Re: tetration and logaritms

first , my thanks to lwalke3 and gottfried for replying to this thread.


lwalke3 wrote :

On Jul 9, 11:59 am, amy666 <tommy1...@xxxxxxxxxxx>
wrote:
we all know the taylor series for log(1+x) which
holds for all real x between -1 and 1.
this series starts with 0 + x - x^2/2 + ...
and from looking at the first terms of this taylor
series we see that we can get series reversion which
of course gives exp(x) - 1.
fractional iterations can be done for exp(x)-1
since its a taylor series with f(0) = 0. ( and its
unique )
and thus also for log(1+x) , either directly since
log(1) = 0 or from the series reversion of the
fractional iterations of exp(x)-1.

...


Hmmm, this looks interesting.

Let us return to the U-tetration notation first
introduced by
Gottfried Helms. In this notation:

exp(x)-1 = U_e(x,1)

I see no reason why we can't write:

log(1+x) = U_e(x,-1)

and start using negative values for the second
parameter of
U_e (which Gottfried calls "h" for "height").

indeed.



Then tommy1729 is really asking for U_e(x,-1/2).

asking ? no giving ;)



So
now I
wonder whether the same matrix that Gottfried would
use to
find, say, U_e(x,1/2) would work for U_e(x,-1/2) as
well.

thats intresting , but i feel your drifting off a bit from the OP.

dont forget we are aiming at normal tetration too !!!

( base e )

you snipped that , and perhaps most importantly , i showed a method that is probably better then gottfriends.

why ?

1 ) unique
2) no oo variables or oo matrices
3) no guessing interpolation but analytic computation
4) convergeance
5) larger radius of convergeance and no problems at
e^(1/e)
6) smooth


I don't have Gottfried's full matrix method available
(he's
explained it but it's a bit too complex to follow),

no offense to you or gottfried , but thats probably why you cant see my method is probably better.

gottfried has not proven uniqueness nor convergeance.

and interpolation is just a guess, as he himself explained with the sine function in this very thread.

also his matrices are oo wich is " clumpsy " to work with , especially when we dont know certainly that it converges and have questions about summability methods.

no complicated summability methods are neccesary for my method.


but here's
a simpler version that only works for base e. (This
is because
Gottfried has discovered that the Taylor coefficients
are a
bivariate polynomial that evaluates to 0/0 when the
base is
either e or 1/e. But L'Hopital's Rule can be used in
this case
to give a polynomial in the single variable h. This
trick simply
gives us this polynomial in h.)

Suppose we only want the first five terms of our
Taylor
series for U_e(x,h). So we begin with a 6x6 matrix
with zeros
everywhere, except on the superdiagonal where it is
one. This
matrix corresponds to U_e(x,0) = x, the identity
function.

To find the matrix for U_e(x,1), we take the Taylor
series for
exp(x)-1 (which is x+x^2/2+x^3/6+x^4/24+x^5/120),
plug in the
previous matrix for x. Naturally, this gives us a
strictly upper
triangular matrix with 1 on the superdiagonal, then
1/2 on the
diagonal above that, then 1/6 on the next diagonal,
and finally
1/24 and 1/120 in the upper-left corner. This tells
us that
U_e(x,1) is x+x^2/2+x^3/6+x^4/24+x^5/120 (but this is
trivial).

We then repeat the process by plugging in the new
matrix for
x in the series for exp(x)-1, to find U_e(x,2). Here
we find that
not only does the superdiagonal contain 1, but so
does the
diagonal above it. The next diagonal has the value
5/6, and
finally 5/8 and 13/30 in the upper-left corner. So we
have that
U_e(x,2) = x+x^2+5x^3/6+5x^4/8+13x^5/120.

this is nothing new to any of us.

though as said , that interpolation is doubtfully correct or unique.



Continuing in this manner, we find:
U_e(x,3) = x+3x^2/2+2x^3+5x^4/2+179x^5/60
U_e(x,4) = x+2x^2+11x^3/3+77x^4/12+163x^5/15

We could continue like this forever, but this is all
we need
to find the polynomials in h. We simply take the
coefficients
for the x, x^2, x^3, ..., terms, view them as
functions of h,
and then take the Lagrange interpolating polynomial.

The x terms are trivial: 1, 1, 1, 1, 1. So we see
that the
linear coefficient is always 1.

The x^2 terms is also easy. Starting with U_e(x,0)
they
form the patten 0, 1/2, 1, 1/2, 2. So the x^2
coefficients
must be h/2.

For x^3 we have 0, 1/6, 5/6, 2, 11/3. The x^3
coefficients
work out to be h^2/4-h/12.

For x^4 we have 0, 1/24, 5/8, 5/2, 77/12. The x^4
coefficients
work out to be h^3/8-5h^2/48+h/48.

For x^5 we have 0, 1/120, 13/30, 179/60, 163/15. The
x^5
work out to be h^4/64-13h^3/144-h^2/24-h/180.

Notice that the polynomial for x^n always works out
to be a
polynomial in h of degree n-1.

Putting this together, we have the Taylor polynomial
U_e(x,h) =
x+(h/2)x^2+(h^2/4-h/12)x^3+(h^3/8-5h^2/48+h/48)x^4

+(h^4/64-13h^3/144-h^2/24-h/180)x^5.

And this seems to work for any _positive_ value of h
that
Gottfried chose. So for h = 1/2:

U_e(x,1/2) = x+x^2/4+x^3/48+x^5/3840

And so there's nothing stopping us from letting h be
negative:

U_e(x,-1) = x-x^2/2+x^3/3-x^4/4+x^5/5

well you can check by taking exp(x) - 1 of it.

will it still be x ?



And this of course is the well-known Taylor series
for log(1+x)
that tommy1729 already mentioned in his post.

indeed , as expected.

but more importantly ( as said above ) you missed more important parts of my OP.

for one that i dont restrict to U-tetration !!


Finally, h=-1/2:

U_e(x,-1/2) = x-x^2/4+5x^3/48-5x^4/96+109x^5/3840


gottfried will be pleased.

but the OP has a lot more potential !!!

like normal ( smooth ) tetration.

regards

tommy1729
.