320 lines
8.4 KiB
Markdown
320 lines
8.4 KiB
Markdown
Page 284
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**Definition**
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If $m$ and $n$ are integers and $m \leq n$, the symbol $\sum_{k=m}^{n}{a_k}$,
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read the **summation from $k$ equals $m$ to $n$ of $a$-sub-$k$**, is the sum of
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all the terms $a_m, a_{m + 1}, a_{m + 2}, \dots, a_n$. We say that
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$a_m + a_{m + 1} + a_{m + 2} + \dots + a_n$ is the **expanded form** of the sum,
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and we write
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$$ \sum_{k=m}^{n}{a_k} = a_m + a_{m + 1} + a_{m + 2} + \dots + a_n $$
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We call $k$ the **index** of the summation, $m$ the **lower limit** of the
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summation, and $n$ the **upper limit** of the summation.
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---
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Page 287
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**Definition**
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If $m$ and $n$ are integers and $m \leq n$, the symbol $\prod_{k = m}^{n}{a_k}$
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read the **product from $k$ equals $m$ to $n$ of $a$-sub-$k$**, is the product
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of all the terms $a_m, a_{m + 1}, a_{m + 2}, \dots, a_n$.
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We write
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$$ \prod_{k = m}^{n}{a_k} = a_m \cdot a_{m + 1} \cdot a_{m + 1} \dots a_n $$
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---
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Page 288
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**Theorem 5.1.1**
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If $a_m, a_{m + 1}, a_{m + 1}, \dots$ and $b_m, b_{m + 1}, b_{m + 1}, \dots$ are
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sequences of real numbers and $c$ is any real number, then the following
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equations hold for any integer $n \geq m$:
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1. $\sum_{k = m}^{n}{a_k} + \sum_{k = m}^{n}{b_k} = \sum_{k = m}^{n}{(a_k + b_k)}$
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2. $c \cdot \sum_{k = m}^{n}{a_k} = \sum_{k = m}^{n}{c \cdot a_k} \quad \text{generalized distributive law}$
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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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---
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Page 291
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**Definition**
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For each positive integer $n$, the quantity **$n$ factorial** denoted $n!$, is
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defined to be the product of all the integers from $1$ to $n$:
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$$ n! = n \cdot (n - 1) \dots 3 \cdot 2 \cdot 1 $$
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**Zero factorial**, denoted $0!$, is defined to be $1$:
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$$ 0! = 1 $$
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---
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Page 292
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**Definition**
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Let $n$ and $r$ be integers with $0 \leq r \leq n$. The symbol
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$$ \binom{n}{r} $$
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is read "**$n$ choose $r$**" and represents the number of subsets of size $r$
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that can be chosen from a set with $n$ elements.
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---
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Page 292
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**Formula for Computing $\dbinom{n}{r}$**
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For all integers $n$ and $r$ with $0 \leq r \leq n$,
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$$ \binom{n}{r} = \frac{n!}{r!(n - r)!} $$
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---
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Page 295
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**Algorithm 5.1.1 Decimal to Binary Conversion Using Repeated Division by $2$**
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_[In Algorithm 5.1.1 the input is a nonnegative integer $a$. The aim of the
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algorithm is to produce a sequence of binary digits $r[0], r[1], r[2], \dots
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r[k] so that the binary representation of $n$ is_
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$$ \left(r[k]r[k - 1] \dots r[2]r[1]r[0]\right)_2 $$
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_That is,_
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$$ a = 2^k \cdot r[k] + 2^{k - 1} \cdot r[k - 1] + \dots + 2^3 \cdot r[2] + 2^1 \cdot r[1] + 2^0 \cdot r[0] $$
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_.]_
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**Input:** $a$ _[a nonegative integer]_
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**Algorithm Body:**
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$q := a, i := 0$
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_[Repeatedly perform the integer division of $q$ by $2$ until $q$ becomes $0$.
