Question

`y=x^x` and y = the Functional Continued Fraction (FCF) developed by `y=x^(x(1+1/y)` are growth functions, for `x>=1`. How do you prove that at `x = 0.26938353091863307 X10^2`, y-ratio is 1 in 100 and curve-direction is vertical, nearly, for both?

Precisely the functions grow from `x >= 1/e`. This x-value is my limit for computer tabulation of 17-sd y and y', in long precision. Is yours, from your 64-bit or ( for longer precision ) 128-bit processor, higher?

Answer 1

Apt semi-log graphs reveal that the ratio is 1, in the neighboring domain. In the question, "1 in 100# has to be changed to "1.000000.....".

Explanation:

As `(x^x)' =x^x(1+ ln x)`, `tan^(-1)(x^x)'`, at x = 26.938353, is

`90^(o).000000...`. The other graph is asymptotic. Note that

`26.938353^26.938353 = 3.40 X 10^38`, nearly. The domains

and ranges for the graphs are befittingly chosen.

See the Semi-`log_ 100` graphs for the rapid growths, with

graduations Y = 0 1 2 3 4 ..., from `Y= log_(100) y`. The y-ratio

100 should appear as Y-difference 1 (nearly). But, here, there is no

difference.

Superimposition would produce a blank graph, if both are the

same. So, I changed y to y + 0.05 for showing one over the other

Here, Y = log y. See the third graph.

Separate graphs, using `Y = log_(100) y`:
graph{100^y-x^x=0[26 28 18 21.5]}
graph{100^y-x^(x(1+100^(-y)))=0[26 28 18 21.5]}

Combined graph, using `Y = log y`:

graph{(10^y-x^x)(10^(y+0.05)-x^(x(1+10^(-y))))=0[26 28 36 43]}