Sigma Notation and Riemann Sums

One strategy for calculating the area of a region is to cut the region into simple shapes, calculate the area of each simple shape, and then add these smaller areas together to get the area of the whole region. We will use that approach, but it is useful to have a notation for adding a lot of values together: the sigma $(\Sigma)$ notation.

The variable (typically $i$, $j$, or $k$) used in the summation is called the counter or index variable. The function to the right of the sigma is called the summand, and the numbers below and above the sigma are called the lower and upper limits of the summation. (Fig. 1)

Practice 1: Write the summation denoted by each of the following:

$(a) \sum_{k=1}^{5} k^{3}$$,        $(b) \sum_{j=2}^{7}(-1)^{j} \frac{1}{j}$,            $(c) \sum_{\mathrm{m}=0}^{4}(2 \mathrm{~m}+1).$

In practice, the sigma notation is frequently used with the standard function notation:

$\sum_{\mathrm{k}=1}^{3} \mathrm{f}(\mathrm{k}+2)=\mathrm{f}(1+2)+\mathrm{f}(2+2)+\mathrm{f}(3+2)=\mathrm{f}(3)+\mathrm{f}(4)+\mathrm{f}(5)$ and

$\sum_{\mathrm{i}=1}^{4} \mathrm{f}\left(\mathrm{x}_{\mathrm{i}}\right)=\mathrm{f}\left(\mathrm{x}_{1}\right)+\mathrm{f}\left(\mathrm{x}_{2}\right)+\mathrm{f}\left(\mathrm{x}_{3}\right)+\mathrm{f}\left(\mathrm{x}_{4}\right)$

Table 1

Example 1: Use the values in Table 1 to evaluate $\sum_{k=2}^{5} 2 \cdot f(k)$ and $\sum_{j=3}^{5}(5+f(j-2))$.

Solution: $\sum_{k=2}^{5} 2 \cdot f(k)=2 \cdot f(2)+2 \cdot f(3)+2 \cdot f(4)+2 \cdot f(5)=2 \cdot(3)+2 \cdot(1)+2 \cdot(0)+2 \cdot(3)=14$

$\begin{gathered} \sum_{j=3}^{5}(5+f(j-2))=(5+f(3-2))+(5+f(4-2))+(5+f(5-2))=(5+f(1))+(5+f(2))+(5+f(3)) \\ =(5+2)+(5+3)+(5+1)=21 \end{gathered}$

Practice 2: Use the values of $f$, $g$ and $h$ in Table 1 to evaluate the following:

(a) $\sum_{k=2}^{5} g(k)$     (b) $\sum_{j=1}^{4} h(j)$     (c) $\sum_{k=3}^{5}(g(k)+f(k-1))$

Example 2: For $f(x)=x^{2}+1$, evalutate $\sum_{\mathrm{k}=0}^{3} \mathrm{f}(\mathrm{k})$.

Solution: $\sum_{k=0}^{3} f(k)=f(0)+f(1)+f(2)+f(3)$
$=\left(0^{2}+1\right)+\left(1^{2}+1\right)+\left(2^{2}+1\right)+\left(3^{2}+1\right)=1+2+5+10=18$

Practice 3: For $g(x)=1 / x$, evaluate $\sum_{k=2}^{4} g(k)$ and $\sum_{\mathrm{k}=1}^{3} \mathrm{~g}(\mathrm{k}+1)$.

The summand does not have to contain the index variable explicitly: a sum from $\mathrm{k}=2$ to $\mathrm{k}=4$ of the constant function $f(k)=5$ can be written as

$\sum_{k=2}^{4} f(k)$ or $\sum_{k=2}^{4} 5=5+5+5=3 \cdot 5=15$. Similarly, $\sum_{k=3}^{7} 2=2+2+2+2+2=52=10$.

