Problems in (0,1/e)
Ioannis Galidakis
a minor flaw in my analysis and I was wondering if anyone could
correctly predict why it happens.
In my analysis on my web page:
<http://www.crosswinds.net/athens/~jgal/Math/Exponents.html>
I consider the function f(z)=-LambertW(-log(z))/log(z), which
essentially gives the fixed points of the iteration for c^z, allowing
for a direct computation of the value of c^c^c^c^... when this
converges.
Upon reading Munafo's page on numbers <http://www.mrob.com/numbers.html>
he very quickly mentions that the interval of convergence of x^x^x^...
is [(1/e)^e,e^(1/e)]. That agrees with my web page initially as far as
the right bound of the interval is concerned, and upon a closer
inspection, I found that my f fails to predict a value of convergence
for values in the interval (0,1/e) [Modified interval because I consider
(x^(1/x))^(x^(1/x))^... etc] I have accordingly slightly modified the
contents to conform with this.
Upon a closer inspection of f, I can tell that it behaves really badly
near 0, (because the Log(z) terms explode), yet it remains a fact that f
is real valued on the interval (0,e^(1/e)]. I am still suspicious of the
fact, thus, that the exponential converges for all values in (0,e^(1/e)]
and not only for [(1/e)^e,e^(1/e)].
Checking this with Maple, it does appear that for very small values of
epsilon, the value of epsilon^(1/epsilon) is very close to 0. [Another
way to look at it would be to look at (1/delta)^delta as delta->+oo. In
fact, the previous tends to zero by looking at the graph of the function
(1/x)^x]. So as far as I can tell from looking at it, my exponentials
[x^(1/x)]^[x^(1/x)]^.. would APPEAR to converge to 1 for epsilon->0.
The question is, since the exponentials [x^(1/x)]^[x^(1/x)]^.. converge
for x in (0,1/e) to 1 why does the function not predict the correct
values in the convergence?
If anyone can shed some light on this I would appreciate it immensely.
Thanx in advance.
--
Ioannis Galidakis jg...@ath.forthnet.gr
<http://www.crosswinds.net/athens/~jgal/main.html>
______________________________________
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Robert Israel
Ioannis Galidakis <jg...@ath.forthnet.gr> writes:
> I consider the function f(z)=-LambertW(-log(z))/log(z), which
> essentially gives the fixed points of the iteration for c^z, allowing
> for a direct computation of the value of c^c^c^c^... when this
> converges.
> Upon reading Munafo's page on numbers <http://www.mrob.com/numbers.html>
> he very quickly mentions that the interval of convergence of x^x^x^...
> is [(1/e)^e,e^(1/e)]. That agrees with my web page initially as far as
> the right bound of the interval is concerned, and upon a closer
> inspection, I found that my f fails to predict a value of convergence
> for values in the interval (0,1/e) [Modified interval because I consider
> (x^(1/x))^(x^(1/x))^... etc] I have accordingly slightly modified the
> contents to conform with this.
> Upon a closer inspection of f, I can tell that it behaves really badly
> near 0, (because the Log(z) terms explode), yet it remains a fact that f
> is real valued on the interval (0,e^(1/e)]. I am still suspicious of the
> fact, thus, that the exponential converges for all values in (0,e^(1/e)]
> and not only for [(1/e)^e,e^(1/e)].
f(x) is a fixed point of g(z) = x^z, i.e. x^(f(x)) = f(x). But for x < 1/e
the fixed point is unstable. Note that g'(f(x)) = -1 at x = (1/e)^e,
and is increasing on this interval. When the fixed point is unstable,
the iterates will not converge to it (unless they happen to hit it exactly).
> Checking this with Maple, it does appear that for very small values of
> epsilon, the value of epsilon^(1/epsilon) is very close to 0. [Another
> way to look at it would be to look at (1/delta)^delta as delta->+oo. In
> fact, the previous tends to zero by looking at the graph of the function
> (1/x)^x]. So as far as I can tell from looking at it, my exponentials
> [x^(1/x)]^[x^(1/x)]^.. would APPEAR to converge to 1 for epsilon->0.
x^(1/x) = exp(ln(x)/x) -> 0 as x -> 0, but if z is near 0 then (x^(1/x))^z
is near 1. What happens when x is small is that the iterates will alternate
between values near 0 and values near 1. For example, with x = .1,
y = x^(1/x) = 10^(-10), we have y^y = .99999999769741,
y^(y^y) = 1.00000005301898 * 10^(-10), y^(y^(y^y)) = .99999999769741 etc.:
it appears to be a stable 2-cycle.
Robert Israel isr...@math.ubc.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia
Vancouver, BC, Canada V6T 1Z2
Ioannis Galidakis
[snip]
> essentially gives the fixed points of the iteration for c^z, allowing
> for a direct computation of the value of c^c^c^c^... when this
> converges.
