# Is there any simpler way of writing 1/(x+1/((x+1)+1/((x+2)+1/((x+3)+ddots, where x=RR?

## For example, 1/(1+1/(2+1/(3+1/(4+ddots

Jun 29, 2017

In terms of notation, sure. If you wanted to denote an infinite continued fraction, i.e.

x + 1/(x+1 + 1/(x+2 + 1/(x + 3 + 1/(x + 4 + ddots))),

you would write

[x; x + 1, x + 2, x + 3, . . . ]

For example, if you represent $\frac{425}{27}$, a finite continued fraction, you could write

$15 + \frac{20}{27}$,

or by multiplying through by $\frac{\frac{1}{20}}{\frac{1}{20}}$, and then $\frac{\frac{1}{7}}{\frac{1}{7}}$ on the inside, etc., until all nested numerators are $1$:

(425)/27 = color(red)(15) + 1/(color(red)(1) + 1/(color(red)(2) + 1/(color(red)(1) + 1/color(red)(6))),

or

[15; 1, 2, 1, 6]

So, for your example, to write

$\frac{1}{1 + \frac{1}{2 + \frac{1}{3 + \frac{1}{4 + \ddots}}}}$,

you could write

[0; 1, 2, 3, 4, . . . ]

Jun 29, 2017

Calling f(x) = 1/(x+1/((x+1)+1/((x+2)+1/((x+3)+ddots we can build the recurrence equation
$f \left(x\right) = \frac{1}{x + f \left(x + 1\right)}$ which gives a nonlinear functional equation as
$f \left(x\right) \left(f \left(x + 1\right) + x\right) - 1 = 0$