# The Unapologetic Mathematician

## Differentiable Exponential Functions

The exponential property is actually a rather stringent condition on a differentiable function $f:\left(0,\infty\right)\rightarrow\mathbb{R}$. Let’s start by assuming that $f$ is a differentiable exponential function and see what happens.

We calculate the derivative as usual by taking the limit of the difference quotient

$\displaystyle f'(x)=\lim\limits_{\Delta x\rightarrow0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

Then the exponential property says that our derivative is

$\displaystyle f'(x)=\lim\limits_{\Delta x\rightarrow0}\frac{f(x)f(\Delta x)-f(0)f(x)}{\Delta x}=f(x)\lim\limits_{\Delta x\rightarrow0}\frac{f(\Delta x)-f(0)}{\Delta x}=f(x)f'(0)$

So we have a tight relationship between the function and its own derivative. Let’s see what happens for the exponential function $\exp$. Since it’s the functional inverse of $\ln$ we can use the chain rule to calculate

$\displaystyle\exp'(y)=\frac{1}{\ln'(\exp(y))}=\frac{1}{\frac{1}{\exp(y)}}=\exp(y)$

Showing that this function is its own derivative. That is, this is the exponential function with $f'(0)=1$.

Since a general (differentiable) exponential function $f$ is a homomorphism from the additive group of reals to the multiplicative group of positive reals, we can follow it by the natural logarithm. This gives a differentiable homomorphism from the additive reals to themselves, which must be multiplication by some constant $C_f$. That is: $\ln(f(x))=C_fx$. How can we calculate this constant? Take derivatives!

$\displaystyle\frac{d}{dx}\ln(f(x))=\frac{1}{f(x)}f'(x)=\frac{f(x)f'(0)}{f(x)}=f'(0)$

So our constant is the derivative $f'(0)$ from before. Of course we could also write

$\ln(f(x))=f'(0)x=\ln(\exp(f'(0)x))$

And since $\ln$ is invertible this tells us that $f(x)=\exp(f'(0)x)$. That is, every differentiable exponential function comes from $\exp$ by taking some constant multiple of the input.

By the usual yoga of inverse functions we can then see that every differentiable logarithmic function (an inverse to some differentiable exponential function) is a constant multiple of the natural logarithm $\ln$. That is, if $g(x)$ satisfies the logarithmic property, then $g(x)=C_g\ln(x)$

April 10, 2008 Posted by | Analysis, Calculus | 7 Comments

## The Exponential Property

We’ve defined the natural logarithm and shown that it is, in fact, a logarithm. That is, it’s a homomorphism from the multiplicative group of positive real numbers to the additive group of all real numbers. Now I assert that this function is in fact an isomorphism.

First off, the derivative of $\ln(x)$ is $\frac{1}{x}$, which is always positive for positive $x$. Thus it’s always strictly increasing. That is, if $0 then $\ln(x)<\ln(y)$. So no two distinct numbers ever have the same natural logarithm, and the function is thus injective.

Flipping this around tells us that we definitely have some nonzero values for the function. For example, we know that $0<\ln(2)$. Now, since the real numbers are an Archimedean field, no matter how big a number $y>0$ we pick, there will be some natural number $n$ so that $y, where the latter inequality follows from the logarithmic property.

That is, no matter how large a number we pick, $\ln$ takes values at least that large. But because $\ln$ is continuous on a connected interval there must be some number $x$ with $\ln(x)=y$. Similarly, if $y<0$ then there will be some $x$ with $\ln(x)=-y$, and thus $\ln(\frac{1}{x})=y$. Thus the natural logarithm is surjective.

So, since our function is one-to-one and onto, it has an inverse function. We will call this function the “exponential” (denoted $\exp$), and define it to be the unique function satisfying

$\exp( \ln(x) )=x$
$\ln(\exp(y))=y$

for all positive real $x$ and all real $y$.

From here it’s straightforward to see that $\exp$ must be the inverse homomorphism. That is, given two real numbers $y_1$ and $y_2$ we know there must be (unique!) positive real numbers $x_1$ and $x_2$ with $\ln(x_i)=y_i$. Then we calculate

$\exp(y_1+y_2)=\exp(\ln(x_1)+\ln(x_2))=\exp(\ln(x_1 x_2))=$
$x_1x_2=\exp(\ln(x_1))\exp(\ln(x_2))=\exp(y_1)\exp(y_2)$

And it’s clear from here that $\exp(0)=1$. A homomorphism from the additive reals to the multiplicative positive reals like this is said to satisfy the “exponential property”, which is just the reverse of the logarithmic property from last time.

April 10, 2008 Posted by | Analysis, Calculus | 8 Comments