5.2: The number e.

The implication

$$\text { For } \quad E(t)=B^{t} \quad \Longrightarrow \quad E^{\prime}(t)=E^{\prime}(0) \times E(t)$$

would be even simpler if we find a value for $B$ so that $E ^{\prime} (0) = 1$. We next find such a value for $B$. It is universally denoted by $e$ and

$$e \quad \text { is approximately } \quad 2.71828182845904523536 ; \quad e \quad \text { is not a rational number. }$$

Then for $E(t) = e ^{t} , ~ E ^{\prime} (0) = 1$ so that $E^{\prime}(t)=E^{\prime}(0) E(t)=1 \times E(t)=e(t)$.

$$\text{Thus} \quad \left[e^{t}\right]^{\prime}=e^{t}, \quad \text{a very important result.}$$

Our goal is to find a base, $e$, so that

$$\left.\left[e^{t}\right]^{\prime}\right|_{t=0}=1$$

Because

$$\left.\left[2^{t}\right]^{\prime}\right|_{t=0} \doteq 0.69341 \quad \text { and }\left.\quad\left[3^{t}\right]^{\prime}\right|_{t=0} \doteq 1.098612,$$

we think the base $e$ that we seek is between 2 and 3 and perhaps closer to 3 than to 2.

Figure $\PageIndex{1}$: Graphs of exponential functions, $y = 2^t$, and $y = 3^t$ and segments of the tangents for each. The tangents have slopes approximately 0.69341 and 1.098612, respectively.

Explore 5.2.1

$$\left.\left[e^{t}\right]^{\prime}\right|_{t=0}=1 \quad \text { and }\left.\quad e^{t}\right|_{t=0}=1$$

and the line of slope 1 through (0,1) is tangent to the graph of $y = e ^t$. Draw a line of slope 1 through the point (0,1). Then draw a graph of an exponential function whose tangent at (0,1) is the line you drew. Compare your graph with those in Figure $\PageIndex{1}$.

Assuming that

$$\left.\left[e^{t}\right]^{\prime}\right|_{t=0}=1, \quad \text { it must be that } \quad \lim _{h \rightarrow 0} \frac{e^{h}-e^{0}}{h}=\lim _{h \rightarrow 0} \frac{e^{h}-1}{h}=1 \text {. }$$

Intuitively then, for $h$ close to zero,

$$\frac{e^{h}-1}{h} \doteq 1, \quad e^{h}-1 \doteq h, \quad e^{h} \doteq 1+h, \quad e \doteq(1+h)^{\frac{1}{h}}$$

We will use this to explore $e$, and in Subsection 5.2.1 we will find that

$$e=\lim _{h \rightarrow 0}(1+h)^{\frac{1}{h}} \label{5.3}$$

Values of $(1 + h) ^{\frac{1}{h}}$ for progressively smaller values of $h$ are shown in Table 5.2. As $h > 0$ decreases toward 0, $(1 + h) ^{\frac{1}{h}}$ increases. We show in Subsection 5.2.1 that $(1 + h) ^{\frac{1}{h}}$ approaches a number as $h$ approaches 0, and we denote that number by $e$.

Explore 5.2.2 Compute $A=(1+h)^{1 / h}$ for $h$ = 0.0001, $h$ = 0.00001, and $h$ = 0.000001.

Your last estimate of $A$ should be approximately 2.71828047, which is an estimate correct to 6 digits of the irrational number $e$ that we are seeking.

Table 5.2: Approximations to the number $e$. As $h$ moves toward 0, $A = (1 + h) ^{1/h}$ increases toward $e$.

$h$	$A = (1 + h) ^{1/h}$
1.000	2.00000
0.500	2.25000
0.200	2.48832
0.100	2.59374
0.050	2.65329
0.010	2.70481
0.005	2.71152
0.001	2.71692
$\downarrow$	$\downarrow$
0	$e$

Definition 5.2.1 The number e

The number $e$ is defined by

$$\lim _{h \rightarrow 0}(1+h)^{\frac{1}{h}}=e$$

Correct to 21 digits, $e$ = 2.71828 18284 59045 23536

We show in Subsection 5.2.1 that if $h = \frac{1}{n}$ for $n$ an integer greater than 2

$$(1+h)^{\frac{1}{h}}<e<\left(1+\frac{h}{1-h}\right)^{\frac{1}{h}}, \label{5.4}$$

and we assume the inequalities are valid for all $0 < h < 1/2$. Using these inequalities it is easy to show that the function $E(t) = e ^t$ has the property that $E ^{\prime} (0) = 1$.

