Re: Bringing back an old tetration curiosity?

Am 22.09.2007 06:42 schrieb lwalke3@xxxxxxxxx:
On Sep 20, 8:35 pm, mike3 <mike4...@xxxxxxxxx> wrote:
Interesting. How smooth is the curve one gets,
when one uses a base of, say, 2? What would
a graph of the tetrational function to the base 2,
with a tetrapower from 0 to 3? Could you show
me a graph? Does the points appear to be outlining
a smooth curve?

Unfortunately, as interesting as this may seem, I know
for a fact that it's _not_ outlining a smooth curve. Indeed,
no tetration function with defines (x^^(1/n))^^n = x (i.e.
that uses tetraroots) can ever be smooth!

Why not? Consider the values:

2^^(1/2), 2^^(1/3), 2^^(1/4), ... 2^^(1/n), ...

defined as the second, third, fourth, ... nth, ... tetraroots
of two. Now what is the limit as n -> infinity? We know
that it must converge to the _infinite_ tetraroot of two,
which is sqrt(2) since sqrt(2)^^(infinity)=2.

But limit as n -> infinity of 1/n is zero, so 2^^(1/n) ought
to approach 2^^0 = 1 as n -> infinity. And thus we must
conclude that the function isn't even _continuous_ at
zero, much less differentiable. And because zero is our
base case for extending tetration to all positive rationals,
we conclude that the function isn't continuous at _any_
of the rationals! (Notice that, unlike Theron's graph in
the other subthread, this function is at least _defined_
for all positive rationals, a countable dense subset.)

This is actually a grave problem indeed. Suppose we
were to use Nelson's superlog (tetralog) algorithm to
find the base-2 tetralog of 3 (so x =3, y = 2). (In the
following, p(x,y) is the base-y tetralog of x and f(x,y)
is the base-y log of x.)

p[x_,y_] :=1+ p[f[x,y],y] /; x >= y

So p(3,2) = 1+p(log(3)/log(2),2)
= 1+p(1.5849625,2)

p[x_,y_] :=1/p[y,x] /; x < y

= 1+1/p(2,1.5849625)
= 1+1/(1+p(1.505007,1.5849625))
= 1+1/(1+1/p(1.5849625,1.505007))
= 1+1/(1+1/(1+p(1.122662, 1.505007)))
= 1+1/(1+1/(1+1/p(1.505007, 1.122662)))

And here's where we get into trouble. For it's
impossible to find p(1.505007,1.122662), which
is the base-1.122662 tetralog of 1.505007, because
1.122662^^(infinity) is only 1.14647, so 1.505007 is
far above the horizontal asymptote. Indeed, if we
attempted to continue the algorithm anyway:

= 1+1/(1+1/(1+1/(1+p(3.428802, 1.122662))))
= 1+1/(1+1/(1+1/(2+p(10.3352, 1.122662))))
= 1+1/(1+1/(1+1/(3+p(19.58955, 1.122662))))
= 1+1/(1+1/(1+1/(4+p(24.95287, 1.122662))))
...
= 1+1/(1+1/(1+1/(16+p(27.9271, 1.122662))))
...

and the x-values converge to 27.9271, when the
algorithm needs it either to reach a base case of zero
or one, or at least drop below y so that we can switch
the values of x and y.

And this happens for almost _every_ value of x and y --
we eventually reach a case where the tetration of y has
an asymptote and x is greater than the asymptote. The
only values for which the algorithm actually works for
base two are hand-picked values such as 0, 1, 2, 4, 16,
65536, 1.55961 (second tetraroot of two), 2^1.55961, etc.

So as promising as this idea may seem, defining
x^^(1/n) as a tetraroot will never give a function that will
be continous. This is why Gottried Helms's (and others')
functions disagree with the tetraroot -- x^^(1/n) must
be less than the nth tetraroot for sufficiently large n in
order for the function to be continous at zero.

Possibly I have an -at least- partial answer.
The problem seems to be in the notation.

Assume base b=sqrt(2).

What my tetration-method does, is implementing
{base,topexponent}^^tetraheight = {b,x}^^h

So we have
b^b^b...^b^x using b h-fold.

Then, it is best to read the notation downwards:

x, x, x , and so on.
b b
b

Now starting with x=1 gives a certain limit; x=2 gives the
limit 2 and x=4 gives the limit 4.

Your consideration goes in the opposite direction: let h->0.
I tried different h and got

for x=1
{b,1}^^1 = 1.4142136...
{b,1}^^0.5 = 1.2436216...
{b,1}^^0.25 = 1.1333935...
{b,1}^^0.125 = 1.0700019...
...
{b,1}^^(2^-10) = 1.0005750...
...
tending to 1.

Using x=2
{b,2}^^1 = 2.0000000...
{b,2}^^0.5 = 2.0000000...
{b,2}^^0.25 = 2.0000000...
{b,2}^^0.125 = 2.0000000...
...
{b,2}^^(2^-10) = 2.0000000...
...
tending to 2.


Using x=sqrt(2)
{b,x}^^1 = 1.6325269...
{b,x}^^0.5 = 1.5388054...
{b,x}^^0.25 = 1.4811284...
{b,x}^^0.125 = 1.4489431 ...
...
{b,x}^^(2^-10) = 1.4144954...
...
tending to sqrt(2), which is also {sqrt(2),sqrt(2)}^^0 = sqrt(2),
what we would expect from the definition of this operation.

So we get in these examples for h->0

lim{h->0} {b,x}^^h = x

Perhaps this explains some discrepancies; if this is
not sufficient however, I possibly still didn't understand
your argument correctly.

Gottfried
--
---

Gottfried Helms, Kassel
.

Relevant Pages

  • Re: Bringing back an old tetration curiosity?
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  • Re: Bringing back an old tetration curiosity?
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