fourier partial sums are given by the Dirichlet kernel

Theorem

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

\begin{aligned} S_{N} f (x) & = \int_{- π}^{π} D_{N} (x - t) f (t) d x \\ D_{N} (x) & = {\begin{cases} \frac{2 N + 1}{2 π} & x = 0 \\ \frac{\sin (N + \frac{1}{2}) x}{2 π \sin \frac{x}{2}} & x \neq 0 \end{cases} \end{aligned}

The function $D_{N}$ is 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 $f \in L^{2} ([- π, π])$ , we know that

\begin{aligned} S_{N} f (x) & = \sum_{| n | \leq N} (\frac{1}{2 π} \int_{- π}^{π} f (t) e^{- i n t} d t) e^{i n x} \\ (*) & = \int_{- π}^{π} f (t) \underset{D_{N} (x - t)}{\underset{⏟}{(\frac{1}{2 π} \sum_{| n | \leq N} e^{i n (x - t)})}} d t \end{aligned}

Where $(*)$ is by linearity of the integral

\begin{aligned} D_{N} (x) & = \frac{1}{2 π} \sum_{| n | \leq N} e^{i n x} \\ = \frac{1}{2 π} e^{- i N x} \sum_{n = 0}^{2 N} e^{i n x} \\ = {\begin{cases} \frac{1}{2 π} e^{- i N x} [\frac{1 - e^{i (2 N - 1) x}}{1 - e^{i x}}], & e^{i x} \neq 1 ⟺ x \neq 0 \\ \frac{2 N + 1}{2 π}, & x = 0 \end{cases} \\ = {\begin{cases} \frac{1}{2 π} [\frac{e^{i (N + 1 / 2) x} - e^{i (N + 1 / 2) x}}{e^{i x / 2} - e^{- i x / 2}}], & x \neq 0 \\ \frac{2 N + 1}{2 π}, & x = 0 \end{cases} \\ (\sin x = \frac{e^{i x} - e^{- i x}}{2 i}) ⟹ & = {\begin{cases} \frac{\sin (N + \frac{1}{2}) x}{2 π \sin \frac{x}{2}}, & x \neq 0 \\ \frac{2 N + 1}{2 π}, & x = 0 \end{cases} \end{aligned}

Which gives us the desired result

\begin{matrix} ◼ \end{matrix}

References

See Also

Mentions

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