We write, for $0 < h < 1/2$,

$$\begin{array}\ &(1+h)^{\frac{1}{h}} & < \quad e & < \quad \left(1+\frac{h}{1-h}\right)^{\frac{1}{h}} \\ &1+h & < \quad e^{h} & < \quad 1+\frac{h}{1-h} \\ &h & < \quad e^{h}-1 & < \quad \frac{h}{1-h} \\ &1 & < \quad \frac{e^{h}-1}{h} & < \quad \frac{1}{1-h} \\ \text { As } h \rightarrow 0 & \downarrow & \quad \downarrow & \quad \downarrow \\ &1 & \leq \quad E ^{\prime}(0) & \leq \quad 1\\ \end{array}$$

Thus for $E(t) = e ^t$, $E ^{\prime} (0) = 1$.

Because of Theorem 5.1.1, $E ^{\prime} (t) = E ^{\prime} (0) \times E(t) = e ^t$, we have another Primary Formula for computing derivatives:

$$\left[e^{t}\right]^{\prime}=e^{t} \quad \text { Exponential Rule } \label{5.5}$$

Strategy for computing derivatives: Now we can use three Primary Formulas (Constant, Power, Exponential) and three Composition Formulas (Sum, Constant Factor, Power Chain) to compute derivatives. In finding derivatives of functions with many terms, students sometimes ask what derivative rule to use first and in subsequent steps. We think of the derivative procedure as peeling the layers off of an onion – outside layer first, etc. For example, to compute the derivative of $F(t) = ( 2 + 3t ^{2} + 5e ^{t} ) ^3$ we write:

$$\left. \begin{array}\ F^{\prime}(t) &=\left[\left(2+3 t^{2}+5 e^{t}\right)^{3}\right]^{\prime} & & \text { Symbolic Identity } \\ &=3\left(2+3 t^{2}+5 e^{t}\right)^{2}\left[2+3 t^{2}+5 e^{t}\right]^{\prime} & & \text { Power Chain Rule } \\ &=3\left(2+3 t^{2}+5 e^{t}\right)^{2}\left([2]^{\prime}+\left[3 t^{2}\right]^{\prime}+\left[5 e^{t}\right]^{\prime}\right) & & \text { Sum Rule } \\ &=3\left(2+3 t^{2}+5 e^{t}\right)^{2}\left(0+\left[3 t^{2}\right]^{\prime}+\left[5 e^{t}\right]^{\prime}\right) & & \text { Constant Rule } \\ &=3\left(2+3 t^{2}+5 e^{t}\right)^{2}\left(3\left[t^{2}\right]^{\prime}+5\left[e^{t}\right]^{\prime}\right) & & \text { Constant Factor Rule } \\ &=3\left(2+3 t^{2}+5 e^{t}\right)^{2}\left(3(2 t)+5\left[e^{t}\right]^{\prime}\right) & & \text { Power Rule } \\ &=3\left(2+3 t^{2}+5 e^{t}\right)^{2}\left(6 t+5 e^{t}\right) & & \text { Exponential Rule } \\ \end{array} \right\} \label{5.6}$$

Think how the expression for $F$ in the previous example, $F(t) = (2 + 3t ^{2} + 5e ^{t} ) ^3$, is evaluated. Given a value for $t$ you would compute $t^2$ and multiply it by 3 and you would compute $e^t$ and multiply it by 5, and then you would sum the three terms. The last step (outside layer) in the evaluation is to cube the sum. The first step in computing the derivative is to 'undo' that cube (use the Power Chain Rule).

The next step is to undo the sum with the Sum Rule. The next step is not uniquely determined; we worked from left to right and chose to evaluate $[2]^{\prime}$.

In evaluating $3t^2$ and $5e^t$, the last step would be to multiply by 3 or 5 – the first step in finding the derivative of $3t^2$ and $5e^t$ is to factor 3 and 5 from the derivative (Constant Factor Rule),

$$\left[3 t^{2}\right]^{\prime}=3\left[t^{2}\right]^{\prime} \quad\left[5 e^{t}\right]^{\prime}=5\left[e^{t}\right]^{\prime}$$

Finally the Primary Formulas for $[t ^{2}] ^{\prime}$ and $[e ^{t}] ^{\prime}$ are used.

