Re: tetration - inf superroot

Gottfried Helms wrote:
Let's define the n'th iterative root ("rit") via

f(x,1) = x f(x,2) = x^x f(x,3) = x^(x^x)

such that

rit(y,3) = x --> f(x,3) = y

and consider the sequence

rit(3,1) , rit(3,2), rit(3,3),..., rit(3,k),... k=1,inf

what is x in

x = lim {k->inf} rit(3,k)

The sequence decreases from 3 down to e^(1/e) + eps
but I think, it cannot fall below.
k x
---------------------
1 3.000000 rit(3,k)
2 1.825455
4 1.563628
8 1.484080
16 1.457948
32 1.449171
64 1.446164
128 1.445135 rit(3,k)
...
->inf -> ?? rit(3,inf)
================================
compare other limits
inf 1.444668 =e^(1/e)
--------------------------------
inf 1.442250 =3^(1/3)

On the other hand, it should arrive at 3^(1/3)...
Do I actually overlook something and the sequence can
indeed cross e^(1/e)?

Hallo Gottfried,

As you correctly found, the function:

f(x) = x^(1/x) is a partial inverse of the power tower.

The final value of the iteration depends on the value of x.

Because the function f(x) has a maximum at x = e, this value (i.e.,
~1.444667861) will be attained ONLY when the the power tower has as base x = e.

Any other iteration for rit(x,k) will converge to the value given by f(x). Plot
the function f(x) from 0 to 5 say, and see the graph. For x = 3, the iteration
will converge to

f(3) ~= 1.442249570

while for x = 2, the iteration will converge to

f(2) = sqrt(2) ~= 1.414213562

Again, since f(x) attains a max at x = e, any value different from e will force
the iteration to converge to something less than e^(1/e), thereby crossing
e^(1/e) from above, but settling on the corresponding value given by f(x), for
which we always have x^(1/x) <= e^(1/e).

<urrks>

Gottfried
--
Ioannis

.

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