3 Sums, products, powers

This unit introduces notation for the elementary arithmetic operations.

3.1 Sums

Definition 3.1 (Summation symbol) It denotes

\[\begin{equation} \sum_{i = 1}^{n} x_i = x_1 + x_{2} + \cdots + x_{n}. \end{equation}\]

Here,

\(\Sigma\) denotes the Greek letter Sigma, mnemonically standing for sum,
the subscript \(i = 1\) for the summation index and the starting index,
the superscript \(n\) denotes the terminal index, and
\(x_1, x_2, ..., x_n\) denote the summands.

The indices of summation must be specified unambiguously, either by a starting and an ending index or by an index set. The name of the summation index, however, is irrelevant to the value of the sum. We have

\[\begin{equation} \sum_{i = 1}^n x_i = \sum_{j = 1}^n x_j. \end{equation}\]

Sometimes the running index is also given as an element of an index set. If, for example, the index set \(I := \{1,5,7\}\) is defined, then

\[\begin{equation} \sum_{i \in I} x_i := x_1 + x_5 + x_7. \end{equation}\]

In the following, we briefly consider a few examples of how the summation symbol is used.

Summation of predefined summands. Let \(x_1 := 2\), \(x_2 := 10\), and \(x_3 := -4\). Then

\[\begin{equation} \sum_{i = 1}^3 x_i = x_1 + x_2 + x_3 = 2 + 10 - 4 = 8. \end{equation}\]

Summation of weighted predefined summands. Again let \(x_1 := 2\), \(x_2 := 10\), and \(x_3 := -4\). In addition, let the weighting coefficients \(a_1 := \frac{1}{2}\), \(a_2 := \frac{1}{5}\), and \(a_3 := 2\) be defined. Then

\[\begin{equation} \sum_{i = 1}^3 a_i x_i = a_1x_1 + a_2x_2 + a_3x_3 = \frac{1}{2} \cdot 2 + \frac{1}{5} \cdot 10 + 2 \cdot (-4) = 1 + 2 - 8 = -5. \end{equation}\]

Expressions of the form \(\sum_{i=1}^n a_i x_i\) are also called linear combinations of \(x_1,...,x_n\) with the coefficients or weighting parameters \(a_1,...,a_n\).

Summation of natural numbers. We have

\[\begin{equation} \sum_{i = 1}^5 i = 1 + 2 + 3 + 4 + 5 = 15. \end{equation}\]

Summation of even natural numbers. We have

\[\begin{equation} \sum_{i = 1}^5 2i = 2 \cdot 1 + 2 \cdot 2 + 2 \cdot 3 + 2 \cdot 4 + 2 \cdot 5 = 2 + 4 + 6 + 8 + 10 = 30. \end{equation}\]

Summation of odd natural numbers. We have

\[\begin{equation} \sum_{i = 1}^5 (2i - 1) = 2 \cdot 1 - 1 + 2 \cdot 2 - 1 + 2 \cdot 3 - 1 + 2 \cdot 4 - 1 + 2 \cdot 5 - 1 = 1 + 3 + 5 + 7 + 9 = 25. \end{equation}\]

Working with the summation symbol can often be simplified by applying the following calculation rules.

Theorem 3.1 (Calculation rules for sums)

Sums of identical summands

\[\begin{equation} \sum_{i = 1}^n x = nx \end{equation}\]

Addition of sums of equal length

\[\begin{equation} \sum_{i = 1}^n x_i + \sum_{i = 1}^n y_i = \sum_{i = 1}^n (x_i + y_i) \end{equation}\]

Distributivity of multiplication by a constant

\[\begin{equation} \sum_{i = 1}^n ax_i = a\sum_{i = 1}^n x_i \end{equation}\]

Splitting sums for \(1 < m < n\)

\[\begin{equation} \sum_{i = 1}^n x_i = \sum_{i = 1}^m x_i + \sum_{i = m+1}^n x_i \end{equation}\]

