Alphabeta Math
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Fourier partial sums of the sawtooth

Example

Assume the Axiom of Countable Choice.

Let f be the one-periodic sawtooth given by f(0)=0 and f(x)=x−12 for 0<x<1. Then f^(0)=0 and, for k≠0,

f^(k)=−12πik.

Hence

SNf(x)=−∑k=1Nsin⁡(2πkx)πk.

At every noninteger x, SNf(x)→f(x), while at every integer x, SNf(x)→0.

Facts & Assumptions

Given: The Axiom of Countable Choice and the one-periodic sawtooth f(0)=0 and f(x)=x−12 for 0<x<1.

[L1]

Fourier coefficients and partial sums are defined by the one-period formulas in Period-one Fourier coefficients, partial sums, and convolution on the torus.

[L2]

Assuming the Axiom of Countable Choice, a one-periodic bounded-variation function converges at each point to the midpoint of its one-sided limits under Fourier partial sums (Dirichlet-Jordan pointwise convergence).

Verification

technique · direct
1.1L1algebra

Since ∫01(x−12) dx=0, [L1] gives f^(0)=0. For k≠0, direct integration gives f^(k)=∫01(x−12)e−2πikx dx=−12πik.

2.1L1step 1.1algebra

Insert the coefficients from step 1.1 into the partial-sum formula [L1]. Pairing the k and −k terms gives SNf(x)=−∑k=1Nsin⁡(2πkx)πk.

3.1L2step 2.1choosealgebra∎

The sawtooth is piecewise C1, hence of bounded variation on one period. If x∉Z, choose the unique integer m with y:=x−m∈(0,1). Because both f and its Fourier partial sums are one-periodic, one has SNf(x)=SNf(y) and f(x)=f(y)=y−12. The one-sided limits at y therefore both equal f(x), so [L2] gives SNf(x)→f(x). At an integer x, the one-sided limits are −1/2 and 1/2, whose midpoint is 0, so [L2] gives SNf(x)→0.

Depends on

Used by

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Dependency tree · two levels

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Sources