# On the scaling power of logarithmic FCF: log_(cf) (x;a;b)=log_b (x+a/log_b(x+a/log_b (x+...))), b in (1, oo), x in (0, oo) and a in (0, oo). How do you prove that log_(cf) ( "trillion"; "trillion"; "trillion" )=1.204647904, nearly?

Aug 8, 2016

Calling $\text{trillion} = \lambda$ and substituting in the main formula
with $C = 1.02464790434503850$ we have

$C = {\log}_{\lambda} \left(\lambda + \frac{\lambda}{C}\right)$ so
${\lambda}^{C} = \left(1 + \frac{1}{C}\right) \lambda$ and
${\lambda}^{C - 1} = \left(1 + \frac{1}{C}\right)$
following with simplifications
lambda = (1+1/C)^{1/(C-1}
finally, computing the value of $\lambda$ gives

$\lambda = 1.0000000000000 \cdot {10}^{12}$

We observe also that

${\lim}_{\lambda \to \infty} {\log}_{\lambda} \left(\lambda + \frac{\lambda}{C}\right) = 1$ for $C > 0$

May 18, 2018

This is my continuation to the nice answer by Cesareo. Graphs for ln, choosing b = e and a = 1, might elucidate the nature of this FCF.

#### Explanation:

Graph of y = log_(cf)(x;1;e) = ln(x + 1/y):

Not bijective for x > 0.
graph{x-2.7183^y+1/y=0 [-10 10 -10 10]}

Graph of y = log_(cf)(-x;1;e) = ln(-x + 1/y):

Not bijective for x < 0.

graph{-x-2.7183^y+1/y=0 [-10 10 -10 10]}

Combined graph:

graph{(x-2.7183^y+1/y)(-x-2.7183^y+1/y)=0 [-10 10 -10 10]}

The two meet at ( 0, 0.567..). See the graph below. All graphs are

attributed to the power of Socratic graphics facility.

graph{x-2.7128^(-y)+y = 0 [-.05 .05 0.55 .59]}

The answer to the question is 1.02... and Cesareo is right.

See the graphical revelation below.

graph{x-y+1+0.03619ln(1+1/y)=0[-.1 .1 1.01 1.04]}