fourier partial sums are given by the Dirichlet kernel

[[concept]]

Topics

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Theorem

For all $f\in L^2([-\pi ,\pi ])$ and all $N\in \mathbb{N}\cup \{0\}$,

S_{N}f(x) &= \int _{-\pi}^\pi D_{N}(x-t) f(t) \, dx \\ D_{N}(x) &= \begin{cases} \frac{2N+1}{2\pi} & x=0 \\ \frac{\sin\left( N+\frac{1}{2} \right)x}{2\pi \sin \frac{x}{2}} & x \neq 0 \end{cases} \end{align}$$ The function $D_{N}$ is [[Concept Wiki/continuous map\|continuous]] and also smooth and is called the **Dirichlet kernel**.

ie the partial Fourier sum can be written as the integral of $f$ with the Dirichlet kernel

Proof (warmup for some calculations)

For any $f\in L^2([-\pi ,\pi ])$, we know that

S_{N}f(x) &= \sum_{\lvert n \rvert \leq N} \left( \frac{1}{2\pi} \int _{-\pi}^\pi f(t) e^{- i n t} \, dt \right)e^{inx} \\ (*)&= \int _{-\pi}^\pi f(t) \underbrace{\left( \frac{1}{2\pi} \sum_{\lvert n \rvert \leq N} e^{i n (x-t)}\right)}_{D_{N}(x-t)}\, dt \\ \end{align}$$ Where $(*)$ is by [[Concept Wiki/lebesgue integral of sum is sum of the integral\|linearity of the integral]] $$\begin{align} D_{N}(x) &= \frac{1}{2\pi} \sum_{\lvert n \rvert \leq N} e^{i n x} \\ &= \frac{1}{2\pi} e^{-i N x} \sum_{n=0}^{2N} e^{i n x} \\ &= \begin{cases} \frac{1}{2\pi} e^{-i N x} \left[ \frac{1 - e^{i(2N-1)x}}{1-e^{ix}} \right] ,& e^{ix}\neq 1 \iff x\neq 0 \\ \frac{2N+1}{2\pi}, & x=0 \end{cases} \\ &= \begin{cases} \frac{1}{2\pi} \left[ \frac{e^{i(N + 1/2)x} - e^{i(N+1/2)x}}{e^{ix/2}-e^{-ix/2}} \right] ,& x\neq 0 \\ \frac{2N+1}{2\pi}, & x=0 \end{cases} \\ \left( \sin x = \frac{e^{ix} - e^{-ix}}{2i} \right) \implies&=\begin{cases} \frac{\sin\left( N+\frac{1}{2} \right)x}{2\pi \sin \frac{x}{2}},& x\neq 0 \\ \frac{2N+1}{2\pi}, & x=0 \end{cases} \end{align}$$ Which gives us the desired result $$\tag*{$\blacksquare$}$$

References

References

See Also

Mentions

Mentions

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