Particle in a Box: Fourier Sine Series Coefficients

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SUMMARY

The discussion focuses on the Fourier sine series coefficients for a particle in a one-dimensional infinite square well of width 'a'. The eigenfunctions are defined as ψₓ(x) = N sin(kₙ x) for 0 < x < a, where kₙ = nπ/a and n = 1, 2, 3, ... The coefficients of the Fourier sine series are established as cₙ = (2/a) ∫₀ᵃ sin(kₙ x) f(x) dx. The period of the function is determined by the width 'a', ensuring that the boundary conditions for the eigenfunctions are satisfied.

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For a particle in a one dimensional infinite squat well of width a, s.t 0<x<a the eignefunctions are given by

[tex]\psi_{x} (x) = N \sin k_{n} x[/tex] for 0 < x < a

where [tex]k_{n} = \frac{n \pi}{a}[/tex] and n = 1,2,3,...

Consider the Fourier sine series for the function f(x) on the interval 0<x<a
[tex]f(x) = \sum_{n=1,2,3,...} c_{n} \psi_{n} (x)[/tex]

Showthat the coefficients of this series are given by
[tex]c_{n} = \frac{2}{a} \int_{0}^{a} \sin (k_{n} x) f(x) dx[/tex]

do i have to PROVE that the coefficients are given by Cn??

isnt the expression by Cn given by the definition of Cn from the Foureir series?? Also why is the persiod a? If n was not 1 then the period would not be a, would it/?
 
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You have a definition of f(x), so just plug that into the integral. All of the terms will drop out except the one with c_n. I don't understand your question about the period.
 
Are you talking about the a in the denominator of the k_n? This puts everything in the right length scale, so the eigenfunction's conditions at the boundary of the box are met. Does that make sense?
 

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