# Tetration Forum

Full Version: "game" - to calculate 1.00...001 ^^ 0.5 = ?
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andydude Wrote:dt := [seq([seq(if j==k then c^k else 0 endif, k=0..n)], j=0..n)];

I got messages "Error, invalid argument sequence" by "if" and "Error, invalid =" by "==". I tried it and I don't run it.
the "endif" isn't the syntax of "if" in maple because "endif" is variable but yes "end if".
you will correct these errors.
I have maple 12 and mathematica 6.
if you don't correct these errors then you do new code for mathematica.
Oh Mathematica I can do
Code:
CarlemanMatrix[series:SeriesData[x_, x0_, _, _, n_, 1]] :=   Table[D[Normal[series]^j, {x, k}]/k! /. {x -> x0}, {j, 0, n}, {k, 0, n}]; RegularIterate[series:SeriesData[x_, x0_, _, _, n_, 1], t_] :=   SeriesData[x, x0, MatrixPower[CarlemanMatrix[series], t][[2]], 0, n, 1];

Then you can do regular tetration with this code:
Code:
tet = c + Normal[RegularIterate[Series[c^((x + c)/c) - c, {x, 0, 3}], y]]/.x->(1-c); tet /. {c -> 2, y -> 0.5}
which would calculate ${}^{0.5}(\sqrt{2})$.
andydude Wrote:Oh Mathematica I can do
Code:
CarlemanMatrix[series:SeriesData[x_, x0_, _, _, n_, 1]] :=   Table[D[Normal[series]^j, {x, k}]/k! /. {x -> x0}, {j, 0, n}, {k, 0, n}]; RegularIterate[series:SeriesData[x_, x0_, _, _, n_, 1], t_] :=   SeriesData[x, x0, MatrixPower[CarlemanMatrix[series], t][[2]], 0, n, 1];
I don't run because I got 2 messages errors (SeriesData::sdatc: "Coefficient specification _ in \
SeriesData[x_,x0_,_,_,n_,1] is not a list.").

andydude Wrote:Then you can do regular tetration with this code:
Code:
tet = c + Normal[RegularIterate[Series[c^((x + c)/c) - c, {x, 0, 3}], y]]/.x->(1-c); tet /. {c -> 2, y -> 0.5}
which would calculate ${}^{0.5}(\sqrt{2})$.
I got calculate w/ only 6 digits. I don't get more than 6 digits after I tried N[tet /. {c -> 2, y -> 0.5},20].
nuninho1980 Wrote:I got calculate w/ only 6 digits. I don't get more than 6 digits after I tried N[tet /. {c -> 2, y -> 0.5},20].

First, you should probably change the 3 to 10 or 15 if you want more accuracy, and second, you're setting precision in the wrong place. If you want more precision in the output, you need more precision in the input:
Code:
tet /. {c -> 2, y -> SetPrecision[0.5, 20]}
and lastly, the errors are harmless, that has to do with the way SeriesData is implemented, and I usually use
Code:
Off[SeriesData::sdatc]
to prevent those errors anways.

Andrew Robbins
andydude Wrote:
nuninho1980 Wrote:I got calculate w/ only 6 digits. I don't get more than 6 digits after I tried N[tet /. {c -> 2, y -> 0.5},20].

First, you should probably change the 3 to 10 or 15 if you want more accuracy, and second, you're setting precision in the wrong place. If you want more precision in the output, you need more precision in the input:
Code:
tet /. {c -> 2, y -> SetPrecision[0.5, 20]}
and lastly, the errors are harmless, that has to do with the way SeriesData is implemented, and I usually use
Code:
Off[SeriesData::sdatc]
to prevent those errors anways.

