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$f:[0,1]\rightarrow\mathbb R$ and $f\in C^1$, then the limit $\lim_{n\rightarrow\infty} n(\int_{0}^{1}f(x)dx-\frac{1}{n}\sum_{k=1}^{n}f(\frac{k-1}{n}))$ exists.

I guess the kernel lies in the sum because then I can write the sum as an integral but I do not know how.

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Every continuous function is integrable an hence all Riemann sums converge to the value of the integral, notice that the sum is just a Riemann sum of a partition of [0,1]. Now you have an estimation of the error which is made by the Riemann sum. Here use that the derivative is continuous too and hence takes a maximum in [0,1] (check your definition of differentiable on a closed interval, if you have defined it differently this might not work).

When you look at the intervalls $I=\left[\frac{i}{n},\frac{i+1}{n}\right]$ you have \[ \left|\int_{\frac{i}{n}}^\frac{i+1}{n} f(x) \; \mathrm{d}x -\frac{1}{n} f\left(\frac{i}{n}\right)\right| \leq \frac{1}{n^2} \cdot \sup_{\xi \in I} f'(\xi)\]

A little more explanation, with the fundamental theorem of calculus we have \[\int_{\frac{i}{n}}^\frac{i+1}{n} f(x)\; \mathrm{d}x= \frac{1}{n} \cdot f(\xi_1)\] with $\xi_1 \in \left(\frac{i}{n}, \frac{i+1}{n}\right)$. So we know that \[\int_{\frac{i}{n}}^\frac{i+1}{n} f(x) \; \mathrm{d}x -\frac{1}{n} f\left(\frac{i}{n}\right) = \frac{1}{n} \left(f(\xi_1)-f(\tfrac{i}{n}) \right) \] With mean value theorem you get \[ \frac{1}{n} \left(f(\xi_1)-f(\tfrac{i}{n}) \right)==\frac{1}{n} \cdot \left(\xi_1-\frac{i}{n} \right) \cdot f'(\xi_2)\leq \frac{1}{n^2} f'(\xi_2)\] with $\xi_2 \in \left( \frac{i}{n} , \xi_i\right)$

Using all this, we have \[ \lim_{n\to \infty} n \cdot \left( \int_0^1 f(x)\; \mathrm{d}x - \frac{1}{n} \sum_{k=1}^n f(\tfrac{i-1}{n})\right) = \lim_{n\to \infty} n \cdot \left( \sum_{i=1}^n \frac{1}{n} \cdot \left( \xi_i-\frac{i}{n}\right) f'(\xi_i) \right)\] Now we look at the absolute values and have \begin{align*} \lim_{n\to \infty} n \cdot \left( \sum_{i=1}^n \frac{1}{n} \cdot \left( \xi_i-\frac{i}{n}\right) f'(\xi_i) \right)& \leq \lim_{n\to \infty} \sum_{i=1}^n \left( \xi_i -\frac{1}{n} \right) |f'(\xi_i)|\\ & \leq \lim_{n\to \infty} \sum_{i=1}^n \frac{1}{n} |f'(\xi_i)| \\ &\leq \sup_{\xi \in [0,1]} |f'(\xi)| \end{align*}

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The right hand side therefore goes to $\int_{0}^{1}f(k) dk$? –  Voyage Mar 31 '13 at 13:35
    
yeah but you can't take the limit in the parenthesis as $n$ isn't bounded. You should have some estimations for the approximation –  Dominic Michaelis Mar 31 '13 at 13:39
    
Could you explain a little bit more your error approximation ? Where comes the $\frac{1}{n^2}$ from? Therefore, as n goes to infinity the whole term goes to 0, right? –  Voyage Mar 31 '13 at 13:57
    
@Voyage made it more explicit, the term goes not to 0 but you can prove it converges –  Dominic Michaelis Mar 31 '13 at 14:14
    
Are you sure you used the mean value theorem in a correct way? I think the equation $\frac{1}{n} \left(f(\xi_1)-f(\tfrac{i}{n}) \right)= \frac{1}{n^2} f'(\xi_2)$ is wrong –  Voyage Apr 1 '13 at 13:40
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