Continued fraction of exponential function

818 Views Asked by At

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.