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Section 8.3 Exponential and Logarithmic Functions

We now construct the exponential function directly from its power series. Uniform convergence on bounded intervals gives a genuine function, and the algebra of absolutely convergent series yields the familiar laws of exponents. The logarithm will then appear naturally as the inverse function.

Subsection 8.3.1

Proof.
Fix \(R \gt 0\text{.}\) If \(|x| \le R\text{,}\) then
\begin{equation*} \left|\frac{x^n}{n!}\right| \le \frac{R^n}{n!}. \end{equation*}
The numerical series \(\sum_{n=0}^\infty R^n/n!\) converges by the ratio test, since
\begin{equation*} \frac{R^{n+1}/(n+1)!}{R^n/n!}=\frac{R}{n+1}\to 0. \end{equation*}
Therefore the Weierstrass M-test (Theorem 8.1.6) applies and gives uniform convergence on \([-R,R]\text{.}\)
We write
\begin{equation*} e^x := E(x)=\sum_{n=0}^\infty \frac{x^n}{n!} \qquad (x \in \R), \end{equation*}
and in particular \(e:=e^1\text{.}\)

Proof.
For fixed \(x,y \in \R\text{,}\) the numerical series \(\sum x^n/n!\) and \(\sum y^n/n!\) converge absolutely. Therefore Mertens’ theorem (Theorem 3.3.1) applies to their product:
\begin{equation*} e^x e^y = \left(\sum_{m=0}^\infty \frac{x^m}{m!}\right) \left(\sum_{n=0}^\infty \frac{y^n}{n!}\right) = \sum_{k=0}^\infty \sum_{j=0}^k \frac{x^j}{j!}\frac{y^{k-j}}{(k-j)!}. \end{equation*}
By the binomial theorem,
\begin{equation*} \sum_{j=0}^k \frac{x^j}{j!}\frac{y^{k-j}}{(k-j)!} = \frac{1}{k!}\sum_{j=0}^k \binom{k}{j}x^j y^{k-j} = \frac{(x+y)^k}{k!}. \end{equation*}
Substituting this into the previous display gives
\begin{equation*} e^x e^y = \sum_{k=0}^\infty \frac{(x+y)^k}{k!} = e^{x+y}. \end{equation*}

Proof.
Fix \(a \in \R\text{.}\) For every \(x \in \R\text{,}\) the addition law gives
\begin{equation*} e^x=e^a e^{x-a} = e^a \sum_{n=0}^\infty \frac{(x-a)^n}{n!}. \end{equation*}
Thus near the point \(a\text{,}\) the function \(e^x\) is represented by a power series in \(x-a\text{.}\) Since \(a\) was arbitrary, \(e^x\) is analytic on \(\R\text{.}\)

Proof.
The power series for \(e^x\) has radius of convergence \(\infty\text{,}\) so by Theorem 6.2.2 it may be differentiated term by term:
\begin{equation*} \frac{d}{dx} e^x = \sum_{n=1}^\infty \frac{n x^{n-1}}{n!} = \sum_{n=1}^\infty \frac{x^{n-1}}{(n-1)!} = \sum_{m=0}^\infty \frac{x^m}{m!} = e^x. \end{equation*}

Proof.
The identity \(e^0=1\) is immediate from the defining series. The addition law with \(y=-x\) gives
\begin{equation*} e^x e^{-x}=e^0=1, \end{equation*}
so \(e^{-x}=1/e^x\) and, in particular, \(e^x \neq 0\) for every \(x\text{.}\)
Because \(e^x\) is continuous and \(e^0=1 \gt 0\text{,}\) the Intermediate Value Theorem implies that \(e^x\) cannot be negative: if it were, continuity would force a zero somewhere in between. Therefore \(e^x \gt 0\) for every \(x\text{.}\)
Finally, Proposition 8.3.4 shows that \((e^x)'=e^x \gt 0\) everywhere. By Proposition 5.4.2, \(e^x\) is strictly increasing on \(\R\text{.}\)