Andrew Robbins

thanks!
but...
1.00000000000000000001^^0.5:
mathematica:
Code:
input - tet = c +     Normal[RegularIterate[Series[c^((x + c)/c) - c, {x, 0, 10}],       y]] /. x -> (1 - c); tet /. {c -> SetPrecision[1.00000000000000000001, 500],   y -> SetPrecision[0.5, 500]} output - 1.00000000000000000000999999999900000000107632355135617795995209920536\ 8238513511966004491298409050071250667500689538361713781158824235061751\ 8804942336518455216635292814415323018925473233062089824544715074650425\ 6359732393835545208872315977093580695663321688857957469520534738177800\ 1693967855762254625860886110633315544031059619857729710726124870557076\ 6594635989115759590552321672637545798115791229033207673798680646642688\ 6796500107272130759792942964123634745456350777468683283970538611734299\ 73796429395
it's correct!??? no.

maple:
Code:
output - 1.00000000000000000000999999999900000000009999999998750000000149999999983645833335 16666666645026041669
it's correct! but you remember I do during 30~60minutes.

maple can be better than mathematica using your method is faster but your code for maple has errors.
nuninho1980 Wrote:it's correct!??? no.

Sorry about that, that code calculates ${}^{y}(c^{1/c})$. So, you input the wrong value. 'c' is the fixed point, not the base. In order to calculate base 1.00000000000000000001 tetration you must use
Code:
H[x_] := -ProductLog[-Log[x]]/Log[x]; base = 1.00000000000000000001500; tet = c + Normal[RegularIterate[Series[c^((x + c)/c) - c, {x, 0, 10}], y]]/.x->(1-c); tet /. {c -> H[base], y -> 0.5500}
Hope that helps.

Andrew Robbins
(05/02/2009, 01:55 AM)andydude Wrote: [ -> ]Sorry about that, that code calculates ${}^{y}(c^{1/c})$. So, you input the wrong value. 'c' is the fixed point, not the base. In order to calculate base 1.00000000000000000001 tetration you must use
Code:
H[x_] := -ProductLog[-Log[x]]/Log[x]; base = 1.00000000000000000001500; tet = c + Normal[RegularIterate[Series[c^((x + c)/c) - c, {x, 0, 10}], y]]/.x->(1-c); tet /. {c -> H[base], y -> 0.5500}
Hope that helps.

Andrew Robbins

Code:
output - 1.00000000000000000001000000000000000000010000000000000000000150000000\ 0000000000023333333333333333333733333333333333333340408333333333333333\ 4633611111111111111135535912698412698413166670634920634920644032949735\ 4497354499150912566137566137601928296321348404682457616448312489979171\ 2269571579076787413095423662162995000357635728649265105119539204243709\ 2522410243792905150688929459341845994846953474372107421754091910061849\ 4939242020663284392349237601427212405552632791142487827221751 + RegularIterate[-1.\ 0000000000000000000150000000000000000002333333333333333333373333333333\ 3333333340408333333333333333463361111111111111113553591269841269841316\ 6670634920634920644032949735449735449915091256613756613760192829632134\ 8404682457616448312489979171226957157907678741309542366216299500035763\ 5728649265105119539204243709252241024379290515068892945934184599484695\ 3474372107421754091910061849493924202066328439234923760140216029716719\ 107236580733179125579896375851214328898*10^-40,   0.500000000000000000000000000000000000000000000000000000000000000000\ 0000000000000000000000000000000000000000000000000000000000000000000000\ 0000000000000000000000000000000000000000000000000000000000000000000000\ 0000000000000000000000000000000000000000000000000000000000000000000000\ 0000000000000000000000000000000000000000000000000000000000000000000000\ 0000000000000000000000000000000000000000000000000000000000000000000000\ 0000000000000000000000000000000000000000000000000000000000000000000000\ 00000000000000]
I got not number only but I use this code only?
(05/02/2009, 12:54 PM)nuninho1980 Wrote: [ -> ]I got not number only but I use this code only?

You have to define CarlemanMatrix and RegularIterate like in my previous post.
(05/03/2009, 08:31 AM)andydude Wrote: [ -> ]
(05/02/2009, 12:54 PM)nuninho1980 Wrote: [ -> ]I got not number only but I use this code only?

You have to define CarlemanMatrix and RegularIterate like in my previous post.

I calculated w/ 500 digits and w/ n=10 (accuracy) during 4 seconds.
thanks!
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