Monday, March 25, 2013

The exponential function

Recall from last class:

Theorem: There exists a function $E(x) : \mathbb R \to \mathbb R$ such that

i) $E'(x) = E(x)$ for all $x \in \mathbb R$
ii) $E(0) = 1$

Proof: Let $E_n = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \ldots + \frac{x^n}{n!}$. Let $A > 0$ be given, then for $|x| \leq A$ and $m > m > 2A$,

$$ |E_m(x) - E_n(s)| = \left| \frac{x^{n+1}}{(n+1)!} + \ldots + \frac{x^m}{m!} \right| \leq \frac{A^{n+1}}{(n+1)!}\left( 1 + \frac{A}{n} + \frac{A^2}{n^2} + \ldots + \left(\frac{A}{n}\right)^{m-n-1} \right) < \frac{2A^{n+1}}{(n+1)!} $$

as $\frac{1}{n} < \frac{1}{2A}$. Now since $\frac{A^n}{n!} \to 0$ as $n \to \infty$ for a fixed $A$, i.e. $E_n$ converges uniformly on $[-A, A]$ in particular, $E_n(x)$ converges for all $x \in \mathbb R$. Let $\lim E_n(x) = E(x)$. For fixed $x_0 \in \mathbb R$, there exists some $A > 0$ such that $x_0 \in [-A, A]$. $E_n \rightrightarrows E$ on $[-A, A]$ so $E$ is continuous at $x_0$. Observe that $E_n(0) = 1$ for all $n \in \mathbb N$ so $E(0) = 1$. Observe as well that since

$$ E_n(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \ldots + \frac{x^n}{n!} $$

it follows that

$$ E_n'(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \ldots + \frac{x^{n-1}}{(n-1)!} = E_{n-1}(x) $$

i.e. $E_n(x)$ is differentiable for all $x \in \mathbb R$ and $E_n'(x) = E_{n-1}(x)$. SInce $E_{n-1}$ converges uniformly on $[-A, A]$, so does $E_n'$ so by the theorem on interchange of limits and differentiability, we get that

$$ \lim E_n'(x) = (\lim E_n(x))' = E'(x) = \lim E_{n-1}(x) = E(x) $$

i.e. $E'(x) = E(x)$ and $E(0) = 1$ for all $x \in [-A, A]$. As $A > 0$ is arbitrary we conclude $E'(x) = E(x)$ for all $x \in \mathbb R$.

Corollary: $E$ has derivatives of all order and $E^{(n)}(x) = E(x)$ for all $n \in \mathbb N$, for all $x \in \mathbb R$.

Corollary: If $x > 0$, $1 + x < E(x)$

Proof: Obvious by series definition of $E(x)$.

Theorem: The function $E : \mathbb R \to \mathbb R$ such that $E'(x) = E(x)$ for all $x \in \mathbb R$ and $E(0) = 1$ is unique.

Proof: Suppose we have two such functions $E(x)$ and $F(x)$ where $E'(x) = E(x)$ and $F'(x) = F(x)$, $E(0) = F(0) = 1$. Consider $G(x) = E(x) - F(x)$. $G'(x) = G(x)$ trivially, and $G^{(n)}(x) = G(x)$. $G(0) = 0$. Let $x \in \mathbb R$ and $I$ be the interval with endpoints 0 and $x$. Since $G$ is continuous on $I$, $G$ is uniformly bounded on $I$ so that $|G(t)| \leq k \in \mathbb R$ for all $t \in I$. By Taylor's theorem applied to $G$ on $I$,

$$ G(x) = G(0) + \frac{G'(0)}{1!}x + \frac{G''(0)}{2!}x^2 + \ldots + \frac{G^{(n-1)}(0)}{(n-1)!}x^{n-1} + \frac{G^{(n)}(C_n)}{n!}x^n = \frac{G^{(n)}(C_n)}{n!}x^n $$

thus

$$ G(x) = \frac{G^{(n)}(C_n)}{n!}x^n = \frac{G(C_n)}{n!}x^n $$

and

$$ |G(x)| \leq \frac{k|x|^n}{n!} $$

for fixed $x$, the RHS above goes to 0 as $n \to \infty$, so by the squeeze theorem $G(x) = 0$. As $x$ is arbitrary we can conclude that $G(x) \equiv 0$ and therefore $E(x) = F(x)$ for all $x \in \mathbb R$.

Definition: The unique function $E : \mathbb R \to \mathbb R$ such that $E'(x) = E(x)$ and $E(0) = 1$ is called the exponential function. $e = E(1)$ is called Euler's number.

Theorem: $E(x)$ satisfies:

i) $E(x) \neq 0$ for all $x \in \mathbb R$
ii) $E(x+y) = E(x)E(y)$
iii) $E(R) = e^R$ for all $R \in \mathbb Q$

Proof:
i) Suppose $E(a) = 0$ for some $a \in \mathbb R$. Let $J$ be the closed and bounded interval with endpoints 0 and $a$. $E(x)$ is continuous on $J$ so there exists some $k \in \mathbb R$ such that $|E(x)| \leq k$ on $J$. By Taylor's theorem,

$$ E(0) = E(a) + \frac{E'(a)}{1!}(-a) + \frac{E''(a)}{2!}(-a)^2 + \ldots + \frac{E^{(n-1)}(a)}{(n-1)!}(-a)^{n-1} + \frac{E^{(n)}(c_n)}{n!}(-a)^n $$

for some $c_n \in J$. By assumption, $E(a) = 0$ so that $E'(a) = 0 = E^{(n)}(a)$ so that

$$ E(0) = \frac{E(c_n)}{n!}(-a)^n $$

so $|E(0)| \leq \frac{k|a|^n}{n!} \to 0 $ which is a contradiction since $E(0) = 1$.

We'll finish the rest next class.