Since the sigma notation is simply a notation for addition, it has all of the familiar properties of addition

Summation Properties

Sum of Constants: $\sum_{k=1}^{\mathrm{n}} \mathrm{C}=\mathrm{C}+\mathrm{C}+\mathrm{C}+\ldots+\mathrm{C}(\mathrm{n} \text { terms })=\mathrm{n} \cdot \mathrm{C}$

Addition: $\sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\mathrm{a}_{\mathrm{k}}+\mathrm{b}_{\mathrm{k}}\right)=\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{k}}+\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{b}_{\mathrm{k}}$

Subtraction: $\sum_{\mathrm{k}=1}^{\mathrm{n}}\left(\mathrm{a}_{\mathrm{k}}-\mathrm{b}_{\mathrm{k}}\right)=\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{k}}-\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{b}_{\mathrm{k}}$

Constant Multiple: $\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{C}^{*} \mathrm{a}_{\mathrm{k}}=\mathrm{C} \cdot \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{a}_{\mathrm{k}}$

Problems 16 and 17 illustrate that similar patterns for sums of products and quotients are not valid.

Source: Dale Hoffman, https://s3.amazonaws.com/saylordotorg-resources/wwwresources/site/wp-content/uploads/2012/12/MA005-5.2-Sigma-Notation-and-Riemann-Sums.pdf
This work is licensed under a Creative Commons Attribution 3.0 License.

In Section 4.2 we will approximate the areas under curves by building rectangles as high as the curve, calculating the area of each rectangle, and then adding the rectangular areas together.

Example 3: Evaluate the sum of the rectangular areas in Fig. 2, and write the sum using the sigma notation.

Solution:

$\begin{aligned} \{\text { sum of the rectangular areas }\}=&\{\text { sum of (base):(height) for each rectangle }\} \\ &=(1) \cdot(1 / 3)+(1) \cdot(1 / 4)+(1)^{*}(1 / 5)=47 / 60 \end{aligned}$

Using the sigma notation,

$(1) \cdot(1 / 3)+(1) \cdot(1 / 4)+(1) \cdot(1 / 5)=\sum_{k=3}^{5} \frac{1}{k}$

Practice 4: Evaluate the sum of the rectangular areas in Fig. 3, and write the sum using the sigma notation.

The bases of the rectangles do not have to be equal. For the rectangular areas in Fig. 4

so the sum of the rectangular areas is $8 + 16 + 50 = 74$.

Example 4: Write the sum of the areas of the rectangles in Fig. 5 using the sigma notation.

Solution: The area of each rectangle is $\text { (base) } \cdot \text { (height)}$.

The area of the K^threctangle is $\left(x_{k}-x_{k-1}\right) \cdot f\left(x_{k}\right)$. and the total area of the rectangles is the sum $\sum_{\mathrm{k}=1}^{3}\left(\mathrm{x}_{\mathrm{k}}-\mathrm{x}_{\mathrm{k}-1}\right) \mathrm{f}\left(\mathrm{x}_{\mathrm{k}}\right)$.

Practice 5: Write the sum of the areas of the shaded rectangles in Fig. 6 using the sigma notation.

Suppose we want to calculate the area between the graph of a positive function $f$ and the interval $[a, b]$ on the x–axis (Fig. 7). The Riemann Sum method is to build several rectangles with bases on the interval $[a, b]$ and sides that reach up to the graph of f (Fig. 8). Then the areas of the rectangles can be calculated and added together to get a number called a Riemann Sum of f on [a, b]. The area of the region formed by the rectangles is an approximation of the area we want.

Example 5: Approximate the area in Fig. 9a between the graph of $f$ and the interval $[2, 5]$ on the x–axis by summing the areas of the rectangles in Fig. 9b.

Solution: The total area of rectangles is $(2)(3) + (1)(5) = 11$ square units.

In order to effectively describe this process, some new vocabulary is helpful: a partition of an interval and the mesh of the partition.

A partition $P$ of a closed interval $[a,b]$ into $n$ subintervals is a set of $n+1$ points $\left\{x_{0}=\mathrm{a}, x_{1}, x_{2}, x_{3}, \ldots, x_{\mathrm{n}-1}, x_{\mathrm{n}}=\mathrm{b}\right\}$ in increasing order, $\mathrm{a}=x_{0}$.

