# Fourier Series of sin(6t)

1. Sep 12, 2010

### Stacks!

1. The problem statement, all variables and given/known data
f(t)=sin(|6t|), −pi<t<pi

with f(t) = f(t+2pi)

2. Relevant equations
Show that the Fourier series for f(t) can be written as (24/pi) time the sum, from n=0 to infinity, of ( 1/( 36 - (2k+1)^2 ) )cos(2k+1)t.

3. The attempt at a solution

I have an answer of a0 being 0, and a guess that an is also zero, however i cannot onfirm this since i don't know how to do the integral of [sin(nt)6cos(6t)].

I run into a similar problem with bn, getting stuck at the integral of [sin(6t)sin(nt)]

any help would be much appreciated

2. Sep 12, 2010

### n.karthick

Hi integral of [sin(nt)6cos(6t)] will be zero because integral [sinA.cosB] is zero over a period 2pi.
Note use the trigonometric formula sinAcosB=sin(A+B)-sin(A-B)

3. Sep 12, 2010

### WrittenStars

Interesting problem hmm, can you show how you got a0 and explain hoe you got upto where you are?

From there I think some trigonometric identities will be useful. But I will put pen to paper once I see where and how you got there!

Seems like a tricky question!

4. Sep 12, 2010

### Stacks!

well a0 is just a straightforward integration of sin(6t) between -pi and pi, which comes out as -(1/6)cos(6t), which comes out as (-1/6) - (-1/6), which = 0.

for an, i started with the equation (1/L)int[ f(t)*cos(nt)].dt, for n>=1. (L is given as pi, and the region is -pi < t < pi.
i am currently working through this but its looking ugly after the sin(nt)cos(6t) integral.

for bn, i'm starting with the same equation as for an, except with a sin term instead of the cos term.
this one i am stuck at the integral of sin(6t)sin(nt)

5. Sep 12, 2010

### WrittenStars

For an just substituting in I get:

(1/L) int [ sin(6t) * cos(nt) ] .dt

Not sure if you started with this....

If you did I am not sure how this led to:

the sin(nt)cos(6t) integral.

6. Sep 12, 2010

### Stacks!

yes, that is what i started with, i wasn't too sure how to proceed with it so i used integration by parts (u=sin(6t), dv/dt = cos(nt) ) and that's how it happened.

7. Sep 12, 2010

### WrittenStars

Usually for integration of something like sinxcosx i use the identity:

sin2x = 2sinxcosx

But not sure how that applies in this case. I am very curious, hopefully someone can answer!

8. Sep 12, 2010

### vela

Staff Emeritus
You're finding a0 for sin 6t, not sin |6t|, which is what the problem asked for.
You can probably get it to work out with integration by parts, but it's simpler using the product-to-sum trig identities.

http://en.wikipedia.org/wiki/List_o...#Product-to-sum_and_sum-to-product_identities

9. Sep 12, 2010

### WrittenStars

thanks for that vela, i looked at the identities prior to you post but I didnt know how to apply them to this case. Would you have any pointers for which one it is? :)

10. Sep 12, 2010

### WrittenStars

You're finding a0 for sin 6t, not sin |6t|, which is what the problem asked for.

For sin |6t| we ignore the - values and we are left with:

1/2pi [-1/6] = -1/12pi

Hope that is right!

11. Sep 12, 2010

### hgfalling

You can verify these yourself:
$$\int_{-L}^{L} \cos \frac{n\pi x}{L} \cos \frac{m\pi x}{L} = \left\{\begin{array}{cc}0,&\mbox{if } n \neq m\\L, & \mbox {if } n=m>0\\2L, & \mbox {if } n=m=0 \end{array}\right$$
$$\int_{-L}^{L} \sin \frac{n\pi x}{L} \sin \frac{m\pi x}{L} = \left\{\begin{array}{cc}0,&\mbox{if } n \neq m\\L, & \mbox {if } n=m>0\end{array}\right$$
$$\int_{-L}^{L} \sin \frac{n\pi x}{L} \cos \frac{m\pi x}{L} = 0$$
(m and n nonnegative integers)

12. Sep 12, 2010

### Stacks!

how does sin(|6t|) change the answer? which are the values you discount. After doing all the integration i'm left with the sum of 2 negative answers

13. Sep 12, 2010

### vela

Staff Emeritus
Which one looks like (sin 6t)(cos nt)?

14. Sep 12, 2010

### vela

Staff Emeritus
No, it's not. sin |6t| is non-negative, so the area under the curve from -pi to pi is positive.

Note f(t)=sin |6t| is an even function, so

$$\int_{-\pi}^\pi \sin |6t| \cos nt \, dt = 2\int_0^\pi \sin |6t| \cos nt \, dt = 2\int_0^\pi \sin 6t \cos nt \, dt$$

You can get rid of the absolute value in the integral over [0,pi] because t will be positive.

15. Sep 12, 2010

### WrittenStars

Thanks for that, but from here I am struggling to find the right identity to simplify equation!

This is not my problem but I cant help be intrigued on how to do it!

16. Sep 12, 2010

### Stacks!

yes i've tried a coupld of times now and i keep ending up at this horrendous thing :

(2/pi){cos(pi(n+6))[-1/(n(n+6)) - 1/(n(n-6)) ] }

ummm, yeh, i'm lost

17. Sep 12, 2010

### vela

Staff Emeritus
You're not integrating correctly. Show your work.

18. Sep 12, 2010

### WrittenStars

Yeah that didnt seem right. The problem I'm having is that very first step! LOL

19. Sep 12, 2010

### Stacks!

i have redone my a0, using a restricted domain of 0 to pi/6, then multiplying it to get the total area under the curve for the actual range.
This came out to be 8/pi.

correct?

20. Sep 12, 2010

### vela

Staff Emeritus
Oops, I led you astray. The function isn't non-negative over the entire interval, and the constant term should indeed be a0=0.

It would really help if you would show your work instead of just posting an answer and asking if it's correct. I know you got a0=0 earlier, but I could also tell from what else you had written, you got that result by accident.