# Homework Help: Fourier Series Analysis

1. Aug 25, 2009

### CSNabeel

1. The problem statement, all variables and given/known data

Show that provided that m and n are arbitrary integers, the two functions f(t) = sin nt and g(t) = cos mt are orthogonal over the interval (0,2$$\pi$$). Explain the significance of this result in Fourier series analysis. Hint: you may find the following trigonometric identity useful 2sin a cos b = sin(a+b) - sin (a-b)

2. Relevant equations

3. The attempt at a solution

$$\int$$ f(t) g(t) dt
$$\int$$ sin nt cos mt dt
$$\int$$ 0.5( sin(n+m)t - sin(n-m)t)dt
0.5[ $$\frac{cos(n+m)t}{n+m}$$ - $$\frac{cos(n-m)t}{n-m}$$ ]

put (0,2$$\pi$$) into (n,m) respectively (sorry not good with code thing).

After subistion the final answer end up being

t over 2$$\pi$$

this is my working out so far but what should I write about it under fourier analysis

2. Aug 25, 2009

### Dick

You don't put (0,2*pi) into (n,m). You set t=2*pi and t=0 into the anti derivative and subtract the results. There's no t in the final answer. What's cos(k*2*pi) where k is an integer?

3. Aug 25, 2009

### CSNabeel

so in that case does it become

[ cos 2*pi*(n+m) / n + m ] - [ cos 2*pi*(n-m) / n - m ]

?

4. Aug 25, 2009

### Dick

It becomes that minus [ cos 0*(n+m) / n + m ] - [ cos 0*(n-m) / n - m ]. Haven't you done definite integrals before?

5. Aug 25, 2009

### CSNabeel

yeah sorry I made a mistake cancelling that because cos 0 = 1 and forget that the denominator where different

so it becomes

[ cos 2*pi*(n+m) / n + m ] - [ cos 2*pi*(n-m) / n - m ] - [1 / n + m] - [ 1 / n - m]

right?

6. Aug 25, 2009

### Dick

I would write that as cos(2*pi*(n+m))/(n+m) - cos(2*pi*(n-m))/(n-m) - [(1/(n+m)-1/(n-m)]. Use enough parentheses and brackets so your meaning is clear. But, yes, I think you mean the right thing. Now what's cos(k*2*pi) where k is an integer?

7. Aug 25, 2009

### CSNabeel

the integral of cos(k*2*pi) is -sin(k*2*pi)/(2*pi)

8. Aug 25, 2009

### Dick

I'm not asking you what the integral is. I'm asking you what the value of cos(k*2*pi) is where k is an 'integer' - not 'integral'. If k=1, that's cos(2*pi), if k=2, that's cos(4*pi), if k=3 that's cos(6*pi) etc etc. I'm asking you this because (n+m) and (n-m) are integers.

9. Aug 25, 2009

### CSNabeel

oh well in that case cos(k*2*pi) = 1

10. Aug 25, 2009

### Dick

Right, so what's the value of the integral now that you know what all the terms are?

11. Aug 25, 2009

### CSNabeel

I think it's [(1/(n+m)-1/(n-m)] - [(1/(n+m)-1/(n-m)] which I think cancels out?

12. Aug 25, 2009

### Dick

Yes, it's zero. That means f(t)=sin(nt) and g(t)=cos(mt) are orthogonal over the interval [0,2*pi] since the integral of f(t)*g(t) over that interval is zero. There is a minor technical point. What happens if n=m or n=(-m)? Then your expression has a zero in the denominator. Can you see why that's not really a problem?

13. Aug 25, 2009

### CSNabeel

yeah I think so, they still cancel out even though 1/0 tends to infinity

14. Aug 25, 2009

### Dick

That's dangerous reasoning. The point is that if e.g. n=m, then sin((n-m)*t)=sin(0*t)=0. The cos((n-m)t)/(n-m) term isn't in the antiderivative. If you are going to be formal about the proof the you might want to make those special cases.

15. Aug 25, 2009

### CSNabeel

ahhh I see. Thanks you been really helpful