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 Incoming links: Fourier partial sum

Fejér sum 2026-09-25

The

n

th Fejér sum is the arithmetic mean of the first

n

σ_{n} (f, x) = \frac{1}{π} \int_{T} (\frac{1}{n} \sum_{k = 0}^{n - 1} D_{k} (x - t)) f (t) d t .

The finite trigonometric sum is

\sum_{k = 0}^{n - 1} sin (k + \frac{1}{2}) u = \frac{s i n ^{2} ( n u /2 )}{s i n ( u /2 )} .

Since

D_{k} (u) = sin ((k + 1/2) u) / (2 sin (u /2))

, the Fejér kernel is therefore

F_{n} (u) = \frac{1}{n} k = 0 \sum n - 1 D_{k} (u) = \frac{1}{2 n} \frac{sin ^{2} ( n u /2 )}{sin ^{2} ( u /2 )}

and

σ_{n} (f, x) = \frac{1}{π} \int_{T} F_{n} (x - t) f (t) d t .

The displayed square shows that

F_{n} \geq 0

σ_{n} (1) = 1

. Substituting

f = 1

\frac{1}{π} \int_{T} F_{n} (t) d t = 1.

Nonnegativity then yields

\frac{1}{π} \int_{T} ∣ F_{n} (t) ∣ d t = 1 .

s_{n} (f, x) = \frac{1}{2 π} \int_{T} f (t) \sum_{k = - n}^{n} e^{ik (x - t)} d t = \frac{1}{π} \int_{T} D_{n} (x - t) f (t) d t,

where

D_{n} (u) = \frac{sin (( n + \frac{1}{2} ) u )}{2 sin ( u /2 )} .