Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I've been trying to solve this problem, but I haven't been able to calculate the exact limit, I've just been able to find some boundaries. I hope you guys can help me with it.

Let $f:[0,1] \to \mathbb{R}$ a differentiable function with a continuous derivative, calculate: $$\lim_{n\to \infty}\left(\sum_{k=1}^nf\left(\frac{k}{n}\right)-n\int_0^1f(x)dx\right) $$ I tried using Mean Value Theorem for derivatives and integrals and I got that $$\lim_{n\to \infty}\left(\sum_{k=1}^nf\left(\frac{k}{n}\right)-n\int_0^1f(x)dx\right)=\lim_{n\to \infty}\left(\sum_{k=1}^nf'\left(x_k**\right)(\frac{k}{n}-x_k*)\right) $$ Where $x_k*\in [\frac{k-1}{n},\frac{k}{n}]$ and $x_k**\in [x_k*,\frac{k}{n}]$, which looks like a Riemann Sum but I'm not sure if it's a Riemann Sum of $f'$ from $0$ to $1$, if this was true I believe the limit is $f(1)-f(0)$ but I'm not really sure about this.

Edit: fixed some typos with the $\frac{k}{n}$.

Edit 2: $k$ starts from $1$ not $0$.

share|improve this question
    
If you try something simple like $f(x) = x+c$ for some constant $c$, the limit is equal to $c+\frac{1}{2}$. Does this agree with your guess of $f(1)-f(0)$? If not, modify your guess. –  JimmyK4542 Aug 30 at 20:21
    
Thanks, just did that and the limit equals to $\frac{1}{2}$, not $c+\frac{1}{2}$, but $\frac{1}{2}=\frac{f(1)-f(0)}{2}$ which leads me to think that the limit might be the halve of the integral, because the last sum I got looks like a Riemann Sum with some parts missing in each interval. Any ideas? –  user142859 Aug 30 at 20:35
    
No, the limit does equal $c + \frac{1}{2}$. Notice that the summation is from $k = 0$ to $n$, so the summation has $n+1$ terms not $n$ terms. Now, that you have edited the problem, the limit is $\frac{1}{2}$. –  JimmyK4542 Aug 30 at 20:38
    
Sorry, my mistake, $k$ starts from $1$ just edited it. –  user142859 Aug 30 at 20:40

3 Answers 3

up vote 4 down vote accepted

Let $$ x_n=\sum_{k=0}^nf\left(\frac{k}{n}\right)-n\int_0^1f(x)dx $$

We will use the following result:

Lemma If $g:[0,1]\to\mathbb{R}$ is a continuously differentiable function. Then $$ \frac{g(0)+g(1)}{2}-\int_0^1g(x)dx= \int_{0}^1\left(x-\frac{1}{2}\right)g'(x)dx. $$ Indeed, this is just integration by parts: $$\eqalign{ \int_{0}^1\left(x-\frac{1}{2}\right)g'(x)dx &=\left.\left(x-\frac{1}{2}\right)g(x)\right]_{x=0}^{x=1} -\int_0^1g(x)dx\cr &=\frac{g(1)+g(0)}{2}-\int_0^1g(x)dx }$$

Now applying this to the functions $x\mapsto f\left(\frac{k+x}{n}\right)$ for $k=0,1,\ldots,n-1$ and adding the resulting inequalities we obtain $$ x_n-\frac{f(0) +f(1)}{2} = \int_0^1\left(x-\frac{1}{2}\right)H_n(x)dx\tag{1} $$ where, $$ H_n(x)=\frac{1}{n}\sum_{k=0}^{n-1}f'\left(\frac{k+x}{n}\right) $$ Clearly for every $x$, $H_n(x)$ is a Riemann sum of the function continuous $f'$, hence $$ \forall\,x\in[0,1],\quad\lim_{n\to\infty}H_n(x)=\int_0^1f'(t)dt $$ Moreover, $| H_n(x)|\leq\sup_{[0,1]}|f'|$. So, taking the limit in $(1)$ and applying the Dominated Convergence Theorem, we obtain $$ \lim_{n\to\infty}\left(x_n-\frac{f(0) +f(1)}{2}\right)= \left(\int_0^1f'(t)dt\right)\int_0^1\left(x-\frac{1}{2}\right)dx=0. $$ This proves that $$ \lim_{n\to\infty}x_n=\frac{f(0) +f(1)}{2} $$

And consequently $$ \lim_{n\to\infty}\left(\sum_{k=\color{red}{1}}^nf\left(\frac{k}{n}\right)-n\int_0^1f(x)dx\right)=\frac{f(1)-f(0)}{2} $$

share|improve this answer
    
Now that the OP changed the summation to start at $k = 1$ instead of $k = 0$, the limit is now $\frac{f(1)-f(0)}{2}$. –  JimmyK4542 Aug 30 at 20:56
    
