LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-04

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Closed form and size bounds for the Dirichlet kernel

Statement

For every integer $N \geq 0$ ,

$D_{N} (t) = 1 + 2 \sum_{k = 1}^{N} \cos (2 π k t) .$

If $t \notin Z$ , then

$D_{N} (t) = \frac{\sin ((2 N + 1) π t)}{\sin (π t)} .$

If $m \in Z$ , then $D_{N} (m) = 2 N + 1$ . In particular $D_{N}$ is even and

$∣ D_{N} (t) ∣ \leq 2 N + 1 (t \in R),$

while for $t \notin Z$ ,

$∣ D_{N} (t) ∣ \leq \frac{1}{∣ \sin (π t) ∣} .$

Facts & Assumptions

Given: An integer $N \geq 0$ and a real $t$ .

[L1]

The Dirichlet kernel is $D_{N} (t) = \sum_{∣ k ∣ \leq N} e^{2 π i k t}$ (Dirichlet and Fejer kernels).

Proof

technique · direct

1.1

Pair the terms with indices $k$ and $- k$ in [L1]. This gives $D_{N} (t) = 1 + \sum_{k = 1}^{N} (e^{2 π i k t} + e^{- 2 π i k t}) = 1 + 2 \sum_{k = 1}^{N} \cos (2 π k t) .$ Hence $D_{N}$ is even. If $t = m \in Z$ , then every exponential equals $1$ , so $D_{N} (m) = 2 N + 1$ .

L1algebra

1.2

Assume $t \notin Z$ and put $z = e^{2 π i t} \neq 1$ . Then [L1, algebra] $D_{N} (t) = z^{- N} \sum_{j = 0}^{2 N} z^{j} = z^{- N} \frac{1 - z^{2 N + 1}}{1 - z} .$ Rewriting numerator and denominator with half-angle factors yields $D_{N} (t) = \frac{e^{- π i (2 N + 1) t} - e^{π i (2 N + 1) t}}{e^{- π i t} - e^{π i t}} = \frac{\sin ((2 N + 1) π t)}{\sin (π t)} .$

L1algebra

2.1

The triangle inequality applied to [L1] gives $∣ D_{N} (t) ∣ \leq 2 N + 1$ for every $t$ . If $t \notin Z$ , step 1.2 and $∣ \sin ((2 N + 1) π t) ∣ \leq 1$ give $∣ D_{N} (t) ∣ \leq \frac{1}{∣ \sin (π t) ∣} .$

L1step 1.2algebra∎

Depends on

Dirichlet and Fejer kernels

Used by

Dependency tree · two levels

2 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Loukas Grafakos, Classical Fourier Analysis, 3rd ed. (standard reference, not scraped)
Richard S. Laugesen, Harmonic Analysis Lecture Notes (standard reference, not scraped)