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

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Homework Help Overview

The discussion revolves around finding the Fourier sine series for the function y(x) = Ax(L-x) within the interval 0 ≤ x ≤ L. Participants are exploring the formulation of the series and the necessary coefficients.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the general form of the Fourier sine series and the coefficients involved. There are attempts to express the coefficients b_n in terms of integrals involving the function Ax(L-x). Some participants express confusion regarding the integration process and the application of the formulas.

Discussion Status

The discussion is ongoing, with participants providing various formulations for the coefficients and questioning the correctness of their expressions. There is a recognition that further steps, such as performing the integral, are necessary to progress.

Contextual Notes

Some participants express uncertainty about the multitude of equations and elements involved in the Fourier series, indicating a potential need for review of foundational concepts and derivations.

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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