We are now defining the Natural Exponential Function, as the unique solution of initial-value problem

There are several propositions based on this definition.

First, let’s show that

is true:

,

and when ,

,

and are solutions of initial-value problem

Hence, by uniqueness theorem, .

As a consequence of (2),

Let’s prove it too:

From (2), we know that .

This shows that is a non-negative function.

But suppose there then

.

This is a contradiction since by definition, .

Therefore, there is no , i.e., .

Next, we have

The fact that

,

and when

and are both solutions of initial-value problem

It follows from above and uniqueness theorem, .

Finally,

A short proof is given below:

By (4), (2), .

The value of can be obtained by solving (1) numerically. For example,

Fig. 1 Compute

An alternative expression for is . Using this notation, propositions (2), (3), (4), (5) are expressed as

,

,

,

respectively.

Let’s conclude this post with some humor from the internet:

* 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 asks “Why are you running away?” The constant answers, “That’s a differential operator. If it acts on me, I’ll disappear.” The says “I’m , I don’t have anything to worry about,” and keeps walking. When he reaches the differential operator, he says “Hi, I’m .” *

*The differential operator responds, “Hi, I’m .”*

*Exercise*: Prove that .

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

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