I recently started studying continuous fractions and came across some decompositions whose origin is unclear to me. For example, for $e^x$: $$e^{x}=1+\frac{2 x}{2 -x+\dfrac{x^2}{6 +\dfrac{x^2}{10 +\dfrac{x^2}{14 +\dfrac{x^2}{18 +\ddots}}}}}\tag 1$$ I also know another decomposition for $e^x$ that has a clear origin using Euler’s identity and Taylor series expansion. $$e^x=\frac{x^0}{0!}+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots=1+\frac{x}{1-\dfrac{1 x}{2+x-\dfrac{2 x}{3+x-\dfrac{3 x}{4+x-\ddots}}}}$$ But I really want to figure out how the continuous fraction $(1)$ is obtained, since it is not reducible to a Taylor series expansion if you reduce it to normal form $$ 1+\dfrac{2x}{2-x+\dfrac{x^2}{6+\dfrac{x^2}{10}}} =\dfrac{x^3+12x^2+60x+120}{-x^3+12x^2-60x+120} $$ Thanks.
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1https://math.stackexchange.com/questions/30283/deriving-a-trivial-continued-fraction-for-the-exponential – Bowei Tang May 29 '25 at 10:57
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Thanks, but only that continuous fraction $e^x$ in the link makes sense to me, the whole question is the fraction (1). – Ivan_Rogers May 29 '25 at 11:10
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Isn't (1) just $e^x$? – lcv May 29 '25 at 11:27
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Yes, but (1) is another decomposition into a continuous fraction that doesn't follow from Taylor series. – Ivan_Rogers May 29 '25 at 11:37
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https://fr.wikipedia.org/wiki/Fraction_continue_et_approximation_diophantienne#Exemple_:_le_nombre_e – Anne Bauval May 29 '25 at 14:43
1 Answers
Here is one approach, based on a summary of Euler's approach on French Wikipedia provided by Anne Bauval in comments above. (Despite this origin, I will largely eschew attempting a historically-accurate description of Euler's actual proof.)
First, consider the Riccati equation $dq/dx+q(x)^2=1$ which we solve by separating variables:
\begin{align} dp &= \frac{dq}{1-q^2}=\frac{dq}{2(q+1)}-\frac{dq}{2(q-1)}\\ \implies p &=\int\frac{dq}{1-q^2}+C=\frac12 \ln\frac{q+1}{q-1}+C \end{align} (Euler in fact considered $a(dq/dp)+q(p)^2=1$, but this amounts to substituting $x=p/a$.) Hence a particular solution is $$e^{2x}=\frac{q+1}{q-1}\implies q(x)=\frac{e^{2x}-1}{e^{2x}+1}=\coth x$$ Euler then develops a continued fraction solution for $q$:
$$q=\coth x=\frac{1}{x}+\frac{1}{3/x+}\frac{1}{5/x+}=\frac{1}{x}\left(1+\frac{x^2}{3+}\frac{x^2}{5+}\frac{x^2}{7+}\right)\cdots$$
(For some details see section 2 of the article here, taking into account the sign change mentioned later at the bottom of page 5.) This readily develops to Lambert's continued fraction for $\tanh$:
$$\tanh x = \frac{1}{1/x+}\frac{1}{3/x+}\frac{1}{5/x+}\cdots=\frac{x}{1+}\frac{x^2}{3+}\frac{x^2}{5+}\cdots$$ But of more interest here is to write \begin{align} e^{2x}=\frac{q+1}{q-1} &=1+\frac{2x}{-x+xq}\\ &= 1+\frac{2x}{1-x+}\frac{x^2}{3+}\frac{x^2}{5+}\\ &= 1+\frac{4x}{2-2x+}\frac{(2x)^2}{6+}\frac{(2x)^2}{10+}\\ \end{align} where the last equality follows via multiplying each 'level' of the continued fraction by $2/2$. Replacing $x\mapsto x/2$ then yields the desired result:
$$e^{x}=1+\frac{2x}{2-x+}\frac{x^2}{6+}\frac{x^2}{10+}\cdots$$
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