If I have a polynomial:
$a(x)= 2 + 5x -3x^2 +x^3$
The Fourier transform for N=4 is the evaluation of this polynomial in ${\omega}^0,{\omega}^1,{\omega}^2,{\omega}^3$ with ${\omega}^h = \cos(2\pi h/4) + i\sin(2\pi h/4) $. So:
${\omega}^0 = 1$
${\omega}^1 = i$
${\omega}^2 = -1$
${\omega}^3 = -i$
I evaluated the polynomial in these 4 points, and the result is $5, 5+4i, 7, 5-4i$.
Now I want to get the inverse transform, and according to the theory it's enough that I compute again the FFT on the resulting vector, but with ${\omega}^{-h}$ instead of ${\omega}^{h}$, and then I should divide the result for N=4.
I get:
${\omega}^0 = 1$
${\omega}^{-1} = -i$
${\omega}^{-2} = -1$
${\omega}^{-3} = i$
For example if I evaluate the resulting polynomial in ${\omega}^0$ I get 22/4, but the result should be 2. Where's the mistake?
Sign error: $a(-1)=-7$. Then the inverse transform gives the correct result $8/4$ for the constant coefficient.