(A partition is a collection of points on the axis and it does not depend on the function in any way.)

The points of the partition $P$ divide the interval into $n$ subintervals (Fig. 10): $\left[x_{0}, x_{1}\right],\left[x_{1}, x_{2}\right],\left[x_{2}, x_{3}\right], \ldots, \text { and }\left[x_{\mathrm{n}-1}, x_{\mathrm{n}}\right]$ with lengths $\Delta x_{1}=x_{1}-x_{0}, \Delta x_{2}=x_{2}-x_{1}, \Delta x_{3}=x_{3}-x_{2}, \ldots$, and $\Delta x_{\mathrm{n}}=x_{\mathrm{n}}-x_{\mathrm{n}-1}$. The points $x_{\mathrm{k}}$ of the partition P are the locations of the vertical lines for the sides of the rectangles, and the bases of the rectangles have lengths $\Delta x_{\mathrm{k}} \text { for } \mathrm{k}=1,2,3, \ldots, \mathrm{n}$.

The mesh or norm of partition $P$ is the length of the longest of the subintervals $\left[x_{\mathrm{k}-1}, x_{\mathrm{k}}\right]$, or, equivalently, the maximum of $\Delta x_{\mathrm{k}} \text { for } \mathrm{k}=1,2,3, \ldots, \mathrm{n}$.

For example, the set $\mathrm{P}=\{2,3,4.6,5.1,6\}$ is a partition of the interval $[2,6]$ (Fig. 11) and divides the interval $[2,6]$ into 4 subintervals with lengths $\Delta x_{1}=1, \Delta x_{2}=1.6, \Delta x_{3}=.5 \text { and } \Delta x_{4}=.9$. The mesh of this partition is 1.6, the maximum of the lengths of the subintervals. (If the mesh of a partition is "small," then the length of each one of the subintervals is the same or smaller.)

Practice 6: $\mathrm{P}=\{3,3.8,4.8,5.3,6.5,7,8\}$ is a partition of what interval? How many subintervals does it create? What is the mesh of the partition? What are the values of $x_{2}$ and $\Delta x_{2}$?

A function, a partition, and a point in each subinterval determine a Riemann sum.

Suppose $f$ is a positive function on the interval $[a,b]$

$\mathrm{P}=\left\{x_{0}=\mathrm{a}, x_{1}, x_{2}, x_{3}, \ldots, x_{\mathrm{n}-1}, x_{\mathrm{n}}=\mathrm{b}\right\}$ is a partition of $[a,b]$.

$\mathbf{c}_{\mathbf{k}}$ is an x–value in the k^th subinterval $\left[x_{\mathrm{k}-1}, x_{\mathrm{k}}\right]: \quad x_{\mathrm{k}-1} \leq \mathrm{c}_{\mathrm{k}} \leq x_{\mathrm{k}}$.

Then the area of the k^th rectangle is $\mathrm{f}\left(\mathrm{c}_{\mathrm{k}}\right) \cdot\left(x_{\mathrm{k}}-x_{\mathrm{k}-1}\right)=\mathrm{f}\left(\mathrm{c}_{\mathrm{k}}\right) \cdot \Delta x_{\mathrm{k}}$. (Figure 12)

Definition: A summation of the form $\sum_{\mathbf{k}=1}^{\mathbf{n}} \mathbf{f}\left(\mathbf{c}_{\mathbf{k}}\right) \cdot \Delta x_{\mathbf{k}}$ is called a Riemann Sum of $f$ for the partition $P$.

This Riemann sum is the total of the areas of the rectangular regions and is an approximation of the area between the graph of f and the x–axis.

Example 6:

Find the Riemann sum for $f(x)=1 / x$ and the partition ${1, 4, 5}$ using the values $\mathrm{c}_{1}=2$ and $c_{2}=5$. (Fig. 13)

Solution: The 2 subintervals are $[1,4]$ and $[4,5]$ so $\Delta x_{1}=3$ and $\Delta x_{2}=1$.

