Little help on Fourier Series of Sin(x/3)

zizou1089
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Homework Statement



Let f be the 2 periodic function defined on [−pi,pi ) by
f(x) = sin(x/3)

Find the Fourier coefficients of f.

Homework Equations



.5[cos(A-B)-cos(A+B)]=sinAsinB

The Attempt at a Solution



After much work using trigonometry and integration by parts I have deduced

(1/2pi) [(3/(1-3n))sin(((1-3n)/3)x) - (3/(1+3n))sin(((1+3n)/3)x)]

the trouble is when i put the limits [-pi,pi) into the equation I get 0 which is not the answer!

and the next part of the question states:

show for all x E (-pi,pi)

sin (x/3) = (((9)(3^.5))/pi)(((-1)^n)n)/(1-9n^2))
 
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One of the coefficients is indeed 0. The other--the one that .5[cos(A-B)-cos(A+B)]=sinAsinB is useful for--is not.
 
zizou1089 said:

Homework Statement



Let f be the 2 periodic function defined on [−pi,pi ) by
f(x) = sin(x/3)

Find the Fourier coefficients of f.

Homework Equations



.5[cos(A-B)-cos(A+B)]=sinAsinB

The Attempt at a Solution



After much work using trigonometry and integration by parts I have deduced

(1/2pi) [(3/(1-3n))sin(((1-3n)/3)x) - (3/(1+3n))sin(((1+3n)/3)x)]

You don't have an equation. What is that supposed to be equal to?

the trouble is when i put the limits [-pi,pi) into the equation I get 0 which is not the answer!

and the next part of the question states:

show for all x E (-pi,pi)

sin (x/3) = (((9)(3^.5))/pi)(((-1)^n)n)/(1-9n^2))

I'm pretty sure that isn't what you are supposed to show since there is no x on the right side and no n on the left.
 
The answer:

bn = (1/2pi) [(3/(1-3n))sin(((1-3n)/3)x) - (3/(1+3n))sin(((1+3n)/3)x)]

is my attempt in integrating the bn coefficient of the Fourier series of sin(x/3) as a0 and an are both equal to zero after using the .5[cos(A-B)-cos(A+B)]=sinAsinB formula.

However the values I get after inserting pi and -pi are 0, which shouldn't be the case for bn.

And the second part of the question only states:

show for all x E (-pi,pi)

sin (x/3) = (((9)(3^.5))/pi)(((-1)^n)n)/(1-9n^2))

Maybe by Dirichlets theorem, I think the part before needs to be answered.
 
Did you even read my post? Your first equation for bn has x in it and your last equation for sin(x/3) has n's in it. Neither is correct. You might want to include appropriate integral signs or sums. And if you expect anyone to help you find any errors in your work you are going to have to show what you did. We can't read your mind.

And yes, Dirichlet's theorem has something to do with it once you write it correctly.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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