The Fejér–Jackson inequality #
Every partial sum ∑_{k=1}^{M} sin(kθ)/k is strictly positive for θ ∈ (0, π).
The proof is the classical interior-minimum argument. The derivative is the shifted Dirichlet
kernel, handled without division: 2 sin(θ/2) ∑_{k=1}^M cos(kθ) = sin((M + ½)θ) - sin(θ/2). At an
interior minimum the derivative vanishes, which forces sin(Mθ) ≥ 0, so the minimum is at least
the previous partial sum, positive by induction; the boundary values are 0.
Main definitions #
Dilog.fejerSum M θ:∑_{k=1}^M sin(kθ)/k.
Main results #
Dilog.dirichlet_telescope: the telescoped Dirichlet kernel.Dilog.fejerSum_pos: the Fejér–Jackson inequality.
References #
- D. Jackson, Über eine trigonometrische Summe, Rend. Circ. Mat. Palermo 32 (1911) 257–262.
The Fejér–Jackson partial sum ∑_{k=1}^{M} sin(kθ)/k (indexed k ∈ range M with
summand sin((k+1)θ)/(k+1)).
Equations
- Dilog.fejerSum M θ = ∑ k ∈ Finset.range M, Real.sin ((↑k + 1) * θ) / (↑k + 1)
Instances For
Termwise differentiation: fejerSum M has derivative the shifted Dirichlet kernel
∑_{k=1}^M cos(kθ).
The multiplicative (division-free) Dirichlet kernel identity:
2 sin(θ/2) · ∑_{k=1}^M cos(kθ) = sin((M+½)θ) − sin(θ/2), by telescoping.
The critical-point dichotomy: at an interior local minimum of fejerSum M,
the "last term" sine sin(Mθ₀) is nonnegative — the two branches of the equality
sin((M+½)θ₀) = sin(θ₀/2) give sin(Mθ₀) = 0 or sin(Mθ₀) = sin θ₀ > 0.