Problem: Consider partitioning $f(x)=x^2$ on $[-1,1]$ into $n$equal sub-intervals. We seek to derive an expression for $L(f, P)$ and $U(f, P)$ in terms of $n$, where $n$ is even.

Now the question seemed a bit vague to me, but non the less this is what I did and arrived at:

Given $P = \{x_0, x_1, ..., x_n \}$

$$L(f, P) := \sum_{i=1}^{n}m_i(x_i - x_{i-1})$$ $$U(f, P) := \sum_{i=1}^{n}M_i(x_i - x_{i-1})$$ where $$m_i = \inf\{f(x): x \in [x_{x-1}, x_i]\}$$ $$M_i = \sup\{f(x): x \in [x_{i-1}, x_{i}]\}$$

Now since we wish to partition $f$ over $[-1, 1]$ into $n$ equal sub-intervals

$$\implies L(f, P) = (x_n - x_{n-1})\sum_{i=1}^{n}m_i$$ $$\implies U(f, P) = (x_n - x_{n-1})\sum_{i=1}^{n}M_i$$

And since we know $n$ is even, $\exists k \ni 2k =n$. This allows us to break our partition $P$ into two separate partitions $P_1$ and $P_2$ over two intervals $[-1, 0]$ and $[0, 1]$. Thus

$$\begin{align}P_1 &= \{x_1, x_2, ...., x_k\} \ & \text{where} && \ x_1 = -1 < x_2 < ... < x_k = 0 \\ P_2 &= \{x_{k}, x_{k+1}, ...., x_{2k}\} \ & \text{where} && \ x_k = 0 < x_{k+1} < ... < x_{2k} = 0 \end{align}$$

Since $f$ is continuous on $[-1,1]$, $m_i = \min(f : [x_{n-1}, x_n] \to \mathbb{R})$ and $M_i = \max(f : [x_{n-1}, x_n] \to \mathbb{R})$.

Therefore for $P_1$: $m_i = f(x_n)$ and $M_i = f(x_{n-1})$, and for $P_2$: $m_i = f(x_{n-1})$ and $M_i = f(x_n)$.

Finally putting all this together allows us to rewrite $L(f, P)$ and $U(f, P)$ as follows:

$$\implies L(f, P) = (x_n - x_{n-1})\left(\sum_{n=1}^{k}f(x_n) + \sum_{n=k}^{2k}f(x_{n-1})\right)$$

$$\implies U(f, P) = (x_n - x_{n-1})\left(\sum_{n=1}^{n}f(x_{n-1}) + \sum_{n=k}^{2k}f(x_{n})\right)$$

But I'm assuming when they say 'derive an expression' they want a closed form solution for the Riemann Sum,and I'm not sure how to convert what I've arrived at into a closed form, or if it is even possible to convert into closed form as the summand is varying.

Can a closed form solution for these lower and upper Riemann Sums be found?

  • 1
    $\begingroup$ Hint: $\sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6}$ $\endgroup$ – florence Aug 15 '16 at 23:05
  • $\begingroup$ First notice that there is a formula for the $x_i$ since they divide $[-1,1]$ into $n$ equal parts (you can ask yourself what is the length of those parts). $\endgroup$ – Joel Cohen Aug 15 '16 at 23:15

$f$ is even and the number of partitions is even (so $n=2m$), so $$L(f,P) = 2\sum_{i=m+1}^n(x_i-x_{i-1})m_i$$ Each subinterval has length $\frac{1}{m}$, and since $f$ is increasing on $[0,\infty)$, $$m_i = x_{i-1}^2 = (\frac{i-m-1}{m})^2 = \frac{(i-m-1)^2}{m^2}$$ So the sum turns into $$L(f,P)=2\sum_{i=m+1}^{n}\frac{1}{m}\frac{(i-m-1)^2}{m^2} = \frac{2}{m^3}\sum_{i=m+1}^n (i-m-1)^2 = \frac{2}{m^3}\sum_{k=0}^{m-1} k^2 = \frac{2}{m^3}\frac{m(m-1)(2m-1)}{6}$$


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