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I'm doing a project on the Dirichlet Kernel and I believe I've solved one of the problems that was assigned to me. If any one here could proof read my answer I would be very grateful.


Insert the formula for the Fourier coecients $c_n$ of the 2-periodic function $f$ into the sum \begin{equation} s_N(x) = \sum _{n = -N} ^N c_n e^{inx} \end{equation}

and hence show that

\begin{equation} s_N(x) = \int _{-\pi} ^{\pi} f(t) D_N (x-t)dt \end{equation}


\begin{equation} D_N = \frac{1}{2\pi} \sum _{n = -N} ^N e^{inx} \end{equation}


According to the lecture notes the Fourier coefficient $c_n$ is given by the formula \begin{equation*} c_n = \frac{1}{2 \pi}\int ^{\pi} _{-\pi} f(t) e ^{-i nt} dt \end{equation*} Now we get \begin{align*} s_N &= \sum ^N _{n = -N} c_n e^{inx} \\ &= \sum ^N _{n = -N} \frac{1}{2 \pi}\int ^{\pi} _{-\pi} f(t) e ^{-i nt} dt \cdot e^{inx} \\ &= \sum ^N _{n = -N} \frac{1}{2 \pi}\int ^{\pi} _{-\pi} f(t) e ^{in(x-t)} dt \\ &= \int ^{\pi} _{-\pi} f(t) \cdot \frac{1}{2 \pi} \sum ^N _{n = -N} e ^{in(x-t)} dt \end{align*} But we know that \begin{equation*} D_N(x)= \frac{1}{2 \pi} \cdot \sum ^N _{n = -N} e^{inx} \end{equation*} so \begin{align*} \int ^{\pi} _{-\pi} f(t) \cdot \frac{1}{2 \pi} \sum ^N _{n = -N} e ^{in(x-t)} dt &= \int ^{\pi} _{-\pi} f(t) \cdot D_N(x-t) dt \end{align*} which completes the proof. $\square$

Thank you!

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Looks good to me. – littleO Nov 26 '12 at 20:13
As formulated here, there's not more to it. The interesting part is writing $D_N(x)$ in closed form, i.e., without a $\sum$-sign. Having this expression in hand you can begin to understand why the inversion formula works. – Christian Blatter Nov 26 '12 at 20:18
Yes, it's very interesting to see what the graph of $D_N$ looks like as $N \to \infty$. It looks more and more like a delta function, which helps to explain why $s_N(x)$ might converge to $f(x)$. – littleO Nov 26 '12 at 20:22
Thanks! I appreciate it. – docjay Nov 27 '12 at 8:45

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