tetration limit ??
#21
so , what are the answers ?
#22
Let's see here.

The fixed point for base (eta + eps) is e + delta(eps), where delta satisfies:

\( [1/e + \ln(1 + \eps/\eta)] (e + \delta(\eps)) = 1 + \ln(1 + \delta(\eps)/e) \)

\( [1/e + \eps/\eta + O(eps^2)] (e + \delta(\eps)) = 1 + \ln(1 + \delta(\eps)/e) \)

\( [\eps/\eta + O(\eps^2)] (e + \delta(\eps)) = -(\delta(\eps)/e)^2/2 + O(\delta(\eps)^3) \)

So \( \delta(\eps) \approx \sqrt{\eps} \) and

\( 0 = e\eps/\eta + \delta(\eps) \eps/\eta + (\delta(\eps)/e)^2/2 + O((\eps)^{3/2} \)

\( \delta(\eps) = \frac{ - \eps/\eta + \sqrt{ \eps^2/\eta^2 - \frac{2 \eps}{e \eta} } } { 1/e^2 } + O(\eps^{3/2}) \)

\( \delta(\eps) = \frac{ - \eps/\eta + \sqrt{\frac{-2 \eps}{e \eta}} [1 - \frac{e \eps}{4 \eta} + O(\eps^2)] } { 1/e^2 } + O(\eps^{3/2}) \)


\( \delta(\eps) = -e^2 \eps/\eta + \sqrt{2/\eta} e^{3/2} \sqrt{-\eps} + O(\eps^{3/2}) \)
#23
BenStandeven Wrote:Let's see here.

The fixed point for base (eta + eps) is e + delta(eps), where delta satisfies:

\( [1/e + \ln(1 + \eps/\eta)] (e + \delta(\eps)) = 1 + \ln(1 + \delta(\eps)/e) \)

\( [1/e + \eps/\eta + O(eps^2)] (e + \delta(\eps)) = 1 + \ln(1 + \delta(\eps)/e) \)

\( [\eps/\eta + O(\eps^2)] (e + \delta(\eps)) = -(\delta(\eps)/e)^2/2 + O(\delta(\eps)^3) \)

So \( \delta(\eps) \approx \sqrt{\eps} \) and

\( 0 = e\eps/\eta + \delta(\eps) \eps/\eta + (\delta(\eps)/e)^2/2 + O((\eps)^{3/2} \)

\( \delta(\eps) = \frac{ - \eps/\eta + \sqrt{ \eps^2/\eta^2 - \frac{2 \eps}{e \eta} } } { 1/e^2 } + O(\eps^{3/2}) \)

\( \delta(\eps) = \frac{ - \eps/\eta + \sqrt{\frac{-2 \eps}{e \eta}} [1 - \frac{e \eps}{4 \eta} + O(\eps^2)] } { 1/e^2 } + O(\eps^{3/2}) \)


\( \delta(\eps) = -e^2 \eps/\eta + \sqrt{2/\eta} e^{3/2} \sqrt{-\eps} + O(\eps^{3/2}) \)

euh

i dont mean to be rude , but euh , where is the limit ?

this thread is about a limit , so whatever you wrote may be brilliant , but i dont know what you mean or if you talk about the same thing ?

regards

tommy1729
#24
BenStandeven Wrote:Let's see here.

The fixed point for base (eta + eps) is e + delta(eps), where delta satisfies:

\( \delta(\eps) = -e^2 \eps/\eta + \sqrt{2/\eta} e^{3/2} \sqrt{-\eps} + O(\eps^{3/2}) \)

Now \( (\eta + \eps)^{e + \Re(\delta(\eps))} = (\eta + \eps)^{e + -e^2 \eps/\eta + O(\eps^{3/2})} = \e( \ln(\eta + \eps) (e + -e^2 \eps/\eta + O(\eps^{3/2}))) \), which is:

\( \e( 1 + -e \eps/\eta + \ln(1 + \eps/\eta) (e + -e^2 \eps/\eta) + O(\eps^{3/2})) = \e( 1 + O(\eps^{3/2})) = e + O(\eps^{3/2}) \).

So:

\( (\eta + \eps)^{e + \Re(\delta(\eps)) + \theta} = (\eta + \eps)^{(e + \Re(\delta(\eps)))(1 + \theta/(e + \Re(\delta(\eps))))} = e^{(1 + O(\eps^{3/2}))(1 + \theta (1 - \Re(\delta(\eps))/e + O(\eps^2))/ e )} \) which for \( |\theta| << 1/\sqrt{\eps} \) is \( e^{1 + \theta (1 + e \eps/\eta)/ e + O(\eps^{3/2}) } = e^{1 + \theta (1 + e \eps/\eta)/e} + O(\eps^{3/2}) \)

To be continued...
#25
BenStandeven Wrote:
BenStandeven Wrote:Let's see here.

