Fourier series(past exam question)

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SUMMARY

This discussion focuses on solving a past exam question involving Fourier series and differential equations. The integral A{n}=2/pi∫(xsinx)sin nx dx is evaluated, revealing that A{n} equals pi^2/4 for n=1, 0 for odd n greater than 1, and -4n/(n^2-1) for even n. The user expresses confusion regarding the classification of the function xsinx as even or odd, ultimately leading to the use of a sine series to satisfy the boundary conditions u(x,t)=u(pi,t)=0.

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  • Understanding of Fourier series and their applications in solving differential equations.
  • Knowledge of integration techniques, specifically integration by parts.
  • Familiarity with the concepts of even and odd functions in mathematical analysis.
  • Basic proficiency in calculus, particularly in evaluating definite integrals.
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  • Study the properties of Fourier series, particularly in relation to boundary value problems.
  • Learn advanced integration techniques, including the use of integration by parts in complex scenarios.
  • Explore the classification of functions as even or odd and their implications in integral calculus.
  • Investigate the application of sine series in fulfilling initial and boundary conditions in differential equations.
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Students preparing for exams in mathematical analysis, particularly those focusing on Fourier series and differential equations, as well as educators seeking to clarify these concepts for their students.

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Homework Statement



This is a past exam question that involves solving a differential equation. I am using a Fourier series. in the solution they simply state that it is obvious that this equals pi^2/4 for n=1, 0 for n is odd>1, and -4n/(n^2-1) for even n.

A{n}=2/pi∫(xsinx)sin nx dx

I always integrate by parts, which for this equation is way too long, especially for an exam. So i am wondering how can i evaluate the value of the integral without integrating? i have tried using the addition formulas, but that seems like just as much work.

Another point of confusion is knowing when a function is even or odd. isn't the function xsinx an even function since f(-X)=-xsin(-X)=-(-sinx)=xsinx and thus an odd function * an even gives an odd, and i thought that the integral would then be zero. But in the solution they specify to use a sine series to fulfill the initial conditions that u(x,t)=u(pi,t)=0.


Homework Equations





The Attempt at a Solution

 
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Well i used the addition formula and i got

-0.5( cos(x(n+1) -cos(x(n-1))

Those seem very easily integrated by parts

INT( xcos(x(n+1))dx = (1/(n+1))*xsin(x(n+1) - (1/(n+1))*INT(sin(x(n+1))dxwhere INT is the integral symbol.
 

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