## Exponential Function

So we threw down a challenge at the end of our discussion of Tangent Lines, lets solve it.

Recall we wanted to find a curve $y=f(x)$ such that for any $(x_{0},f(x_{0}))$ we have the line tangent to that point $t_{0}(x)$ satisfy $t_{0}(x_{0}-1)=0$.

(Historical note: Debeaune asked Descartes this problem in a letter in 1638; Leibniz solved this problem in 1684 using “infinitesimals”. We will use modern calculus to solve the problem!)

So what is $f(x)=$? Spoiler below!

Well, we set up the equation describing the tangent line
(1)$\displaystyle0=f(x_{0})+f'(x_{0})\cdot(x_{0}-1-x_{0})$
We can rewrite this to be
(2)$\displaystyle 0=f(x_{0})-f'(x_{0}).$
Or equivalently $f'(x_{0})=f(x_{0})$ for any $x_{0}$.

Lets try using the tangent line as an approximation for this curve. So for small $h$ we write
(3)$\displaystyle f(x+h)\approx f(x)+f'(x)\cdot h.$
But since $f'(x)=f(x)$ we rewrite the right hand side as
(3′)$\displaystyle f(x+h)\approx f(x)\left(1+\frac{f'(x)}{f(x)}\cdot h\right)=f(x)\cdot(1+h).$
What to do?

Assumption: Lets write $f(0)=1$.

We can have a coarse approximation
$\displaystyle f(x+h)=f(x)\cdot(1+h).$
So for small $h$ we have
$\displaystyle f(h)=f(0)\cdot(1+h)=(1+h).$
We could also break this down into two steps:
(4)$\displaystyle f(h)=f(h/2)\cdot\left(1+\frac{h}{2}\right)=\left(1+\frac{h}{2}\right)^{2}$.
Why can we say this? Well, because
$\displaystyle f(h/2)=(1+h/2)$
and plugging this into (4) gives the correct relation.

We could get a better approximation iterating this three times:
(4a)$\displaystyle f(h)=\left(1+\frac{h}{3}\right)^{3}$.
We could do this $n$ times:
(4b)$\displaystyle f(h)=\left(1+\frac{h}{n}\right)^{n}$.
Taking the limit $n\to\infty$ we obtain the definition
(5)$\displaystyle f(h)=\lim_{n\to\infty}\left(1+\frac{h}{n}\right)^{n}$.
For sufficiently large $n$, we have for any $x$ the quantity $x/n$ be really small. So Equation (5) works for any real number.

What is this $f(x)$? It is precisely the exponential function, usually written $\exp(x)=\mathrm{e}^{x}$.

Exercise 1. Prove $\exp(a)\exp(b)=\exp(a+b)$.

Exercise 2. What if we drop our assumption and allow $f(0)$ be some nonzero real number? What if we let it be zero?

Exercise 3. Prove $\exp(-a)=1/\exp(a)$ for any positive real number $a$.

Exercise 4. Calculate $\exp(1)=\mathrm{e}$ to 7 digits of precision. (This mathematical constant “$\mathrm{e}$” is called “Euler’s constant”.) [Hint: we need for $k$ digits of precision to evaluate the limit with some $n\geq 4\cdot10^{k}$.]

Exercise 5. Lets prove some inequalities involving $\exp(x)$! Prove the following:
(a) $\exp(x)>1+x$
(b) $\displaystyle \exp(x)<\frac{1}{1-x}$ for $x<1$
(c) $\displaystyle \frac{x}{1+x}<1-\exp(-x) for $-1
(d) $\displaystyle x<\exp(x)-1<\frac{x}{1-x}$ for $x\exp\bigl(x/[1+x]\bigr)$ for $x>-1$
(f) $\exp(x)>x^{p}/p!$ for $x>0$ and $p\in\mathbb{N}$ (i.e., $p$ is a positive integer)
(g) $\displaystyle \exp(x)>\left(1+\frac{x}{y}\right)^{y}>\exp\bigl(xy/[x+y]\bigr)$ for $x>0$, $y>0$
(h) $\displaystyle \exp(x)<1-\frac{x}{2}$ for $0.