The fixed point for base (eta + eps) is e + delta(eps), where delta satisfies:

\( \delta(\eps) = -e^2 \eps/\eta + \sqrt{2/\eta} e^{3/2} \sqrt{-\eps} + O(\eps^{3/2}) \)

\( (\eta + \eps)^{e + \Re(\delta(\eps)) + \theta} = e^{1 + \theta (1 + e \eps/\eta)/e} + O(\eps^{3/2}) \)

Now if \( \theta \) is on the order of \( \sqrt \eps \), we have \( (\eta + \eps)^{e + \Re(\delta(\eps)) + \theta} = e + \theta + \theta^2/2e + O(\eps^{3/2}) \), so the effect of an additional level of tetration is to add \( -\Re(\delta(\eps)) + \theta^2/2e \) to the exponent. To cross this region of length \( 2 \sqrt \eps \) would require between \( 2 / (\sqrt \eps (e^2/\eta + 1/2e)) \) and \( 2 / (\sqrt \eps (e^2/\eta)) \) steps.

But if \( \theta \) is of a larger order, the epsilon-dependent terms may be neglected, and we get that \( (\eta + \eps)^{e + \theta} = \eta^{e + \theta} + O(\eps)} \).

So it takes roughly \( slog_{\eta}(e - \sqrt\eps) \) tetration levels to reach \( e - \sqrt \eps \).

To be continued...
#26
tommy1729 Wrote:euh

i dont mean to be rude , but euh , where is the limit ?

this thread is about a limit , so whatever you wrote may be brilliant , but i dont know what you mean or if you talk about the same thing ?

regards

tommy1729

Sorry, I'm trying to find the limit in several steps; I should have said so from the start. I think it should only take one or two more posts to finish it.
#27
BenStandeven Wrote:
BenStandeven Wrote:
BenStandeven Wrote:Let's see here.

The fixed point for base (eta + eps) is e + delta(eps), where delta satisfies:

\( \delta(\eps) = -e^2 \eps/\eta + \sqrt{2/\eta} e^{3/2} \sqrt{-\eps} + O(\eps^{3/2}) \)

\( (\eta + \eps)^{e + \Re(\delta(\eps)) + \theta} = e^{1 + \theta (1 + e \eps/\eta)/e} + O(\eps^{3/2}) \)

Now if \( \theta \) is on the order of \( \sqrt \eps \), we have \( (\eta + \eps)^{e + \Re(\delta(\eps)) + \theta} = e + \theta + \theta^2/2e + O(\eps^{3/2}) \), so the effect of an additional level of tetration is to add \( -\Re(\delta(\eps)) + \theta^2/2e \) to the exponent. To cross this region of length \( 2 \sqrt \eps \) would require between \( 2 / (\sqrt \eps (e^2/\eta + 1/2e)) \) and \( 2 / (\sqrt \eps (e^2/\eta)) \) steps.

But if \( \theta \) is of a larger order, the epsilon-dependent terms may be neglected, and we get that \( (\eta + \eps)^{e + \theta} = \eta^{e + \theta} + O(\eps)} \).

So it takes roughly \( slog_{\eta}(e - \sqrt\eps) \) tetration levels to reach \( e - \sqrt \eps \).

To be continued...

Now \( slog_{\eta} ( e - \sqrt\eps) = 2e/\sqrt\eps + O(\eps) \). So we see that \( \lim_{\eps \to 0} {}^{C/\sqrt\eps}(\eta + \eps) \) is e for any constant from \( 2e \) to \( e^2/\eta \). Also, it is e for any constant less than 2e, since \( {}^{C/\sqrt\eps}(\eta) = e - 2e/{C/\sqrt\eps} + O(\eps) \). Similarly, \( \lim_{\eps \to 0} {}^{C \eps^{-1/2+\sigma}}(\eta + \eps) = e \) for any positive sigma.

But assuming positive sigma again, \( \lim_{\eps \to 0} {}^{C \eps^{-1/2-\sigma}}(\eta + \eps) \) would be \( \eta + O(\sqrt\eps) \) exponentiated \( C \eps^{-1/2-\sigma/2} - D \eps^{-1/2} \) times and then another \( C \eps^{-1/2-\sigma} - C \eps^{-1/2-\sigma/2} \); the first operation is enough to move the exponent up to at least \( e + C \eps^{-\sigma/2} \), while the next would take it from there to \( e + C/2e \eps^{-\sigma} \) and each additional step would square the excess over e again. So we would get at least \( e + C' \eps^{-\sigma 2^{C \eps^{-1/2-\sigma} - C \eps^{-1/2-\sigma/2} - 2}} \), which clearly tends to infinity as epsilon approaches zero.

