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.
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
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 0.249998407To 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