# Continued fraction of exponential function

There is this $$\exp(x)=1+\cfrac{x}{1-\cfrac{x/2}{1+x/2-\cfrac{x/3}{1+x/3-\cfrac{x/4}{1+x/4-\dots}}}}$$ and there is $$\exp(1)=1+\cfrac{1}{0+\cfrac{1}{1+\cfrac{1}{1+\cfrac{1}{2+\cfrac{1}{1+\cfrac{1}{\cdots}}}}}}$$

Is there a way to rewrite the second equation, replacing some numbers with expressions $x$, so that we get something like the first equation, valid in a neighborhood of $1$, and such that, plugging in $x$, recovers the second equation?

I tried something like $$1+x+x^2/2+\cdots=1+\cfrac{x}{0+\cfrac{1}{?}}$$ as well as a few other guesses, but found it difficult.

• This - en.wikipedia.org/wiki/… - will help. You can get this by the Euler's formula. – rookie Jun 16 '17 at 10:29
• No. I know how to derive the first equation. I wanted to extend the second equation. – MaudPieTheRocktorate Jun 16 '17 at 10:37
• @md2perpe Yes! The first and second equation seem to be very different. I want to find a third equation, that looks kind of like the first equation, with $exp(x)=$ some continued fraction, such that when plugging $x=1$, gets the second equation. – MaudPieTheRocktorate Jun 16 '17 at 18:01
• The coefficients of $x^{-1}$ are $1,3,5,\ldots$: $$e^x= 1 +\cfrac{1}{x^{-1}-1 +\cfrac{1}{1 +\cfrac{1}{1 +\cfrac{1}{3x^{-1}-1 +\cfrac{1}{1 +\cfrac{1}{\cdots}}}}}}$$ I think this answers the question? I'm not sure it converges just above $1$ though – GPhys Jun 19 '17 at 13:17
• You can guess it from the continued fractions of $\exp(1/n)$. As I recall it can be proved as a special case of a for the continued fraction of a quotient of Bessel functions along the lines of dlmf.nist.gov/10.33 – Noam D. Elkies Jun 24 '17 at 3:55