In Euler's paper "De Fractionibus Continuis Dissertatio" (English Translation) he proves that the constant $e\approx 2.71828$ is irrational.1 One step in the proof threw me for a loop, though. In the middle of paragraph 29, Euler solves the ODE

$$ a\,\mathrm{d}q+q^2\,\mathrm{d}p=\mathrm{d}p $$

by rearranging and integrating to get

$$\begin{align} \frac{a}{1-q^2}\mathrm{d}q&=\mathrm{d}p\\ \frac{a}{2} \log\left(\frac{1+q}{1-q}\right)&=p+C. \end{align}$$

So far, so good. But now Euler says (quoting from Wyman's translation) "the constant ought to be determined from this equation by setting $q=\infty$ when $p=0$. Wherefore there follows \begin{equation} \frac a2\log\left(\frac{q+1}{q-1}\right)=p." \end{equation}

This is the part I feel icky about.

My current interpretation of this is that when we choose the initial condition $q=\infty$ and $p=0$, we get $C=\frac a2\log(-1)$, so that

$$\begin{align} \frac{a}{2} \log\left(\frac{1+q}{1-q}\right)&=p+\frac a2\log(-1)\\ \frac{a}{2} \left(\log\left(\frac{1+q}{1-q}\right)+\log(-1)\right)&=p\\ \frac{a}{2} \log\left(\frac{1+q}{1-q}(-1)\right)&=p\\ \frac{a}{2} \log\left(\frac{q+1}{q-1}\right)&=p. \end{align}$$ This interpretation leaves more than a little to be desired. For one thing, $\log(-1)$ doesn't even make sense in a purely real context, but if we decide to work in $\mathbb{C}$ then we can't just freely use the sum-to-product rule \begin{equation} \log(zw)=\log(z)+\log(w). \end{equation}

How should I understand this step in Euler's proof?

1) Ed Sandifer claims that this is the first rigorous proof that $e$ is irrational (Link. Note that the linked PDF didn't display correctly in the two browsers I tried, but was fine when I downloaded it.)

  • $\begingroup$ I think you wanted $q-1$ instead of $1-q$; otherwise you are right that there is a problem. Which one of the two is correct depends on where the initial value for $q$ lies on the number line: $(-\infty,-1),(-1,1)$ or $(1,\infty)$. $\endgroup$ – Ian Jan 23 '18 at 1:47
  • 1
    $\begingroup$ When you solve the DE, you do partial fractions and then integrate the two pieces on the left. Properly, there should be absolute value bars in both logs, and hence in the single log when you combine them. I looked at the Latin version of the paper, and the notation is not clear. Perhaps the absolute value is implicit in Euler's notation(?) $\endgroup$ – B. Goddard Jan 23 '18 at 12:32

$$p = \frac{a}{2} \log \left(\frac{q+1}{q-1}\right)$$ is certainly a solution of the differential equation with $p \to 0$ as $q \to +\infty$, as can be verified directly.

  • $\begingroup$ +1 for thinking to solve the differential equation by differentiating Euler's answer. I'm kicking myself for not thinking of it. I'll leave the question open for the moment, though, in case someone wants to try to expand upon B. Goddard's comment. $\endgroup$ – A. Howells Jan 23 '18 at 14:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.