Re: Extending tetration to the reals: linear approximation on an interval

On Aug 3, 1:17 am, Gottfried Helms <he...@xxxxxxxxxxxxx> wrote:
Am 02.08.2007 22:51 schrieb jayd...@xxxxxxxxx:
However, try using a linear approximation in the interval [0,1], and
you'll notice that all the bows disappear, and that the equation now
looks "right" when graphed.

f(z=1.76322..., x) =
{
1+(z-1)*frac(x) if x in [0,1],
power(z, f(z,x-1)) if n > 1,
log_z(f(z, x+1)) if n in (-2, 0)
}

What we have done is taken a good first order approximation of "the"
function x^^n, and made it a good *second* order approximation. The
first derivative is now continuous.

Try the graph for f(z,n)=z^^x, z=1.60107515392372877932706476774783545
(z^(z^z) = z^^3 = e, for the curious)

I tried my matrix-approach as described in earlier threads,
using the eigensystem-decomposition, and then exponentiating
the eigenvalues.

So to say, using your z as s in my notation:
s = z = 1.60107515392372877932706476774783545
[snip]

Since it is only a rough approximation to use a finite
submatrix of Bs of dimension 20, 24 or 28 I give the table
whose entries are the results, using this three dimension.

The table is

x s^^x (dim=20) s^^x (dim=24) s^^x (dim=28)
-------------------------------------------------------
1 1.60107515392 1.60107515392 1.60107515392
1.1 1.65421216949 1.65421216924 1.65421216919
1.2 1.70681266678 1.70681266636 1.70681266628
1.3 1.75901977647 1.75901977595 1.75901977587
1.4 1.81097069279 1.81097069225 1.81097069216
1.5 1.86279825502 1.86279825451 1.86279825443
1.6 1.91463239939 1.91463239895 1.91463239889
1.7 1.96660151842 1.96660151809 1.96660151804
1.8 2.01883375998 2.01883375976 2.01883375973
1.9 2.07145829533 2.07145829522 2.07145829521
2 2.12460658362 2.12460658362 2.12460658362
2.1 2.17841365921 2.17841365929 2.17841365929
2.2 2.23301946840 2.23301946852 2.23301946852
2.3 2.28857028340 2.28857028351 2.28857028350
2.4 2.34522022304 2.34522022310 2.34522022307
2.5 2.40313291318 2.40313291315 2.40313291310
2.6 2.46248332386 2.46248332372 2.46248332365
2.7 2.52345982592 2.52345982568 2.52345982560
2.8 2.58626651726 2.58626651700 2.58626651691
2.9 2.65112587825 2.65112587820 2.65112587814
3 2.71828182777 2.71828182846 2.71828182846
------------------------------------------------------

and dim=24 and dim=28 agree already up to the 8'th to 10'th
digit.

I could not yet describe the numerical conditions in
all details, but I think that the theoretical approach,
to use the eigenvalue-decomposition (or an approximation
to it) is the best and most coherent one to define the
continuous version of tetration.

Gottfried Helms

Fascinating. I'll need to study your method more, to see what it's
doing. I'm surprised you're able to get an answer with a matrix of 28
dimensions, but perhaps it works differently from what I'm picturing
it. I tried my hand at a matrix method for finding the taylor series
for e^^x on the interval [-1,0], and I hit a steep wall at degree 11.
I ended up having to solve a degree 10 equation, and manually tweak
the 11th coefficient until I closed in on the solution. The 11th
degree matrix was extremely unstable, but then again, I'm pretty sure
this is because the coefficients for degree 1 through 5 feed back into
the matrix. (It's at this point that I admit that I'm using Excel with
a multiprecision library called X-Numbers, not having access to
Mathematica or any other proper mathematical tool.)

What I really liked about seeing your solution is that it confirmed a
suspicion of mine that there is an inflection point very close to the
center of the unit interval I gave, which would explain why the
solution can be so closely approximated by a linear equation on that
interval. Using the values in your table at 1.4, 1.5, and 1.6, one can
verify that the secant approximation of the second derivative very
close to 0, and is about 18-20 times smaller than the second
derivative at 1.4 and 1.6, with a sign change occurring just before
1.5.

Actually, if f(z,a)=e, then I'm wondering how closely to f(z,a-1.5)
that inflection point occurs. Has anyone studied this particular
question?

Anyway, this would seem to confirm my analysis, at least for use as a
very quick approximation, especially for small z. But for more
precision, or for larger z, I'm not sure what direction to head, so
I'll start by studying your method.

.

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