>
> Upon reading Munafo's page on numbers <http://www.mrob.com/numbers.html>
> he very quickly mentions that the interval of convergence of x^x^x^...
> is [(1/e)^e,e^(1/e)]. That agrees with my web page initially as far as
> the right bound of the interval is concerned, and upon a closer
> inspection, I found that my f fails to predict a value of convergence
> for values in the interval (0,1/e) [Modified interval because I consider
> (x^(1/x))^(x^(1/x))^... etc] I have accordingly slightly modified the
> contents to conform with this.
>
> Upon a closer inspection of f, I can tell that it behaves really badly
> near 0, (because the Log(z) terms explode), yet it remains a fact that f
> is real valued on the interval (0,e^(1/e)]. I am still suspicious of the
> fact, thus, that the exponential converges for all values in (0,e^(1/e)]
> and not only for [(1/e)^e,e^(1/e)].
>
> epsilon, the value of epsilon^(1/epsilon) is very close to 0. [Another
> way to look at it would be to look at (1/delta)^delta as delta->+oo. In
> fact, the previous tends to zero by looking at the graph of the function
> (1/x)^x]. So as far as I can tell from looking at it, my exponentials
> [x^(1/x)]^[x^(1/x)]^.. would APPEAR to converge to 1 for epsilon->0.
>
> for x in (0,1/e) to 1 why does the function not predict the correct
> values in the convergence?
Well, after dwelling in the arms of Morpheus for 14 hours, I figured it
out.
Turns out that the symbol c^c^c^... does NOT converge for values close
to 0. (No wonder f in this case is so badly behaved).
To show this, let's assume the following:
let s=epsilon.
let w be a value very close to 0.
then s^w will be very close to 1.
AND
let w be a value very close to 1
then s^w will be very close to 0.
One only needs to check that indeed, say, if on starts with s=0.00001
close to zero, the successive exponentials do indeed alternate between
values close to 0 and 1, as the sequence s, s^s, s^s^s, s^s^s^s,... is
indeed alternating. So the limit cannot exist.
To conclude: The fixed point in the orbit of c^z for small c exists, but
it's repellant.
Man, I almost fell into a coma to draw the resources for this. I will
update my page ASAP.
Ioannis Galidakis
[snip]
> One only needs to check that indeed, say, if on starts with s=0.00001
> close to zero, the successive exponentials do indeed alternate between
> values close to 0 and 1, as the sequence s, s^s, s^s^s, s^s^s^s,... is
> indeed alternating. So the limit cannot exist.
I need to say more, of course. Since for values >1/e the sequence still
alternates but the sup and the inf converge to the same number, the
fixed point which in this case is attractive.
Chiao
Ioannis Galidakis
> [snip]
> x^(1/x) = exp(ln(x)/x) -> 0 as x -> 0, but if z is near 0 then (x^(1/x))^z
> is near 1. What happens when x is small is that the iterates will alternate
> between values near 0 and values near 1. For example, with x = .1,
> y = x^(1/x) = 10^(-10), we have y^y = .99999999769741,
> y^(y^y) = 1.00000005301898 * 10^(-10), y^(y^(y^y)) = .99999999769741 etc.:
> it appears to be a stable 2-cycle.
Correct, but 0.1 won't do though, because it's >1/e=.06598803588. For
0.1 it converges actually. This was probably a typo. You probably meant
some smaller value, like 0.01 or 0.001
Thanks however, I figured it out just some time before you sent the
message. Here's what seems to be happening:
for small values of d, (i.e. d very close to 0), name two sequences, the
sequences of odd and even terms as follows:
a(n)={d^d, d^d^d^d, d^d^d^d^d^d,...}
and
b(n)={d, d^d^d, d^d^d^d^d,...}
then *IF* a(n) and b(n) are to converge (and they do) then their limits
will respectivelly be:
limit for a(n): solution to lim1=d^d^lim1 (1)
and limit for b(n): lim2=d^lim1 (2)
Now, *IF* the cycle converges altogether, we need to have:
lim1=lim2 which gives:
d^d^lim=d^lim, and alas, the solution to this is:
-LambertW(-log(d))/log(d).
I was hoping that there would be some indication on the graph of
-LambertW(-log(z))/log(z) which would predict the point 1/e, but I don't
think there is.
If you see such behaviour please post or email. I need to resolve this
asap, cause I can't sleep. :*)
One could manually check whether lim1=-LambertW(-log(d))/log(d), and if
they are equal, the cycle converges, but I don't like this. I was hoping
that I could get the damn 1/e point from some bad property of my f, but
it seems unlikely. However, f is really strange near zero. If I could
get Maple to show some more detail on its behaviour, perhaps I could
deduce something.