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Store successive remainders in a one-dimensional array
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$r[0], r[1], r[2], \dots r[k]$. Even if the initial-value of $q$ equals $0$, the
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loop should execute one time (so that $r[0]$ is computed). Thus the guard
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condition for the **while** loop is $i = 0$ or $q \neq 0$.]_
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$\text{\textbf{while }}(i = 0 \text{ or } q \neq 0)\\ \ \ r[i] := q \mod 2\\ \ \ q := q \text{ div } 2\\ \ \ \text{[r[i] and q can be obtained by calling the division algorithm.]}\\ \ \ i := i + 1\\ \text{\textbf{end while}}$
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_[After execution of this step, the values of $r[0], r[1], \dots, r[i - 1]$ are
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all $0$'s and $1$'s, and
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$a = \left(r[i - 1]r[i - 2] \dots r[2]r[1]r[0]\right)_2$.]_
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**Output:** $r[0], r[1], r[2], \dots, r[i - 1]$ _[a sequence of integers]_
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---
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Page 300
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**Principle of Mathematical Induction**
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Let $P(n)$ be a property that is defined for integers $n$, and let $a$ be a
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fixed integer. Suppose the following two statements are true:
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1. $P(a)$ is true.
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2. For every integer $k \geq a$, if $P(k)$ is true then $P(k + 1)$ is true.
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Then the statement
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$$ \text{for every integer } n \geq a, P(a) $$
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is true.
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---
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Page 301
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**Method of Proof by Mathematical Induction**
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Consider the statement of the form, "For every integer $n \geq a$, a property
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$P(n)$ is true." To prove such a statement, perform the following two steps:
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**Step 1 (basis step):**
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Show that $P(a)$ is true.
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**Step 2 (inductive step):**
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Show that for every integer $k \geq a$, if $P(k)$ is true then $P(k + 1)$ is
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true. To perform this step,
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**suppose** that $P(k)$ is true, where $k$ is any particular but arbitrarily
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chosen integer with $k \geq a$. _[This supposition is called the **inductive
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hypothesis**.]_
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Then **show** that $P(k + 1)$ is true.
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---
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Page 303
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**Theorem 5.2.1 Sum of the First $n$ Integers**
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For every integer $n \geq 1$,
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$$ 1 + 2 + \dots + n = \frac{n(n + 1)}{2} $$
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**Proof (by mathematical induction):**
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Let the property $P(n)$ be the equation
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$$ 1 + 2 + 3 + \dots + n = \frac{n(n + 1)}{2} $$
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_Show that $P(1)$ is true:_
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To establish $P(1)$, we must show that
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$$ 1 = \frac{1(1 + 1)}{2} $$
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But the left-hand side of this equation is $1$ and the right-hand side is
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$$ \frac{1(1 + 1)}{2} = \frac{2}{2} = 1 $$
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also. Hence $P(1)$ is true.
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_Show that for every integer $k \geq 1$, if $P(k)$ is true then $P(k + 1)$ is
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also true:_
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_[Suppose that $P(k)$ is true for a particular but arbitrarily chosen integer
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$k \geq 1$. That is:]_
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Suppose that $k$ is any integer with $k \geq 1$ such that
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$$ 1 + 2 + 3 + \dots + k = \frac{k(k + 1)}{2} $$
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_[We must show that $P(k + 1)$ is true. That is:]_ We must show that
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$$ 1 + 2 + 3 + \dots + (k + 1) = \frac{(k + 1)[(k + 1) + 1]}{2} $$
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or, equivalently, that
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$$ 1 + 2 + 3 + \dots + (k + 1) = \frac{(k + 1)(k + 2)}{2} $$
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_[We will show that the left-hand side and the right-hand side of $P(k + 1)$ are
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equal to the same quantity and thus are equal to each other.]_
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The left-hand side of $P(k + 1)$ is
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$$ 1 + 2 + 3 + \dots + (k + 1) $$
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$$ = 1 + 2 + 3 + \dots + k + (k + 1) $$
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$$ = \frac{k(k + 1)}{2} + (k + 1) $$
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$$ = \frac{k(k + 1)}{2} + \frac{2(k + 1)}{2} $$
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$$ = \frac{k^2 + k}{2} + \frac{2k + 2}{2} $$
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$$ = \frac{k^2 + 3k + 2}{2} $$
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And the right-hand side of $P(k + 1)$ is
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$$ \frac{(k + 1)(k + 2)}{2} = \frac{k^2 + 3k + 2}{2} $$
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Thus the two sides of $P(k + 1)$ are equal to the same quantity and so they are
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equal to each other. Therefore, the equation $P(k + 1)$ is true _[as was to be
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shown]_.
