# Showing property for the derivative partial_x T of a trigonometric polynomial

Showing property for the derivative ${\mathrm{\partial }}_{x}T$ of a trigonometric polynomial
Let be
$T=\sum _{n=0}^{N}\stackrel{^}{T}\left(n\right){e}^{inx}$
a trigonometric polynomial of grade N without negative frequencies.
I wanna show that
${\mathrm{\partial }}_{x}T=-iN\left({F}_{N}\ast T-T\right)$
Where ${F}_{N}\ast T$ meas the convolution of the Fejer Kernel and T.
might be easy..but I just can't work out the right conversion for this property..
SO
${\mathrm{\partial }}_{x}T=\sum _{n=0}^{N}\stackrel{^}{T}\left(n\right)in{e}^{inx}$
$=\sum _{n=0}^{N}\left(\frac{1}{2\pi }{\int }_{-\pi }^{\pi }T\left(y\right){e}^{-iny}dy\right){e}^{inx}in$
$=\frac{1}{2\pi }{\int }_{-\pi }^{\pi }T\left(y\right)\left(\sum _{n=0}^{N}{e}^{in\left(x-y\right)}\right)in$
from there on I get carried away in the wrong direction. is the derivative right?
Also..the Fejer Kernel can be expressed as the mean arithmetic value of the dirichtlet kernel so:
${F}_{N}=\frac{1}{n+1}\sum _{k=0}^{n}{D}_{k}\left(x\right)$
Where ${D}_{k}=\sum _{n=-k}^{k}{e}^{inx}$ is the Dirichtlet Kernel
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Step 1
1). The last equality you wrote for the derivative is weird, since you have the in outside a term that involves summing in n, so that's a mistake.
2). The definition of ${F}_{N}$ should involved 'N' rather than 'n'.
3). Either FN should be defined to be $\frac{1}{N}\sum _{k=0}^{N-1}{D}_{k}$ or the result you wish to prove should be ${\mathrm{\partial }}_{x}T=-i\left(N+1\right)\left({F}_{N}\ast T-T\right)$.
I will keep your definition of ${F}_{N}$ (with point (2) taken into account of course) and prove ∂xT=−i(N+1)(FN∗T−T). We show the fourier transform of both sides is the same -- this suffices.
Step 2
Note $\left[-i\left(N+1\right)\left({F}_{N}\ast T-T\right)\right]\stackrel{^}{}\left(m\right)=-i\left(N+1\right)\left[\stackrel{^}{{F}_{N}}\left(m\right)\stackrel{^}{T}\left(m\right)-\stackrel{^}{T}\left(m\right)\right]$ (the latter you can see from your first equality for the derivative). So, immediately, if $m<0$ or if $m\ge N+1$, both sides are 0 and we're good. So suppose $0\le m\le N$. Then, $\stackrel{^}{{F}_{N}}\left(m\right)=\frac{1}{N+1}\sum _{k=0}^{N+1}\stackrel{^}{{D}_{k}}\left(m\right)=\frac{1}{N+1}\sum _{k=0}^{N+1}{1}_{m\le k}=\frac{1}{N+1}\left[N+1-m\right]$, so we get $\left[-i\left(N+1\right)\left({F}_{N}\ast T-T\right)\right]\stackrel{^}{}\left(m\right)=-i\left(N+1\right)\stackrel{^}{T}\left(m\right)\frac{-m}{N+1}=im\stackrel{^}{T}\left(m\right)$, as desired.