# Two Peas in a Pod, Part 2 We are now defining the Natural Exponential Function, $exp(x)$ as the unique solution of initial-value problem $\begin{cases} {d \over dx} f(x) = f(x), \\ f(0)=1\end{cases}\quad\quad\quad(1)$

There are several propositions based on this definition.

First, let’s show that $exp(x)exp(y)=exp(x+y)\quad\quad\quad(2)$

is true: ${d \over dx}(exp(x)exp(y)) = exp(y){d \over dx}exp(x)=exp(x)exp(y)$, ${d \over dx} exp(x+y) = exp(x+y)\cdot 1 = exp(x+y)$

and when $x=0$, $exp(x)exp(y)=exp(0)exp(y)=1\cdot exp(y) = exp(y)$, $exp(x+y)=exp(0+y)=exp(y)$ $\implies exp(x)exp(y)$ and $exp(x+y)$  are solutions of initial-value problem $\begin{cases} {d \over dx} f(x) = f(x), \\ f(0)=exp(y)\end{cases}$

Hence, by uniqueness theorem, $exp(x)exp(y) \equiv exp(x+y)$.

As a consequence of (2), $\forall x, exp(x) > 0\quad\quad\quad(3)$

Let’s prove it too:

From (2), we know that $exp(x) = exp({x \over 2} + {x \over 2}) = (exp({x \over 2}))^2 \geq 0$.

This shows that $exp(x)$ is a non-negative function.

But suppose there $\exists x_* \ni exp(x_*) =0$ then $\forall x, exp(x)=exp(x+x_* - x_*)=exp(x_*)exp(x-x_*)=0\cdot exp(x-x_*) =0$ $\implies \forall x, exp(x) = 0$.

This is a contradiction since by definition, $exp(0)=1 \neq 0$.

Therefore, there is no $x_* \ni exp(x_*) = 0$, i.e., $\forall x, exp(x) > 0$.

Next, we have $exp(-x)= {1 \over exp(x)}\quad\quad\quad(4)$

The fact that ${d \over dx} exp(-x) = exp(-x)\cdot -1 = -exp(-x)$, ${d \over dx} ({1 \over exp(x)}) ={-1\cdot exp(x) \over (exp(x))^2}= -{1 \over exp(x)}$

and when $x=0, exp(-x) = exp(0)=1, {1 \over exp(x)} = {1 \over exp(0)} = 1$ $\implies exp(-x)$ and ${1 \over exp(x)}$ are both solutions of initial-value problem $\begin{cases} {d \over dx} f(x) = -f(x), \\ f(0)=1\end{cases}$

It follows from above and uniqueness theorem, $ex(-x) \equiv {1 \over exp(x)}$.

Finally, ${exp(x) \over exp(y) }= exp(x-y)\quad\quad\quad(5)$

A short proof is given below:

By (4), (2), ${exp(x) \over exp(y)} = exp(x)\cdot {1 \over exp(y)} = exp(x) \cdot exp(-y) = exp(x-y)$.

The value of $exp(x)$ can be obtained by solving (1) numerically. For example, Fig. 1 Compute $exp(1)$

An alternative expression for $exp(x)$ is $e^x$. Using this notation, propositions (2), (3), (4), (5) are expressed as $e^x e^y=e^{x+y}$, $\forall x,e^x>0$, $e^{-x}={1 \over e^x}$, ${e^x \over e^y} =e^{x-y}$

respectively.

Let’s conclude this post with some humor from the internet: $e^x$ and a constant are walking down the street together when the constant sees a  differential operator coming their way. He starts to run away, and $e^x$ asks “Why are you running away?” The constant answers, “That’s a differential operator. If it acts on me, I’ll disappear.” The $e^x$ says “I’m $e^x$, I don’t have anything to worry about,” and keeps walking. When he reaches the differential operator, he says “Hi, I’m $e^x$.”

The differential operator responds, “Hi, I’m ${d \over dy}$.”

Exercise:  Prove that $e^{n\cdot x} = (e^x)^n, n \in N$.

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## 2 thoughts on “Two Peas in a Pod, Part 2”

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