Noncircular construction of $e$ and $\ln$ for the real line

Question

Could anyone direct me to (or possibly detail) a construction of $e$ and $\ln$ along the reals?

For example, they can define $e=\lim_{n\rightarrow\infty}(1+\frac{1}{n})^n$ but from this definition how do they prove:

• It converges!
• $\frac{d}{dx}e^x=e^x$!
• etc.!

Then if we know $e^x$ injective from $\mathbb{R} \rightarrow\mathbb{R^+}$, we can call $\ln(x)$ the inverse of it. If we can prove $\ln$ is differentiable on its domain, then we can say: $$1=\frac{d}{dx}x=\frac{d}{dx}e^{\ln(x)}=e^{\ln(x)}\cdot\ln(x)'=x\cdot\ln(x)\Rightarrow\ln(x)'=x^{-1}$$ but this all depends on the above.

Answer 1

There are many possible paths through the exposition graph.

• select a definition one or both of $e^x$ and $\log x$ for real $x$

• select a definition of $e$

• prove all standard compatibilities between these definitions (such as $e = \exp(1) = \log^{-1}(1) = \lim (1+\frac{1}{x})^x = \sum \frac{1}{n!}$ ), equivalence to other standard definitions, and basic properties of the functions such as functional equations and power series developments.

Almost every book on real analysis, or on "rigorous calculus" (Spivak, Apostol or similar), and many ordinary calculus books that do not claim complete rigor but achieve it in this part of the exposition, will choose at least one path through the maze that accomplishes what you want.

Answer 2

Lang does this in A First Course in Calculus, based on $\log x = \int_1^x \frac{1}{t} \, dt$.

Answer 3

We can define $ln(x)=\int_{1}^{x}\frac{dt}{t}$ and define $e^x$ as its inverse.

However, even the apprach you suggested works. Showing that the above sequence converges is not very hard. You can show that the sequence is increasing.

Also, you can show that the "power series" for $e^x$ converges absolutely.

Answer 4

Spivak's Calculus has a beautiful presentation of these ideas, starting from the integral definition of the logarithm.

Answer 5

Let $\exp(x) = \sum_{k=0}^\infty {x^n \over n!}$. Show that the series has an infinite radius of convergence, hence well defined everywhere. Furthermore, $\exp$ is smooth, $\exp' = \exp$, $\exp(0) = 1$ and $\lim_{x \to \infty} \exp(x) = \infty$.

Show $\exp(a)\exp(b) = \exp(a+b)$, from which we get $\exp(-x) = {1 \over \exp(x)}$ and $\exp(x) >0$ for all $x$. Hence $\exp$ is strictly increasing, $\lim_{x \to \infty} \exp(-x) = 0$ and so $\exp:\mathbb{R} \to (0,\infty)$ is a bijection, and we let $\ln = \exp^{-1}$ (the $\ ^{-1}$ meaning the function inverse). The inverse function theorem shows that $\ln$ is differentiable, and $\ln'(x) = { 1 \over x}$, which is smooth, and hence $\ln x = \int_1^x {dt \over t}$.

Answer 6

Bartle and Sherbert's Introduction to Real Analysis 4th edition does precisely this on pp253-259. Specifically they show that there is a unique function $E:\mathbb{R}\rightarrow\mathbb{R}$ where:

$\text{(i)}\quad E'(x)=E(x)\ \ \forall x\in\mathbb{R}$

$\text{(ii)}\quad E(0)=1$

then show the sequence of functions: $E_1(x)=1+x,\ E_{n+1}(x)=1+\int_{0}^{x}E_n(t)dt$ converges uniformly to $E(x)$ on an interval domain of arbitrary length, and derive $\ln$ as the inverse of $E$.