This task comes from my text book, can someone help me solve it, please?
Let $$R_{n} = \sum_{i = 1}^{n} f(\xi_{n})\Delta x_{i} = > f(\xi_{1})\Delta x_{1} + f(\xi_{2})\Delta x_{2} + f(\xi_{3})\Delta x_{3} + ...+f(\xi_{n})\Delta x_{n}$$ be a Riemann sum of a function $f$. Explain the notation and provide the assumptions for:
- $\Delta x_{i}$ -
- $\xi_{i}$ -
- $f(xi_{i})$ -
- $n$ -
- $f$ is -
Can someone help me solve this theory task?
My attempt following the same order:
- $\Delta x_{i} = x_{i} - x_{i - 1}$ - partition of equal size, the length of subinterval
- $\xi_{i} \in [x_{i-1}, x_{i}]$
- $f(xi_{i}) - $ no idea
- $n $ is a number of subintervals
- f is continuous (?)
Can someone help me fix it? The text book left little space to fill, so I expect something basic is required from me.
Thank you.
No, the $Δx_i$ need not have equal length. The partition of the interval $[a,b]$ is given by the points $x_i$ with $$ a=x_0<x_1<...<x_n=b. $$ Yes to $\xi_i\in [x_{i-1},x_i]$ and $f(\xi_i)$ being the function values.
There is nothing more to this sum, and all these properties come from the "Riemann" in "Riemann sum".
If you want the discussion to extend to the Riemann integral, then you need to add the bounded total variation of $f$ and that the maximal interval length of the subdivision goes to zero.
As to the $\xi_i$, Riemann starts his exploration of the integral with a continuously differentiable function $F$ and the mean value theorem $$ \frac{F(x_i)-F(x_{i-1})}{x_i-x_{i-1}}=F'(\xi_i) $$ for some $\xi_i\in (x_{i-1},x_i)$. In reversing this he found $$ F(b)-F(a)=\sum_{i=1}^n F'(\xi_i)Δx_i. $$ Then he turned this around and asked for the condition that, replacing $F'$ with $f$, these sums for infinitesimal partitions will all give the same result in the limit of the partition sub-interval lengths going to zero.
See chapter 4 in Bernhard Riemann (1867): "Ueber die Darstellbarkeit einer Function durch eine trigonometrische Reihe."