I want to give the complex trigonometric polynomial to the function $f(x)=x$ with support $x_k=\dfrac{\pi k}{2}$ for $k=0,1,2,3$ using the Fast Fourier Transformation.
For the trigonometric polynomial
$p(x)=\sum_{k=0}^3 \alpha_ke^{ikx}$ with $p(x_k)=\dfrac{\pi k}{2}$ the coefficients $\alpha_j$ are given by:
$\alpha_j=\displaystyle{\frac14\sum_{k=0}^3 \frac{\pi k}{2}}e^{\displaystyle{{-ijk\pi/4}}}$
for $j=0,1,2,3$.
Now I split the sum into even and odd parts.
Then
$\alpha_j=\displaystyle{\frac14\left(\sum_{k=0}^2 \frac{2k\pi}{2}e^{\displaystyle{-ij(2k)(2\pi)/4}}\right)}+e^{\displaystyle{-ij\pi/2}}\left(\frac14\sum_{k=0}^2 \frac{(2k+1)\pi}{2}e^{\displaystyle{-ij(2k+1)(2\pi)/2}}\right)$
We get:
$\alpha_0=\pi$
$\alpha_1=-\dfrac34\pi+\dfrac14i\pi$
$\alpha_2=-\dfrac12\pi$
$\alpha_3=-\dfrac34\pi-\dfrac14i\pi$
So $p(x)=\pi+(-\dfrac34\pi+\dfrac14i\pi)e^{ix}-\dfrac12\pi e^{2ix}+(-\dfrac34\pi-\dfrac14i\pi)e^{3ix}$
But if I plot the real part of this function, I get:
Which does not look very accurate.
Have I done a mistake, or can you tell me what I am supposed to do?
Can you give the correct solution to the FFT here, so I can try again?
Do you know how to perfor this specific FFT with a software like Python for example.
Thanks in advance.