Důkaz. Pomocí Taylorova rozvoje: $$\begin{array}{rl}{f}_2& =f\left(x_0\right)+2h{f}^{\prime }\left(x_0\right)+2h^2{f}^{\prime \prime }\left(x_0\right)+\frac{4h^3}{3}{f}^{\prime \prime \prime }\left(x_0\right)+\frac{2h^4}{3}{f}^{\left(4\right)}\left(x_0\right)+\frac{4h^5}{15}{f}^{\left(5\right)}\left(x_0\right)+𝒪\left(h^6\right)\\ f_1& =f\left(x_0\right)+h{f}^{\prime }\left(x_0\right)+\frac{h^2}{2}{f}^{\prime \prime }\left(x_0\right)+\frac{h^3}{6}{f}^{\prime \prime \prime }\left(x_0\right)+\frac{h^4}{24}{f}^{\left(4\right)}\left(x_0\right)+\frac{h^5}{120}{f}^{\left(5\right)}\left(x_0\right)+𝒪\left(h^6\right)\\ f_{-1}& =f\left(x_0\right)-h{f}^{\prime }\left(x_0\right)+\frac{h^2}{2}{f}^{\prime \prime }\left(x_0\right)-\frac{h^3}{6}{f}^{\prime \prime \prime }\left(x_0\right)+\frac{h^4}{24}{f}^{\left(4\right)}\left(x_0\right)-\frac{h^5}{120}{f}^{\left(5\right)}\left(x_0\right)+𝒪\left(h^6\right)\\ f_{-2}& =f\left(x_0\right)-2h{f}^{\prime }\left(x_0\right)+2h^2{f}^{\prime \prime }\left(x_0\right)-\frac{4h^3}{3}{f}^{\prime \prime \prime }\left(x_0\right)+\frac{2h^4}{3}{f}^{\left(4\right)}\left(x_0\right)-\frac{4h^5}{15}{f}^{\left(5\right)}\left(x_0\right)+𝒪\left(h^6\right)\end{array}$$ Danou lineární kombinací dostáváme $$\begin{array}{rl}{f}_2-4f_1+6f_0-4f_{-1}+f_{-2}& =h^4{f}^{\left(4\right)}\left(x_0\right)+𝒪\left(h^6\right)\end{array}$$ Po vydělení $h^4$ dostaneme znění věty.