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I have been working on this problem for 3 days and I still have no idea where to start with it. I previously proved an equation: $$hf'(x) = \frac{1}{2}(f(x+h)-f(x-h)) - \frac{h^2}{12}(f''(x+h)-f''(x-h)) + O(h^5)$$ and I am asked to prove that, given $x_i = a + ih$ and $h = \frac{b-a}{2N}$ (I assume 2N is the number of points), and $R(h) = 2h\Sigma_{i=0}^{N-1} f(x_{2i+1})$ $$\int_a^bf(s)ds -R(h) = \frac{h^2}{6} [f'(b)-f'(a)]+ O(h^4) $$

My prof. gave us a hint to apply the first equation to $\int^xf(s)ds$ but I have no idea how that helps. I have toyed around with it using Taylor series, some messy algebra, and working backwards, but I can't connect the first equation with the equation to be proven.

Can someone give me a nudge in the right direction? My biggest problem here is that I don't know how to start or connect the equations. Thanks!

user728757
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1 Answers1

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The number of points is actually $2N+1$, $2N$ is the number of sub-intervals. If you follow your professor's suggestion , you can obtain

$$ \int_{x-h}^{x+h} f(t) dt = 2h f(x) + \frac{h^2}{6}(f'(x+h)-f'(x-h)) + O(h^5). $$

and, in particular,

$$ \int_{x_{2i+1}-h}^{x_{2i+1}+h} f(t) dt = 2h f(x_{2i+1}) + \frac{h^2}{6}(f'(x_{2i+1}+h)-f'(x_{2i+1}-h)) + O(h^5). $$

If you now add up these equations, you'll get what you need.

PierreCarre
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