Usual course of events.

In computing derivatives, use the Composition Formulas first and the Primary Formulas last.

 

5.2.1 Proof that lim $\lim _{h \rightarrow 0}(1+h)^{1 / h}$ exists.

Our goal is to prove that

$$\lim _{h \rightarrow 0}(1+h)^{1 / h} \quad \text { exists. }$$

We only consider the values of $h$ for $h = \frac{1}{n}$ where $n$ a positive integer, however, and show that

$$\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n} \quad \text { exists. }$$

You may accept this information without proof and proceed to the section exercises. The material of this subsection is very important, however, and deserves your substantial independent effort.

Explore 5.2.3 Do this! Explain the MATLAB program and the graph it creates and their relevance to our goal.

Code $\PageIndex{1}$ (MATLAB):

close all;clc;clear

h=[0.01:0.01:1];

y=(1+h).^(1./h);

plot(h, y, 'k', 'linewidth', 2)

hold('on')

For n = 1:8

z=(1+1/n)^n;

plot(1/n,z,’ks’,’linewidth’,2)

end

axis([0 1.1 1.8 3])

The graph in Explore 5.2.3 may convince you that $\lim _{h \rightarrow 0}(1+h)^{1 / h}$ exists and that there is nothing left to prove. But it is the existence of a number that is the limit that is in question and that existence requires a crucial property of the number system called completeness.

Axiom $\PageIndex{1}$ Completeness property of the number system.

If $S_1$ and $S_2$ are two sets of numbers and

1. Every number belongs to either $S_1$ or $S_2$, and
2. Every number of $S_1$ is less than every number of $S_2$,

then there is a number $C$ such that $C$ is either the largest number of $S_1$ or $C$ is the least number of $S_2$.

The statement of the Completeness Axiom by Richard Dedekind in 1872 greatly increased our understanding of the number system. In this text, the word 'set' means a nonempty set.

As noted above, we only consider values of $h = 1/n$ where $n$ is a positive integer and we define

$$s_{n}=\left(1+\frac{1}{n}\right)^{n} \quad \text { and prove that } \lim _{n \rightarrow \infty} s_{n} \quad \text { exists. }$$

That a sequence of numbers $s_{1}, s_{2}, s_{3} \cdots$ is bounded means that an open interval $(A, B)$¹ contains every number in $s_{1}, s_{2}, s_{3} \cdots$; $A$ is called a lower bound and $B$ is called an upper bound of $s_{1}, s_{2}, s_{3} \cdots$.

The sequence {$1, 4, 9, 16, 25, \cdots$} has no upper bound and is not bounded. The sequence {$1, \frac{1}{2} , \frac{1}{3} , \frac{1}{4} , \cdots$} is bounded by the numbers 0 and 2. The sequence {$1, -2, 3, -4, 5, -6, \cdots$} has neither a lower bound nor an upper bound and is not bounded.

For this section we prove:

Theorem 5.2.1 If $s_{1} \leq s_{2} \leq s_{3} \leq \cdots$ is a bounded nondecreasing sequence of numbers there is a number $s$ such that $\lim _{n \rightarrow \infty} s_{n}=s$.

To prove Theorem 5.2.1 we need a clear definition of limit of a sequence.

Definition 5.2.2 A number $s$ is the limit of a number sequence $s_{1}, s_{2}, s_{3} \cdots$ means that if $(u, v)$ is an open interval containing $s$ there is a positive integer $N$ such that if $n$ is an integer greater than $N, ~ s_n$ is in $(u, v)$. It is sometimes said that $s_{1}, s_{2}, s_{3} \cdots$ converges to $s$ or approaches $s$.

Example 5.2.1 The limit of the sequence {$1, \frac{1}{2} , \frac{1}{3} , \frac{1}{4} , \cdots$} is zero. If $(u, v)$ contains 0, $v$ is greater than zero and there is² a positive integer $N$ that is greater than $1/v$. Therefore, if $n > N, ~ n > 1/v$ so that $1/n < v$, and $u < 0 < 1/n < v$, so $\frac{1}{n}$ is in $(u, v)$).

