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How to prove the following continued fraction of $e^{x/y}$

$${\displaystyle e^{x/y}=1+{\cfrac {2x}{2y-x+{\cfrac {x^{2}}{6y+{\cfrac {x^{2}}{10y+{\cfrac {x^{2}}{14y+{\cfrac {x^{2}}{18y+\ddots }}}}}}}}}}}$$

Since $a_i \geq b_i$ for all $i \geq 1$. By the condition of irrationality of generalized continued fraction, proving this directly proves that $e^{x/y}$ will be an irrational number!

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  • $\begingroup$ I can’t make much sense of the question. What are $a_i$ and $b_i$, and how do they relate to $x$ and $y$? Also, there clearly are values of $x$ and $y$ for which $e^{x/y}$ is rational, such as $x=\log2$, $y=1$. $\endgroup$
    – Emil Jeřábek
    Commented Feb 6, 2021 at 10:25
  • $\begingroup$ I think it's clear from context that the OP means $x,y\in\mathbb{Z}$. $\endgroup$
    – gmvh
    Commented Feb 6, 2021 at 10:44
  • $\begingroup$ I think Euler himself derived this. see e.g. sciencedirect.com/science/article/pii/S0315086014000044 $\endgroup$
    – Carl Witthoft
    Commented Feb 25, 2022 at 14:54

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I think this might be a solution.

The Continued Fraction Expansion of the hyperbolic tanh function discovered by Gauss is

$$\tanh z = \frac{z}{1 + \frac{z^2}{3 + \frac{z^2}{5 + \frac{z^2}{...}}}} \\\\$$

We also know that the hyperbolic tanh function is related to the exponential function with the following formula

$$\tanh z=\frac{e^z-e^{-z}}{e^z+e^{-z}}$$

Putting $\frac{x}{y}$ in place of $z$ in the previous equation we get $$\frac{e^{\frac{x}{y}}-e^{-{\frac{x}{y}}}}{e^{\frac{x}{y}}+e^{-{\frac{x}{y}}}}= \frac{{\frac{x}{y}}}{1 + \frac{{\frac{x}{y}}^2}{3 + \frac{{\frac{x}{y}}^2}{5 + \frac{{\frac{x}{y}}^2}{...}}}} \\\\$$

This continued fraction can be simplified into

$$\frac{e^{\frac{x}{y}}-e^{-{\frac{x}{y}}}}{e^{\frac{x}{y}}+e^{-{\frac{x}{y}}}}= \frac{x}{y + \frac{x^2}{3y + \frac{x^2}{5y + \frac{x^2}{...}}}} \\\\$$

This equation can be further be simplified as

$$1+\frac{2}{e^{\frac{2x}{y}}-1}=y + \frac{x^2}{3y + \frac{x^2}{5y + \frac{x^2}{...}}} \\\\$$

$$\frac{e^{\frac{2x}{y}}-1}{2}=\frac{1}{(\frac{y}{x}-1) + \frac{x}{3y + \frac{x^2}{5y + \frac{x^2}{...}}}}$$

From this equation after some algebraic manipulation, we finally get the continued fraction expansion of $e^{x/y}$ as

$${\displaystyle e^{x/y}=1+{\frac {2x}{2y-x+{\frac {x^{2}}{6y+{\frac {x^{2}}{10y+{\frac {x^{2}}{14y+{\frac {x^{2}}{18y+\ddots }}}}}}}}}}}$$

This is an infinite generalized continued fraction. We will now state the necessary and sufficient condition for the continued fraction proved by Lagrange given in Corollary 3, on page 495, in chapter XXXIV on "General Continued Fractions" of Chrystal's Algebra Vol.II to converge into an irrational number.

\begin{theorem}The necessary and sufficient condition that the continued fraction $$\frac{b_1}{a_1 + \frac{b_2}{a_2 + \frac{b_3}{a_3 + \dots}}}$$

is irrational is that the values $a_{i}, b_{i}$ are all positive integers, and if we have $|a_i| > |b_i|$ for all $i$ greater than some $n$ \end{theorem}

In the continued fraction of $e^{x/y}-1$ as we have derived $a_{i}, b_{i}$ are equals to $2(2i-1),x^2$ except $i=1$.Therefore we have $|a_i| > |b_i|$ for all $i>\frac{\frac{x^2}{2}+1}{2}$. Hence we have proved that $e^{x/y}-1$ is irrational which in turn means $e^{x/y}$ is irrational.

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