Proof.
We already know from Proposition 8.3.5 that \(e^x \gt 0\) for all \(x\) and that \(e^x\) is strictly increasing, so it is injective.
Also,
\begin{equation*} e=e^1=1+1+\frac{1}{2!}+\frac{1}{3!}+\cdots \gt 2. \end{equation*}
Therefore
\begin{equation*} e^n \ge 2^n \to \infty \qquad \text{as } n \to \infty, \end{equation*}
and hence \(e^x\) is unbounded above. Likewise,
\begin{equation*} e^{-n}=\frac{1}{e^n} \to 0, \end{equation*}
so the values of \(e^x\) come arbitrarily close to \(0\text{.}\)
Since \(e^x\) is continuous and strictly increasing, its image is an interval. The previous paragraph shows that this interval contains arbitrarily large positive numbers and also positive numbers arbitrarily close to \(0\text{.}\) Therefore the image is exactly \((0,\infty)\text{.}\)
Because \(e^x\) is a continuous bijection from the interval \(\R\) onto the interval \((0,\infty)\text{,}\) it has a continuous inverse by Proposition 4.4.14. We call this inverse the natural logarithm and write
\begin{equation*} \ln \colon (0,\infty) \to \R. \end{equation*}
Since \(\ln\) is the inverse of a continuous strictly increasing function, it is itself continuous and strictly increasing on \((0,\infty)\text{.}\)

Proof.
Since \(e^0=1\) and \(\ln\) is the inverse of \(e^x\text{,}\) we have \(\ln 1=0\text{.}\)
Let \(u=\ln x\) and \(v=\ln y\text{.}\) Then \(x=e^u\) and \(y=e^v\text{.}\) Hence
\begin{equation*} \begin{aligned} xy &= e^u e^v\\ &= e^{u+v}. \end{aligned} \end{equation*}
and applying \(\ln\) to both sides gives
\begin{equation*} \ln(xy)=u+v=\ln x+\ln y. \end{equation*}
Taking \(y=1/x\) in the previous identity yields
\begin{equation*} 0=\ln 1=\ln x+\ln(1/x), \end{equation*}
so \(\ln(1/x)=-\ln x\text{.}\) The quotient rule for logarithms follows by combining the product rule with the reciprocal rule:
\begin{equation*} \ln(x/y)=\ln x+\ln(1/y)=\ln x-\ln y. \end{equation*}

Proof.
Apply Proposition 5.1.9 to the continuous bijection \(e^x \colon \R \to (0,\infty)\text{.}\) Since \((e^x)'=e^x \neq 0\) for every \(x\text{,}\) the inverse function \(\ln\) is differentiable, and
\begin{equation*} (\ln)'(y)=\frac{1}{(e^x)'\big|_{x=\ln y}} = \frac{1}{e^{\ln y}} = \frac{1}{y} \end{equation*}
for every \(y \gt 0\text{.}\)

Proof.
Let
\begin{equation*} F(x)=\int_1^x \frac{dt}{t} \qquad (x \gt 0). \end{equation*}
By the Fundamental Theorem of Calculus, \(F'(x)=1/x\text{.}\) By Proposition 8.3.8, the same is true for \(\ln x\text{.}\) Therefore
\begin{equation*} \frac{d}{dx}\bigl(F(x)-\ln x\bigr)=0 \end{equation*}
on \((0,\infty)\text{.}\) By Corollary 5.4.5, the difference \(F-\ln\) is constant. Evaluating at \(x=1\) gives
\begin{equation*} F(1)-\ln 1 = 0-0 = 0, \end{equation*}
so the constant is \(0\text{.}\)
Finally, if \(a \gt 0\text{,}\) one may define
\begin{equation*} a^x := e^{x\ln a} \qquad (x \in \R). \end{equation*}
The addition law for the exponential function then gives the usual rules for general real powers as well.
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