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Q: Let $f$ be a continuous function on [0, 1] and its derivative $f'$ is continuous on [0, 1]. Show that$$|\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{k}{n})-\int_0^1f(x)dx|\leq\frac{M}{2n}$$ where M is the maximum value of $|f'|$ on [0, 1].

I find this inequality very tricky to understand. Because the max $|f'|$ just came out of nowhere... I tried to draw a graph of $\frac{1}{n}\sum_{k=0}^{n-1}f(\frac{k}{n})$ but didn't get much information. Could somebody give me a hint or the direction of the solution?

LM2357
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user379423
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  • Do you know Riemann sums? – TMM Mar 27 '17 at 09:41
  • Yes but the Riemann sum should be sufficiently close to its integral... Not like M/2n. – user379423 Mar 27 '17 at 09:47
  • Why is that? You are approximating the integral by just evaluating at a bunch of points. If the derivative can have very large values, then you may miss out on a portion of the integral just because the peaks are in-between your evaluation points. This should hold for Riemann sums as well. (And the quantity you have looks exactly like a Riemann sum.) – TMM Mar 27 '17 at 09:51

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