# Inverse Discrete Fourier Transform proof help

1. Jan 28, 2009

### emnaki

I'm reading the text Understanding Digital Signal Processing Second Edition and in the text they give the IDFT without any proof and so I tried to do a quick proof, but I have not been able do it here is my attempted steps:
$$X(m)=\sum^{N-1}_{n=0}x(n)e^{-i2\pi mn/N} \\$$
Considering x(1),
$$x(1)e^{-i2\pi m/N}=X(m)-\sum^{N-1}_{n=0,n\neq 1}x(n)e^{-i2\pi mn/N} \\$$
$$x(1)=X(m)e^{i2\pi m/N}-\sum^{N-1}_{n=0,n\neq 1}x(n)e^{-i2\pi mn/N} e^{i2\pi m/N} \\$$
$$x(1)=X(m)e^{i2\pi m/N}-\sum^{N-1}_{n=0,n\neq 1}x(n)e^{i2\pi m(1-n)/N} \\$$
$$\mbox{Add up for all m} \\$$
$$Nx(1)=\sum^N_{m=1}X(m)e^{i2\pi m/N}-\sum^N_{m=1}\sum^{N-1}_{n=0,n\neq 1}x(n)e^{i2\pi m(1-n)/N} \\$$
$$x(1)=\frac{1}{N}(\sum^N_{m=1}X(m)e^{i2\pi m/N}-\sum^N_{m=1}\sum^{N-1}_{n=0,n\neq 1}x(n)e^{i2\pi m(1-n)/N}) \\$$

So it looks like I got the first term, but how do I remove the second term with the double summation? Is my working wrong? Does the second term actually cancel out?