# Tetration Forum

Full Version: "game" - to calculate 1.00...001 ^^ 0.5 = ?
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bo198214 Wrote:No, thatswhy I wrote "is a method that works for arbitrary bases $b>1$"!!!
ok but I don't mistake.

Quote:I would get a result. But actually its not clear why these values are interesting.

Quote:So why dont you use it then!?
I don't want quote.

you:
1. ----------
2. ----------
...

I:
1. ---------
...
because it's smaller and easier than quotes. do you understand well?
nuninho1980 Wrote:you:
1. ----------
2. ----------
...

I:
1. ---------
...
because it's smaller and easier than quotes. do you understand well?

No *lol*
I'll have a go at it.

1.00000000000000000001^^(1/2) =
1.000000000000000000009999999999000000000099999999987500000001499999999836458333351666666664502604166916666666638173719621422222221817020670621277777771935930347661211970809173335777978054727401267322532195532504077689101941770151
andydude Wrote:I'll have a go at it.

1.00000000000000000001^^(1/2) =
1.000000000000000000009999999999000000000099999999987500000001499999999836458333351666666664502604166916666666638173719621422222221817020670621277777771935930347661211970809173335777978054727401267322532195532504077689101941770151

it's correct!! you get w/ 225 cases decimal. but you calculated still too time to result?
how many digits and many exponents w/ iterations (= n_1^n_2^...^n_k^x) for to evaluate better? do you use maple?

I am 7000 digits and 199 exponents for evaluate, using maple, in only 100 cases decimal.
andydude Wrote:I'll have a go at it.

Wah, Andy my saviour! *ggg*
nuninho1980 Wrote:it's correct!! you get w/ 225 cases decimal. but you calculated still too time to result?
how many digits and many exponents w/ iterations (= n_1^n_2^...^n_k^x) for to evaluate better? do you use maple?

It didn't take that much time at all. The base is very close to 1, which means regular iteration works much better (The natural superlogarithm is very erratic for bases between 1 and eta). Also, I used Mathematica. So I used a 10-term and 15-term regular iteration series (polynomial), compared them, and each took about 3 seconds to evaluate. When they were identical, I assumed all the digits were "significant".

Andrew Robbins
andydude Wrote:It didn't take that much time at all. The base is very close to 1, which means regular iteration works much better (The natural superlogarithm is very erratic for bases between 1 and eta). Also, I used Mathematica. So I used a 10-term and 15-term regular iteration series (polynomial), compared them, and each took about 3 seconds to evaluate. When they were identical, I assumed all the digits were "significant".

Andrew Robbins

slog_b (x) = y
I didn't convert from natural slog to tetration using maple but I tried many times to add new each one of cases decimal of x since x=1 when the result is very close to y=0.5. these tentations last 30-60 minutes.
do you convert from natural slog to tetration?
nuninho1980 Wrote:do you convert from natural slog to tetration?

No, I used regular iteration from the fixed-point p =
1.000000000000000000010000000000000000000100000000000000000001500000000000000000023333333333333333333733333333333333333340408333333333333333463361111111111111113553591269841269841316667063492063492064403294973544973544991509125661375661376019282963213484046824576164483124899791712269571579076787413095423662162995000357635728649265105119539204243709252241024379290515068892945934184599484695347437210742175409191006184949392420206632843923492376014272124055526327911424878272
in the series
${}^{x}a = p + \ln(p)^x(1 - p) + \frac{p\ln(a)^2(1 - \ln(p)^x)\ln(p)^{x-1}}{2(1 - \ln(p))}(1 - p)^2 + \cdots$
where a = 1.00000000000000000001

Andrew Robbins
andydude Wrote:I used regular iteration from the fixed-point p =
1.000000000000000000010000000000000000000100000000000000000001500000000000000000023333333333333333333733333333333333333340408333333333333333463361111111111111113553591269841269841316667063492063492064403294973544973544991509125661375661376019282963213484046824576164483124899791712269571579076787413095423662162995000357635728649265105119539204243709252241024379290515068892945934184599484695347437210742175409191006184949392420206632843923492376014272124055526327911424878272
in the series
${}^{x}a = p + \ln(p)^x(1 - p) + \frac{p\ln(a)^2(1 - \ln(p)^x)\ln(p)^{x-1}}{2(1 - \ln(p))}(1 - p)^2 + \cdots$
where a = 1.00000000000000000001

Andrew Robbins

this method is much faster. thanks!

p+ ... (1-p)^2+ ...
what do you add new last terms, if you want 15 or more terms?
I know you use maple, so I've tried to make this Maple-ish, but I do not have Maple, so I'm programming blind. Let me know if this works:

Code:
carleman_matrix := proc(expr, x, p, n)   [seq([seq(subs(diff((expr-p)^j, x, k), x=p), k=0..n)], j=0..n)]; end proc; iterate := proc(m, p, t)   c := m[1][1];   n := length(m);   ev := eigenvects(m);   sm := [seq(ev[3*j + 1], j=0..n)];   dt := [seq([seq(if j==k then c^k else 0 endif, k=0..n)], j=0..n)];   mt := evalm(inverse(sm) &* dt &* sm);   p + sum(mt[1][k]*(1 - p)^k, k=1..n); end proc; # if this works, you can use it like this: a := 1.0001; # the base p := 1.000999; # the fixed point mat := carleman_matrix(a^x, x, p, 20); flo := iterate(mat, p, 0.5);

Andrew Robbins
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