# #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?

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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?

Precisely the functions grow from

##### 1 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

and ranges for the graphs are befittingly chosen.

See the Semi-

graduations Y = 0 1 2 3 4 ..., from

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

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

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