tiny limit-curiosity [ from ratio b^^(2j) / b^^j ]

[Gottfried]

Hi

I played a bit around with the behave of approximation when the tetrate progresses to its fixpoint and the stepwith for iteration is increasing, say exponentially, for instance at step k=3 the iteration-height for x_3 is j=2^3=8 and at step k=4 the iteration-height for x_4 is 2^4=16 and so on. Then how do the x_(k+1)/x_k -ratios behave?

Clearly this seems to approximate 1; but I modified the criterion a bit; I used

\( \hspace{24} \log(\frac{ x_{\small k+1}}{x_{\small k}}) ^{\frac1{\small 2^k}} \)

and find, that I get the log of the fixpoint with this, at least for some tested bases.

Formally:
with \( b=t^{\frac1t} \), t in the range 1<t<exp(1)

it seems that

\( \hspace{24} \lim_{k->\infty, \ j=2^k} \hspace{24} \log( \frac{b\^\^ ^{2j}}{b\^\^ ^j })^{\frac1j} -> \log(t) \)

Perhaps there is an "obvious" reason, which I overlooked...

Gottfried

[tommy1729]

Gottfried wrote :

Hi

I played a bit around with the behave of approximation when the tetrate progresses to its fixpoint and the stepwith for iteration is increasing, say exponentially, for instance at step k=3 the iteration-height for x_3 is j=2^3=8 and at step k=4 the iteration-height for x_4 is 2^4=16 and so on. Then how do the x_(k+1)/x_k -ratios behave?

Clearly this seems to approximate 1; but I modified the criterion a bit; I used

\( \hspace{24} \log(\frac{ x_{\small k+1}}{x_{\small k}}) ^{\frac1{\small 2^k}} \)

and find, that I get the log of the fixpoint with this, at least for some tested bases.

Formally:
with \( b=t^{\frac1t} \), t in the range 1<t<exp(1)

it seems that

\( \hspace{24} \lim_{k->\infty, \ j=2^k} \hspace{24} \log( \frac{b\^\^ ^{2j}}{b\^\^ ^j })^{\frac1j} -> \log(t) \)

Perhaps there is an "obvious" reason, which I overlooked...

Gottfried

i think you can reduce to :

\( \hspace{24} \lim_{k->\infty, \ j=2^k} \hspace{24} \log( \frac{h(b)}{b\^\^ ^j })^{\frac1j} -> \log(t) \)


it seems to converge faster and to the same value.

for instance dont you get the same result for the ratio b^^3j / b^^ j as the ratio b^^2j / b^^ j ?

it seems so at first sight , without using many iterations though.

im not sure im right here , but you have the power to check it.

btw i mainly tried base sqrt(2) for convenience.

high regards

tommy1729

[tommy1729]

i think you can reduce to :

\( \hspace{24} \lim_{k->\infty, \ j=2^k} \hspace{24} \log( \frac{h(b)}{b\^\^ ^j })^{\frac1j} -> \log(t) \)


it seems to converge faster and to the same value.

for instance dont you get the same result for the ratio b^^3j / b^^ j as the ratio b^^2j / b^^ j ?

it seems so at first sight , without using many iterations though.

im not sure im right here , but you have the power to check it.

btw i mainly tried base sqrt(2) for convenience.

high regards

tommy1729

not ?

Moderator's note: Please quote only as much as is really needed (I shortened your quote)

[Gottfried]

Here is a list of values.

I used as base for the tetration b=t^(1/t)
where t=1.5, u=log(t), l=u/t, b=exp(l) = t^(1/t)

In the second column we have

err(k) = u - log( b^^j2 / b^^j1) ^(1/j1) // using j1=2^k; j2=2^(k+1)

Because we see, that the error err(k) approx halves each step, column q1 contains their exact quotients

Then these seem to approximate the constant value 0.5 but with some halving in the following digits
So in the next column d1 is that value with that constant 0.5 removed.
And so on.

Code:

.
    
| k:      err(k)                q1=         d1=q1-0.5      q2=         d2 =q2-0.5      q3            d3           q4             d4              q5
|                          =err(k+1)/err(k)           =d1(k+1)/d1(k)
| ------------------------------------------------------------------------------------------------------------------------------------------------------
|  1:      0.197795491256     ***  
|  2:      0.107205167894    0.542000059   0.042000059    ***  
|  3:     0.0568456435509    0.530250963   0.030250963   0.720260023   0.220260023    ***  
|  4:     0.0294828108981    0.518646796   0.018646796   0.616403381   0.116403381   0.528481650   0.278481650    ***  
|  5:     0.0150195849019    0.509435310   0.009435310   0.506001659   0.006001659   0.051559145  -0.198440855  -0.712581439  -0.962581439    ***
|  6:    0.00758065719662    0.504718156   0.004718156   0.500053120   0.000053120   0.008850813  -0.241149187   1.215219452   0.965219452  -1.002740561
|  7:    0.00380821235831    0.502359131   0.002359131   0.500011131   0.000011131   0.209546225  -0.040453775   0.167754143  -0.082245857  -0.085209490
|  8:    0.00190859823922    0.501179572   0.001179572   0.500002783   0.000002783   0.250002575   0.000002575  -0.000063646  -0.250063646   3.040440637
|  9:   0.000955424785547    0.500589787   0.000589787   0.500000696   0.000000696   0.250000696   0.000000696   0.270204466   0.020204466  -0.080797295
| 10:   0.000477994141307    0.500294893   0.000294893   0.500000174   0.000000174   0.250000174   0.000000174   0.250000043   0.000000043   0.000002152
| 11:   0.000239067549336    0.500147447   0.000147447   0.500000043   0.000000043   0.250000043   0.000000043   0.250000011   0.000000011   0.249998407

To compute the last row and last column (refering to b^^4096) I needed 1200 digits precision (I'll try to find the true required precision here).
It would be nice if we could develop this to something like the operation of dividing in the height-parameter, b^^a = f(b^^(a/2)).


