Showing property for the derivative ${\mathrm{\partial}}_{x}T$ of a trigonometric polynomial

Let be

$T=\sum _{n=0}^{N}\hat{T}(n){e}^{inx}$

a trigonometric polynomial of grade N without negative frequencies.

I wanna show that

${\mathrm{\partial}}_{x}T=-iN({F}_{N}\ast T-T)$

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

${\mathrm{\partial}}_{x}T=\sum _{n=0}^{N}\hat{T}(n)in{e}^{inx}$

$=\sum _{n=0}^{N}(\frac{1}{2\pi}{\int}_{-\pi}^{\pi}T(y){e}^{-iny}dy){e}^{inx}in$

$=\frac{1}{2\pi}{\int}_{-\pi}^{\pi}T(y)(\sum _{n=0}^{N}{e}^{in(x-y)})in$

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:

${F}_{N}=\frac{1}{n+1}\sum _{k=0}^{n}{D}_{k}(x)$

Where ${D}_{k}=\sum _{n=-k}^{k}{e}^{inx}$ is the Dirichtlet Kernel

Let be

$T=\sum _{n=0}^{N}\hat{T}(n){e}^{inx}$

a trigonometric polynomial of grade N without negative frequencies.

I wanna show that

${\mathrm{\partial}}_{x}T=-iN({F}_{N}\ast T-T)$

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

${\mathrm{\partial}}_{x}T=\sum _{n=0}^{N}\hat{T}(n)in{e}^{inx}$

$=\sum _{n=0}^{N}(\frac{1}{2\pi}{\int}_{-\pi}^{\pi}T(y){e}^{-iny}dy){e}^{inx}in$

$=\frac{1}{2\pi}{\int}_{-\pi}^{\pi}T(y)(\sum _{n=0}^{N}{e}^{in(x-y)})in$

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:

${F}_{N}=\frac{1}{n+1}\sum _{k=0}^{n}{D}_{k}(x)$

Where ${D}_{k}=\sum _{n=-k}^{k}{e}^{inx}$ is the Dirichtlet Kernel