Two Peas in a Pod, Part 2

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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,

Screen Shot 2018-01-11 at 8.59.41 PM.png

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

  1. Pingback: Two Peas in a Pod, Part 3 | Vroom

  2. Pingback: What goes up fast, will slow down, Part One | Vroom

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