# Help on Fourier Series of Sin(x/3)

• Apr 24th 2010, 02:54 AM
zizou1089
Help on Fourier Series of Sin(x/3)
THE QUESTION STATES:

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

Find the Fourier coefficients of f.

My attempt at the question,

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

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

Now I only need to put the limits in. 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 think I may need to use Dirichlets theorem however I need the answer to the part of the question before!
• Apr 24th 2010, 03:11 AM
HallsofIvy
Quote:

Originally Posted by zizou1089
THE QUESTION STATES:

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

Find the Fourier coefficients of f.

My attempt at the question,

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

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

This makes no sense. bn cannot depend on x. Do you mean this is the integral of sin(nx)sin(x/3) before you evaluate at the limits of integration?

Quote:

Now I only need to put the limits in. The trouble is when i put the limits [-pi,pi) into the equation I get 0 which is not the answer!
How do you get that? When n= 1, for example, that gives
$\displaystyle \frac{1}{2\pi}[(3/2) sin((2/3)\pi)- (3/2)sin((4/3)\pi))= \frac{1}{2\pi}(\sqrt{3}{2}- (-\sqrt{3}{2})= \frac{\sqrt{3}}{2\pi}$ at $\displaystyle x= \pi$ and evaluating at $\displaystyle x= -\pi$ gives the negative of that so the difference is $\displaystyle \frac{\sqrt{3}}{\pi}$.

Quote:

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))
and this can't be true! The left side depends on x and the right side doesn't!

Quote:

I think I may need to use Dirichlets theorem however I need the answer to the part of the question before!
• Apr 24th 2010, 03:41 AM
zizou1089
For the first bit yes i meant this is the integral of sin(nx)sin(x/3) is
(1/2pi) [(3/(1-3n))sin(((1-3n)/3)x) - (3/(1+3n))sin(((1+3n)/3)x)]

before you evaluating at the limits of integration. I just dont know how to simplify the answer after putting the limits in

For the second part, the question only states:

show for all x E (-pi,pi)
.....................................infinity
sin (x/3) = (((9)(3^.5))/pi) SUM (((-1)^n)n)/(1-9n^2))
......................................n=1
• Apr 24th 2010, 06:50 AM
zizou1089
Quote:

Originally Posted by HallsofIvy
This makes no sense. bn cannot depend on x. Do you mean this is the integral of sin(nx)sin(x/3) before you evaluate at the limits of integration?

How do you get that? When n= 1, for example, that gives
$\displaystyle \frac{1}{2\pi}[(3/2) sin((2/3)\pi)- (3/2)sin((4/3)\pi))= \frac{1}{2\pi}(\sqrt{3}{2}- (-\sqrt{3}{2})= \frac{\sqrt{3}}{2\pi}$ at $\displaystyle x= \pi$ and evaluating at $\displaystyle x= -\pi$ gives the negative of that so the difference is $\displaystyle \frac{\sqrt{3}}{\pi}$.

and this can't be true! The left side depends on x and the right side doesn't!

For the first bit yes i meant this is the integral of sin(nx)sin(x/3) is
(1/2pi) [(3/(1-3n))sin(((1-3n)/3)x) - (3/(1+3n))sin(((1+3n)/3)x)]

before you evaluating at the limits of integration. I just dont know how to simplify the answer after putting the limits in

For the second part, the question only states:

show for all x E (-pi,pi)
.....................................infinity
sin (x/3) = (((9)(3^.5))/pi) SUM (((-1)^n)n)/(1-9n^2))
......................................n=1