@JimmyK4542, I hope I am as fast as he is in this game ! –  Omran Kouba Aug 30 at 20:58
    
Wow thanks, very nice solution. I forgot to say this, but this problem was presented in a context where we only know Mean Value Theorem for Integrals and the Fundamental Theorem of Calculus. Do you think there is a simpler way to approach to this problem only using only the definitions of Riemann Sums andsimpler theorems? –  user142859 Aug 30 at 21:19
    
It looks like an application of Abel-Plana formula. –  Felix Marin Aug 31 at 1:07

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$ $\ds{\lim_{n\ \to\ \infty}\bracks{\sum_{k = 1}^{n}\fermi\pars{k \over n} -n\int_{0}^{1}\fermi\pars{x}\,\dd x}:\ {\large ?}}$.

$\ds{\tt\mbox{This is an application of}}$ Abel-Plana formula:

\begin{align}&\color{#c00000}{\sum_{k = 1}^{n}\fermi\pars{k \over n}} =\sum_{k = 0}^{n - 1}\fermi\pars{k + 1 \over n} =\sum_{k = 0}^{\infty}\bracks{\fermi\pars{k + 1 \over n} -\fermi\pars{k + n + 1 \over n}} \\[5mm]&=\int_{0}^{\infty} \bracks{\fermi\pars{x + 1 \over n} - \fermi\pars{x + n + 1 \over n}}\,\dd x +\half\bracks{\fermi\pars{1 \over n} - \fermi\pars{n + 1 \over n}} \\[3mm]&+\color{#00f}{\ic\int_{0}^{\infty} \bracks{\fermi\pars{\ic x + 1 \over n} - \fermi\pars{\ic x + n + 1 \over n} -\fermi\pars{-\ic x + 1 \over n} + \fermi\pars{-\ic x + n + 1 \over n}}\times} \\[3mm]&\color{#00f}{\dd x \over \expo{2\pi x} - 1} \\[5mm]&=n\int_{1/n}^{\infty}\fermi\pars{x}\,\dd x -\int_{1 + 1/n}^{\infty}\fermi\pars{x}\,\dd x +\half\bracks{\fermi\pars{1 \over n} - \fermi\pars{n + 1 \over n}} + \color{#00f}{"\mbox{the blue term}"} \\[3mm]&=n\int_{1/n}^{1 + 1/n}\fermi\pars{x}\,\dd x +\half\bracks{\fermi\pars{1 \over n} - \fermi\pars{n + 1 \over n}} + \color{#00f}{"\mbox{the blue term}"} \\[3mm]&=n\int_{0}^{1}\fermi\pars{x}\,\dd x -n\int_{0}^{1/n}\fermi\pars{x}\,\dd x + n\int_{1}^{1 + 1/n}\fermi\pars{x}\,\dd x +\half\bracks{\fermi\pars{1 \over n} - \fermi\pars{n + 1 \over n}} \\[3mm]&\mbox{}+ \color{#00f}{"\mbox{the blue term}"} \end{align}

Since $\ds{\lim_{n\ \to\ \infty}\color{#00f}{\pars{"\mbox{the blue term}"}} = 0}$ and $\ds{\lim_{n\ \to\ \infty}n\int_{0}^{1/n}\fermi\pars{x}\,\dd x = \fermi\pars{0}}$ and $\ds{\lim_{n\ \to\ \infty}n\int_{1}^{1 + 1/n}\fermi\pars{x}\,\dd x = \fermi\pars{1}}$:

\begin{align} &\color{#66f}{\large\lim_{n\ \to\ \infty}\bracks{% \sum_{k = 1}^{n}\fermi\pars{k \over n} - n\int_{0}^{1}\fermi\pars{x}\,\dd x}} \\[3mm]&=-\fermi\pars{0} + \fermi\pars{1} +\half\bracks{\fermi\pars{0} - \fermi\pars{1}} =\color{#66f}{\large{\fermi\pars{1} - \fermi\pars{0} \over 2}} \end{align}

share|improve this answer

From the error analysis paragraph in the Wikipedia page for the Trapezoidal rule we have that the limit is: $$\frac{f(0)+f(1)}{2}.$$

share|improve this answer
    
I believe that tells me that the limit is: $$\lim_{n\to \infty}n*error + \frac{f(a)+f(b)}{2}$$ Am I doing anything wrong? Edit: I got it now, $n*error$ still tends to $0$. Thanks! –  user142859 Aug 30 at 20:52

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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