notation sorry.

Thanks twice.

Cyrus

Last edited: Nov 10, 2005
5. Nov 11, 2005

quasar987

Isn't it more like

$$cos(x) = \sum_{n=0}^{\infty}(-1)^n x^{2n} / (2n!)$$.

Your question is strangely worded. What do you mean by reduced?

6. Nov 11, 2005

Cyrus

$$cos(n \pi) = (-1) ^n$$