discrete_mathematics_with_applications/chapter_5/notes.md

Page 284

Definition

If m and n are integers and m \leq n, the symbol \sum_{k=m}^{n}{a_k}, read the summation from k equals m to n of $a$-sub-$k$, is the sum of all the terms a_m, a_{m + 1}, a_{m + 2}, \dots, a_n. We say that a_m + a_{m + 1} + a_{m + 2} + \dots + a_n is the expanded form of the sum, and we write

 \sum_{k=m}^{n}{a_k} = a_m + a_{m + 1} + a_{m + 2} + \dots + a_n 

We call k the index of the summation, m the lower limit of the summation, and n the upper limit of the summation.

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Page 287

Definition

If m and n are integers and m \leq n, the symbol \prod_{k = m}^{n}{a_k} read the product from k equals m to n of $a$-sub-$k$, is the product of all the terms a_m, a_{m + 1}, a_{m + 2}, \dots, a_n.

We write

 \prod_{k = m}^{n}{a_k} = a_m \cdot a_{m + 1} \cdot a_{m + 1} \dots a_n 

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Page 288

Theorem 5.1.1

If a_m, a_{m + 1}, a_{m + 1}, \dots and b_m, b_{m + 1}, b_{m + 1}, \dots are sequences of real numbers and c is any real number, then the following equations hold for any integer n \geq m:

1. \sum_{k = m}^{n}{a_k} + \sum_{k = m}^{n}{b_k} = \sum_{k = m}^{n}{(a_k + b_k)}

2. c \cdot \sum_{k = m}^{n}{a_k} = \sum_{k = m}^{n}{c \cdot a_k} \quad \text{generalized distributive law}

3. \left(\prod_{k = m}^{n}{a_k}\right) \cdot \left(\prod_{k = m}^{n}{b_k}\right) = \prod_{k = m}^{n}{(a_k \cdot b_k)}

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Page 291

Definition

For each positive integer n, the quantity n factorial denoted n!, is defined to be the product of all the integers from 1 to n:

 n! = n \cdot (n - 1) \dots 3 \cdot 2 \cdot 1 

Zero factorial, denoted 0!, is defined to be 1:

 0! = 1 

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Page 292

Definition

Let n and r be integers with 0 \leq r \leq n. The symbol

 \binom{n}{r} 

is read "n choose $r$" and represents the number of subsets of size r that can be chosen from a set with n elements.

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Page 292

Formula for Computing $\dbinom{n}{r}$

For all integers n and r with 0 \leq r \leq n,

 \binom{n}{r} = \frac{n!}{r!(n - r)!} 

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Page 295

Algorithm 5.1.1 Decimal to Binary Conversion Using Repeated Division by $2$

[In Algorithm 5.1.1 the input is a nonnegative integer a. The aim of the algorithm is to produce a sequence of binary digits $r[0], r[1], r[2], \dots r[k] so that the binary representation of n is

 \left(r[k]r[k - 1] \dots r[2]r[1]r[0]\right)_2 

That is,

 a = 2^k \cdot r[k] + 2^{k - 1} \cdot r[k - 1] + \dots + 2^3 \cdot r[2] + 2^1 \cdot r[1] + 2^0 \cdot r[0] 

.]

Input: a [a nonegative integer]

Algorithm Body:

q := a, i := 0

[Repeatedly perform the integer division of q by 2 until q becomes 0. Store successive remainders in a one-dimensional array r[0], r[1], r[2], \dots r[k]. Even if the initial-value of q equals 0, the loop should execute one time (so that r[0] is computed). Thus the guard condition for the while loop is i = 0 or q \neq 0.]

\text{\textbf{while }}(i = 0 \text{ or } q \neq 0)\\ \ \ r[i] := q \mod 2\\ \ \ q := q \text{ div } 2\\ \ \ \text{[r[i] and q can be obtained by calling the division algorithm.]}\\ \ \ i := i + 1\\ \text{\textbf{end while}}

[After execution of this step, the values of r[0], r[1], \dots, r[i - 1] are all $0$'s and $1$'s, and a = \left(r[i - 1]r[i - 2] \dots r[2]r[1]r[0]\right)_2.]

Output: r[0], r[1], r[2], \dots, r[i - 1] [a sequence of integers]

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Page 300

Principle of Mathematical Induction

Let P(n) be a property that is defined for integers n, and let a be a fixed integer. Suppose the following two statements are true:

1. P(a) is true.

2. For every integer k \geq a, if P(k) is true then P(k + 1) is true.

Then the statement

 \text{for every integer } n \geq a, P(a) 

is true.

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Page 301

Method of Proof by Mathematical Induction

Consider the statement of the form, "For every integer n \geq a, a property P(n) is true." To prove such a statement, perform the following two steps:

Step 1 (basis step):

Show that $P(a)$ is true.

Step 2 (inductive step):

Show that for every integer k \geq a, if P(k) is true then P(k + 1) is true. To perform this step,

suppose that P(k) is true, where k is any particular but arbitrarily chosen integer with k \geq a. [This supposition is called the inductive hypothesis.]

