5.7 Properties of the exponential function

At this stage we have developed a rich theory of differentiation, with highlights including the inverse function theorem and the first derivative test. We now apply these tools to study one specific example: the exponential function $\exp\colon\mathbb{R}\to\mathbb{R}$ , introduced in Definition 4.5.

Differential properties of the exponential function

The exponential function has some very special properties and plays a fundamental role in the theory of differentiation. Recall from Lemma 5.12 that $\exp$ is differentiable and satisfies $\exp^{\prime}=\exp$ and $\exp(0)=1$ . Our first goal is to show that $\exp$ is in fact the only function with this property!

Proposition 5.55 (Uniqueness of the exponential).

Suppose $F\colon\mathbb{R}\to\mathbb{R}$ is differentiable and

(5.24) (5.24)

\left\{\begin{array}[]{@{}rclcl@{}}F^{\prime}(x)&=&F(x)&&\text{for all $x\in% \mathbb{R}$,}\\ F(0)&=&1.&&\end{array}\right.

Then $F=\exp$ .

Remark 5.56.

The conditions in (5.24) on $F$ provide a simple example of a differential equation. In particular, Proposition 5.55 determines which functions $F$ satisfy the equation (5.24) involving the derivative (it shows there is only one such function: the exponential). Differential equations are a central topic in both pure and applied mathematics, and are fundamental to mathematical modelling of physical processes (and therefore indispensable to engineering, physics and other sciences). You shall explore the fascinating world of differential equations in detail if you take the year 2 course Modelling and Computing.

To prove the proposition, we first show the following lemma.

Lemma 5.57.

The exponential function is non-vanishing: $\exp(x)\neq 0$ for all $x\in\mathbb{R}$ .

We discussed this basic property of the exponential in Chapter 4, but only justified it for $x\geq 0$ (where it follows directly from Definition 4.5). We can now use our theory of differentiation to give a satisfying and complete proof.

Proof (of Lemma 5.57).

Consider the function $G\colon\mathbb{R}\to\mathbb{R}$ given by $G(x):=\exp(x)\exp(-x)$ for all $x\in\mathbb{R}$ . By using the product and chain rules, we have

$\displaystyle G^{\prime}(x)$	$\displaystyle=\exp^{\prime}(x)\exp(-x)-\exp(x)\exp^{\prime}(-x)$
	$\displaystyle=\exp(x)\exp(-x)-\exp(x)\exp(-x)=0\qquad\text{for all $x\in% \mathbb{R}$.}$

Hence, by the first derivative test, $G$ is constant. If we fix $x\in\mathbb{R}$ , then $G(x)=G(0)=\exp(0)^{2}=1$ and so $\exp(x)\exp(-x)=1$ . This implies that $\exp(x)\neq 0$ . ∎

Exercise 5.58.

Using the intermediate value theorem, strengthen the conclusion of Lemma 5.57 by showing that $\exp(x)>0$ for all $x\in\mathbb{R}$ .

We now have all the tools we need to prove Proposition 5.55.

Proof (of Proposition 5.55).

By Lemma 5.57, the reciprocal $\exp(x)^{-1}$ is defined for all $x\in\mathbb{R}$ . Let

H\colon\mathbb{R}\to\mathbb{R},\qquad H(x):=F(x)\exp(x)^{-1}\qquad\text{for % all $x\in\mathbb{R}$,}

and note that $H$ is differentiable by the quotient rule. Moreover,

H^{\prime}(x)=\frac{F^{\prime}(x)\exp(x)-F(x)\exp^{\prime}(x)}{\exp(x)^{2}}=% \frac{F(x)\exp(x)-F(x)\exp(x)}{\exp(x)^{2}}=0\qquad\text{for all $x\in\mathbb{% R}$.}

Hence, by the first derivative test, $H$ is constant. Thus, $H(x)=H(0)=F(0)\exp(0)^{-1}=1$ and so $F(x)=\exp(x)$ for all $x\in\mathbb{R}$ . ∎

Algebraic properties of the exponential function

Surprisingly, we can use the special differential properties of the exponential function to show that is satisfies a very special algebraic property!

Theorem 5.59 (Multiplicative identity).

$\exp(x+y)=\exp(x)\exp(y)$ for all $x$ , $y\in\mathbb{R}$ .

Proof.

Fix $y\in\mathbb{R}$ and consider the function

F\colon\mathbb{R}\to\mathbb{R},\qquad F(x):=\frac{\exp(x+y)}{\exp(y)}\qquad% \text{for all $x\in\mathbb{R}$.}

One can check that $F$ is differentiable, $F^{\prime}(x)=F(x)$ for all $x\in\mathbb{R}$ and $F(0)=1$ (see Exercise 5.60; remember $y$ is fixed here and not a variable).

