Appendix B. The Natural Exponential

Let's look at exponential functions in greater detail so that we can explain the differential law for the natural exponential.

Theorem.

A function \( z \mathrel{\overset{f}{\leftarrow}} x \) is exponential if it satisfies the rules:

\( f \of 0 = 1 \).
\( f (m + n) = f \of m \mult f \of n \) for any numbers \( m : \Number \) and \( n : \Number \).
The function has positive output, \( z \gt 0 \), only.

Why?

Let's start by defining the constant \( b \) as the output corresponding to the input \( x = 1 \).

\[ \begin{align} b &= f \of 1 \end{align} \]

We'd like to show that our function takes the form

\[ \begin{align} z &= b^x. \end{align} \]

Let's check a few \( x \)-values.

\[ \begin{align} f \of 0 &= 1 = b^0 \\ f \of 1 &= b = b^1 \\ f \of 2 &= f (1 + 1) = (f \of 1) \mult (f \of 1) = b^2 \\ f \of 3 &= f (2 + 1) = (f \of 2) \mult (f \of 1) = (f \of 1) \mult (f \of 1) \mult (f \of 1) = b^3 \end{align} \]

Sure enough, using the assumed rules, we're finding

\[ \begin{align} f \of x &= b^x. \end{align} \]

You may wish to check that the formula holds for other inputs, say \( x = -1 \), or \( x = 1 / 2 \), or \( x = 1 / 3 \).

We can use the previous theorem to give a new formula for the natural exponential.

Formula.

We may write the natural exponential function as the sum of infinitely-many terms

\[ \begin{align} \exp x &= 1 + \frac{1}{1!} \mult x + \frac{1}{2!} \mult x^2 + \frac{1}{3!} \mult x^3 + \cdots \end{align} \]

where \( j! \) is \( j \) factorial.

\[ \begin{align} j! &= 1 \mult 2 \mult 3 \cdots j \end{align} \]

Why?

Let's define a new function \( f \) which is equal to the right hand side of the above equation.

\[ \begin{align} z &\mathrel{\overset{f}{\leftarrow}} x \\ z &= 1 + \frac{1}{1!} \mult x + \frac{1}{2!} \mult x^2 + \frac{1}{3!} \mult x^3 + \cdots \end{align} \]

We wish to show that \( f \) is equal to the natural exponential. To do this, we'll apply our theorem. Let's check the three assumptions.

We compute.

\[ \begin{align} f \of 0 &= 1 + \frac{1}{1!} \mult 0 + \frac{1}{2!} \mult 0^2 + \frac{1}{3!} \mult 0^3 + \cdots \\ &= 1 \end{align} \]
Let's check that \( f \) takes a sum of inputs to a product of outputs.

\[ \begin{align} f (m + n) &= f \of m \mult f \of n \end{align} \]

We'll begin by looking at the left hand side.

\[ \begin{align} f (m + n) &= 1 + \frac{(m + n)}{1} + \frac{(m + n)^2}{2} + \frac{(m + n)^3}{6} + \cdots \\ &= 1 + m + n + \frac{m^2}{2} + m n + \frac{n^2}{2} + \frac{m^3}{6} + \frac{m^2 n}{2} + \frac{m n^2}{2} + \frac{n^3}{6} + \cdots \end{align} \]

Now let's look at the right hand side.

\[ \begin{align} f \of m \mult f \of n &= \biggl(1 + \frac{m}{1} + \frac{m^2}{2} + \frac{m^3}{6} + \cdots\biggr) \mult \biggl(1 + \frac{n}{1} + \frac{n^2}{2} + \frac{n^3}{6} + \cdots\biggr) \end{align} \]

When we multiply these out, we find the terms:

\[ \begin{align} &1 \mult 1 & &1 \mult \frac{n}{1} & &1 \mult \frac{n^2}{2} & &1 \mult \frac{n^3}{6} & &\cdots \\ &\frac{m}{1} \mult 1 & &\frac{m}{1} \mult \frac{n}{1} & &\frac{m}{1} \mult \frac{n^2}{2} & &\ddots & & \\ &\frac{m^2}{2} \mult 1 & &\frac{m^2}{2} \mult \frac{n}{1} & &\ddots & & & & \\ &\frac{m^3}{6} \mult 1 & &\ddots & & & & & & \\ &\vdots & & & & & & & & \end{align} \]

Sure enough, these match the terms for the left hand side!
Let's check that our function \( f \) has positive output for any input. If the input is positive, \( x \gt 0 \), then the output is positive because it is the sum of positive terms.

\[ \begin{align} f \of x &= 1 + \frac{x}{1} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \end{align} \]

If the input is negative, it is not at all obvious from the definition that the output will be positive. Instead, notice that the calculation

\[ \begin{align} f \of x \mult f (- x) &= f (x - x) = f \of 0 = 1 \end{align} \]

shows that \( f \of x \) and \( f (- x) \) are reciprocals. Since our output is positive for \( x \gt 0 \), it must also be positive for \( x \lt 0 \).

We've shown that our function \( f \) meets all three of the assumptions for the theorem, so we may conclude that \( f \) is exponential. We find the base by evaluating our function at one.

\[ \begin{align} f \of 1 &= 1 + \frac{1}{1!} + \frac{1}{2!} + \frac{1}{3!} + \cdots \\ &= \naturalbase \end{align} \]

We've found the natural base!

We can now calculate the differential law for the natural exponential.

Law.

The differential law for the natural exponential function states

\[ \begin{align} \diff (\exp u) &= \exp u \mult \diff u \end{align} \]

where \( u : \Number \) is a number.

Why?

Let's calculate the \( u \)-derivative.

\[ \begin{align} \overset{u}{\diff} (\exp u) &= \overset{u}{\diff} \biggl( 1 + \frac{1}{1!} \mult u + \frac{1}{2!} \mult u^2 + \frac{1}{3!} \mult u^3 + \frac{1}{4!} \mult u^4 + \cdots \biggr) \\ &= \overset{u}{\diff} (1) + \overset{u}{\diff} \biggl(\frac{1}{1!} \mult u\biggr) + \overset{u}{\diff} \biggl(\frac{1}{2!} \mult u^2\biggr) + \overset{u}{\diff} \biggl(\frac{1}{3!} \mult u^3\biggr) + \overset{u}{\diff} \biggl(\frac{1}{4!} \mult u^4\biggr) + \cdots \\ &= \frac{1}{1!} \mult \overset{u}{\diff} (u) + \frac{1}{2!} \mult \overset{u}{\diff} \bigl(u^2\bigr) + \frac{1}{3!} \mult \overset{u}{\diff} \bigl(u^3\bigr) + \frac{1}{4!} \mult \overset{u}{\diff} \bigl(u^4\bigr) + \cdots \\ &= 1 + \frac{1}{2!} \mult 2 u + \frac{1}{3!} \mult 3 u^2 + \frac{1}{4!} \mult 4 u^3 + \cdots \\ &= 1 + \frac{1}{1!} \mult u + \frac{1}{2!} \mult u^2 + \frac{1}{3!} \mult u^3 + \cdots \\ &= \exp u \end{align} \]

The differential law follows.