ギブズ現象
f ^ ( n ) := 1 L ∫ 0 L f ( x ) e − 2 π i n x / L d x a n := 2 L ∫ 0 L f ( x ) cos  ( 2 π n x / L ) d x b n := 2 L ∫ 0 L f ( x ) sin  ( 2 π n x / L ) d x {\displaystyle {\begin{aligned}{\hat {f}}(n)&:={\frac {1}{L}}\int _{0}^{L}f(x)e^{-2\pi inx/L}\ dx\\a_{n}&:={\frac {2}{L}}\int _{0}^{L}f(x)\cos(2\pi nx/L)\ dx\\b_{n}&:={\frac {2}{L}}\int _{0}^{L}f(x)\sin(2\pi nx/L)\ dx\end{aligned}}}
従って、次の式が得られる:
lim N → ∞ S N f ( x 0 + L / 2 N ) = f ( x 0 + ) + a ⋅ 0.089490 ... lim N → ∞ S N f ( x 0 − L / 2 N ) = f ( x 0 − ) − a ⋅ 0.089490 ... {\displaystyle {\begin{aligned}\lim _{N\to \infty }S_{N}f(x_{0}+L/2N)&=f(x_{0}^{+})+a\cdot 0.089490\dots \\\lim _{N\to \infty }S_{N}f(x_{0}-L/2N)&=f(x_{0}^{-})-a\cdot 0.