Problems with Fourier transformation: jump at discontinuity

[kahn10]

Homework Statement

Consider a particle in an infinite square well of width
= 1. The particle is in a state
given by:

Phi=A(1/2-|x-1/2)

a=1

b) Find the general form of the expansion coefficients (the Fourier coefficients, right?)
for expanding the function in terms of the square-well basis set.
c) On the same graph, plot the wave function and the expansion in terms of square well
functions (take the sum to at least five terms or so) to verify the notion of expanding
the wave function in eigenstates of the well.

Homework Equations

a_n=1/L *Integrate[f[x]*Cos(n*Pi*x/L),{x,-L,L}]
b_n=1/L *Integrate[f[x]*Sin(n*Pi*x/L),{x,-L,L}]

f[x]=a_0/2+Sum[a_n*Cos(n*Pi*x/L)+b_n*Sin(n*Pi*x/L),{n,1,Infinity}]

The Attempt at a Solution

I ran through the generalized form for a_n and b_n and got the following values:

a_n=2*(-(1/(2 n^2 \[Pi]^2)) + (-1)^n/(2 n^2 \[Pi]^2)) Cos[2 n \[Pi] x]

b_n=2*((-1)^n Sin[2 n \[Pi] x])/(2 n \[Pi])

I got these by splitting the transformation at 1/2 and rewriting 0 to 1/2 as A(1/2-(1/2-x)=x and 1/2 to 1 as A(1/2-(x-1/2)=(1-x). When :I graph my final transform it lines up nicely with the right side of my phi but not the left. I know I'm missing an a_0 but I can't explain why it is so off for tho 0 to 1/2 portion.

 

[vela]

You might want to take the time to learn how to write the equations in LaTeX. I found your post incredibly hard to read.

Homework Statement

Consider a particle in an infinite square well of width
= 1. The particle is in a state
given by:

Phi=A(1/2-|x-1/2)

a=1

What does a represent?

b) Find the general form of the expansion coefficients (the Fourier coefficients, right?)
for expanding the function in terms of the square-well basis set.
c) On the same graph, plot the wave function and the expansion in terms of square well
functions (take the sum to at least five terms or so) to verify the notion of expanding
the wave function in eigenstates of the well.

Homework Equations

a_n=1/L *Integrate[f[x]*Cos(n*Pi*x/L),{x,-L,L}]
b_n=1/L *Integrate[f[x]*Sin(n*Pi*x/L),{x,-L,L}]

f[x]=a_0/2+Sum[a_n*Cos(n*Pi*x/L)+b_n*Sin(n*Pi*x/L),{n,1,Infinity}]

The Attempt at a Solution

I ran through the generalized form for a_n and b_n and got the following values:

a_n=2*(-(1/(2 n^2 \[Pi]^2)) + (-1)^n/(2 n^2 \[Pi]^2)) Cos[2 n \[Pi] x]

b_n=2*((-1)^n Sin[2 n \[Pi] x])/(2 n \[Pi])

I got these by splitting the transformation at 1/2 and rewriting 0 to 1/2 as A(1/2-(1/2-x)=x and 1/2 to 1 as A(1/2-(x-1/2)=(1-x). When :I graph my final transform it lines up nicely with the right side of my phi but not the left. I know I'm missing an a_0 but I can't explain why it is so off for tho 0 to 1/2 portion.

Right idea, but you need to think about what the eigenstates for the square well are. It's not the same as the Fourier basis.