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# 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 $x \ge \frac{1}{e}$. 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?

Then teach the underlying concepts
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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
Jun 18, 2018

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 $\left({x}^{x}\right) ' = {x}^{x} \left(1 + \ln x\right)$, ${\tan}^{- 1} \left({x}^{x}\right) '$, at x = 26.938353, is

${90}^{o} .000000 \ldots$. 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]}

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