exponential – Sine of Madness

For

$\begin{aligned} x&>0 \\ \frac{e^x}{x^n}&=\frac{1+x+\frac{x^2}{2}+...}{x^n} \\ &> \displaystyle\frac{\frac{x^m}{m!}}{x^n} \end{aligned}$

for any possible integer m we care to choose.

Choose $m_0>n$ then $\displaystyle \frac{e^x}{x^n} > \frac{x^{m_0-n}}{m_0!}$ .

But $\displaystyle \frac{x^{m_0-n}}{m_0!} \to \infty$ as $x \to \infty$ .

So, $\displaystyle \frac{e^x}{x^n}\to\infty$ as $x\to\infty$ .

So it can be said that $e^x$ increases more rapidly than any power of x.

Also (for $n>0$ ).

$\begin{aligned} \displaystyle \lim_{x\to\infty} \frac{\ln{x}}{x^n} &= \lim_{y\to\infty}\frac{y}{e^{ny}} \\&= \lim_{z\to\infty} \frac{z}{ne^z} = 0 \;\mbox{from above.} \end{aligned}$

Where $x=e^y$ as $x\to\infty, y\to\infty$ , and $z=ny$ as $y\to\infty, z\to\infty$ as $n>0$ .

So it can be said that $\ln{x}$ increases more slowly than any positive power of x.

Also (for $n>0$ ).

$\begin{aligned} \displaystyle \lim_{x\to 0^+} x^n\cdot\ln{x} &= \lim_{y\to\infty}\frac{1}{y^n}\cdot\ln{\frac{1}{y}} \\&= \lim_{y\to\infty}\frac{-\ln{y}}{y^n} = 0 \;\mbox{from above.} \end{aligned}$

Where $x=\frac{1}{y}, y=\frac{1}{x}, x\to0^+, y\to\infty$ .

Examples

$\begin{aligned} \displaystyle \lim_{x\to\infty}(e^x-x^4) &= \lim_{x\to\infty}(e^4(\frac{e^x}{x^4}-1)) \\&= \infty \end{aligned}$

$\begin{aligned} \displaystyle \lim_{x\to\infty}\frac{e^x(-2x^2+3)}{3x^4+2x^2+6} &= \lim_{x\to\infty}\frac{e^x(-2+\frac{3}{x^2})}{3x^2+2+\frac{6}{x^2}} \\&= -\infty \end{aligned}$

$\begin{aligned} \displaystyle \lim_{x\to\infty}\frac{2e^{2x}+x^2+2}{3e^{3x}+7} &= \lim_{x\to\infty}\frac{2e^{-x}+\frac{x^2}{e^{3x}}+\frac{2}{e^{3x}}}{3+\frac{7}{e^{3x}}} \\&= 0 \end{aligned}$

$\begin{aligned} \displaystyle \lim_{x\to\infty} \frac{(x-1)\ln{x^2}}{x^2} &= \lim_{y\to\infty}\frac{(y^\frac{1}{2}-1)\ln{y}}{y} \\&= \lim_{y\to\infty}(\frac{\ln{y}}{y^\frac{1}{2}}-\frac{\ln{y}}{y}) \\&= 0 \end{aligned}$

$x^2=y, x\to\infty, y\to\infty$ .

We define

$\exp{x} = 1 +x+\frac{x^2}{2!}+\frac{x^3}{3!}+...+\frac{x^n}{n!}+...$ which in fact converges $\forall x \in \mathbb{R}$ .

$\begin{aligned} \exp{x} \cdot \exp{y} &= (1 +x+\frac{x^2}{2!}+...)(1 +y+\frac{y^2}{2!}+...) \\&= 1 +x+y+\frac{x^2}{2!}+xy+\frac{y^2}{2!}+...+\frac{x^n}{n!}+\frac{x^{n-1}y}{n-1!}+\frac{x^{n-2}y^2}{(n-2)!2!}+...+\frac{y^2}{2!}+... \\&= 1 +(x+y)+\frac{(x+y)^2}{2!}+...+\frac{(x+y)^n}{n!}+... \\&= \exp{(x+y)} \end{aligned}$

We have assumed that we can multiply infinite series as if they were finite algebraic expressions. This is not always true, but it is for convergent power series.

The relationship $\exp{x}\cdot\exp{y} = \exp{(x+y)}$ leads us to write $\exp{x}=e^x$ so that $e^x\cdot e^y = e^{x+y}$ .

Note, $\exp{1} = e^1 = e = 1+1+\frac{1}{2!}+\frac{1}{3!}+...$

On the assumption that a power series can be differentiated term by term $\frac{d e^x}{dx} = 0 + 1 +\frac{x^2}{2!}+...=e^x$ .

Properties of exp

The graph of $y=e^x$ is as shown, (with series definition this is not easy to see), but there is some evidence from the property $e^{x_1}\cdot e^{x_2}=e^{x_1+x_2}$

Observe from the graph the properties:

$\displaystyle \lim_{x\to\infty}e^x=\infty$ .
$\displaystyle \lim_{x\to-\infty}e^x=0$ .
$\forall x, e^x >0$ .
$e^x$ is continuous everywhere.
$D(e^x)=\mathbb{R}$ .
$e^x$ is an injection (or one to one).
ie. $x_1 \neq x_2 \Rightarrow f(x_1) \neq f(x_2)$ .

This means there will be an inverse function. We call it the log function. We have $\log{y}$ equal to the unique x such that $y=e^x$ .

$y=e^x$ and $x=\ln{y}$ mean of course the same thing. The graph of $y=\ln{x}$ will be a reflection of $y=e^x$ in the line $y=x$ .

Note that the domain of ln = range of exp = $\mathbb{R}^+$ .

Properties of ln

If $y=\ln{x}$ then $x=e^y$ , so $\frac{dx}{dy} = e^y$ .
Therefore, $\frac{dy}{dx} = \frac{1}{\frac{dx}{dy}} = \frac{1}{e^y} = \frac{1}{x}$ .
$\ln{x}$ is an injection.
Proof:
Let $\ln{x}=\ln{y}$ .
Put $\ln{x}=p$ and $\ln{y}=p$ .
So, $e^p=x$ and $e^p=y$ .
Whence $x=y$ .
$\forall x, x=e^{\ln{x}}$ .
Proof:
Let $y=e^{ln{x}}$ .
Then $\ln{y} = \ln{x}$ .
Whence $x=y$ by 2.
So, $x=e^{\ln{x}}$ .
$\ln{xy}=\ln{x}+\ln{y}$ .
Proof:
$\begin{aligned} \ln{xy}&=\ln{e^{\ln{x}}\cdot e^{\ln{y}}} \\&= \ln{e^{\ln{x}+\ln{y}}} \\ xy &= e^{ln{x}+\ln{y}} \\ \mbox{Hence}\; \ln{x}+\ln{y} &= \ln{xy} \end{aligned}$ .
$\forall a \in \mathbb{R}^+$
(i) if m in an integer then $a^m = m\ln{a}$ .
(ii) if $\frac{m}{n}$ is in $\mathbb{Q}$ , then equally we have $\ln{a\frac{m}{n}}=\frac{m}{n}\ln{a}$ . ( $a^{\frac{m}{n}} \mbox{ means } \sqrt[n]{a^m}$ .