12.4 The Exponential Function

12.22   Example. Suppose we had a complex function $E$ such that $E$ is everywhere differentable and

$$% latex2html id marker 19578E'= E, \mbox{ and } E(0)=1. \index{differential equation (\ref{diffEE})}$$ (12.23)

Let $H(z)=E(z)E(-z)$ for all $z\in\mbox{{\bf C}}$. By the chain and product rules,

$$H'(z)=E'(z)E(-z)+E(z)[-E'(-z)]=E(z)E(-z)-E(z)E(-z)=0$$

on $\mbox{{\bf C}}$, so $H'$ is constant. Since $H(0)=E(0)E(0)=1$, we conclude

$$E(z)E(-z)=1\mbox{ for all } z\in\mbox{{\bf C}}.$$ (12.24)

In particular $E(z)$ is never $0$, and

$$E(-z)=\left(E(z)\right)^{-1} \mbox{ for all }z \in \mbox{{\bf C}}.$$

Now let $a\in\mbox{{\bf C}}$ and define a function $H_a\colon\mbox{{\bf C}}\to\mbox{{\bf C}}$ by

$$H_a(z)=E(z+a)E(-z).$$

We have

$$\begin{eqnarray*} H_a'(z)&=&E'(z+a)E(-z)+E(z+a)[-E'(-z)]\\ &=& E(z+a)E(-z) - E(z+a)E(-z) =0 \end{eqnarray*}$$

for all $z\in\mbox{{\bf C}}$, so $H_a$ is constant, and $H_a(0)=E(a)E(0)=E(a)$. Thus

$$E(z+a)E(-z)=E(a)\mbox{ for all }z\in\mbox{{\bf C}},\; a\in\mbox{{\bf C}},$$

and by (12.24),

$$E(z+a)=E(a)E(z)\mbox{ for all }z\in\mbox{{\bf C}},\;a\in\mbox{{\bf C}}.$$ (12.25)

Next suppose you know some function $e\colon\mbox{{\bf R}}\to\mbox{{\bf R}}$ such that $e'(t)=e(t)$ for all $t\in\mbox{{\bf R}}$ and $e(0)=1$. (You do know such a function from your previous calculus course.) Let

$$K(t)=E(-t)e(t)\mbox{ for all }t\in\mbox{{\bf R}}.$$

Then by the product and chain rules,

$$K'(t)=[-E(-t)]e(t)+E(-t)e(t)=0\mbox{ for all } t\in\mbox{{\bf R}},$$

so $K$ is constant on $\mbox{{\bf R}}$, and since $K(0)=E(0)e(0)=1$, we have $E(-t)e(t)=1$. By (12.24),

$$e(t)=E(t)\mbox{ for all } t\in\mbox{{\bf R}}.$$

Now I will try to construct a function $E$ satisfying the differential equation (12.23) by hoping that $E$ is given by a power series. Suppose

$$\begin{eqnarray*} E(z)&=&a_0+a_1z+a_2z^2+a_3z^3+a_4z^4\cdots\mbox{ for all }z\in\mbox{{\bf C}}. \\ E(0)&=&a_0+0+0+\cdots. \end{eqnarray*}$$

Since $E(0)=1$, we must have $a_0=1$, and

$$E(z)=1+a_1z+a_2z^2+a_3z^3+a_4z^4+\cdots.$$

By the differentiation theorem,

$$E'(z)=a_1+2a_2z+3a_3z^2+4a_4z^3+\cdots,$$

and

$$a_1=E'(0)=E(0)=1.$$

By the differentiation theorem again,

$$E'(z)=2\cdot 1 a_2+3\cdot 2a_3z+4\cdot 3a_4z^2+\cdots,$$

so

$$2\cdot 1a_2=E'(0)=E(0)=1 \mbox{ and } a_2={1\over 2\cdot 1}.$$

Hence

$$E(z)=E'(z)=1+3\cdot 2a_3z+4\cdot 3a_4z^2+\cdots.$$

Repeating the process, we get

$$E'(z)=3\cdot 2\cdot 1a_3+4\cdot 3\cdot 2a_4z+\cdots,$$

so

$$3\cdot 2 \cdot 1 a_3=E'(0)=E(0)=1\mbox{ and }a_3={1\over 3\cdot 2\cdot 1}.$$

I see a pattern here: $${a_n={1\over {n!}}}$$.