Proof of Theorem 5.2.1. Suppose $s_{1}, s_{2}, s_{3} \cdots$ is a nondecreasing sequence bounded by $A$ and $B$. Let $S_1$ be the set of numbers $x$ for which some number of $s_{1}, s_{2}, s_{3} \cdots$ is greater than $x$. $A$ is a member of $S_1$ and $B$ is not a member of $S_1$. Let $S_2$ denote all of the numbers not in $S_1$ Clearly every number is in either $S_1$ or $S_2$.

To use the Completeness Axiom we must show that every number of $S_1$ is less than every number of $S_2$. Suppose not; suppose there is a number $x$ of $S_1$ that is greater than a number $y$ of $S_2$. Then $x$ has the property that some number, $s_n$, say of \(s_{1}, s_{2), s_{3} \cdots\) is greater than $x$ and $y$ is not in $S_1$ so no number of $s_{1}, s_{2}, s_{3} \cdots$ is greater than $y$. However, the supposition that $y < x$ and $x < s_n$, leads to $y < s_n$ which is a contradiction. Therefore every number of $S_1$ is less than every number of $S_2$.

By the Completeness Axiom, either $S_1$ has a largest number or $S_2$ has a least number. Suppose $S_1$ has a largest number, $L_1$. By definition of $S_1$ there is a member $s_n$ of $s_{1}, s_{2}, s_{3} \cdots$ greater than $L_1$. Now $\frac{L_{1} + s_{n}}{2}$ is less than $s_n$ so $\frac{L_{1} + s_{n}}{2}$ is in $L_1$. But $\frac{L_{1} + s_{n}}{2}$ is greater than $L_1$ so $L_1$ is not the largest member of $S_1$, which is a contradiction; $S_1$ does not have a largest number. Therefore $S_2$ must have a least number, $L_2$.

We prove that $s_{1}, s_{2}, s_{3} \cdots$ approaches $L_2$. Suppose ($u, v$) is an open interval containing $L_2$. Then $u$ is less than $L_2$ so belongs to $S_1$ and there is a number, $S_N$ in $s_{1}, s_{2}, s_{3} \cdots$ that is greater than $u$. Because $s_{1}, s_{2}, s_{3} \cdots$ is increasing, if $n$ is greater than $N$, $S_{N} \leq s_{n}$. No number of $s_{1}, s_{2}, s_{3} \cdots$ is greater than $L_2$. Therefore if $n$ is greater than $N$,

$$u<s_{N} \leq s_{n}<L_{2}<v$$

and the definition that $s_{1}, s_{2}, s_{3} \cdots$ approaches $L_2$ is satisfied.

End of proof.

Example 5.2.2 We can use Theorem 5.2.1 to show a useful result, that if $a$ is a number and $0 < a < 1$, then the sequence $x_{n} = a ^{n}$ converges to zero.

Proof. We assume the alternate version of Theorem 5.2.1 that If $s_{1} \geq s_{2} \geq s_{3} \geq \cdots$ is a bounded nonincreasing sequence of numbers there is a number $s$ such that $\lim _{n \rightarrow \infty} s_{n}=s$

Because $0 < a < 1$ and $x_{n} = a ^n$, $x_{n+1}=a \cdot x_{n}$ and $x_{n+1} < x_n$, and it follows that $x_n$ is a nonincreasing (actually decreasing) sequence. Then {$x_n$} approaches the greatest lower bound, $s$ of {$x_n$}.

If $s < 0$, then ($2s, 0$) is an open interval containing $s$ and there is a number, $x_m$ in {$x_n$} in $(2s, 0)$. Then $x_{m} = a ^{m}$ is negative which is a contradiction.

Suppose $s > 0$. Because $0 < a < 1$), the number $s/a$ is greater than $s$ and there is a number, $x_m$ in {$x_n$} such that $x_{m} < s_{a}$. Because $s$ is a lower bound on {$x_n$}, $s \leq x_m$. Then

$$s \leq x_{m}<s / a, \quad s \leq a^{m}<s / a, \quad s \cdot a<a^{m+1}<s, \quad x_{m+1}<s$$

But this contradicts the condition that $s$ is a lower bound on {$x_n$}.

We conclude that $s$, the greatest lower bound on {$x_n$}, is zero and {$x_n$} converges to 0.

