Find the order of $f'(x)\approx \frac{1}{12h}[-f(x+2h)+8f(x+h)-8f(x-h)+f(x-2h)]$

approximationderivativesnumerical methodsreal-analysis

  1. Find the order of approximation of
    $$f'(x)\approx \frac{1}{12h}[-f(x+2h)+8f(x+h)-8f(x-h)+f(x-2h)]$$

  2. Use the expression to find approximation for $f''(x)$

  1. $$-f(x+2h)=-(f(x)+2hf'(x)+\frac{4h^2}{2}f''(x)+o(h^3)=\\=-f(x)-2hf'(x)-\frac{4h^2}{2}f''(x)-o(h^3)$$

$$f(x-2h)=f(x)-2hf'(x)+\frac{4h^2}{2}f''(x)-o(h^3)$$

$$8f(x+h)=8(f(x)+hf'(x)+\frac{h^2}{2}f''(x)+o(h^3)=8f(x)+8hf'(x)+8\frac{h^2}{2}f''(x)+8o(h^3)$$

$$-8f(x-h)=-8(f(x)-hf'(x)+\frac{h^2}{2}f''(x)-o(h^3)=-8f(x)+8hf'(x)-8\frac{h^2}{2}f''(x)+8o(h^3)$$

So

$$|f'(x)-\frac{1}{12h}[-f(x+2h)+8f(x+h)-8f(x-h)+f(x-2h)]|=\\=|f'(x)-\frac{1}{12h}[-4hf'(x)-2o(h^3)+16hf'(x)+16o(h^3)]|=\\=|f'(x)-f'(x)-\frac{14o(h^3)}{12}|=o(h^3)?$$

2.

We just plug $x+2h,x+h,x-h,x-2h$ to the approximation and get:

$$ f''(x)\approx \frac{[-f(x+4h)-16f(x+3h)+16f(x+h)-130f(x)+64f(x+2h)+64f(x-h)+64f(x-2h)-64f(x-3h)]}{144h^2}$$

Is it correct?

Best Answer

Define $u(h)=f(x+h)-f(x-h)$, then we get the Taylor-MacLaurin expansion $$ u(h)=2f'(x)h+\frac13f'''(x)h^3+\frac1{60}f^{(5)}(x)h^5+... $$ Then apply Richardson extrapolation to eliminate the third degree term \begin{align} u(2h)-2^3u(h)&=2f'(x)[(2h)-2^3h]+\frac13f'''(x)[(2h)^3-2^3h^3]+\frac1{60}f^{(5)}(x)[(2h)^5-2^3h^5]+...\\ 8u(h)-u(2h)&=12f'(x)h-\frac25f^{(5)}(x)h^5+... \end{align} so that indeed you get $$ \frac{-f(x+2h)+8f(x+h)-8f(x-h)+f(x-2h)}{12h}=f'(x)-\frac1{30}f^{(5)}(x)h^4+O(h^6) $$

Now you can do the same with $$v(h)=f(x+h)-2f(x)+f(x-h)=2f''(x)h^2+\frac1{12}f^{(4)}(x)h^4+\frac{2}{6!}f^{(6)}(x)h^6+...$$

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