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Symmetric difference formula for Fourier partial sums

Statement

Let f be a one-period integrable function, let x,sR, and let N0. Then

SNf(x)s=01/2(f(x+t)+f(xt)2s)DN(t)dt.

Equivalently,

SNf(x)s=01/2(f(x+t)+f(xt)2s)sin((2N+1)πt)sin(πt)dt.

Facts & Assumptions

Given: A one-period integrable function f, reals x,s, and an integer N0.

[L1]

Fourier partial sums are Dirichlet convolutions: SNf(x)=01f(xt)DN(t)dt (Fourier partial sums are Dirichlet convolutions).

[L2]

The Dirichlet kernel is even and 01DN(t)dt=1 (Dirichlet and Fejer kernels).

[L3]

For tZ, DN(t)=sin((2N+1)πt)/sin(πt) (Closed form and size bounds for the Dirichlet kernel).

Proof

technique · direct
1.1

By [L1], SNf(x)s=01(f(xt)s)DN(t)dt, because [L2] gives 01DN(t)dt=1.

L1L2algebra
2.1

Split the integral in step 1.1 at 1/2 and substitute u=1t on [1/2,1]. Since f is one-periodic and DN(1u)=DN(u)=DN(u) by [L2], this yields SNf(x)s=01/2(f(xt)+f(x+t)2s)DN(t)dt.

step 1.1L2algebra
3.1

The first displayed formula is step 2.1 with the two summands reordered. Replacing DN(t) by the closed form from [L3] gives the second displayed formula.

step 2.1L3algebra

Depends on

Used by

Dependency tree · two levels

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Sources