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 $$ \partial_x T= -iN(F_N \ast T-T) $$
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
$ \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
Very thankful for any help !
Best Answer
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 $F_N$ should be defined to be $\frac{1}{N}\sum_{k=0}^{N-1} D_k$ or the result you wish to prove should be $\partial_x T = -i(N+1)(F_N*T-T)$.
I will keep your definition of $F_N$ (with point (2) taken into account of course) and prove $\partial_x T = -i(N+1)(F_N*T-T)$. We show the fourier transform of both sides is the same -- this suffices.
Note $[-i(N+1)(F_N*T-T)]\hat{}(m) = -i(N+1)[\widehat{F_N}(m)\widehat{T}(m)-\widehat{T}(m)]$ and $\widehat{\partial_x T}(m) = im\widehat{T}(m)$ (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, $\widehat{F_N}(m) = \frac{1}{N+1}\sum_{k=0}^{N+1}\widehat{D_k}(m) = \frac{1}{N+1}\sum_{k=0}^{N+1} 1_{m \le k} = \frac{1}{N+1}[N+1-m]$, so we get $[-i(N+1)(F_N*T-T)]\hat{}(m) = -i(N+1)\widehat{T}(m)\frac{-m}{N+1} = im\widehat{T}(m)$, as desired.