#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?

1 Answer
Jun 17, 2018

Answer:

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]}