How Do I Solve for the Fourier Series of sin(6t) with f(t)=f(t+2pi)?

[Stacks!]

Homework Statement

f(t)=sin(|6t|), −pi<t<pi

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

Homework 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.

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

 

[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)

 

[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!

 

[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)

 

[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.

 

[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.

 

[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!

 

[vela]

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.

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

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)

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

 

[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? :)

 

[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!

 

[hgfalling]

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

 

[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

 

[vela]

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? :)

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

 

[vela]

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!

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.

 

[WrittenStars]

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.

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

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

 

[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

 

[vela]

You're not integrating correctly. Show your work.

 

[WrittenStars]

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

 

[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?

 

[vela]

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.

 

[Stacks!]

its ok, i have since done a0 again and am confident i got 0 and can understand that bit.

i think that this question may be just too long to post all the working up, especially when there's a half page of working to type up.

i've got 3 a values being a1 = 24/35pi, a2=0, a3=8/9pi, a4=0

again, i appologise for not putting working, but I'm not sure how to do it on this forum so it has an acceptable level of readability

 

[vela]

Do those results match the series you were given?

 

[Stacks!]

they match, wow, i think my whole issue was treating the function as an odd function, my integration was indeed correct and the series match up. the function must be treated as an even function, and then its all good

thank you so much for your time guys, really appreciate the help