I have an initial function u(x,0) = -sin(x) and I want to derive the FFT coefficients for an odd-parity solution in the form of u(x,t) = $\sum_{k \geq 1} a_{k} sin (kx)$. I tried using the normal expansion of the function in terms of $\exp{ikx}$ but it adds some error to the solution.
Can anyone suggest me the procedure of how to filter the Fourier coefficients which remains odd throughout the solution using numpy.fft.fft ?
If the function is inherently odd (like the sine functions) then only the imaginary part of the fft function will be non-zero. I think your problem is that your function is not periodic as it should be, you should exclude the last point:
import numpy as np
x=np.linspace(-np.pi,np.pi,50,endpoint=False)
y=-np.sin(x)
yf=np.fft.fft(y)
even_part=yf.real
odd_part=yf.imag
Here only odd_part[1] is non-zero.
If your function is not odd and you want to force it, you can either use sdt as I mentioned in the comments, or add the inverse of your function on left side then use fft.
Another point, if your input is not complex, then it's faster and more time efficient to use rfft