Then show that P(k + 1) is true.

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Page 303

Theorem 5.2.1 Sum of the First n Integers

For every integer n \geq 1,

 1 + 2 + \dots + n = \frac{n(n + 1)}{2} 

Proof (by mathematical induction):

Let the property P(n) be the equation

 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2} 

Show that P(1) is true:

To establish P(1), we must show that

 1 = \frac{1(1 + 1)}{2} 

But the left-hand side of this equation is 1 and the right-hand side is

 \frac{1(1 + 1)}{2} = \frac{2}{2} = 1 

also. Hence P(1) is true.

Show that for every integer k \geq 1, if P(k) is true then P(k + 1) is also true:

[Suppose that P(k) is true for a particular but arbitrarily chosen integer k \geq 1. That is:]

Suppose that k is any integer with k \geq 1 such that

 1 + 2 + 3 + \dots + k = \frac{k(k + 1)}{2} 

[We must show that P(k + 1) is true. That is:] We must show that

 1 + 2 + 3 + \dots + (k + 1) = \frac{(k + 1)[(k + 1) + 1]}{2} 

or, equivalently, that

 1 + 2 + 3 + \dots + (k + 1) = \frac{(k + 1)(k + 2)}{2} 

[We will show that the left-hand side and the right-hand side of P(k + 1) are equal to the same quantity and thus are equal to each other.]

The left-hand side of P(k + 1) is

 1 + 2 + 3 + \dots + (k + 1) 

 = 1 + 2 + 3 + \dots + k + (k + 1) 

 = \frac{k(k + 1)}{2} + (k + 1) 

 = \frac{k(k + 1)}{2} + \frac{2(k + 1)}{2} 

 = \frac{k^2 + k}{2} + \frac{2k + 2}{2} 

 = \frac{k^2 + 3k + 2}{2} 

And the right-hand side of P(k + 1) is

 \frac{(k + 1)(k + 2)}{2} = \frac{k^2 + 3k + 2}{2} 

Thus the two sides of P(k + 1) are equal to the same quantity and so they are equal to each other. Therefore, the equation P(k + 1) is true [as was to be shown].

[Since we have proved both the basis step and the inductive step, we conclude that the theorem is true.]

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Page 304

Definition

If a sum with a variable number of terms is shown to equal an expression that does not contain an ellipsis or a summation symbol, we say that the sum is written in closed form.

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Page 306

Theorem 5.2.2 Sum of a Geometric Sequence

For any real number r except 1, and any integer n \geq 0,

 \sum_{i = 0}^{n}{r^i} = \frac{r^{n + 1} - 1}{r - 1} 

Proof (by mathematical induction):

Suppose r is a particular but arbitrarily chosen real number that is not equal to 1, and let the property P(n) be the equation

 \sum_{i = 0}^{n}{r^i} = \frac{r^{n + 1} - 1}{r - 1} 

We must show that P(n) is true for every integer n \geq 0. We do this by mathematical induction on n.

Show that P(0) is true:

To establish P(0), we must show that

 \sum_{i = 0}^{0}{r^i} = \frac{r^{0 + 1} - 1}{r - 1} 

The left-hand side of this equation is r^0 = 1 and the right-hand side is

 \frac{r^{0 + 1} - 1}{r - 1} = \frac{r - 1}{r - 1} = 1 

also because r^1 = r and, since r \neq 1, r - 1 \neq 0. Hence P(0) is true.

Show that for every integer k \geq 0, if P(k) is true then P(k + 1) is also true:

[Suppose that P(k) is true for a particular but arbitrarily chosen integer k \geq 0. That is:]

Let k be any integer with k \geq 0, and suppose that

 \sum_{i = 0}^{k}{r^j} = \frac{r^{k + 1} - 1}{r - 1} 

[We must show that P(k + 1) is true. That is:] We must show that

 \sum_{i = 0}^{k + 1}{r^j} = \frac{r^{(k + 1) + 1} - 1}{r - 1} 

or, equivalently, that

 \sum_{i = 0}^{k + 1}{r^j} = \frac{r^{k + 2} - 1}{r - 1} 

[We will show that the left-hand side of P(k + 1) equals the right-hand side.]

The left-hand side of P(k + 1) is

 \sum_{i = 0}^{k + 1}{r^j} = \sum_{i = 0}^{k}{r^i + r^{k + 1}} 

 = \frac{r^{k + 1} - 1}{r - 1} + r^{k + 1} 

 = \frac{r^{k + 1} - 1}{r - 1} + \frac{r^{k + 1}(r - 1)}{r - 1} 

 = \frac{(r^{k + 1} - 1) + r^{k + 1}(r - 1)}{r - 1} 

 = \frac{r^{k + 1} - 1 + r^{k + 2} - r^{k + 1}}{r - 1} 

 = \frac{r^{k + 2} - 1}{r - 1} 

which is the right-hand side of P(k + 1) [as was to be shown].

[Since we have proved the basis step and the inductive step, we conclude that the theorem is true.]