So \( \lim_{\eps \to 0} {}^{C \eps^{-1/2+\sigma}}(\eta + \eps) \) is e if sigma is 0 or positive (probably independent of C), and infinite if sigma is negative.
#28
(04/07/2009, 01:35 AM)nuninho1980 Wrote:
bo198214 Wrote:Oh, you mean we have an upper fixed point of the tetrational for \( b\le 1.6353244967 \) and the fixed point can then be computed by \( \operatorname{slog}_b(x)=x \) or \( {^x b} = x \). Ya, interesting. I dont know whether we even have a thread on the forum that dealt with the topic of the fixed point of tetrationals.
Of course there maybe always the dependency of the values from the chosen method of tetration.
slog_b (x) = x <=> b ^^ x = x => b ^^^ oo = x
yeah! Wink

to remember:
one - 1^oo = 1
Euler - (e^(1/e)) ^^ oo = e

now new fixed point
(1.63532...) ^^^ oo ~= (3.08855...) what is this new result? it's a Super-Euler?

bo198214 Wrote:So how big is the difference between both methods, with respect to the computed fixed point?

to remember - you follow "regular slog" - http://en.wikipedia.org/wiki/Talk:Tetrat...on_methods lol Wink
I calculated Nuninho's constant, to 32 decimal digits of precision using my latest kneser.gp program. For bases around 1.6, it takes about 70 seconds to generate sexp accurate to 32 decimal digits. Then we generate the Taylor series, centered around 3.0. Then we generate the Taylor series for sexp'(x), centered around 3.0. Calculate when sexp'(x)=1. Now we have the 'x' value for the local minimum of sexp(x)-x. For that x, calculate sexp(x)-x. If sexp(x)-x>0, then the current base is bigger than Nuinho's constant; if sexp(x)-x<0, then the current base is smaller than Nuinho's constant. Do a binary search .... Here's the result, accurate to 32 decimal digits.
Code:
base                    =  1.6353244967152763993453446183062
upfixed                 =  3.0885322718067176544821807826411
sexp'(upfixed)          =  1.0000000000000000000000000000000
sexp(upfixed)-upfixed   = -1.1371391135491644200632572659231 E-32
lowfixed                = -1.6408725757165933485612321510790
sexp(lowfixed)-lowfixed =  0.0000000000000000000000000000000
sexp'(lowfixed)         =  4.8060057543017516963843938970331
Here is a graph, showing sexp(x), and the line f(x)=x. The two graphs intersect each other at the lower fixed point, and at the upper fixed point. At the lower fixed point, the slope>1, so regular iteration is well defined. The slope at the upper fixed point=1, so this is a parabolic fixed point, much like eta.
- Sheldon
   

sexp taylor series, centered at 0, accurate to 32 digits
Code:
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3.2442551084741376457891294676456 E-32

regular pentation generated from the lower fixed point, via pentation.gp code, pentation taylor series, centered at 0, accurate to ~21 digits
Code:
base 1.6353244967152763993453446183062
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sexp taylor series, centered at the upper fixed point. Parabolic regular iteration, since sexp(upfixed)-upfixed=0, and the derivative=1. However, the pentation series above was developed from the lower fixed point. It might also be interesting to develop the pentation from the upper fixed point.
Code:
base 1.6353244967152763993453446183062
upper fixed point 3.0885322718067176544821807826411
3.0885322718067176544821807826411
1.0000000000000000000000000000000
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#29
@sheldonison - it's correct!! Smile

\( 1.635324496715 \uparrow\uparrow\uparrow \infty \approx 3.088535321467 \) (13 digits corrects)
\( 1.635324496716^-\uparrow\uparrow\uparrow \infty \approx 3.088532271794 \) (" " " )
\( 1.63532449671527639\uparrow\uparrow\uparrow \infty \approx 3.088532271806 \) (18 " " )

\( 1.63532449671527639\uparrow\uparrow\uparrow \infty \approx 3.08853225407356834 \) (18 " " )
\( 1.63532449671527640^-\uparrow\uparrow\uparrow \infty \approx 3.08853227180671764 \) (" " " )
\( 1.63532449671527639934534\uparrow\uparrow\uparrow \infty \approx 3.08853227181918334 \) (23 " " )

\( 1.63532449671527639934535^-\uparrow\uparrow\uparrow \infty \approx 3.0885322718067176544821 \) (23 " " )

base - \( 1.635324496716^- \) = base -
Code:
init(1.635324496716)
(by kneser.gp pari/gp) but \( b^- \) it's for to avoid infinite (result).

Smile
#30
(04/29/2009, 01:08 PM)tommy1729 Wrote: so , what are the answers ?

using logarithmic semi operators (lol)
{q : 0 <= q <= 1 q E R}, if S(x) is the identity function, q:ln(x) = exp^[-q](x):

\( \lim_{h\to\0}\,(S(1-q)\, \{-q\}\, q:ln(h))\,\{2-q\}\,\frac{1}{h} = q:ln(e) \)



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