Then the Riemann sum for this partition is

$\begin{aligned} \sum_{k=1}^{n} f\left(c_{k}\right) \cdot \Delta x_{k} &=\sum_{k=1}^{2} \mathrm{f}\left(c_{k}\right) \cdot \Delta x_{k}=f\left(c_{1}\right) \cdot \Delta x_{1}+f\left(c_{2}\right) \cdot \Delta x_{2}=f(2) \cdot(3)+f(5) \cdot(1) \\ &=\frac{1}{2}(3)+\frac{1}{5}(1)=1.7 \end{aligned}$.

Practice 7:

Practice 8: What is the smallest value a Riemann sum for $\mathrm{f}(x)=1 / x$ and the partition ${1, 4, 5}$ can have? (You will need to select values for $\mathrm{c}_{1}$ and $\mathrm{c}_{2}$.) What is the largest value a Riemann sum can have for this function and partition?

Table 2 shows the results of a computer program that calculated Riemann sums for the function $f(x)=1 / x$ with different numbers of subintervals and different ways of selecting the points $\mathrm{c}_{\mathrm{i}}$ in each subinterval.

When the mesh of the partition is small (and the number of subintervals large), all of the ways of selecting the $\mathrm{c}_{\mathrm{i}}$ lead to approximately the same number for the Riemann sums. For this decreasing function, using the left endpoint of the subinterval always resulted in a sun that was larger than the area. Choosing the right end point gave a value smaller that the area. Why?

Table 2: Riemann sums for $f(x)=1 / x$ on the interval $[1,5]$

Values of the Riemann sum for different choices of $\mathrm{c}_{\mathrm{k}}$

As the mesh gets smaller, all of the Riemann Sums seem to be approaching the same value, approximately $\text { 1.609. }(\ln 5=1.609437912)$.

Example 7: Find the Riemann sum for the function $f(x)=\sin (x)$ on the interval $[0, \pi]$ using the partition $\{0, \pi / 4, \pi / 2, \pi\}$ with $\mathrm{c}_{1}=\pi / 4, \mathrm{c}_{2}=\pi / 2, \mathrm{c}_{3}=3 \pi / 4$.

Solution: The 3 subintervals (Fig. 14) are $[0, \pi / 4],[\pi / 4, \pi / 2], \text { and }[\pi / 2, \pi]$, and $[\pi / 2, \pi]$ so $\Delta \mathrm{x}_{1}=\pi / 4, \Delta \mathrm{x}_{2}=\pi / 4$ and $\Delta \mathrm{x}_{3}=\pi / 2$. The Riemann sum for this partition is

$\begin{aligned} \sum_{\mathrm{k}=1}^{3} \mathrm{f}\left(\mathrm{c}_{\mathrm{k}}\right) \cdot \Delta \mathrm{x}_{\mathrm{k}} &=\sin (\pi / 4) \cdot(\pi / 4)+\sin (\pi / 2) \cdot(\pi / 4)+\sin (3 \pi / 4) \cdot(\pi / 2) \\ &=\frac{\sqrt{2}}{2} \cdot \frac{\pi}{4}+1 \cdot \frac{\pi}{4}+\frac{\sqrt{2}}{2} \cdot \frac{\pi}{2} \approx 2.45148 . \end{aligned}$.

Practice 9: Find the Riemann sum for the function and partition in the previous example, but use $\mathrm{c}_{1}=0, \mathrm{c}_{2}=\pi / 2, \mathrm{c}_{3}=\pi / 2$.

Two particular Riemann sums are of special interest because they represent the extreme possibilities for Riemann sums for a given partition.

Definition: Suppose $f$ is a positive function on $[a,b]$, and $P$ is a partition of $[a,b]$. Let $\mathbf{m}_{\mathbf{k}}$ be the x–value in the kth subinterval so that $\mathrm{f}\left(\mathbf{m}_{\mathbf{k}}\right)$ is the minimum value of $f$ in that interval, and let $\mathbf{M}_{\mathbf{k}}$ be the x–value in the kth subinterval so that $\mathrm{f}\left(\mathbf{M}_{\mathbf{k}}\right)$ is the maximum value of $f$ in that interval.