Now to the number, $e$. We are to show that $\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n}$ exists. We will show that

1. The sequence $s_{n}=\left(1+\frac{1}{n}\right)^{n}$ is an increasing sequence.
2. The sequence $t_{n}=\left(1+\frac{1}{n-1}\right)^{n}, n>1$, is a decreasing sequence.
3. For every $n > 1, s_{n} < t_{n}$.
4. As $n$ increases without bound, $t_{n} − s_{n}$ approaches 0.

Conditions 1 and 3 show that $s_{1} < s_{2} < s_{3} < \cdots$ is a bounded increasing sequence and therefore approaches a number $s$. That number $s$ is the number we denote by $e$. Conditions 2 and 3 show that $t_{1} > t_{2} > t_{3} > \cdots$ is a bounded decreasing sequence and it follows from Theorem 5.2.13 that there is a number $t$ such that $t_{1} > t_{2} > t_{3} > \cdots$ converges to $t$ By condition 4 above, $t = s$.

The proofs of conditions 3 and 4 above are left as exercises. Our argument for conditions 1 and 2 is that of N. S. Mendelsohn⁴ based on the following theorem. The theorem is of general interest and we prove it in Subsection 8.1.2, independently of the work in this section.

Theorem 5.2.2 If $a_{1}, a_{2}, \cdots$, $a_n$ is a sequence of $n$ positive numbers then

$$\frac{a_{1}+a_{2}+\cdots a_{n}}{n} \geq \sqrt[n]{a_{1} a_{2} \cdots a_{n}} \label{5.7}$$

with equality only when $a_{1} = a_{2} = \cdots = a_{n}$.

The left side of inequality 5.7 is the arithmetic mean and the right side is the geometric mean of $a_{1}, a_{2}, \cdots, a_{n}$. The theorem states that the arithmetic mean is greater than or equal to the geometric mean.

1. Proof that $\left(1+\frac{1}{n}\right)^{n}$ is increasing. Consider the set of $n + 1$ numbers $$1, 1+\frac{1}{n}, 1+\frac{1}{n} \cdots 1+\frac{1}{n}$$  They are not all equal and they have an arithmetic mean of $1 + 1/(n + 1)$ and a geometric mean of $\left(1 \times(1+1 / n)^{n}\right)^{1 /(n+1)}$. Then $$1+\frac{1}{n+1}>\left((1+1 / n)^{n}\right)^{1 /(n+1)} \quad \text { or } \quad\left(1+\frac{1}{n+1}\right)^{n+1}>(1+1 / n)^{n}$$  Hence $s_{n+1} > s_{n}$ and the sequence $s_{1}, s_{2}, \cdots$ is increasing.
2. Proof that $\left(1+\frac{1}{n-1}\right)^{n}$ is decreasing. Consider the set of $n + 1$ numbers $$1, 1-\frac{1}{n}, 1-\frac{1}{n} \cdots 1-\frac{1}{n}$$  They have an arithmetic mean of $n/(n + 1)$ and a geometric mean of $\left(1 \times((n-1) / n)^{n}\right)^{1 /(n+1)}$ and $$\frac{n}{n+1}>\left(\frac{n-1}{n}\right)^{n /(n+1)}$$  By taking reciprocals this becomes $$\frac{n+1}{n}<\left(\frac{n}{n-1}\right)^{n /(n+1)} \quad \text { or } \quad\left(1+\frac{1}{(n+1)-1}\right)^{n+1}<\left(1+\frac{1}{n-1}\right)^{n}$$  It follows that $t_{n+1} < t_{n}$ and $t_{1}, t_{2}, \cdots$ is decreasing. End of proof.

We have shown that $\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{n}=e$ We claim the more general result that $\lim _{h \rightarrow 0}(1+h)^{1 / h}=e$. We are encouraged to this claim by an identification of $h = 1/n$, but there are other values of $h$ to consider that are not reciprocals of integers. You will show in Exercise 12.2.5 that the slope of the graph in Explore 5.2.3 is negative and this provides a way to complete the proof that $\lim _{h \rightarrow 0}(1+h)^{1 / h}=e$. In the suggested solution to that exercise, $[\ln u(x)]^{\prime}$ is used. $\log _{10} u(x)$ could be used equally well, so that the argument is not dependent on the number e.

Explore 5.2.4 Explain the MATLAB program and the graph it creates and their relevance to the previous paragraph.

