Show that if $f$ is continuously differentiable on $[0,1]$, then $$\lim\limits_{n\rightarrow\infty}n\left(\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n}\right)-\int_{0}^{1}f(x)dx\right)=\frac{f(1)-f(0)}{2}$$

Observe that \begin{align*} \lim\limits_{n\rightarrow\infty}n\left(\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n}\right)-\int_{0}^{1}f(x)dx\right)&=\lim\limits_{n\rightarrow\infty}n\left(\frac{1}{n}\sum_{i=1}^{n}f\left(\frac{i}{n}\right)-\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}f(x)dx\right)\\ &=\lim\limits_{n\rightarrow\infty}n\left(\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}\left[f\left(\frac{i}{n}\right)-f(x)\right]dx\right)\\ &=\lim\limits_{n\rightarrow\infty}n\left(\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}f'(c_i)\left[\left(\frac{i}{n}\right)-x\right]dx\right) \end{align*} where the last equality follows from the Mean Value Theorem.

Let $m_i=\inf\{f'(x):x\in[(i-1)/n,i/n]\}$ and $M_i=\sup\{f'(x):x\in[(i-1)/n,i/n]\}$, then we have the follow inequality: $$m_i\int_{(i-1)/n}^{i/n}\left[\left(\frac{i}{n}\right)-x\right]dx\leq\int_{(i-1)/n}^{i/n}f'(c_i)\left[\left(\frac{i}{n}\right)-x\right]dx\leq M_i\int_{(i-1)/n}^{i/n}\left[\left(\frac{i}{n}\right)-x\right]dx$$

Consequently $$\frac{1}{2n}\sum_{i=1}^{n}m_i\leq n\sum_{i=1}^{n}\int_{(i-1)/n}^{i/n}f'(c_i)\left[\left(\frac{i}{n}\right)-x\right]dx\leq\frac{1}{2n}\sum_{i=1}^{n} M_i$$ where $\int_{(i-1)/n}^{i/n}\left[\left(\frac{i}{n}\right)-x\right]dx=\frac{1}{2n^2}$.

I stuck at this step. And it seems not right because when I take the limit both sides, I have $0$. Can someone give me a hint or suggestion. Thanks in advanced.

  • $\begingroup$ You can't just write $f'(c_i)$, since $c_i$ depends on $x$ - it is a different value for each $x$. $\endgroup$ – Thomas Andrews Dec 23 '15 at 0:39
  • $\begingroup$ @ThomasAndrews that equality wrong? $\endgroup$ – Simple Dec 23 '15 at 0:52
  • $\begingroup$ Yes, your application of the mean value theorem is wrong - otherwise, every function is linear if there is a $c\in (a,b)$ such that $f(x)-f(a)=f'(c)(x-a)$ for all $x\in(a,b)$. $\endgroup$ – Thomas Andrews Dec 23 '15 at 0:55
  • $\begingroup$ @ThomasAndrews Can you give me a suggestion to modify it? $\endgroup$ – Simple Dec 23 '15 at 1:07
  • 2
    $\begingroup$ Possible duplicate of Limit with Integral and Sigma $\endgroup$ – Paramanand Singh Jun 13 '16 at 5:49

Your bounds

$$ \frac{1}{2n} \sum_{i=1}^{n} m_i \quad \text{and} \quad \frac{1}{2n} \sum_{i=1}^{n} M_i $$

converge to the same quantity, namely $\frac{1}{2}(f(1) - f(0))$. This is essentially because they are Riemann sums for $\frac{1}{2}f'(x)$.

Proof using Taylor Theorem. Let $x_i = i/n$ for brevity and consider $F(x) = \int_{0}^{x} f(t) \, dt$. Then we may write

$$ n \left( \frac{1}{n} \sum_{i=1}^{n} f(x_i) - \int_{0}^{1} f(x) \, dx \right) = n \sum_{i=1}^{n} (F(x_{i-1}) - F(x_i) + \tfrac{1}{n}F'(x_i)). $$

By Taylor Theorem, we can pick $c_i \in [x_{i-1}, x_i]$ such that

$$ F(x_{i-1}) = F(x_i - \tfrac{1}{n}) = F(x_i) - \tfrac{1}{n}F'(x_i) + \tfrac{1}{2n^2}F''(c_i). $$

Plugging this back, we have

$$ n \left( \frac{1}{n} \sum_{i=1}^{n} f(x_i) - \int_{0}^{1} f(x) \, dx \right) = \frac{1}{2n} \sum_{i=1}^{n} f'(c_i). $$

Taking $n \to \infty$, this converges to $\frac{1}{2}(f(1) - f(0))$ as desired.


Hint: Use Taylor's theorem and write the following: $$f(x) = f(\frac{i}{n}) +f'(\frac{i}{n})(x-\frac{i}{n})+h_i(x)(x-\frac{i}{n}) $$, where $h_i(x)$ is a continuous function that has a limit 0 at $x = \frac{i}{n}$. The rest should be a matter of computation.

  • $\begingroup$ I don't know the Taylow's theorem yet $\endgroup$ – Simple Dec 23 '15 at 1:08
  • $\begingroup$ Then, I think the answer by @sangchul lee serves the purpose. $\endgroup$ – dezdichado Dec 23 '15 at 3:12

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Leftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ Indeed, it involves the application of Bernoulli Numbers $\braces{B_{i}}$: \begin{align} \sum_{i = 1}^{n}\mathrm{f}\pars{{i \over n}} & = \int_{0}^{n}\mathrm{f}\pars{{x \over n}}\,\dd x + \sum_{i = 1}^{m}{B_{i} \over i!\,n^{i - 1}}\bracks{% \mathrm{f}^{\pars{i - 1}}\pars{1} - \mathrm{f}^{\pars{i - 1}}\pars{0}} + R_{+}\pars{\mathrm{f},m} \\[3mm] & = n\int_{0}^{1}\mathrm{f}\pars{x}\,\dd x\ +\ \overbrace{B_{1}}^{\ds{\half}}\ \bracks{\mathrm{f}\pars{1} - \mathrm{f}\pars{0}} + {1 \over 2n}\ \overbrace{B_{2}}^{\ds{{1 \over 6}}}\bracks{\mathrm{f}'\pars{1} - \mathrm{f}'\pars{0}} + R_{+}\pars{\mathrm{f},2} \end{align}

which yields \begin{align} &n\bracks{{1 \over n}\sum_{i = 1}^{n}\mathrm{f}\pars{{i \over n}} - \int_{0}^{n}\mathrm{f}\pars{{x \over n}}\,\dd x} = {\mathrm{f}\pars{1} - \mathrm{f}\pars{0} \over 2} + {\mathrm{f}'\pars{1} - \mathrm{f}'\pars{0} \over 12n} + R_{+}\pars{\mathrm{f},2} \end{align}

Then, \begin{align} \color{#f00}{\lim_{n \to \infty}n\bracks{% {1 \over n}\sum_{i = 1}^{n}\mathrm{f}\pars{{i \over n}} - \int_{0}^{1}\mathrm{f}\pars{x}\,\dd x}} & = \color{#f00}{{\mathrm{f}\pars{1} - \mathrm{f}\pars{0} \over 2}} \end{align}

whenever the remaining term $R_{+}\pars{\mathrm{f},2}$ vanishes out ( see the above cited link ) in the $n \to \infty$ limit.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.