What is the value of #1+1/(2+1/(3+1/(4+1/(5\cdots# ?
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"Suppose that I don't have a formula for #g(x)# but I know that #g(1)
= 3# and #g'(x) = sqrt(x^2+15)# for all x. How do I use a linear approximation to estimate #g(0.9)# and #g(1.1)#?"
1 Answer
Aug 14, 2018
It does not seem to have a name, though it is known and probably transcendental.
Explanation:
Some well known transcendental numbers have regular continued fractions, e.g.:
#e = [1;1,2,1,1,4,1,1,6,1,1,8,1,...]#
In the case of the given expression:
#[1;2,3,4,5,6,...] = 1+1/(2+1/(3+1/(4+1/(5+1/(6+...)))))#
we can approximate the value by truncating early to find:
#[1;2,3,4,5,6,...] ~~ 1.433127426722311758#
This is almost certainly a transcendental number, but seems to have no great significance and no particular name. (Most real numbers are transcendental). A Google search for this value finds a couple of references, but nothing more.
