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

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Homework Help Overview

The problem involves finding the Fourier coefficients of the periodic function defined by f(x) = sin(x/3) over the interval [-π, π). Participants are exploring the implications of their calculations and the relationships between the coefficients.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants discuss the derivation of the Fourier coefficients, with one noting that their integration results in zero, which raises questions about the validity of their approach. Others suggest that one of the coefficients is indeed zero while the other is not, prompting further exploration of the equations involved.

Discussion Status

The discussion is ongoing, with participants questioning the correctness of the derived equations and the assumptions made. Some guidance has been offered regarding the need for clarity in the equations presented, and there is an acknowledgment of the relevance of Dirichlet's theorem in the context of the problem.

Contextual Notes

There are indications of confusion regarding the relationship between the variables x and n in the equations, as well as the need for proper notation in the expressions used. Participants are also grappling with the implications of the periodic nature of the function and the limits of integration.

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.
 

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