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?

2 Answers
Cesareo R.
Aug 8, 2016

Calling #"trillion" = lambda # and substituting in the main formula
with #C = 1.02464790434503850# we have

#C = log_{lambda}(lambda+lambda/C)# so
#lambda^C = (1+1/C)lambda# and
#lambda^{C-1} = (1+1/C)#
following with simplifications
#lambda = (1+1/C)^{1/(C-1}#
finally, computing the value of #lambda# gives

#lambda=1.0000000000000*10^12#

We observe also that

#lim_{lambda->oo}log_{lambda}(lambda+lambda/C) = 1# for #C > 0#

A. S. Adikesavan
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]}

Related questions
See all questions in Applications of Exponential Functions
Impact of this question
1865 views around the world
You can reuse this answer
Creative Commons License