Thus, by Proposition 5.55, we have $F(x)=\exp(x)$ for all $x\in\mathbb{R}$ . Rearranging this identity gives the desired result. ∎

Exercise 5.60.

Fill in the details of the proof of Theorem 5.59 by showing $F$ is differentiable, $F^{\prime}(x)=F(x)$ for all $x\in\mathbb{R}$ and $F(0)=1$ .

Irrational exponentiation

Going right back to the beginning of the course, in 0.5 we asked how to make sense of irrational exponents, such as $10^{\sqrt{2}}$ or $2^{\pi}$ . The multiplicative identity for the exponential function is key to understanding this problem. Before we get to that, however, we first explore some more direct consequence of Theorem 5.59.

Exercise 5.61.

Verify that $\exp\colon\mathbb{R}\to\mathbb{R}$ satisfies the following properties:

(i)

$\exp(-x)=\exp(x)^{-1}$ for all $x\in\mathbb{R}$ ;
(ii)

$\exp(kx)=\exp(x)^{k}$ for all $x\in\mathbb{R}$ and $k\in\mathbb{N}$ ;
(iii)

$\exp(x/k)=\exp(x)^{1/k}$ for all $x\in\mathbb{R}$ and $k\in\mathbb{N}$ ;
(iv)

$\exp(qx)=\exp(x)^{q}$ for all $x\in\mathbb{R}$ and $q\in\mathbb{Q}$ ;
(v)

In particular, $\exp(q)=\exp(1)^{q}$ for all $q\in\mathbb{Q}$ .

Hint: for (i), recall the proof of Lemma 5.57.

Exercise 5.61 (v) suggests the number $\exp(1)$ plays an important role for the exponential function.

Definition 5.62 (Euler’s constant).

We define $e:=\exp(1)\in\mathbb{R}$ .

From Exercise 5.61 (v), we see that $\exp(q)=e^{q}$ for all $q\in\mathbb{Q}$ . However, $\exp(x)$ is defined for all values of $x\in\mathbb{R}$ (both rational and irrational), so we can use these observations to introduce a notion of irrational exponentiation.

Definition 5.63.

We define $e^{x}:=\exp(x)\in\mathbb{R}$ for $x\in\mathbb{R}$ .

To reiterate: the idea behind Definition 5.63 is that $\exp(x)$ agrees with our usual algebraic notion of an exponential $e^{x}$ whenever $x\in\mathbb{Q}$ (as expressed by the properties in Exercise 5.61). However, we can use the function $\exp(x)$ to extend the definition of $e^{x}$ to irrational powers. This idea is intuitive if we plot the values $e^{q}$ for $q\in\mathbb{Q}$ , as in Figure 5.14. We see that the values $e^{q}$ for $q\in\mathbb{Q}$ all lie along the graph of $\exp$ , but they leave lots of ‘holes’ at irrational values of $x$ . By defining $e^{x}:=\exp(x)$ , we fill in the holes.

Figure 5.14: The graph of the exponential function

\exp

together with a scatter plot of

e^{q}

for a finite sample of rational values

q\in\mathbb{Q}

So far we have only defined irrational exponentiation for the special base $e$ . Our next goal is to extend this to all bases: that is, to define $a^{x}$ for all $a>0$ and all $x\in\mathbb{R}$ .

Let $a>0$ and $q\in\mathbb{Q}$ . Using the result of Exercise 5.61 (iv), we see that

e^{q\log a}=\exp(q\log a)=\left(\exp(\log a)\right)^{q}=a^{q},

where we have used the fact that $\log$ is (by definition) the inverse of $\exp$ . This shows us that the function $x\mapsto e^{x\log a}$ agrees with the familiar notion of exponentiation $a^{x}$ whenever $x\in\mathbb{Q}$ . However, the function $x\mapsto e^{x\log a}$ is defined for all values of $x\in\mathbb{R}$ (both rational and irrational), so we can use it to extend the definition of exponentiation to irrational powers.

Definition 5.64 (Real exponentiation).

Given $a>0$ and $x\in\mathbb{R}$ , define $a^{x}:=e^{x\log a}$ .

This finally answers 0.5 of how to define irrational powers!

We can also use this definition to show that the formula for differentiating rational powers $x^{1/n}$ (established in Exercise 5.32) can be extended to all real powers, $x^{p}$ .

Exercise 5.65.

Fix $p\in\mathbb{R}$ . Define $f:(0,\infty)\to\mathbb{R}$ by $f(x)=x^{p}:=e^{p\log x}$ . Prove that $f$ is differentiable on $(0,\infty)$ and that $f^{\prime}(x)=px^{p-1}$ .