Re: Tetration again!

Am 19.12.2007 09:55 schrieb mike3:
On Dec 19, 12:35 am, Gottfried Helms <he...@xxxxxxxxxxxxx> wrote:
Am 19.12.2007 05:55 schrieb mike3:

On Dec 18, 3:13 am, Gottfried Helms <he...@xxxxxxxxxxxxx> wrote:
Am 18.12.2007 08:14 schrieb mike3:
PS. "Tower" here (as opposed to "base") is what you call
"height" -- I'm using the term "Tower" by analogy with
"power" -- (T)etrational p(OWER) -> TOWER, and as a pun.
:-))
Ahh, yes! I got it.
Heh.
Anyway, what about the tetration of an "easy"
base, like the base e that I discussed?
I meant "easy" for bases 1/e^e + eps < base < e^(1/e)-eps
where the powerseries converge for each height/infinite
height = according power of tetration-matrix



I noticed something too. The "linear" tetration
given seems to be continuously differentiable
when the base is e. This makes sense, since it
says that at each integer the derivative multiples
by the natural log of the base, and the natural
log of e is 1, so no change -- it's continuous.
Is that something interesting?
Maybe... Someone else may consider this... sorry, no
idea currently.


For the approach via an interpolated powerseries
for continuous tetration U_b(x,h) = b^x-1 I have an
idea like the following, if b = exp(1).

In the following I omit the "_b"-signature, since it refers
to the log(b) and this is 1 in this case.

The U-tetration can be computed by the exponential-
series

U (x,1) = 1/1!*x + 1/2!*x^2 + 1/3!*x^3 + ...

Rewrite this, also analoguously with the series in the
following, as the product of two vectors:

V(x) = [1,x,x^2,x^3,....] and U1 = [0,1/1!,1/2!,1/3!,...]

such that

U(x,1) = V(x)~ * U1

-------------------------------------------------


Now, what is U(x,2) of height 2? We need a vector of coefficients
U2 = [?,?,?,?...] to build

U(x,2) = V(x)~ * U2.

In Abramowitsch/Stegun we read, that the Stirling-numbers 2'nd
kind perform x-> e^x-1, x->(e^x-1)^2, x->(e^x-1)^3 ... column-
wise, if arranged as columns in a Stir2-matrix and rescaled
by factorials.
Let Stir2 =
1 0 0 0 ...
0 1 0 0 ...
0 1 1 0 ...
0 1 3 1 ...
....
and F=diag(0!,1!,2!,3!,...) then the scaled version is

S2 = F^-1 * Stir2 * F

and is a matrix, whose columns contain the coefficients for the
required powerseries to compute
U0: x -> 1
U1: x -> e^x -1
U2: x -> (e^x -1)^2
U3: x -> (e^x -1)^3
...

So
V(x)~ * S2 = [1, e^x-1, (e^x-1)^2, (e^x-1)^3,....]
= [U(x,1)^0, U(x,1)^1, U(x,1)^2,...]
= V(U(x,1))~

and the second column U1 gives our relevant result.

-------------------------------------------------------



Now repeat this and get

V(U(x,1))~ * S2 = V(U(x,2))~
= (V(x)~*S2) * S2
= V(x)~ * S2^2

Since S2 is triangular, we can compute the exact (truncated)
matrix for the x->U(x,2) transformation by the second column
of S2^2.

The top-left segment of S2^2 is
1 . . . . .
0 1 . . . .
0 1 1 . . .
0 5/6 2 1 . .
0 5/8 8/3 3 1 .
0 13/30 35/12 11/2 4 1

The interesting coefficients are again in the second column, call
it U1_2. Similarly we get the column U1_3 of S2^3, which gives
the coefficients for the powerseries in x to compute U(x,3).

---------------------------------------------------------

Assume, we write a list of the subsequent vectors U1_0,U1,U1_2,U1_3,...
as new matrix

List L =
0 0 0 0 0 0 ...
1 1 1 1 1 1
0 1/2 1 3/2 2 5/2
0 1/6 5/6 2 11/3 35/6
0 1/24 5/8 5/2 77/12 105/8
0 1/120 13/30 179/60 163/15 691/24 ...
... ...
and each column gives the coefficients for the powerseries in x
to compute U(x,h) for h=0..n, where n+1 is the number of columns.

Next step is, to observe, that each row has its own progression,
and may be expressed as a polynomial; I give the result computed
by Pari/Gp
polynomials = PY =
( 0 ) / 0!
( 1 ) / 1!
( 1*h +0 ) / 2!
( 3/2*h^2-1/2*h +0 ) / 3!
( 3*h^3-5/2*h^2+1/2*h +0 ) / 4!
( 15/2*h^4-65/6*h^3+5*h^2-2/3*h +0 ) / 5!
...
where each polynomial allows to compute the entries of
the according row in L by inserting the desired h-value
h=0,1,2,3,...
That means, for instance, for h=1 we get the vector of results

U1 = [ 0 1 1/2 1/6 1/24 1/120 ... ]~ \\ read it as column-vector

and for h=2
U1_2 = [ 0 1 1 5/6 5/8 13/30 ...]~

and so on, as expected.



For negative h we get
h = -1
U1_(-1) = [ 0 1 -1/2 1/3 -1/4 1/5 ....]

which is just the series for the natural log, such that

V(x)~ * U1_(-1) = log(1+x)

(for x in the range of convergence), the inverse of the U-trans-
formation or simply the U-transformation of negative height h=-1.

-------------------------------------------------------

Now, since we have polynomials to determine the coefficients
for the powerseries in x for the computation of U(x,h),
we may be able to interpolate to fractional h, say h=1/2.
We get the following vector of coefficients

U1_(1/2) = [ 0 1 1/4 1/48 0 1/3840 ...]~

which is the same vector, which I get in the second
column of S2^(1/2)
1 . . . . .
0 1 . . . .
0 1/4 1 . . .
0 1/48 1/2 1 . .
0 0 5/48 3/4 1 .
0 1/3840 1/96 1/4 1 1

which I get, if I compute the half-iterate by S2^(1/2)
via matrix-logarithm and ~ exponentiation

S2^(1/2) = MExp( 1/2 * MLog(S2))

and thus the interpolation seems to be usable.


By this I get the values

U(1 ,1/2) ~ 1.27103...

and V(1.27103)* S2^(1/2) = V(1.71828)~ = V(exp(1)-1)~

U(1.27103,1/2) ~ 1.71828

----------------------------------------------------

These polynomials seem to allow to "interpolate" w.r.t. the
height parameter, such that we get coefficients for a powerseries
in x, which performs then U(x,h) for negative and even for
fractional h.

Gottfried


--
---

Gottfried Helms, Kassel
.

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