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 tiny limit-curiosity [ from ratio b^^(2j) / b^^j ] Gottfried Ultimate Fellow Posts: 753 Threads: 114 Joined: Aug 2007 07/24/2009, 01:19 PM (This post was last modified: 07/24/2009, 08:48 PM by 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\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 Gottfried Helms, Kassel tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 07/26/2009, 10:02 PM Gottfried wrote : (07/24/2009, 01:19 PM)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\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 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 12/13/2009, 06:38 PM (This post was last modified: 12/13/2009, 08:01 PM by bo198214.) (07/26/2009, 10:02 PM)tommy1729 Wrote: 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 Ultimate Fellow Posts: 753 Threads: 114 Joined: Aug 2007 07/28/2010, 07:52 AM (This post was last modified: 07/28/2010, 09:56 AM by 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 Helms, Kassel Gottfried Ultimate Fellow Posts: 753 Threads: 114 Joined: Aug 2007 07/30/2010, 05:05 AM (This post was last modified: 07/30/2010, 06:51 AM by 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 Gottfried Helms, Kassel tommy1729 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 08/16/2010, 11:28 PM with $b=t^{\frac1t}$, t in the range 1\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 Ultimate Fellow Posts: 1,354 Threads: 328 Joined: Feb 2009 03/28/2014, 12:35 AM Is that correct ? regards tommy1729 « Next Oldest | Next Newest »

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