Question

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 )?

Answer 1

The curves intersect without osculating.

Explanation:

Considering

`y=x^(x(1+1/y)) equiv y^(y/(y+1))-x^x` we have

`f_1(x,y)= y^(y/(y+1))-x^x = 0` and
`f_2(x,y)=y-x^x=0`

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

Now the tangency at this point obeys

`((dy)/(dx))_1 = -(f_1)_x/((f_1)_y) = ( (1 + y)^2 (1 + Log_e(x)))/(1 + y + Log_e(y)) = 2`
`((dy)/(dx))_2 = -(f_2)_x/((f_2)_y) =x^x (1 + Log_e(x)) = 1`

So they intersect without osculating.