Re: tetration - inf superroot

Gottfried Helms wrote:
[snip]

Hi Ionannis -

nice to read from you!

But I can't follow one argument:

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

But this would mean, that the base

x_k = 1.44466... (=e^(1/e) = ί)

must be crossed while "k is not yet infinite", sloppily said, if we assume,

No. That's where the mistake lies. The function f(x) = x^(1/x) is a partial
inverse of the *infinite powertower* x^x^....

Hence, you won't ever get the result you are looking for, for any _finite_ k :-)

You can get only approximations, when k is finite.

Try it the other way around: Choose a value x in [(1/e)^e, e^(1/e)].

Calculate y = x^(1/x). Then,

lim_{n->oo} y^y^... = x.

For any finite iteration, you won't be able to reach x.


that x_1 = 3,
x_2 = 1.825
x_3 =
... decreasing for increasing k
x_k = 1.44466... = ί // "for some finite k"
!? ...
and in the
limit x_inf = 1.4422... (=3^(1/3)) (for k->inf)

or, as you say:
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

But for which *finite* k will then the inverse function f(ί,k) = 3 ?

Or do we assume that the trajectory of x_k misses ί and goes
through the complex plane?
[snip]

Gottfried
--
Ioannis

.

Relevant Pages

  • Re: tetration - inf superroot
    ... For any finite iteration, you won't be able to reach x. ...
    (sci.math)
  • Re: tetration - inf superroot
    ... f= x^is a partial inverse of the power tower. ... Any other iteration for ritwill converge to the value given by f. ... Here I have two plots, ...
    (sci.math)
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