[update] Ah, I should mention, that doing things in steps of j1=3^k,j2=3*j1 instead of j1=2^k,j2=2*j1 we get 0.3333 instead of 0.5 in q1 . Didn't check yet whether this proceeds analoguously...[/update]

Gottfried

[Gottfried]

I could improve the results up to iteration-height h ~ 2^22 which means values of b^^4194304 (which is very near the fixpoint... )

Let's denote b^^h as T(h), t the lower fixpoint, and ln(t) = u

Then with the limit-function

\( \hspace{48} d(h) = u - \ln\( \frac { T(2*h)} { T( h ) } \)^{\frac1h} \)

we seem to get approximation down to zero with increasing height h.



Moreover, if we use the ratio of two such limits in steps of h = 2^k , or even more general in h = c^k with some exponential step-base c :

\( \hspace{48} d( k,c ) = u - \ln\( \frac { T(c^{k+1})} { T( c^k ) } \)^{\frac1{c^k}} \)

or even better for computation

\( \hspace{48} d( k,c ) = u - \ln\(\( T(c^{k+1}-1) - T(c^k-1)\)*\ln(b)\)^{\frac1{c^k}} \)

and then

\( \hspace{48} q(k,c\) = \frac{ d(k-1,c) }{ d( k,c ) } \)

then we seem to arrive at the following

\( \text{Conjecture :} \\
\\
\hspace{48} \lim_{k\to\infty} q( k , c ) \to c \)

Heuristic:

Code:

.
(Pari/GP, float-prec 800 , regular tetration, base b=sqrt(2),t=fp0=2, u=ln(2)

   [c=2 , k, q(k,c)]
% 99 = [2, 15, 1.9999760334888695166]
%100 = [2, 20, 1.9999992510378326004]
%101 = [2, 21, 1.9999996255188461821]
%102 = [2, 22, 1.9999998127594055616]

   [c=3 , k, q(k,c)]
%103 = [3,  7, 2.9989230328032237024]
%104 = [3, 12, 2.9999955667153233567]
%105 = [3, 13, 2.9999985222372279067]
%106 = [3, 14, 2.9999995074122745008]

   [c=5 , k, q(k,c)]
%111 = [5, 4, 4.9874581459928721151]
%112 = [5, 7, 4.9998994774776347240]
%113 = [5, 8, 4.9999798952530082216]
%114 = [5, 9, 4.9999959790409007914]

The extremely high iteration-heights can be achieved in regular tetration when we use the schroeder-function: with this the powertower-iteration need not be done by h-times exponentiation but can be found by raising u to the h'th power. Denote S_a(x) the schroeder-function for the fixpoint-shifted argument and S_b(x) the inverse such that

\( \hspace{48} T(h) = S_b\( u^h * S_a\( \frac1t-1\) \) *t + t \hspace{120} \) // regular tetration

Different T(h) can then be computed conveniently after the value of the schroederfunction for x=1, s = S_a(1/t-1), was saved in a constant s . Also the numerator in the second d-formula can be simplified, since the difference of two T()'s also cancel the numerically uneasy addition of t, and finally ln(b)=u/t where then also t cancels, so

\( \hspace{48} d( k,c ) = u- \ln \( \( S_b(s*u^{c*c^k-1}) - S_b(s*u^{c^k-1})\)*u\)^{\frac1{c^k}}
\)

[Update: don't know: am I the only one who get's the vague feeling that one of these formulae really sucks ? ]

Gottfried

[tommy1729]

with \( b=t^{\frac1t} \), t in the range 1<t<exp(1)

it seems that

\( \hspace{24} \lim_{k->\infty, \ j=2^k} \hspace{24} \log( \frac{h(b)}{b\^\^ ^j })^{\frac1j} -> \log(t) \)

i think i can show this to be correct as well.

that would be 2 proofs in 1 day

i believe the above is a koenigs like form in disguise.

if we take blog instead of log we get :

\( \hspace{24} \lim_{k->\infty, \ j=2^k} \hspace{24} \log( \frac{h(b)}{b\^\^ ^j })^{\frac1j} -> \log(t) \)

yes , exactly the same , because the blog = log / log(b) and the log(b) gets canceled by the power ^(1/j).

then we can reduce the fraction to a difference !

that difference grows like the (de)nominator of the koenigs function does.

hence log(t) is simply the absolute value of the derivate of the fixpoint in the analogue koenigs function.

and thus the identity is explained and proved by koenigs function.


Q.E.D.

regards

tommy1729

[tommy1729]

Is that correct ?

regards

tommy1729