12.26   Definition (Exponential function.) Let $E$ denote the power series $${\sum\left\{ {{z^n}\over{n!}}\right\} =\left\{ 1+z+{{z^2}\over {2!}} +\cdots + {z^n \over n!}\right\}}$$. We will show in exercise 12.31 that $E$ has infinite radius of convergence. We write

$$\displaystyle { E(z) = \exp(z)=\sum_{n=0}^\infty{{z^n}\over {n!}}} \mbox{ for all }z\in\mbox{{\bf C}}.$$

12.27   Theorem. $\exp'=\exp$ and $\exp(0)=1$.

Proof: It is clear that $\exp(0)=1$. The formal derivative of $E$ is

$$DE=\sum\left\{ {{nz^{n-1}}\over {n!}}\right\}=\sum\left\{ {{(... ...over {(n+1)!}}\right\}=\sum\left\{ {{z^n}\over {n!}}\right\}=E,$$

so the [still unproved] differentiation theorem says that $\exp'=\exp$. It follows from our discussion above that $\exp(z)$ is never $0$,

$$\exp (-z)=\left(\exp(z)\right)^{-1}\mbox{ for all }z\in\mbox{{\bf C}},$$ (12.28)

and

$$\exp(a+z)=\exp(a)\exp(z)\mbox{ for all }z\in\mbox{{\bf C}}.$$ (12.29)

It is clear that $\exp(z)$ is real for all $z\in\mbox{{\bf R}}$. In fact, we must have $\exp(z)\in\mbox{${\mbox{{\bf R}}}^{+}$}$ for all $z\in\mbox{{\bf R}}$, since $\exp$ is continuous (differentiable functions are continuous) and if $\exp(t)<0$ for some $z$, the intermediate value theorem would say $\exp (y)=0$ for some $y$ between $0$ and $t$. Since $\exp'(t)=\exp(t)>0$ on $\mbox{{\bf R}}$, $\exp$ is strictly increasing on $\mbox{{\bf R}}$. $\mid\!\mid\!\mid$

12.30   Definition ($e$.) We define $e$ to be the number $\exp (1)$; i.e., $${e=\sum_{n=0}^\infty {1\over {n!}}}$$.

12.31   Exercise. Show that $${\sum\left\{ {{z^n}\over {n!}}\right\}}$$ has infinite radius of convergence.

12.32   Exercise. Use the definition of $e$ to show that $e>2.718$.

12.33   Exercise. A $\quad$

a) Show that $\exp(nz)=\left(\exp(z)\right)^n$ for all $n\in\mbox{{\bf N}}$, $z\in\mbox{{\bf C}}$.

b) Show that $\exp(nz)=\left(\exp(z)\right)^n$ for all $n\in\mbox{{\bf Z}}$, $z\in\mbox{{\bf C}}$.

12.34   Exercise. A From the previous exercise, it follows that

$$\exp(nz)=\left(\exp(z)\right)^n\mbox{ for all }z\in\mbox{{\bf C}},\; n\in\mbox{{\bf Z}}.$$

Use this to prove that

$$\exp\left( {p\over q}t\right)=\left(\exp(t)\right)^{{p\over q... ...R}},\; p\in\mbox{{\bf Z}},\; q\in\mbox{${\mbox{{\bf Z}}}^{+}$};$$

i.e.,

$$\exp(rt)=\left(\exp(t)\right)^r\mbox{ for all }t\in\mbox{{\bf R}},\; r\in\mbox{{\bf Q}}.$$

(Note that for $t=1$, this says

$$\exp(r)=\left(\exp(1)\right)^r=e^r.$$

12.35   Notation ($e^z$.) Another notation for $\exp(z)$ is $e^z$. This notation is motivated by the previous exercise. With this notation, we have

$$\begin{eqnarray*} e^{z+a}&=&e^ze^a\mbox{ for all }z,a\in\mbox{{\bf C}}. \\ (e^z... ...{(tr)}\mbox{ for all } t\in\mbox{{\bf R}},\; r\in\mbox{{\bf Q}}. \end{eqnarray*}$$

12.36   Theorem. Every number $t\in\mbox{${\mbox{{\bf R}}}^{+}$}$ can be written as $t=\exp(s)$ for a unique $s\in\mbox{{\bf R}}$.

Proof: The uniqueness of $s$ follows from the fact that $\exp$ is strictly increasing on $\mbox{{\bf R}}$. Let $t\in(1,\infty)$. From the expansion $${\exp(t)=1+t+{{t^2}\over {2!}}+\cdots}$$, we see that $\exp(t)>t$. Since $\exp$ is continuous, we can apply the intermediate value theorem to $\exp$ on $[0,t]$ to conclude $t=\exp(s)$ for some $s\in(0,t)$. If $t\in(0,1)$, then $${ {1\over t}\in(1,\infty)}$$, so $${ {1\over t}=e^s}$$ for some $s\in (0,\infty)$, and $t=e^{-s}$ where $-s\in(-\infty,0)$. Since $1=e^0$, the theorem has been proved in all cases. $\mid\!\mid\!\mid$