Exp

[!math|{“type”:“definition”,“number”:“auto”,“setAsNoteMathLink”:false,“title”:“exp”,“label”:“exp”,“_index”:0}] Definition 1 (exp). The inverse function of $\log$ from $\mathbb{R}\to {\mathbb{R}}^{+}$ is $\exp (x)$.

Properties §

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:1}] Proposition 2. $\exp (0)=1$

\begin{proof}$\exp (\log (y))=y,\log (1)=0$. \end{proof}

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:2}] Proposition 3. $\frac{d}{dx}\exp =\exp$

\begin{proof}Since $\frac{d}{dx}\log x=\frac{1}{x}\ne 0$, using the formula for the derivative of an inverse,

$$\frac{d}{dx}\exp (x)=\frac{1}{\frac{d}{dx}\log (\exp (x))}=\frac{1}{\frac{1}{\exp (x)}}=\exp (x)$$

\end{proof}

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:3}] Proposition 4. For $x,y\in \mathbb{R}$, $\exp (x+y)=\exp (x)\exp (y)$

\begin{proof} For $x,y\in \mathbb{R}$, exists unique $u,v\in {\mathbb{R}}^{+}$ s.t. $x=\log u,y=\log v$.

$$\begin{array}{rl}\exp (x+y)& =\exp (\log u+\log v)\\ & =\exp (\log uv)\\ & =uv\\ & =\exp (x)\exp (y).\end{array}$$

\end{proof}

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:4}] Proposition 5. $\exp$ is strictly increasing.

\begin{proof}If $x>y$, let $x=\log (u),y=\log v$. $\log$ is strictly increasing, so $u>v$, and thus $u=\exp (x)>\exp (y)=v$. \end{proof}

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:5}] Proposition 6. ${\lim }_{x\to \infty }\exp (x)=\infty$, ${\lim }_{x\to -\infty }(x)=0$

\begin{proof} For the first limit, we must show $\exp (x)>M$ when $x>L$, or equivalently $\log x>\log L⟹x>\log M$, which is possible since we choose $L$. Similarly, for the second limit, we must show $x<L⟹\exp (x)<ϵ$ or $\log x<\log L⟹x<\log ϵ$.\end{proof}

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:6}] Proposition 7. $\int \exp (x)dx=\exp (x)+C$

\begin{proof}Since $\exp$ is differentiable, it is continuous. Thus, it is integrable on finite subintervals. We also know $\frac{d}{dx}\exp =\exp$.\end{proof}

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:7}] Proposition 8. ${\lim }_{x\to \infty }\frac{p(x)}{\exp (x)}=0$ and ${\lim }_{x\to -\infty }p(x)\exp (x)=0$ for polynomials $p$.

\begin{proof}Start with ${\lim }_{x\to \infty }\frac{x^j}{\exp x}$. We know that $\frac{\exp x}{x^j}\to \infty$ since the $\log$ of the previous term is $x-j\log x=x(1-\frac{j\log x}{x})\to \infty$.

For ${\lim }_{x\to -\infty }{x}^j\exp x$, check $\frac{1}{x\exp x}\to -\infty$, which is true since $\log \left(-\frac{1}{x\exp x}\right)=\log \left(-\frac{1}{x}\right)-x=x(\frac{\log x}{-x}-1)$. \end{proof}

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:8}] Proposition 9. ${\int }_{-\infty }^b\exp (x)dx=\exp (b)$

\begin{proof}We need to show

$$L=\underset{x\to -\infty }{\lim }{\int }_x^b\exp (t)dt=\underset{x\to -\infty }{\lim }\exp b-\exp x.$$

[!math|{“type”:“lemma”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:9}] Lemma 10. If $f$ is differentiable on $(a,b)$ and $f^{\prime }(x)=f(x)\forall x\in (a,b)$, then $\exists c$ s.t. $f(x)=c\exp x$ and if $f(0)=1$, $c=1$.

We can prove this lemma is true by differentiating $h(x)=\frac{f(x)}{\exp (x)}$ using quotient rule to find that $h$ is constant, say $c$. \end{proof}

e §

The number $e$ is defined as $e=\exp (1)$.

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:10}] Proposition 11. $\exp (x)=e^x$.

\begin{proof}Apply $\log$ to get $x=\log (e^x)$.

Now, we will prove this is true for $\mathbb{Z},\mathbb{Q},\mathbb{R}$ in that order.

We know

$$1=\log e=\log ((e^{1/n})n)=n\log (e^{1/n})⟹\log (e^{1/n})=\frac{1}{n}$$

and

$$\log (e^n)=n\log (e)=n.$$

Then,

$$\log (e^{m/n})=m\log (e^{1/n})=\frac{m}{n}.$$

Then, for $x\in \mathbb{R}$, take ${\alpha }_n\in \mathbb{Q}\to x$, so

$$\log (e^x)=\log (e^{\lim {\alpha }^{{}_n}})=\log (\lim e^{\alpha n})=\lim \log (e^{\alpha n})=\lim {\alpha }_n=x.$$

\end{proof}

Now, for every $a>0,a^x=e^{x\log n}$, this is monotone so it has an inverse, all it ${\log }_a$. It is easy to see that

[!math|{“type”:“proposition”,“number”:“auto”,“setAsNoteMathLink”:false,“_index”:11}] Proposition 12. $\frac{d}{dx}{a}^x=a^x\log a$.

Hyperbolic Functions §

$e$ is key to the hyperbolic trig functions:

$$\sinh (x)=\frac{e^x-e^{-x}}{2},\cosh (x)=\frac{e^x+e^{-x}}{2},\tanh (x)=\frac{\sinh }{\cosh }.$$

Cool Properties §

1. $\frac{d}{dx}\sinh =\cosh$
2. $\frac{d}{dx}\cosh =\sinh$
3. like Pythagorean theorem but with diff sign
4. model hyperbolas