Hi Folks,
Got a question regarding how to calculate the Fourier coefficents a0, an and bn.

I have a problem with the orthogonal trig identities which are used to identify the particular co-efficent. I know it works and I can just take that and accept, however I would like to see a proof of it. The identitiy is as follows:
I dont now how to show the integral sign so I will use (int instead) sorry!

2*pi
int cos mx. cos nx dx = 0 ( if m does not equal n)
0 = 1 ( if m does equal n(therefore identifying term))

Identity used

cos mx. cos nx = = 1/2 (cos(m+n)x + cos(m-n)x)

substuting this identity into the integral above.

2*pi
1/2 int [(cos(m+n)x + cos(m-n)x)]dx
0

= | sin(m+n)x sin(m-n)x | 2*pi
| ---------- + ---------- |
| (m+n) (m-n) |0

I finally get to my question
According to the orthogonal feature of this, the result will be 1 only when m=n
all other values of m and n will result in the reslut bieng 0.

The thing I cant understand is taht in each part of the integral the sin term is been multiplied by a multiple of 2*pi or 0, surly any multiple of that result in a zero output. So how is it possible to get a result of 1 when m equals n.

I know I am wrong but I would appreciate it if someone will tell me where I am going wrong. I have tried doing it on my calculator and it always works out to be 0.

Jag.

 

Jagjit,
I'll represent the integral you started with in this way:
int[cos(mx)cos(nx)dx;0;2*pi]

If m = n, the integral simplifies to
int[(cos(mx)^2)dx; 0; 2*pi], which comes out for me as pi, not 1 as you stated.

If m <> n, the original integral becomes
1/2 * int[ (cos(m+n)x +cos(m-n)x)dx; 0; 2*pi]

which is 1/2 * [ sin(m+n)x/(m+n) + sin(m-n)x/(m-n), evaluated at 2*pi and 0.
Both sine terms are zero for integer multiples of 2pi and for x=0, so that integral's value is zero.

In summary, the integral you're working with is nonzero only if m = n. This shows that the functions cos(mx) and cos(nx) are orthogonal when m and n are different.
Mark

 

Cheers Mark,
I can see where I went wrong.
Jag.