Reindexing

\[\begin{equation} \sum_{i = 0}^n x_i = \sum_{j = m}^{n+m} x_{j - m} \end{equation}\]

Proof. These rules can be verified by expanding the sums and applying the rules of addition and multiplication. As examples, we show the addition of sums of equal length and the distributivity of multiplication by a constant. For the former, we have

\[\begin{align} \begin{split} \sum_{i = 1}^n x_i + \sum_{i = 1}^n y_i & = x_1 + x_2 + \cdots + x_n + y_1 + y_2 + \cdots + y_n \\ & = x_1 + y_1 + x_2 + y_2 + \cdots + x_n + y_n \\ & = \sum_{i = 1}^n (x_i + y_i). \end{split} \end{align}\]

For the latter, we have

\[\begin{align} \begin{split} \sum_{i = 1}^n ax_i & = ax_1 + ax_2 + \cdots + ax_n \\ & = a(x_1 + x_2 + \cdots + x_n) \\ & = a\sum_{i = 1}^n x_i. \end{split} \end{align}\]

Examples

As a first example of applying the calculation rules recorded in Theorem 3.1, we consider the evaluation of a mean (sometimes also called an average). Let \(x_1, x_2,...,x_n\) be real numbers. The mean of these numbers is the sum of \(x_1, x_2,...,x_n\) divided by the number of numbers \(n\). By statement (3) of Theorem 3.1, it is irrelevant whether the numbers are first added up and the resulting sum is then divided by \(n\), or whether the individual numbers are each divided by \(n\) and the corresponding results are then added up. More precisely, by applying Theorem 3.1 (3) with \(a = 1/n\), we obtain

\[\begin{equation} \frac{1}{n}\sum_{i = 1}^n x_i = \sum_{i = 1}^n \frac{x_i}{n}. \end{equation}\]

For example, the mean of \(x_1 := 1, x_2 := 4, x_3 := 2\), and \(x_4 := 1\) is given by

\[\begin{equation} \frac{1}{4}\sum_{i = 1}^4 x_i = \frac{1}{4}(1 + 4 + 2 + 1) = \frac{8}{4} = 2 = \frac{8}{4} = \frac{1}{4} + \frac{4}{4} + \frac{2}{4} + \frac{1}{4} = \sum_{i = 1}^4 \frac{x_i}{4}. \end{equation}\]

As a second example, we consider the reindexing rule recorded in Theorem 3.1 (5). Let \(n := 3\) and \(m := 2\), and let \(x_0 := 2\), \(x_1 := 3\), \(x_2 := 5\), and \(x_3 := 10\). Then, evidently,

\[\begin{align} \begin{split} \sum_{i = 0}^3 x_i & = x_0 + x_1 + x_2 + x_3 \\ & = 2 + 3 + 5 + 10 \\ & = 20. \end{split} \end{align}\]

But we also have

\[\begin{align} \begin{split} \sum_{j = m}^{n + m} x_{j-m} & = \sum_{j = 2}^{3 + 2} x_{j-2} \\ & = \sum_{j = 2}^{5} x_{j-2} \\ & = x_{2-2} + x_{3-2} + x_{4-2} + x_{5-2} \\ & = x_{0} + x_{1} + x_{2} + x_{3} \\ & = 2 + 3 + 5 + 10 \\ & = 20. \end{split} \end{align}\]

Double Sums

In applications, one often encounters expressions that carry out several summations one after another. For each of these sums, the definitions and calculation rules listed above apply. The meaning of double sums is best made clear by writing them out from the inside to the outside, while noting that the running index of the outer sum remains constant during each iteration of the inner sum.

The following examples may illustrate this.

