🚧 Setup for 5.2

chapter_5/notes.md

@@ -118,3 +118,203 @@ 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]_
+
+---
+
+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.
+
+---
+
+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.
+
+---
+
+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.]_
+
+---
+
+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.**
+
+---
+
+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.]_