Tetration of the Square Root of Two

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Lode Vandevenne

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Jan 3, 2004, 6:55:12 PM1/3/04
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Hello,

I noticed that tetration (also called hyper4, hyperexponentials, powertower,
...) hasn't really been defined for real hyperexponents yet. So I've done
an attempt myself that might be pretty original. The only disadvantage is
that it only works for tetration of the square root of two. So in fact,
I've tried to find the function

sqrt(2)^^x

With x a real number. If you don't know about tetration, do a google
search. I'll use the ^^ notation, call it the tetration operator, and call
x the hyperexponent.

I'm going to make a lot of assumptions though, and I don't think there's a
lot of mathematical "truth" in here, but hey, that what hasn't been defined
yet can be played with.

Disclaimer: I use the notation oo for infinity. In fact I mean the limit of
n going to infinity, but I'll just use oo to avoid limit notations.

A. Recursion
--------------

Tetration is defined using recursion as follows:

x^^1=x
x^^n = x^(x^^(n-1)) with n > 1

I'll immediatly assume that this also works for real n and negative n.

The following two handy recursive formulas are derived from the definition:

x^^(n-1) = ln(x^^n)/ln(x)

and

x^^(n+1) = x^(x^^n)

Examples:

2^^3 = 2^(2^2) = 16
3^^2 = 3^3 = 27
sqrt(2)^^4 = sqrt(2) ^ (sqrt(2) ^ (sqrt(2) ^ sqrt(2)) = 1.84091087

B. Horizontal Asymptote
-------------------------

a^^n is convergent for n going to oo (infinity) if a is positive and a <
e^(1/e). (I'm ignoring negative a here)

e^(1/e) = 1.44466786
sqrt(2) = 1.41421356

So it's a close one, but sqrt(2) < e^(1/e) and thus sqrt(2)^n is convergent.
By checking it out, it appears to go to 2, which can be understood because
sqrt(2)^2 = 2.
In short I'll notate this as:

sqrt(2)^^oo = 2

And this means that the function sqrt(2)^^x has a horizontal asymptote at
+2.

C. Vertical Asymptote
-----------------------

sqrt(2) will also appear to have a vertical asymptote.

With the recursion formula x^^(n-1) = ln(x^^n)/ln(x), we can extend
sqrt(2)^^n for negative n:

sqrt(2)^^1 = sqrt(2)
==> sqrt(2)^^0 = ln(sqrt(2))/ln(sqrt(2)) = 1
==> sqrt(2)^^(-1) = ln(1)/ln(sqrt(2)) = 0
==> sqrt(2)^^(-2) = ln(0)/ln(sqrt(2)) = -oo
Here it stops though, unless someone would want to work with ln(-oo)

This can also be generalized:

x^^2 = x^x
x^^1 = x
x^^0 = 1
x^^(-1) = 0
x^^(-2) = -oo

Either way, this means sqrt(2)^^x has a vertical asymptote in -2 because
sqrt(2)^^(-2) = -oo

D. A Symmetrical Plot!
-----------------------

Now we have some values for sqrt(2)^^x, and two assymptotes

These values can easily be calculated:

sqrt(2)^^1 = 1.4142
sqrt(2)^^2 = 1.6325
sqrt(2)^^3 = 1.7608
sqrt(2)^^4 = 1.8409

These values were assumed/defined under C.

sqrt(2)^^0 = 1
sqrt(2)^^(-1) = 0

And these are the two asymptotes:

sqrt(2)^^oo = 2
sqrt(2)^^(-2) = -oo

Study the last 4 values (the two asymptotes and the ones that become 1 and
0).
It appears that these values are symmetrical around the line y = -x
(the -45° line).

If you plot the points out on a paper and draw the asymptotes, it's clearly
visible.

Looking at that, I'll make a very big assumption:

That the function isn't only symmetrical around y=-x for those 4 values, but
also for all the rest, or in other words:

*******************************
sqrt(2)^^x = y <==> sqrt(2)^^(-y) = -x
*******************************

This is true for

sqrt(2)^^0 = 1
sqrt(2)^^(-1) = 0

and

sqrt(2)^^oo = 2
sqrt(2)^^(-2) = -oo

But using that formula and assuming it works for any x, it's now easy to
find some values for sqrt(2)^^x with real x! Namely take:

sqrt(2)^^1 = 1.4142
sqrt(2)^^2 = 1.6325
sqrt(2)^^3 = 1.7608
sqrt(2)^^4 = 1.8409

Then we get:

sqrt(2)^^(-1.4142) = -1
sqrt(2)^^(-1.6325) = -2
sqrt(2)^^(-1.7608) = -3
sqrt(2)^^(-1.8409) = -4

E. Further extending it
----------------------

Now we can get more values with the recursion formula x^^(n+1) = x^(x^^n),
for example:

sqrt(2)^^(-1.4142)=-1, so:
sqrt(2)^^(-0.4142)=0.7071
sqrt(2)^^(0.5858)=1.2777
etc...