Example (1)

\[\begin{align} \begin{split} \sum_{i = 1}^2 \sum_{j = 1}^3 (i+j) & = \sum_{i = 1}^2 (i + 1 + i + 2 + i + 3) \\ & = (1 + 1 + 1 + 2 + 1 + 3) + (2 + 1 + 2 + 2 + 2 + 3) \\ & = 9 + 12 \\ & = 21 \end{split} \end{align}\]

Example (2)

\[\begin{align} \begin{split} \sum_{i = 1}^2 \sum_{j = 1}^3 (x_i + y_j) & = \sum_{i = 1}^2 (x_i + y_1 + x_i + y_2 + x_i + y_3) \\ & = (x_1 + y_1 + x_1 + y_2 + x_1 + y_3) + (x_2 + y_1 + x_2 + y_2 + x_2 + y_3) \end{split} \end{align}\]

Example (3)

\[\begin{align} \begin{split} \sum_{i = 1}^3 \sum_{j = 1}^2 (x_iy_j) & = \sum_{i = 1}^3 (x_iy_1 + x_iy_2) \\ & = (x_1y_1 + x_1y_2) + (x_2y_1 + x_2y_2) + (x_3y_1 + x_3y_2) \end{split} \end{align}\]

3.2 Products

The product symbol provides notation for products that is analogous to the summation symbol.

Definition 3.2 (Product symbol) It denotes

\[\begin{equation} \prod_{i = 1}^{n} x_i = x_1 \cdot x_{2} \cdot \cdots \cdot x_{n}. \end{equation}\]

Here,

\(\prod\) denotes the Greek letter Pi, mnemonically standing for product,
the subscript \(i = 1\) for the product index and the starting index,
the superscript \(n\) denotes the terminal index, and
\(x_1, x_2, ..., x_n\) denote the product terms.

The indices of multiplication must be specified unambiguously, either by a starting and an ending index or by an index set. Again, the name of the product index is irrelevant. We have

\[\begin{equation} \prod_{i = 1}^n x_i = \prod_{j = 1}^n x_j. \end{equation}\]

Here, too, the running index is in rare cases given as an element of an index set. If, for example, the index set \(J := \mathbb{N}_2^0\) is defined, then

\[\begin{equation} \prod_{j \in J} x_j := x_0 \cdot x_1 \cdot x_2. \end{equation}\]

One example of the use of the product symbol is the definition of the factorial of a natural number \(n\) by

\[\begin{equation} n! := \prod_{i = 1}^n i. \end{equation}\]

For example,

\[\begin{equation} 4! = \prod_{i = 1}^4 i = 1 \cdot 2 \cdot 3 \cdot 4 = 24. \end{equation}\]

There are also a number of calculation rules for products that often simplify working with them. We list some in the following theorem. In doing so, we make anticipatory use of the notation of powers defined in Section 3.3.

Theorem 3.2 (Calculation rules for products)

Products of identical factors

\[\begin{equation} \prod_{i = 1}^n x = x^n \end{equation}\]

Raising constants to powers

\[\begin{equation} \prod_{i = 1}^n ax_i = a^n\prod_{i = 1}^n x_i \end{equation}\]

Splitting products for \(1 < m < n\)

\[\begin{equation} \prod_{i = 1}^n x_i = \prod_{i = 1}^m x_i \prod_{j = m+1}^n x_j \end{equation}\]

Product of products

\[\begin{equation} \prod_{i = 1}^n x_iy_i = \prod_{i = 1}^n x_i \prod_{i = 1}^n y_i \end{equation}\]

Double products

As examples, we consider the following double products.

Example (1)

\[\begin{align} \begin{split} \prod_{i = 1}^2 \prod_{j = 1}^3 (i+j) & = \prod_{i = 1}^2 (i+1)(i+2)(i+3) \\ & = (1+1)(1+2)(1+3)\cdot(2+1)(2+2)(2+3) \\ & = 24\cdot 60 \\ & = 1440. \end{split} \end{align}\]

Example (2)

\[\begin{align} \begin{split} \prod_{i = 1}^2 \prod_{j = 1}^3 (x_i+y_j) & = \prod_{i = 1}^2 (x_i+y_1)(x_i+y_2)(x_i+y_3) \\ & = (x_1+y_1)(x_1+y_2)(x_1+y_3)\cdot(x_2+y_1)(x_2+y_2)(x_2+y_3). \end{split} \end{align}\]

