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I have been working on this homework problem for 2 hours, having thought it was pretty simple, and its really annoying me.

I am given two operators and a function f(x) that is five times differentiable on some interval. $$\alpha f(x)=f(x+h/2) - f(x-h/2)$$ And $$\beta f(x) = 1/2(f(x+h/2) + f(x-h/2))$$ I am asked to prove that $$\beta\alpha^3f(x) = h^2/2(f''(x+h) - f''(x-h)) + O(h^5))$$

I believe we are working with equispaced points, where $\delta x = h$. Possibly relevant equations: At a fixed point $$(1) f''(x) ~ 1/4h^2(\alpha^2)f(x)) = f''(x) + O(h^2)$$ $$ (2) f''(x_v) ~ \frac{f_{v+1}(x) - 2f_v(x) + f_{v-1}(x)}{h^2} = f'(x_v) + O(h^2)$$ (Where $f_v(x) = f(x_0 + vh)$)

Attempted Solution: $$\alpha f(x) = f(x + h/2) - f(x-h/2)$$ $$\alpha^2f(x) = f(x + h) - f(x) - f(x) +f(x-h)$$ $$\alpha^3f(x) = f(x+3h/2) -2f(x + h/2) + f(x-h/2) -f(x+h/2) +2f(x-h/2) - f(x - 3h/2)$$ $$\beta\alpha^3f(x) = 1/2( [f(x+2h) -2f(x +h) + f(x) -f(x+h) +2f(x) -f(x-h)] - [f(x+h) -2f(x) + f(x-h) -f(x) + 2f(x-h) -f(x-2h)]) = 1/2(f(x+2h) -5f(x+h) +5f(x) -5f(x-h) -f(x-2h))$$ I may be a bit off with the expression--every time I work through it, I seem to get different solution.

After expanding it out, I can't seem to get it to fit the equation Im proving. The main reason is 1. The formula (1) for f''(x+h) requires a term f(x+3h). You cannot get it from the operators. Formula (2) you do not need that term, but I cannot get it to work out.

  1. How on earth do I get orders of $h^5$ as the error? I understand getting $O(h^4)$--that follows directly from factoring out $h^2$ when substituting in f''(x).

1 Answers1

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Define the translation operator $$(\tau f)(x)=f(x+h/2)\implies(\tau^n f)(x)=f(x+nh/2).$$ Then $α=τ-τ^{-1}$ and $β=\frac12(τ+τ^{-1})$ so that by the binomial theorem \begin{align} βα^3&=(αβ)α^2=\frac12(τ^2-τ^{-2})α^2\\ \end{align}

By Taylor expansion $τ=\exp(hD/2)$, where $D$ is the differentiation operator, so that $α=2\sinh(hD/2)$. Thus $$ α^2=4\sinh^2(hD/2)=2(\cosh(hD)-1)=h^2D^2+\frac1{24}h^4D^4+... $$ so that finally $$ (βα^3f)(x)=\frac{h^2}2[f''(x+h)-f''(x-h)]+\frac{h^4}{48}[f^{(4)}(x+h)-f^{(4)}(x-h)]+O(h^6) $$ The second term is $O(h^5)$ by the mean value theorem.

Lutz Lehmann
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