11.14 The exponential function

Example of theorem

Consider \(x \in \mathbb{R}\), \(\sum_{n=0}^{\infty} \frac{x^n}{n!}\) is A.C. \(\forall x\).(from)

From previous theorem: \(\sum_{k=0}^{\infty} \frac{x^k}{k!} \sum_{k=0}^{\infty} \frac{y^k}{k!} = \sum_{n=0}^{\infty} C_n\)

\(C_{n}=\sum_{k=0}^{n}a_{k}\cdot b_{n-k}=\sum_{k=0}^{n}\frac{x^{k}}{k!}\cdot\frac{y^{n-k}}{(n-k)!}\)

= \(\frac{1}{n!}\sum_{k=0}^{n}\frac{n!}{k!(n-k)!}x^{k}y^{n-k}=\frac{1}{n!}\sum_{k=0}^{n}\binom{n}{k}x^{k}y^{n-k}\) \(\Rightarrow C_n = \frac{(x + y)^n}{n!}\)

Therefore we have proved

Theorem: \(\left( \sum_{n=0}^{\infty} \frac{x^n}{n!} \right) \left( \sum_{n=0}^{\infty} \frac{y^n}{n!} \right) = \sum_{n=0}^{\infty} \frac{(x + y)^n}{n!}\)

The exponential function

Let us call \(\exp(x) := \sum_{n=0}^{\infty} \frac{x^n}{n!}, \quad \forall x \in \mathbb{R}\).

We have proved: \(\exp(x + y) = \exp(x) \exp(y), \quad \forall x, y \in \mathbb{R}\).

We will see that \(\exp(q) = e^q, \quad \forall q \in \mathbb{Q}\).

Example

Recall that: \(e = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = 2.71 \ldots\)

Parenthesis: \(2^x\) with \(x \in \mathbb{R}\)?

\(2^{5/2} = \sqrt{2^5} = \sqrt{32} = (2^{1/2})^5\)

\(2^{3/2} = \sqrt{2^3} = \sqrt{8} = (2^{1/2})^3\)

\(2^{\sqrt{2}} \approx 2^{1.414} = \sqrt[1000]{2^{1414}} = \left(\sqrt[1000]{2}\right)^{1414}\)

Theorem

\(\exp(x) = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n\)

#Method tool: Eliminating variable with limit#

Proof

We want to prove \(\lim_{n \to \infty} \sum_{k=0}^{n} \frac{x^k}{k!} = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n\)

From the expansion: \(\left(1 + \frac{x}{n}\right)^n = \sum_{k=0}^{n} \binom{n}{k} \left(\frac{x}{n}\right)^k 1^{n-k}\)

\(= \sum_{k=0}^{n} \frac{x^k}{k!} \cdot \frac{n!}{(n-k)! \, n^k}\) \(= \sum_{k=0}^{n} \frac{x^k}{k!} \cdot \frac{n (n-1) (n-2) \cdots (n-k+1)}{n^k}\)

\(= \sum_{k=0}^{n} \frac{x^k}{k!} \cdot 1 \cdot \left(1 - \frac{1}{n}\right) \left(1 - \frac{2}{n}\right) \cdots \left(1 - \frac{k-1}{n}\right)\).

This implies

\(\left(1+\frac{x}{n}\right)^{n}\leq\sum_{k=0}^{n}\frac{x^{k}}{k!}\Rightarrow\widetilde{\exp}(x)\leq\exp(x)\). (take limit)
Consider \(n > m\) arbitrary: \(\left(1+\frac{x}{n}\right)^{n}\geq\sum_{k=0}^{m}\frac{x^{k}}{k!}\left(1-\frac{k-1}{n}\right)^{k}\) where \((1 - \frac{k-1}{n}) > (1 - \frac{m}{n})\) for \(0 < k \leq m\).

Thus \(\geq(1-\frac{m}{n})^{m}\sum_{k=0}^{m}\frac{x^{k}}{k!}\)

Consider \(m\) fixed and take \(\lim_{n \to \infty}\): \(\widetilde{\exp}(x) \geq \sum_{k=0}^{m} \frac{x^k}{k!} \quad \forall m\)

Now \(\lim_{m \to \infty}\): \(\widetilde{\exp}(x) \geq \sum_{k=0}^{\infty} \frac{x^k}{k!} = \exp(x)\)

From (1) and (2): \(\widetilde{\exp}(x) = \exp(x)\)

Proposition

\(\exp(q) = e^q \quad \forall q \in \mathbb{Q}\)

Proof

We want to prove \(\lim_{n \to \infty} \left(1 + \frac{q}{n}\right)^n = \left(\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n\right)^q = e^q\)

Assume \(q \in \mathbb{N}\), \(\lim_{n \to \infty} \left(1 + \frac{q}{n}\right)^n\)

