01/15/2015, 03:09 PM
First, thank, you, Tommy, for your interest in that article.
I try to answer your questions, although I do not understand some notations.
and your slog is equivalent of my ate.
The natural pentation pen in my book and in my article is defined through the following way:
Some integer \( M>1 \) is chosen.
\( \varepsilon=\exp(kz) \)
\( f(z)=L_{e,4,0} + \sum_{m=0}^{M-1} a_m \varepsilon^m \)
Where \( L_{e,4,0}\approx -1.850354529 \) is zeroth fixed point of the 4th Ackermann to base e, id est, real solution \( x \) of equation \( \mathrm{tet}(x)=x \).
Increment \( k \) and coefficients \( a \)
are determined by substitution of the asymptotic supertetration
\( F(z)=f(z)+O(\varepsilon^M) \)
into the transfer equation
\( F(z+1)=\mathrm{tet}(F(z)) \)
I define
\( F_n(z)=\mathrm{tet}^n(f(z-n)) \)
\( F(z)=\lim_{n\rightarrow \infty} F_n \)
\( \mathrm{pen}(z)=F(x_1+z) \)
where
\( x_1 \) is real solution of equation \( F(x_1)=1 \)
My claim is, that resulting pen does not depend on the \( M \) chosen.
I define it in my way, because it seems to work.
\( \mathrm{tet}(z)=\mathrm{pen}(1+\mathrm{ape}(z)) \)
where \( \mathrm{ape}=\mathrm{pen}^{-1} \)
Is it that you mean?
After to construct the approximation, the researcher should substitute it into the equation, calculate the residual and plot the map of the agreement. Then we'll see, how many decimal digits may it keep for any given argument. If the numerical test passes, we may ask Henryk to write the long and complicated proof for some "Aequationes Mathematicae".
I doubt if these equations are useful for the evaluation of tetration.
Usually, such a limits are very slow to converge; I have not enough patience "to press a key, to have a tea".
There is no reason to compare acceleration of a car to the acceleration of the unmovable rock. The only, we may say, that the car moves, while the rock does not. However, one can refer to the theory of relativity, quantum mechanics, etc., but this does not look serious.
I repeat my old statement:
Until now, nobody could calculate and plot complex maps of tetration (nor pentation) with any algorithm, faster or simpler than those I had presented.
As soon, as you or anybody else present any algorithm, that does the same, we can plot the maps of the agreement and compare the algorithms.
Tommy, if you do not have much time, then, do not collect many questions in a bunch. Ask as soon as you have formulated one first question.
For me, it will be also easier to answer. Now, the preview does not fit one screen..
Best regard, Dima.
I try to answer your questions, although I do not understand some notations.
(01/15/2015, 01:12 PM)tommy1729 Wrote: Why isnt pentation defined asI think, your sexp is equivalent of my tet,
\( pent(z) = sexp^{[z]}(x_0) = \) lim \( sexp^{[n]}( L' ^z slog^{[n]}(x_0)) \)
and your slog is equivalent of my ate.
The natural pentation pen in my book and in my article is defined through the following way:
Some integer \( M>1 \) is chosen.
\( \varepsilon=\exp(kz) \)
\( f(z)=L_{e,4,0} + \sum_{m=0}^{M-1} a_m \varepsilon^m \)
Where \( L_{e,4,0}\approx -1.850354529 \) is zeroth fixed point of the 4th Ackermann to base e, id est, real solution \( x \) of equation \( \mathrm{tet}(x)=x \).
Increment \( k \) and coefficients \( a \)
are determined by substitution of the asymptotic supertetration
\( F(z)=f(z)+O(\varepsilon^M) \)
into the transfer equation
\( F(z+1)=\mathrm{tet}(F(z)) \)
I define
\( F_n(z)=\mathrm{tet}^n(f(z-n)) \)
\( F(z)=\lim_{n\rightarrow \infty} F_n \)
\( \mathrm{pen}(z)=F(x_1+z) \)
where
\( x_1 \) is real solution of equation \( F(x_1)=1 \)
My claim is, that resulting pen does not depend on the \( M \) chosen.
I define it in my way, because it seems to work.
Quote:Do you really believe lim \( sexp^{[n]}(f(L,L',z)) = \) lim \( sexp^{[n]}( L' ^z slog^{[n]}(x_0)) \) where \( f \) is a simple elementary function ?No. I do not understand the notation. I would say,
\( \mathrm{tet}(z)=\mathrm{pen}(1+\mathrm{ape}(z)) \)
where \( \mathrm{ape}=\mathrm{pen}^{-1} \)
Is it that you mean?
Quote:Afterall approximating slog^[n] with an elementary function seems wrong/divergent ?It depends on the skills of the colleague who makes the approximation.
After to construct the approximation, the researcher should substitute it into the equation, calculate the residual and plot the map of the agreement. Then we'll see, how many decimal digits may it keep for any given argument. If the numerical test passes, we may ask Henryk to write the long and complicated proof for some "Aequationes Mathematicae".
Quote:Lim \( sexp^{[n]}(x+y) \) is usually very different from lim \( sexp^{[n]}(x) \) even if \( y \) is small or getting smaller with growing \( n \).What is L' ?
Also its not defined as lim \( sexp^{[n]}(L' ^{z-n}) \).
I doubt if these equations are useful for the evaluation of tetration.
Usually, such a limits are very slow to converge; I have not enough patience "to press a key, to have a tea".
Quote:I assume its (your def of pentation in the paper) meant as an acceleration of lim \( sexp^{[n]}(L' ^{z-n}) \).I do not see any equivalent of this formula in my article. Nor in my book..
If not that would appeal weird and dubious to me.
Quote: Is that acceleration really a big improvement ?Hmm... As soon, as anybody reproduces the complex maps from my Book or from my articles with any alternative algorithms, it will be possible to compare the efficiency.
There is no reason to compare acceleration of a car to the acceleration of the unmovable rock. The only, we may say, that the car moves, while the rock does not. However, one can refer to the theory of relativity, quantum mechanics, etc., but this does not look serious.
I repeat my old statement:
Until now, nobody could calculate and plot complex maps of tetration (nor pentation) with any algorithm, faster or simpler than those I had presented.
As soon, as you or anybody else present any algorithm, that does the same, we can plot the maps of the agreement and compare the algorithms.
Tommy, if you do not have much time, then, do not collect many questions in a bunch. Ask as soon as you have formulated one first question.
For me, it will be also easier to answer. Now, the preview does not fit one screen..
Best regard, Dima.