$\mathrm{LS}_{\mathrm{P}}: \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}\left(\mathbf{m}_{\mathbf{k}}\right) \cdot \Delta x_{\mathrm{k}}$ is the lower sum of f for the partition $P$.

$\mathrm{US}_{\mathrm{P}}: \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}\left(\mathbf{M}_{\mathbf{k}}\right) \cdot \Delta x_{\mathrm{k}}$ is the upper sum of f for the partition $P$.

Geometrically, the lower sum comes from building rectangles under the graph of $f$ (Fig. 15a), and the lower sum (every lower sum) is less than or equal to the exact area $\text { A: } L S_{P} \leq A$ for every partition $P$. The upper sum comes from building rectangles over the graph of f (Fig. 15b), and the upper sum (every upper sum) is greater than or equal to the exact area $A$:

$\mathrm{US}_{\mathrm{P}} \geq \mathrm{A}$ for every partition $P$. The lower and upper sums provide bounds on the size of the exact area:

$\mathrm{LS}_{\mathrm{P}} \leq \mathrm{A} \leq \mathrm{US}_{\mathrm{P}}$

For any $\mathrm{c}_{\mathrm{k}}$ value in the kth subinterval, $\mathrm{f}\left(\mathbf{m}_{\mathbf{k}}\right) \leq \mathrm{f}\left(\mathrm{c}_{\mathrm{k}}\right) \leq \mathrm{f}\left(\mathbf{M}_{\mathbf{k}}\right)$, so, for any choice of the $\mathrm{c}_{\mathrm{k}}$ values, the Riemann sum $\mathrm{RS}_{\mathrm{P}}=\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}\left(\mathrm{c}_{\mathrm{k}}\right) \cdot \Delta x_{\mathrm{k}}$ satisfies

$\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}\left(\mathbf{m}_{\mathbf{k}}\right) \cdot \Delta x_{\mathrm{k}} \leq \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}\left(\mathrm{c}_{\mathrm{k}}\right) \cdot \Delta x_{\mathrm{k}} \leq \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{f}\left(\mathrm{M}_{\mathbf{k}}\right) \cdot \Delta x_{\mathrm{k}}$ or, equivalently, $\mathrm{LS}_{\mathrm{P}} \leq \mathrm{RS}_{\mathrm{P}} \leq \mathrm{US}_{\mathrm{P}}$.

The lower and upper sums provide bounds on the size of all Riemann sums. The exact area $A$ and every Riemann sum $\mathrm{RS}_{\mathrm{P}}$ for partition $P$ both lie between the lower sum and the upper sum for $P$ (Fig. 16). Therefore, if the lower and upper sums are close together then the area and any Riemann sum for $P$ must also be close together. If we know that the upper and lower sums for a partition $P$ are within 0.001 units of each other, then we can be sure that every Riemann sum for partition $P$ is within 0.001 units of the exact area.

Unfortunately, finding minimums and maximums can be a time–consuming business, and it is usually not practical to determine lower and upper sums for "wiggly" functions. If f is monotonic, however, then it is easy to find the values for $\mathbf{m}_{\mathbf{k}}$ and $\mathbf{M}_{\mathbf{k}}$ , and sometimes we can explicitly calculate the limits of the lower and upper sums.

For a monotonic bounded function we can guarantee that a Riemann sum is within a certain distance of the exact value of the area it is approximating.

Theorem: If $f$ is a positive, montonically increasing, bounded function on $[a,b]$, then for any partition $P$ and any Riemann sum for $P$,

$\left\{\begin{array}{l} \text { distance between the } \\ \text { Riemann sum and } \\ \text { the exact area } \end{array}\right\} \leq\left\{\begin{array}{l} \text { distance between the } \\ \text { upper sum and } \\ \text { the lower sum } \end{array}\right\} \leq\{\mathrm{f}(\mathrm{b})-\mathrm{f}(\mathrm{a})\} \cdot(\text { mesh of } \mathrm{P})$.