> Robert Israel isr...@math.ubc.ca
> Department of Mathematics http://www.math.ubc.ca/~israel
> University of British Columbia
> Vancouver, BC, Canada V6T 1Z2
--
Dave L. Renfro
the four sci.math postings by Ioannis Galidakis
I found this morning at
<http://forum.swarthmore.edu/epigone/sci.math/ranbrelbrel>
that deal with the convergence of iterated exponentiation.
It is well known that the sequence
x, x^x, x^(x^x), ...
converges for all exp(-e) <= x <= exp(1/e), and diverges
for all other positive values of x. This sequence
diverges to infinity for each x > exp(1/e) and it
diverges in an oscillatory fashion for each
0 < x < exp(-e). The graph of y = x^(x^(x^...))
on its interval of convergence is an upward sloping
curvilinear segment with endpoints
x = exp(-e), y = 1/e and x = exp(1/e), y = e.
Let f_1(x) = x, f_2(x) = x^x, f_3(x) = x^(x^x),
f_4(x) = x^(x^(x^x)), etc. Then for odd values of
n, limit (x --> 0+) of f_n(x) equals 0, and for even
values of n, limit (x --> 0+) of f_n(x) equals 1. This
can be proved by induction. [Let "statement 1"
be the result for f_1(x) AND for f_2(x), and let
"statement k" be the result for f_(2k-1)(x) AND for
f_(2k)(x). It is not difficult to prove (a) "statement 1"
and (b) "statement k" implies "statement k+1".]
Moreover, it is known that for each 0 < x <= exp(1/e),
f_1(x), f_3(x), f_5(x), ... converges and
f_2(x), f_4(x), f_6(x), ... converges.
However, these two sequences have different limits
for each 0 < x < exp(-e). Thus, the graph of
y = x^(x^(x^...)) bifurcates into two branches for
0 < x < exp(-e). Or, another way to put it,
x^(x^(x^...)) is a double-valued function (to use
older terminology) when 0 < x < exp(-e). The branch
associated with y = limit (n odd, n --> inf) of f_n(x)
begins at x = 0, y = 0 (in the sense of having a
continuous extension there) and increases towards
x = exp(-e), y = 1/e. The branch associated with
y = limit (n even, n --> inf) of f_n(x) begins at
x = 0, y = 1 (in the sense of having a continuous
extension there) and decreases towards
x = exp(-e), y = 1/e.
Three early proofs of these results were published
by Euler (1778), Eisenstein (1844), and Seidel (1873).
In fact, Euler actually obtained explicit equations
for these two branches. I *think* they are what I
have below, but be warned that I'm not completely
sure about this.
(odd branch) y = u^[1/(1-u)]
u = (e^e)*x
0 < x < exp(-e)
(even branch) y = u^[u/(1-u)]
I STRONGLY recommend that you look at the following paper.
R. Arthur Knoebel, "Exponentials Reiterated",
The American Mathematical Monthly 88(4) (April 1981),
pages 235-252.
There are 125 references in its bibliography, by
my count. Among the topics mentioned in Knoebel's
survey paper are ----->>>
Several references to an application in biochemistry.
(Section 2 title: "An Application to Biochemistry".)
A discussion of what is known for the convergence
of f_n(z) for complex values of z. [Knoebel, pp. 245-246]
You might also want to look at the paper
A. J. Macintyre, "Convergence of i^(i^(i^...))",
Proc. Amer. Math. Soc. 17 (1966), 67.
Knoebel makes some remarks about the analytic
continuation of z^(z^(z^...)) on p. 246.
References to iterated exponentials with unequal powers:
x1, x1^x2, x1^(x2^x3), ... [Knoebel, p. 248]
The issue of iterated TETRATION for positive real
numbers is discussed in E. M. Wright, "Iteration
of the exponential function", The Quarterly Journal
of Mathematics (Oxford) 18 (1947), 228-235.
I don't believe much, if anything, is known regarding
the existence of a mathematically satisfying
analytic continuation for the tetration operation.
Knoebel brings this matter up on p. 247. On this
page he also poses the question to what extent the
Ackermann function can be analytically continued.
Ioannis Galidakis
[snip]
>
> Note that g'(f(x)) = -1 at x = (1/e)^e,
> and is increasing on this interval.
Robert, could you share some insight on what this actually means? Why
are you considering g'? and was you deduction of 1/e a backwards
calculation (i.e. knowing in advance about 1/e) or does the result have
some significance as it stands?
Sorry, but it's been 14 hours without sleep and I may be missing the
obvious.
Thanx
Ioannis Galidakis
>
> These following comments are in response to
> the four sci.math postings by Ioannis Galidakis
> I found this morning at
> <http://forum.swarthmore.edu/epigone/sci.math/ranbrelbrel>
> that deal with the convergence of iterated exponentiation.
>
> It is well known that the sequence
>
> x, x^x, x^(x^x), ...
Thanx a billion Dave. I will plunge into the refs asap.