# FCS y = exp_(fcs)(x; 1) = e^(x+e^(x+e^(x+... ))). How do you prove that, as y to oo, y + x to C < oo?

Jul 2, 2018

See graphs and explanation.

#### Explanation:

The FCF generator is $y = {e}^{x + y}$.

Inversely,

$x = \ln y - y$.

Easily, $\frac{\mathrm{dx}}{\mathrm{dy}}$ = 1/y - 1 = 0, at y = 1.

As $y \to \infty , \frac{\mathrm{dx}}{\mathrm{dy}} = - 1 = \frac{\mathrm{dy}}{\mathrm{dx}}$.

( -1 1 ) is the turning point.

See this plot, where the asymptote x + y = 0 cuts the FCS, near O.

So, the graph turns towards this direction of x+ y = 0, to align with

all such linear and non-linear curves.

So do the parallel lines x + y = C,

in the direction polar $\theta = \frac{3 \pi}{4}$.

All are asymptotic, to one another.

Graph of FCS y = exp_(fcs)( x; 1 ) = e^(x+y), near origin:
graph{(x-ln y + y)(x+y -3)(x+y+3)(x+y -4)(x+y+4)(x+y)((x+1)^2+(y-1)^2-0.01) = 0}

Same graph far away from origin, for march to $\infty$:
graph{(x-ln y + y)(x+y -3)(x+y+3)(x+y -4)(x+y+4)(x+y) = 0[-8000 -5000 5000 6500]}

Observe scaling, in contracting large rectangle to screen

dimensions..