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_[Since we have proved both the basis step and the inductive step, we conclude
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that the theorem is true.]_
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---
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Page 304
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**Definition**
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If a sum with a variable number of terms is shown to equal an expression that
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does not contain an ellipsis or a summation symbol, we say that the sum is
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written **in closed form.**
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---
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Page 306
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**Theorem 5.2.2 Sum of a Geometric Sequence**
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For any real number $r$ except $1$, and any integer $n \geq 0$,
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$$ \sum_{i = 0}^{n}{r^i} = \frac{r^{n + 1} - 1}{r - 1} $$
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**Proof (by mathematical induction):**
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Suppose $r$ is a particular but arbitrarily chosen real number that is not equal
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to $1$, and let the property $P(n)$ be the equation
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$$ \sum_{i = 0}^{n}{r^i} = \frac{r^{n + 1} - 1}{r - 1} $$
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We must show that $P(n)$ is true for every integer $n \geq 0$. We do this by
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mathematical induction on $n$.
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_Show that $P(0)$ is true:_
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To establish $P(0)$, we must show that
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$$ \sum_{i = 0}^{0}{r^i} = \frac{r^{0 + 1} - 1}{r - 1} $$
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The left-hand side of this equation is $r^0 = 1$ and the right-hand side is
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$$ \frac{r^{0 + 1} - 1}{r - 1} = \frac{r - 1}{r - 1} = 1 $$
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also because $r^1 = r$ and, since $r \neq 1$, $r - 1 \neq 0$. Hence $P(0)$ is
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true.
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_Show that for every integer $k \geq 0$, if $P(k)$ is true then $P(k + 1)$ is
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also true:_
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_[Suppose that $P(k)$ is true for a particular but arbitrarily chosen integer
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$k \geq 0$. That is:]_
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Let $k$ be any integer with $k \geq 0$, and suppose that
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$$ \sum_{i = 0}^{k}{r^j} = \frac{r^{k + 1} - 1}{r - 1} $$
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_[We must show that $P(k + 1)$ is true. That is:]_ We must show that
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$$ \sum_{i = 0}^{k + 1}{r^j} = \frac{r^{(k + 1) + 1} - 1}{r - 1} $$
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or, equivalently, that
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$$ \sum_{i = 0}^{k + 1}{r^j} = \frac{r^{k + 2} - 1}{r - 1} $$
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_[We will show that the left-hand side of $P(k + 1)$ equals the right-hand
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side.]_
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The left-hand side of $P(k + 1)$ is
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$$ \sum_{i = 0}^{k + 1}{r^j} = \sum_{i = 0}^{k}{r^i + r^{k + 1}} $$
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$$ = \frac{r^{k + 1} - 1}{r - 1} + r^{k + 1} $$
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$$ = \frac{r^{k + 1} - 1}{r - 1} + \frac{r^{k + 1}(r - 1)}{r - 1} $$
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$$ = \frac{(r^{k + 1} - 1) + r^{k + 1}(r - 1)}{r - 1} $$
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$$ = \frac{r^{k + 1} - 1 + r^{k + 2} - r^{k + 1}}{r - 1} $$
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$$ = \frac{r^{k + 2} - 1}{r - 1} $$
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which is the right-hand side of $P(k + 1)$ _[as was to be shown]._
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_[Since we have proved the basis step and the inductive step, we conclude that
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the theorem is true.]_
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