Fourier series

  • Thread starter struggles
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  • #1
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Homework Statement


Find the Fourier series defined in the interval (-π,π) and sketch its sum over several periods.
i) f(x) = 0 (-π < x < 1/2π) f(x) = 1 (1/2π < x < π)

2. Homework Equations

ao/2 + ∑(ancos(nx) + bnsin(nx))
a0= 1/π∫f(x)dx
an = 1/π ∫f(x)cos(nx) dx
bn = 1/π ∫f(x) sin(nx)

The Attempt at a Solution


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I've worked out the a0 by splitting the limits and integrating individually for -π < x < 1/2π and 1/2π < x < π. When I did this i got a0 = ½

for an = 1/π ∫(upper limit π, lower limit 1/2π) cos(nx) dx
= 1/π[1/n sin(nx)]
= 1/π ((1/n. sin(πn) - 1/n.sin(πn/2))
= 1/nπ(0- sin(πn/2)

Here is where i get stuck as sin(πn/2) is 0 for even values of n and alternates between 1, -1 for odd values.
Can i leave this written in sin form of the Fourier series as every other example i've changed the value of sin/cos to either 0 or (-1)n.

Thanks for any help!
 

Answers and Replies

  • #2
Buzz Bloom
Gold Member
2,405
441
Hi struggles:

You have already noticed that the integrals for the coefficients can be limited to {π/2,π}. That's a good start.

I think it might be helpful to next separate the function f(x) into two parts:
f(x) = S(x) + A(x)
where S(x) is symmetric, S(x) = S(-x)
and
A(x) is anti-symmertric, A(x) = -A(-x).​
Then think about the coefficients for A(x) and S(x) separately.

You may also want to think about the general form of the integrals and try a substitution
y = nx, dx = dy/n,​
including the lower and upper bounds of the integral.

Hope this helps.

Regards,
Buzz
 

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