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We apply composite Trapezoidal rule over $[0,1]$. Let $h$ be fixed and $f$ smooth such that $$\int_{0}^{1}f(x)dx=\frac{h}{2}[f(0)+2\sum_{r=1}^{n-1}f(rh)+f(1)]+\sum_{i=1}^{N}K_ih^{2i}$$ where $K_i$ depends only on $f$. Write: $$R(h):=\frac{h}{2}[f(0)+2\sum_{r=1}^{n-1}f(rh)+f(1)]$$ Check that $$\int_{0}^{1}f(x)dx=R(\frac{h}{2^n})+\frac{1}{3}[R(\frac{h}{2^n})-R(\frac{h}{2^{n-1}})]-\frac{1}{2^{4n-2}}K_2h^4+O(h^6)$$

I am unsure how to check the last part is true. It seems that routine computation is overcomplicated and I can't get it right. Any idea?

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  • $\begingroup$ In the first formulas, obviously $nh=1$. How does the $n$ of the last formula relate to that? $\endgroup$ – LutzL Nov 21 '16 at 15:21
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You have \begin{align} R(h)&=I+K_1h^2+K_2h^4+O(h^6)\\ R(2h)&=I+K_1\,4h^2+K_2\,16h^4+O(h^6)\\ \end{align} which you can combine to eliminate $K_1$.

Then replace $h$ by $\frac{h}{2^m}$ to get the desired result.

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