Do this for all other points and you'll find that they all lie beautifully
on the plot.

By keeping to use mirroring and recursion, you can (after many many steps)
find sqrt(2)^^x for any x. If x<-2, the result appears to become complex
(mirroring that might also give an idea of what some complex hyperexponents
can give: values higher than 2).

But as a final example, let's find the value of sqrt(2)^^0.5

1) sqrt(2)^^2=1.6325
2) mirroring: sqrt(2)^^-1.6325 = -2
3) recursion: sqrt(2)^^-0.6325 = 0.5
4) mirroring: sqrt(2)^^(-0.5) = 0.6325
5) recursion: sqrt(2)^^0.5 = 1.2451

Or to be more exact: sqrt(2)^^0.5 = sqrt(2)^((sqrt(2)^sqrt(2)-1)

F. In Conclusion
-----------------

So hereby, sqrt(2)^^x is defined for every x. When plotted, you get a plot
with a vertical and a horizontal asymptote, that is symmetrical around the
line

y=-x. In ASCII-art it looks somewhat like this (paste in something like
notepad to see a not-screwed up version)


^
+ | sqrt(2)^^x
+ |
* + |
* + |
*+ |
++++++++++++2+++++++++++++++++++++++++++
+* | ....@.¨¨¨¨¨
+ * | .@....@
+ * | .@¨¨¨
+ * |.¨
+ * 1@
+ * . |
+ . |
+ . * |
+ . *|
---+----@----+----+----+----+----+----->
-2 -1 0|* 1 2 3 4 x
+ . | *
+ . | *
+ . | *
+ M -1+ *
+ . | *
+ . | *
+ . | *
+ . | *
+ M -2+ *
+ . | *
+ . | *
+ . | *
+ . | *
+ M -3+ *
+. | *
+. | *
+. | *
+. | *
+. |

Legend:

----------- = axes
+++++++++++ = asymptotes
*********** = the y=-x line

@ = original points
M = mirrored points
. = points addad after more mirror/recursion cycles

But this was all based on the assumtion that

*******************************
sqrt(2)^^x = y <==> sqrt(2)^^(-y) = -x
*******************************

It looks like a quite analytical function in ]-2,oo[ to the eye.

Extending this to tetration of other values A than sqrt(2) looks quite hard
though, since if A > e^(1/e) there's no horizontal asymptote at all, and
else the horizontal asymptote is not at +2 so you can't use the mirroring.
Maybe this is possible after a suitable transformation, then mirror, then
transform back...

This concludes my attempt to define sqrt(2)^^x for real x, did it make any
sense?

A N Niel

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Jan 4, 2004, 5:55:26 AM1/4/04
to
In article <2058e9de.04010...@posting.google.com>, Lode
Vandevenne <lo...@planetunreal.com> wrote:

> a^^n is convergent for n going to oo (infinity) if a is positive and a <
> e^(1/e). (I'm ignoring negative a here)

You need to recheck a close to zero. For example, 0.1^^n does not
converge, but gets close to a cycle of period 2.

> This concludes my attempt to define sqrt(2)^^x for real x, did it make
> any
> sense?

Here is a mathematical problem it suggests.

Describe all functions g:(-2,oo) -> (-oo,2) such that:
g(x+1)=sqrt(2)^g(x)
g(x)=y <==> g(-y)=-x
g(0)=1

In particular, is there a unique continuous solution?

Phil Carmody

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Jan 4, 2004, 5:37:33 PM1/4/04
to
lo...@planetunreal.com (Lode Vandevenne) writes:
> This concludes my attempt to define sqrt(2)^^x for real x, did it make any
> sense?

Absolutely.
What happens with 3^(1/3)=1.442249570307408382321638311?

Phil
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Ioannis

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Jan 4, 2004, 6:04:28 PM1/4/04
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Ο "Lode Vandevenne" <lo...@planetunreal.com> έγραψε στο μήνυμα
news:2058e9de.04010...@posting.google.com...
>
> Hello,
[snip]

> It looks like a quite analytical function in ]-2,oo[ to the eye.
>
> Extending this to tetration of other values A than sqrt(2) looks quite
hard
> though, since if A > e^(1/e) there's no horizontal asymptote at all, and
> else the horizontal asymptote is not at +2 so you can't use the mirroring.
> Maybe this is possible after a suitable transformation, then mirror, then
> transform back...

The exact bounds where the (real) hyperpower function converges are well
known: [(1/e)^e,e^(1/e)]

As far as extending the hyperpower to reals, you might find Munafo's pages
quite interesting and a good place to start.
http://home.earthlink.net/~mrob/pub/math/ln-notes1.html#real-hyper4
--
Ioannis Galidakis
http://users.forthnet.gr/ath/jgal/
------------------------------------------
Eventually, _everything_ is understandable

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