Code $\PageIndex{1}$ (MATLAB):

close all;clc;clear

h=[0.01:0.01:1];

N=[10, 11, 13,17,20,25,33,50,100];

F=0*h;

for i = 1:9

F(N(i))=1;

end

plot(h, F, 'kd', 'linewidth', 2)

axis([0 1.1 -0.1 1.1])

Exercises for Section 5.2, The number e.

Exercise 5.2.1 Derivatives of functions are computed below. Identify the rule used in each step. In a few steps the rule is an algebraic rule of exponents and not a derivative rule.

1. $$\begin{aligned} &{\left[5 t^{4}-7 e^{t}\right]^{\prime}} \\ &{\left[5 t^{4}\right]^{\prime}-\left[7 e^{t}\right]^{\prime}} \\ &5\left[t^{4}\right]^{\prime}-7\left[e^{t}\right]^{\prime} \\ &5 \times 4 t^{3}-7\left[e^{t}\right]^{\prime} \\ &5 \times 4 t^{3}-7 \times e^{t} \end{aligned}$$
2. $$\begin{aligned} &{\left[\left(1+e^{t}\right)^{8}\right]^{\prime}} \\ &8\left(1+e^{t}\right)^{7}\left[1+e^{t}\right]^{\prime} \\ &8\left(1+e^{t}\right)^{7}\left([1]^{\prime}+\left[e^{t}\right]^{\prime}\right) \\ &8\left(1+e^{t}\right)^{7}\left(0+\left[e^{t}\right]^{\prime}\right) \\ &8\left(1+e^{t}\right)^{7}\left(0+e^{t}\right) \\ &8 e^{t}\left(1+e^{t}\right)^{7} \end{aligned}$$
3. $$\begin{aligned} &{\left[e^{3 t}\right]^{\prime}} \\ &{\left[\left(e^{t}\right)^{3}\right]^{\prime}} \\ &3\left(e^{t}\right)^{2}\left[e^{t}\right]^{\prime} \\ &3\left(e^{t}\right)^{2} \times e^{t} \\ &3 e^{2 t} \times e^{t} \\ &3 e^{3 t} \end{aligned}$$

Exercise 5.2.2 Differentiate (means compute the derivative of) P. Use one rule for each step and identify the rule as, C (Constant Rule), $t ^n$ ($t ^n$ Rule), S (Sum Rule), CF (Constant Factor Rule), PC (Power Chain Rule), or E (Exponential Rule). For example,

$$\begin{array}{rll} {\left[\pi t^{-2}-5\left(e^{t}\right)^{7}\right]^{\prime}} & =\left[\pi t^{-2}\right]^{\prime}-\left[5\left(e^{t}\right)^{7}\right]^{\prime} & \mathrm{S} \\ & =\pi\left[t^{-2}\right]^{\prime}-5\left[\left(e^{t}\right)^{7}\right]^{\prime} & \mathrm{CF} \\ & =\pi \times(-2) t^{-3}-5\left[\left(e^{t}\right)^{7}\right]^{\prime} & t^{n} \\ & =-2 \pi t^{-3}-5(7)\left(e^{t}\right)^{6}\left[e^{t}\right]^{\prime} & \mathrm{PC} \\ & =-2 \pi t^{-3}-35\left(e^{t}\right)^{6} \times e^{t} & \mathrm{E} \\ & =-2 \pi t^{-3}-35\left(e^{t}\right)^{7} &  \text { Algebra } \end{array}$$

1. $P(t)=5 t^{2}+32 e^{t}$
2. $P(t)=3\left(e^{t}\right)^{5}-6 t^{5}$
3. $P(t)=7-8\left(e^{t}\right)^{-1}$
4. $P(t)=\frac{2}{5}+\frac{t}{3}$
5. $P(t)=t^{25}+3 e$
6. $P(t)=\frac{4}{e}+\frac{t^{5}}{7}$
7. $P(t)=1+t+\frac{t^{2}}{2}-e^{t}$
8. $P(t)=\frac{\left(e^{t}\right)^{2}}{2}+\frac{t^{3}}{3}$

Careful, the next two are easier than you might first think.

9. $P(t)=5 e^{3}$
10. $P(t)=10 \times \pi^{2}$

Exercise 5.2.3 Draw the graphs of

$$y_{1}(t)=e^{t} \quad y_{2}(t)=t^{2} \quad y_{3}(t)=t^{3} \quad y_{4}(t)=t^{4} \quad-1 \leq t \leq 5$$

The graphs are close together near $t = 0$ and increase as t increases. Which one grows the most as t increases? Expand the domain and range to $-1 \leq t \leq 10, 0 \leq y \leq 25,000$, and answer the same question.