Trick: to cancel \(q\), consider \(n_k = q \cdot k\)

\(= \lim_{k \to \infty} \left(1 + \frac{q}{n_k}\right)^{n_k}\)\(= \lim_{k \to \infty} \left(1 + \frac{q}{q \cdot k}\right)^{q \cdot k}\)\(= \lim_{k \to \infty} \left(1 + \frac{1}{k}\right)^{q \cdot k}\)

\(= \left(\lim_{k \to \infty} \left(1 + \frac{1}{k}\right)^k\right)^q\)\(= e^q\)
Now let \(q = \frac{a}{b}\) with \(a, b \in \mathbb{N}\). We want to prove:

\(\lim_{n \to \infty} \left(1 + \frac{a/b}{n}\right)^n = \left[\lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n\right]^{a/b}\)

\(\;\iff\;\left(\lim_{n\to\infty}\left(1+\frac{\frac{a}{b}}{n}\right)^{n}\right)^{b}=\left(\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{n}\right)^{a}\)

Since \(a,b\in\N\), we already know in 1

\(\;\iff\;\lim_{n\to\infty}\left(1+\frac{\frac{a}{b}\cdot b}{n}\right)^{n}=\lim_{n\to\infty}\left(1+\frac{a}{n}\right)^{n}\)

This is true.

It remains \(q < 0\) and is left as exercise.

Summarizing

\(\exp(x) := \sum_{k=0}^{\infty} \frac{x^k}{k!}\) by definition

Theorem: \(\exp(x) = \lim_{n \to \infty} \left(1 + \frac{x}{n}\right)^n\)

Also: \(\exp(x + y) = \exp(x) \cdot \exp(y)\)

\(\exp(q) = e^q \quad \forall q \in \mathbb{Q}\)

\(\exp(1) = e\)

\(\exp(0) = 1\)

These define the exponential function.

Def: If \(x \in \mathbb{R}\), define \(e^x = \exp(x)\).

By definition, \(e = \exp(1) = \sum_{k=0}^{\infty} \frac{1^k}{k!} = \lim_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = 2.71 \ldots\)

Topology

Definition

Let \(X\subseteq \R\) and \(x_0\in \R\), we say that

\(x_0\) is interior (inner point) of \(X\) if \(\exists\delta>0:(x_0-\delta,x_0+\delta)\subseteq X\)

Example: \(X=(3,5]\), \(x_0=4.6\) is interior and \(x_0=5\) is not

The set of inner points of \(X\) is denoted by \(\mathring{X}\)
\(x_0\) is an exterior point if \(\exists \delta > 0\) such that \((x_0-\delta,x_0+\delta)\cap X=\emptyset\)

Or if \(x_0\) is the interior of the \(X^c\) which is \((x_0-\delta,x_0+\delta)\subset X^{c}\Rightarrow(x_0-\delta,x_0+\delta)\cap X=\emptyset\)
\(x_0\) is limit point of \(X\) if \(\forall\delta>0:\left\lbrack(x_0-\delta,x_0)\cup(x_0,x_0+\delta)\right\rbrack\cap X\neq\emptyset\)

or \(:0<|x-x_0|<\delta\)

Example: \(x_0=5\) is limit point of \(X=(3,5]\)

\(x_0=3\) is limit point of \(X=(3,5]\)

\(x_0=5\) is not a limit point of \(X=(1,2)\cup \{5\}\cup(6,7)\)
\(x_0\) is a boundary point(extreme point) if \(\forall \delta > 0\), \(\exists y\in X,z\notin X\) with \(|x_0-y|<\delta\), \(|x_0-z|<\delta\)

The set of boundary points is denoted by \(\partial A\)

Equivalence: \(x_0\) is a boundary point of \(A\) if \(\forall \delta > 0 \quad (x_0 - \delta, x_0 + \delta) \cap A \neq \emptyset\) and \((x_0 - \delta, x_0 + \delta) \cap A^c \neq \emptyset\)
\(x_0\) is isolated point if \(x_0\in X\) and \(\exists \delta>0\) such that \((x_0-\delta,x_0+\delta)\cap X=\left\lbrace x_0\right\rbrace\)

Example

\(A = \left\{ \frac{1}{n} : n \in \mathbb{N} \right\}\)

\(0 \notin A\)

Is \(0\) a limit point? Yes

\(\forall \delta > 0, \, \exists \, \frac{1}{N}\) such that \(\frac{1}{N} \in (0 - \delta, 0 + \delta) \cap A\)

Is \(1\) a limit point? No
Inner points

\(\mathring{A}=\emptyset\)
Exterior points

\((-\infty,0)\cup(1,+\infty)\cup\left(\frac{1}{n+1},\frac{1}{n}\right)_{n\in\mathbb{N}}\)
Boundary of A

\(\partial A=A\cup\{0\}\)