Proof: The Riemann sum and the exact area are both between the upper and lower sums so the distance between the Riemann sum and the exact area is less than or equal to the distance between the upper and lower sums. Since $f$ is monotonically increasing, the areas representing the difference of the upper and lower sums can be slid into a rectangle (Fig. 17) whose height equals $f(b)-f(a)$ and whose base equals the mesh of $P$. Then the total difference of the upper and lower sums is less than or equal to the area of the rectangle, $\{\mathrm{f}(\mathrm{b})-f(a)\} \cdot(\text { mesh of } P)$.

Practice 1:

(a) $\sum_{\mathrm{k}=1}^{5} \mathrm{k}^{3}=1+8+27+64+125$.

(b) $\sum_{j=2}^{7}(-1)^{j} \frac{1}{j}=\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}+\frac{1}{6}-\frac{1}{7}$.

(c) $\sum_{\mathrm{m}=0}^{4}(2 \mathrm{~m}+1)=1+3+5+7+9$.

Practice 2:

(a) $\sum_{\mathrm{k}=2}^{5} \mathrm{~g}(\mathrm{k})=\mathrm{g}(2)+\mathrm{g}(3)+\mathrm{g}(4)+\mathrm{g}(5)=1+(-2)+3+5=7$.

(b) $\sum_{j=1}^{4} h(j)=h(1)+h(2)+h(3)+h(4)=3+3+3+3=12$.

(c) $\sum_{\mathrm{k}=3}^{5}(\mathrm{~g}(\mathrm{k})+\mathrm{f}(\mathrm{k}-1))=(\mathrm{g}(3)+\mathrm{f}(2))+(\mathrm{g}(4)+\mathrm{f}(3))+(\mathrm{g}(5)+\mathrm{f}(4))=(-2+3)+(3+1)+(5+0)=10$.

Practice 3: For $\mathrm{g}(x)=1 / x, \sum_{\mathrm{k}=2}^{4} \mathrm{~g}(\mathrm{k})=\mathrm{g}(2)+\mathrm{g}(3)+\mathrm{g}(4)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{13}{12}$ and $\sum_{\mathrm{k}=1}^{3} \mathrm{~g}(\mathrm{k}+1)=\mathrm{g}(2)+\mathrm{g}(3)+\mathrm{g}(4)=\frac{13}{12}$.

Practice 4: Rectangular areas = $1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}=\frac{25}{12}=\sum_{j=1}^{4} \frac{1}{j}$.

Practice 5: $\mathrm{f}\left(\mathrm{x}_{0}\right)\left(\mathrm{x}_{1}-\mathrm{x}_{0}\right)+\mathrm{f}\left(\mathrm{x}_{1}\right)\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)+\mathrm{f}\left(\mathrm{x}_{2}\right)\left(\mathrm{x}_{3}-\mathrm{x}_{2}\right)$
$=\sum_{j=1}^{3} f\left(x_{j-1}\right)\left(x_{j}-x_{j-1}\right) \text { or } \sum_{k=0}^{2} f\left(x_{k}\right)\left(x_{k+1}-x_{k}\right)$

Practice 6: Interval is $[3, 8]$. 6 subintervals. mesh = length of longest subinterval = 1.2.
$\mathrm{x}_{2}=4.8 \text { and } \Delta \mathrm{x}_{2}=\mathrm{x}_{2}-\mathrm{x}_{1}=4.8-3.8=1$.

Practice 7: $\mathrm{RS}=(3)\left(\frac{1}{3}\right)+(1)\left(\frac{1}{4}\right)=1.25$.

Practice 8: smallest RS = $(3)\left(\frac{1}{4}\right)+(1)\left(\frac{1}{5}\right)=0.95$     largest RS = $(3)(1)+(1)\left(\frac{1}{4}\right)=3.25$.

Practice 9: $\mathrm{RS}=(0)\left(\frac{\pi}{4}\right)+(1)\left(\frac{\pi}{4}\right)+(1)\left(\frac{\pi}{2}\right) \approx 2.356$.