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#

2 Answers
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 #425/27#, a finite continued fraction, you could write

#15 + 20/27#,

or by multiplying through by #(1/20)/(1/20)#, and then #(1/7)/(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

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

you could write

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

Jun 29, 2017

Answer:

See below.

Explanation:

Calling #f(x) = 1/(x+1/((x+1)+1/((x+2)+1/((x+3)+ddots# we can build the recurrence equation

#f(x) = 1/(x + f(x+1))# which gives a nonlinear functional equation as

#f(x)(f(x+1)+x)-1 = 0#