# How do you prove that the curves of y=x^x and y = the Functional Continued Fraction (FCF) generated by y=x^(x(1+1/y)) touch y=x, at their common point ( 1, 1 )?

Aug 27, 2016

The curves intersect without osculating.

#### Explanation:

Considering

$y = {x}^{x \left(1 + \frac{1}{y}\right)} \equiv {y}^{\frac{y}{y + 1}} - {x}^{x}$ we have

${f}_{1} \left(x , y\right) = {y}^{\frac{y}{y + 1}} - {x}^{x} = 0$ and
${f}_{2} \left(x , y\right) = y - {x}^{x} = 0$

Those functions intersect at clearly at $\left\{x = 1 , y = 1\right\}$. This can be verified by simple substitution.

Now the tangency at this point obeys

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{1} = - {\left({f}_{1}\right)}_{x} / \left({\left({f}_{1}\right)}_{y}\right) = \frac{{\left(1 + y\right)}^{2} \left(1 + L o {g}_{e} \left(x\right)\right)}{1 + y + L o {g}_{e} \left(y\right)} = 2$
${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{2} = - {\left({f}_{2}\right)}_{x} / \left({\left({f}_{2}\right)}_{y}\right) = {x}^{x} \left(1 + L o {g}_{e} \left(x\right)\right) = 1$

So they intersect without osculating.