Solving Fourier Coefficients: Hints for Finding a_n

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Homework Help Overview

The discussion revolves around finding the Fourier series for the function f(t) defined as 0 for -π < t < 0 and sin(t) for 0 ≤ t ≤ π. Participants are exploring the calculation of Fourier coefficients, specifically a_n and b_n, and are encountering difficulties with their integral evaluations.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss splitting the integral for a_n and express confusion over the results, questioning whether certain coefficients should be zero. There is also mention of needing to consider the b_n coefficients and the implications of the function's evenness or oddness.

Discussion Status

The discussion is ongoing, with participants providing hints and prompting each other to show their work. Some have suggested looking at specific cases, such as n = 1, to clarify the calculations. There is an acknowledgment of a potential mistake regarding the b_1 coefficient.

Contextual Notes

Participants are working under the constraints of homework guidelines, which may limit the amount of direct assistance they can receive. There is a focus on understanding the Fourier series expansion rather than simply obtaining the answer.

errordude
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Homework Statement


Hi i would just like some fast hints, I'm doing the integrals wrong, I am splitting up the integral below and get the wrong answer.

well it's about finding the Fourier series for f(t)={0 for -π<t<0 and sint for 0≤t≤π}

Homework Equations


a_{n} = \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos(nt) dt , n\in Z_{+}


The Attempt at a Solution



well i split the integral up in finding a_{n} like

\frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\cos(nt) dt = \frac{1}{\pi}\int_{-\pi}^{0} 0· dt+\frac{1}{\pi}\int_{0}^{\pi}\sin(t)\cos(nt) dt
Both of these elementary, but it fails to produce the right series.
Hints anyone?
 
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Did you forget the bn?
 
LCKurtz said:
Did you forget the bn?

no but that just get to zero
 
errordude said:
no but that just get to zero

They can't be zero because the function you are expanding is not an even function.
 
LCKurtz said:
They can't be zero because the function you are expanding is not an odd function.

That's what i was thinking


<br /> \frac{1}{\pi}\int_{-\pi}^{\pi}f(t)\sin(nt) dt = \frac{1}{\pi}\int_{-\pi}^{0} 0· dt+\frac{1}{\pi}\int_{0}^{\pi}\sin(t)\sin(nt) dt<br />

but the above is zero!

i'm doing something wrong.
 
I have to run now. You didn't show your work but I'm guessing you need to look what happens when n = 1.
 
halloo??

any1 who knows this Fourier series

f(t)={0 for -π<t<0 and sint for 0≤t≤π}
 
Nobody is going to just give you the answer. Show us your work for the an and bn and we will help you find the mistake.
 
LCKurtz said:
Nobody is going to just give you the answer. Show us your work for the an and bn and we will help you find the mistake.

Hey man chill.

b_1=1/2 that was the problem.
 
  • #10
errordude said:
Hey man chill.

b_1=1/2 that was the problem.

Chill?? Surely you mean "Thanks for the suggestion, eh?"
 
  • #11
LCKurtz said:
Chill?? Surely you mean "Thanks for the suggestion, eh?"

you were right LC, b_1 was the crucial step.

thanx
 

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