Re: tetration and logaritms

On Jul 10, 1:20 pm, amy666 <tommy1...@xxxxxxxxxxx> wrote:
dont forget we are aiming at normal tetration too !!!
( base e )

Here tommy1729 is reminding us that he is aiming at
normal tetration base e -- that is, what we normally
think of as tetration. Gottfried calls this T-tetration.

In other words, the main reason that tommy1729 is
concerned with U_e(x,1/2) is his claim that the function
is more easily convertible to normal T-tetration, then
the method that Gottfried uses.

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)

I assume that in 5) tommy1729 is referring to the fact
that the method Gottfried uses to convert from
U-tetration back to T-tetration, which Gottfried has
himself posted in this thread:

Gottfried Helms:

"The fixpoint-substitution, which relates T and U-tetration
is, for a base b=t^(1/t)
T_b°h(x) = (U_t°h (x/t-1) + 1)*t"

But, as Gottfried himself has pointed out, this method fails
when b > e^(1/e) (= eta). Even if he uses a complex
fixed point of the exponential function, the power series
ends up diverging.

So tommy1729 is hoping that his method of converting from
U-tetration back to T-tetration will work for base e. Now we
may return to the original post.

We have just calculated the Taylor series:

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

and this function has the property that f(f(x)) = log(x+1). From
the OP, we have:

set v = x+1 for v between 0 and 2
(*) beware f(x) is a taylor series in x but you need to convert to a taylor series in x+1 !!
use the substitution x = v-1 and the binomial theorem to compute it.
[two typos corrected]

Better yet, we can use the matrix method again here to find
the series in v. We take a 6x6 matrix with -1 on the diagonal
and 1 on the superdiagonal and then evaluate the series for
f at the matrix. If I did it right, I obtained:

f(v) = -5509/3840+1661v/768-445v^2/384+229v^3/384
-149v^4/768+109v^5/3840

According to tommy1729, f(f(v)) should now be log(v) -- that
is to say, we have now found f(v) = T_e(v,-1/2).

It's hard for me to determine whether this works or not. I took a
graphing calculator and tried graphing f(f(v)) on the interval
[0,2] (since this is the only interval on which tommy1729
claimed this would work). The resulting function appears similar
to a logarithmic function, but it stays negative on the whole
interval [0,2] and doesn't become positive until v is slightly
more than 2. Of course, we only used six terms from the
power series, so we don't expect perfection at all. I don't have
a symbolic algebra system, so I don't have the ability to check
with more terms of the series.
.