Exercise 5.2.4 Draw the graphs of

$$y(t)=e^{t} \quad \text { and } \quad p(t)=1+t+\frac{t^{2}}{2}+\frac{t^{3}}{6}+\frac{t^{4}}{24}$$

Set the domain and range to $-1 \leq t \leq 2$. and $0 \leq y \leq 8$.

Exercise 5.2.5 We found a base e so that $E(t) = e ^t$ has the property that the rate of change of $E$ at 0 is 1. Suppose we had searched for a number $B$ so that the average rate of change of $E_{B}(t)=B^{t}$ on [0, 0.01] is 1:

$$m_{0,0.01}=\frac{E_{B}(0.01)-E_{B}(0)}{0.01}=\frac{B^{0.01}-B^{0}}{0.01}=\frac{B^{0.01}-1}{0.01}=1$$

1. Solve the last equation for $B$.
2. Solve for B in each of the equations: $$\frac{B^{0.001}-1}{0.001}=1 \quad \frac{B^{0.00001}-1}{0.00001}=1 \quad \frac{B^{0.0000001}-1}{0.0000001}=1$$

Exercise 5.2.6 On a popular television business news channel, an analyst exclaimed that ”XXX stock has gone parabolic.” Is there some sense in which this exclamation is more exuberant than ”XXX stock has gone exponential?”

Exercise 5.2.7 Let $y(x) = e ^x$. Compute $y^{\prime}(x), y^{\prime \prime}(x)=\left(y^{\prime}\right)^{\prime}$, and $y^{\prime \prime \prime}(x)$.

Exercise 5.2.8 We introduced the power chain rule $\left[(u(x))^{n}\right]^{\prime}=n(u(x))^{n-1}[u(x)]^{\prime}$ for fractional and negative exponents, $n$, in Section 4.3.1 (see Exercises 4.3.3 and 4.3.4). Use these rules when necessary in the following exercise.

Compute $y^{\prime}(x)$ and $y^{\prime \prime}(x)$ for

1. $y(x)=x^{2}+e^{x}$
2. $y(x)=3 x^{2}+2 e^{x}$
3. $y(x)=\left(1+e^{x}\right)^{2}$
4. $y(x)=\left(e^{x}\right)^{2}$
5. $y(x)=e^{2 x}=\left(e^{x}\right)^{2}$
6. $y(x)=e^{-x}=\left(e^{x}\right)^{-1}$
7. $y(x)=e^{3 x}=\left(e^{x}\right)^{3}$
8. $y(x)=e^{x} \times e^{2 x}$
9. $y(x)=\left(5+e^{x}\right)^{3}$
10. $y(x)=\frac{1}{1+e^{x}}=\left(1+e^{x}\right)^{-1}$
11. $y(x)=\sqrt{e^{x}}=\left(e^{x}\right)^{\frac{1}{2}}$
12. $y(x)=e^{\frac{1}{2} x}$
13. $y(x)=e^{0.6 x}=\left(e^{x}\right)^{0.6}$
14. $y(x)=e^{-0.005 x}$

Exercise 5.2.9 Identify the errors in the following derivative computations.

1. $$\begin{aligned} &{\left[\left(t^{4}+t^{-1}\right)^{7}\right]^{\prime}} \\ &7\left(t^{4}+t^{-1}\right)^{6}\left[t^{4}+t^{-1}\right]^{\prime} \\ &7\left(t^{4}+t^{-1}\right)^{6}\left[t^{4}\right]^{\prime}+\left[t^{-1}\right]^{\prime} \\ &7\left(t^{4}+t^{-1}\right)^{6} 4 t^{3}+(-1) t^{-2} \\ &28 t^{3}\left(t^{4}+t^{-1}\right)^{6}-t^{-2} \end{aligned}$$
2. $$\begin{aligned} &{\left[5 t^{7}+7 t^{-5}\right]^{\prime}} \\ &{\left[5 t^{7}\right]^{\prime}+\left[7 t^{-5}\right]^{\prime}} \\ &5\left[t^{7}\right]^{\prime}+7\left[t^{-5}\right]^{\prime} \\ &5 \times 7 t^{6}+7 \times(-5) t^{-4} \\ &35\left(t^{6}-t^{-4}\right) \end{aligned}$$
3. $$\begin{aligned} &{\left[10 t^{8}+8 e^{5 t}\right]^{\prime}} \\ &{\left[10 t^{8}\right]^{\prime}+\left[8 e^{5 t}\right]^{\prime}} \\ &10\left[t^{8}\right]^{\prime}+8\left[e^{5 t}\right]^{\prime} \\ &10 \times 8 t^{7}+8 \times 5 e^{4 t} \\ &40\left(2 t^{7}+e^{4 t}\right) \end{aligned}$$