Example (3)

\[\begin{align} \begin{split} \prod_{i = 1}^3 \prod_{j = 1}^2 (x_i y_j) & = \prod_{i = 1}^3 (x_i y_1)(x_i y_2) \\ & = (x_1y_1)(x_1y_2)\cdot(x_2y_1)(x_2y_2)\cdot(x_3y_1)(x_3y_2). \end{split} \end{align}\]

3.3 Powers

Products of numbers with themselves can be abbreviated using power notation.

Definition 3.3 (Power) For \(a \in \mathbb{R}\) and \(n \in \mathbb{N}^0\), the \(n\)-th power of \(a\) is defined by

\[\begin{equation} a^0 := 1 \mbox{ and } a^{n+1} := a^n \cdot a. \end{equation}\]

Furthermore, for \(a\in \mathbb{R} \setminus \{0\}\) and \(n \in \mathbb{N}^0\), the negative \(n\)-th power of \(a\) is defined by

\[\begin{equation} a^{-n} := (a^n)^{-1} := \frac{1}{a^n}. \end{equation}\]

\(a\) is called the base, and \(n\) is called the exponent.

Defining \(a^{n+1}\) in terms of the power \(a^n\), as above, is called a recursive definition. The definition \(a^0 := 1\) is called the base case. It makes the recursive definition of \(a^{n+1}\) possible. The definition \(a^{n+1} := a^n \cdot a\) is also called the recursive step. The following rules simplify calculations with powers.

Theorem 3.3 (Calculation rules for powers) For \(a,b \in \mathbb{R}\) and \(n,m \in \mathbb{Z}\), the following rules hold whenever all powers involved are defined. For negative exponents, the corresponding bases must be nonzero. The quotient rule additionally requires \(b \neq 0\).

\[\begin{align} a^n a^m & = a^{n+m}, \\ (ab)^n & = a^n b^n, \\ \left(\frac{a}{b}\right)^n & = \frac{a^n}{b^n}, \\ (a^n)^m & = a^{nm}. \end{align}\]

Instead of a proof, we consider the following four examples.

(1)

\[\begin{equation} 2^2 \cdot 2^3 = (2 \cdot 2) \cdot (2 \cdot 2 \cdot 2) = 2^5 = 2^{2 + 3} \end{equation}\]

(2)

\[\begin{equation} (2 \cdot 4)^2 = (2 \cdot 4) \cdot (2 \cdot 4) = (2 \cdot 2) \cdot (4 \cdot 4) = 2^2 \cdot 4^2 \end{equation}\]

(3)

\[\begin{equation} \left(\frac{2}{3}\right)^3 = \frac{2}{3} \cdot \frac{2}{3} \cdot \frac{2}{3} = \frac{2 \cdot 2 \cdot 2}{3 \cdot 3 \cdot 3} = \frac{2^3}{3^3} \end{equation}\]

(4)

\[\begin{equation} (3^2)^3 = (3 \cdot 3)^3 = (3 \cdot 3) \cdot (3 \cdot 3) \cdot (3 \cdot 3) = 3^6 = 3^{2 \cdot 3} \end{equation}\]

Closely related to powers is the definition of the \(n\)-th root:

Definition 3.4 (\(n\)-th root) For \(a \in \mathbb{R}_{\ge 0}\) and \(n \in \mathbb{N}\), the \(n\)-th root of \(a\) is defined as the nonnegative real number \(r\) with

\[\begin{equation} r^n = a. \end{equation}\]

When computing with roots, power notation for roots is often helpful because it allows the calculation rules for powers to be applied directly.

