How to Find Fourier Series for a Given Function Using Sine Series?

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To find the Fourier sine series for the function y(x) = Ax(L-x) over the interval 0 ≤ x ≤ L, the coefficients b_n are calculated using the formula b_n = (2/L) * integral from 0 to L of (Ax(L-x) * sin(n*pi*x/L) dx). The discussion highlights the confusion surrounding the integration process and the various equations involved. It suggests that understanding the derivation of Fourier series formulas can clarify the process and reduce confusion. Ultimately, performing the integral is essential to obtain the correct coefficients for the series.
RJLiberator
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Homework Statement



Find the Fourier series for the following function (0 ≤ x ≤ L):
y(x) = Ax(L-x)

Homework Equations

The Attempt at a Solution



1. We start with the sum from n to infinity of A_n*sin(n*pi*x/L) where An = B_n*Ax(l-x)

2. We have the integral from 0 to L of f(x)*sin(m*pi*x/L) dx

I really have no idea what to do, I am francticlly looking through notes and websites. I understand the Fourier sine series should be pretty easy to find, it's just plugging in values, but there are so many different equations/elements.

Let me try this solution:

f(x) = L/pi(sum from n = 1 to infinity of sin(n*pi*x/L)

Ah?
 
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When you are asked the for the Fourier series of a function, your answer should be the coefficients appearing in the sum.
 
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From my notes, would:

A_m = 2/L integral from 0 to L f(x)*sin(m*pi*x/L) dx be the answer then?

where f(x) = Ax(L-X)
 
Okay, let me say this:

We start with the Fourier Since series:
sum from 1 to infinity of (b_n*sin(n*pi*x/L))

where b_n = 2/L integral from 0 to L (f(x)*sin(n*pi*x/L))

where f(x) = the function in question, namely Ax(L-x)

All together we have

The sum from n=1 to infinity of 2/L integral from 0 to L of (Ax(L-x))*sin^2(n*pi*x/L) dx
 
While formally correct, this doesn't answer the question. You have to find an expression for the b_n.
 
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b_n = 2/L integral from 0 to L (f(x)*sin(n*pi*x/L))
where f(x) = the function in question, namely Ax(L-x)

so b_n = 2/L integral from 0 to L (Ax(L-x)*sin(n*pi*x/L))
Is that incorrect for b_n?
 
RJLiberator said:
so b_n = 2/L integral from 0 to L (Ax(L-x)*sin(n*pi*x/L))
Is that incorrect for b_n?
It's a start. Now you have to perform the integral.
 
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RJLiberator said:
I really have no idea what to do, I am francticlly looking through notes and websites. I understand the Fourier sine series should be pretty easy to find, it's just plugging in values, but there are so many different equations/elements.
Now that you've been working with Fourier series for a while, it wouldn't hurt to go back and review the derivation of the various formulas (using one reference). If you understand the basics, all the variations/conventions will make more sense and won't seem so confusing.
 
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