Exercise 5.2.10 Locate the point (0, e) on the graph in Explore 5.2.3

Exercise 5.2.11 Use the Completeness Axiom 5.2.1 to show that the positive integers do not have an upper bound. (This is called the Archimedean Axiom).

Exercise 5.2.12 Argue that if $S_1$ and $S_2$ are two sets of numbers and every number is in either $S_1$ or $S_2$ and every number in $S_1$ is less than every number in $S_2$ then it is not true that there are numbers $L_1$ and $L_2$ such that $L_1$ is the greatest number in $S_1$ and $L_2$ is the least number in $S_2$. Is this a contradiction to the Completeness Axiom?

Exercise 5.2.13 Let $S_2$ denote the points of the X-axis that have positive x-coordinate and $S_1$ denote the points of the X-axis that do not belong to $S_2$. Does $S_2$ have a left most point?

Exercise 5.2.14 Suppose $S_2$ is the set of numbers to which x belongs if and only if x is positive and $x ^{2} > 2$ and $S_1$ consists of all of the other numbers.

1. Give an example of a number in $S_2$.
2. Give an example of a number in $S_1$.
3. Argue that every number in $S_1$ is less than every number in $S_2$.
4. Which of the following two statements is true?
   1. There is a number C which is the largest number in $S_1$.
   2. There is a number C which is the least number in $S_2$.
5. Identify the number C in the correct statement of the previous part.

Exercise 5.2.15 Suppose your number system is that of Early Greek mathematicians and includes only rational numbers. Does it satisfy the Axiom of Completion?

Exercise 5.2.16 Show that if $n \geq 2$ and $s_{n}=\left(1+\frac{1}{n}\right)^{n}$ and $t_{n}=\left(1+\frac{1}{n-1}\right)^{n}$ then

1. For every $n$ $s_{n} < t_{n}$.
2. Justify the steps (i) to (iv) in $$\begin{aligned} t_{n}-s_{n} &=\left(1+\frac{1}{n-1}\right)^{n}-\left(1+\frac{1}{n}\right)^{n} & (i)\\ &=\left[\left(1+\frac{1}{n-1}\right)-\left(1+\frac{1}{n}\right)\right] \times & (ii)\\ {\left[\left(1+\frac{1}{n-1}\right)^{n-1}\right.} &\left.+\left(1+\frac{1}{n-1}\right)^{n-2}\left(1+\frac{1}{n}\right)+\cdots+\left(1+\frac{1}{n-1}\right)\left(1+\frac{1}{n}\right)^{n-2}+\left(1+\frac{1}{n}\right)^{n-1}\right] \\ &=\leq \frac{1}{n(n-1)} \times n \times\left(1+\frac{1}{n-1}\right)^{n-1} & (iii)\\ &=\leq \frac{1}{n-1} \times 4 & (iv) \end{aligned}$$
3. As $n$ increases without bound, $t_{n} − s_{n}$ approaches zero.

 

 

¹ If $A$ and $B$ are numbers with $A < B$, the open interval $(A, B)$ consists of all the numbers between $A$ and $B$; the closed interval $[A, B]$ consists of $A, ~ B$, and all of the numbers between $A$ and $B$.

² This is called the Archimedean property of the integers and may be treated as an axiom also. It is, however, a consequence of the Completeness Axiom, Exercise 5.2.11

³ Observe that $-t_{1} < -t_{2} < -t_{3} < \cdots$ is a bounded increasing sequence and there is a number $-t$ such that $-t_{1} < -t_{2} < -t_{3} < \cdots$ approaches $-t$.

⁴ American Mathematical Monthly, 58 (1951) p. 563.