Let #a=root(2016)(2016)#. Which of the following two numbers is greater #2016# or #a^(a^(a^(a^(vdots^a))))#?

3 Answers
Oct 30, 2016

Answer:

#2016#

Explanation:

If #a = root(2016)(2016)# then:

#a^(a^(a^(a^(a^(a^(a^...)))))) ~~ 1.00379575316558#

Note that both this value and #2016# are fixed points of the function:

#x -> (root(2016)(2016))^x#

but #2016# is an unstable fixed point, while #1.00379575...# is a stable one.

Oct 30, 2016

Answer:

#2016 > a^(a^(a^(a^(vdots^a))))#

Explanation:

Calling #f(x)=a^x# and #2016=m# if #a=root(m)(m)# then

#f(m)=a^m=m# so #m# is a fixed point for #f(x)# and then

#f(f(f(cdotsf(m))))=m#

Now, #a = root(m)(m)=e^((log_em)/m) > 1+log_em/m > 1#

so #f(x) = a^x# is a strict increasing function.

This is a consequence of the inequality #e^x > 1+x#.

so is true the relationship

#x > y iff f(f(f(cdotsf(x)))) > f(f(f(cdotsf(y))))#

but

#m > a => f^m(m) > f^m(a) =>m > a^(a^(a^(a^(vdots^a))))#

Finally

#2016 > a^(a^(a^(a^(vdots^a))))#

Oct 30, 2016

Incredible but true. The 17-SD value of the functional ladder-in-

exponentiation, of any order, ad infinitum,

#(((2016)^(1/2016))^((2016^(1/2016))^((2016^((1/2016)))^...)))#

#=1.0037957531655815#.

For my iterations, with starter as a in double precision, this accuracy

was reached in the 7th iterate,

The difference from 2016 is 2014.9962, nearly, for any rise in the

exponential ladder, beyond the seventh step. This difference is not

fixed. There are limitations in computation, for realizing this beyond

about 34-sd..