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
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 are both solutions of initial-value problem
It follows from above and uniqueness theorem, .
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
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 .