Definition 3.5 (Power notation for the \(n\)-th root) Let \(a \in \mathbb{R}_{\ge 0}\), \(n \in \mathbb{N}\), and let \(r\) be the \(n\)-th root of \(a\). We define

\[\begin{equation} a^{\frac{1}{n}} := r. \end{equation}\]

Computing with square roots is greatly facilitated by the power notation

\[\begin{equation} \sqrt{x} = x^{\frac{1}{2}} \end{equation}\]

For example,

\[\begin{equation} \frac{2\pi}{\sqrt{2\pi}} = \frac{2\pi}{(2\pi)^{\frac{1}{2}}} = (2\pi)^{1} \cdot (2\pi)^{-\frac{1}{2}} = (2\pi)^{1-\frac{1}{2}} = (2\pi)^{\frac{1}{2}} = \sqrt{2\pi}. \end{equation}\]

The following relation between square, root, and absolute value is used quite often.

Theorem 3.4 (Root and absolute value) Let \(x \in \mathbb{R}\), and let

\[\begin{equation} |x| := \begin{cases} x & \mbox{for } x \ge 0 \\ -x & \mbox{for } x < 0 \end{cases} \end{equation}\]

denote the absolute value of \(x\). Then

\[\begin{equation} \sqrt{x^2} = |x|. \end{equation}\]

Proof. We consider the cases \(x \ge 0\) and \(x < 0\). First, let \(x \ge 0\). Then

\[\begin{equation} x \ge 0 \Rightarrow x^2 \ge 0 \Rightarrow \sqrt{x^2} = x = |x|. \end{equation}\]

Now let \(x < 0\). Then

\[\begin{equation} x < 0 \Rightarrow x^2 > 0 \Rightarrow \sqrt{x^2} = -x = |x|. \end{equation}\]

Study questions

State the definition of summation notation.
Calculate \(\sum_{i = 1}^3 2\), \(\sum_{i = 1}^3 i^2\), and \(\sum_{i = 1}^3 \frac{2}{3}i\).
Write \(1 + 3 + 5 + 7 + 9 + 11\) using summation notation.
Write \(0 + 2 + 4 + 6 + 8 + 10\) using summation notation.
State the definition of product notation.
State the definition of the \(n\)-th power of \(a \in \mathbb{R}\).
Calculate \(2^2 \cdot 2^3\) and \(2^5\), and state the corresponding rule for powers.
Calculate \(6^2\) and \(2^2 \cdot 3^2\), and state the corresponding rule for powers.
State the definition of the power notation \(a^{\frac{1}{n}}\) for \(a \ge 0\) and \(n \in \mathbb{N}\).

Study question answers

See Definition 3.1.
We have

\[\begin{equation} \begin{aligned} \sum_{i = 1}^3 2 & = 2 + 2 + 2 = 6 \\ \sum_{i = 1}^3 i^2 & = 1 + 4 + 9 = 14 \\ \sum_{i = 1}^3 \frac{2}{3}i & = \frac{2}{3}(1 + 2 + 3) = 4. \end{aligned} \end{equation}\]

We have

\[\begin{equation} 1 + 3 + 5 + 7 + 9 + 11 = \sum_{i = 1}^6 (2i - 1). \end{equation}\]

We have

\[\begin{equation} 0 + 2 + 4 + 6 + 8 + 10 = \sum_{i = 0}^5 2i. \end{equation}\]

See Definition 3.2.
See Definition 3.3.
We have \(a^n \cdot a^m = a^{n + m}\), hence here

\[\begin{equation} 2^2 \cdot 2^3 = 4 \cdot 8 = 32 = 2^5. \end{equation}\]

See Theorem 3.3.

We have \((ab)^n = a^n b^n\), hence here

\[\begin{equation} 6^2 = 36 = 4 \cdot 9 = 2^2 \cdot 3^2. \end{equation}\]

See Theorem 3.3.

For \(a \ge 0\) and \(n \in \mathbb{N}\), we define \(a^{\frac{1}{n}} := r\), where \(r\) is the unique nonnegative real number satisfying \(r^n = a\). See Definition 3.5.