HailstoneMatrix

2246 days ago by calkin

A=matrix(6,6,[1,0,0,1,0,0,0,0,0,0,1/3,0 ,0,1,0,0,1,0 ,0,0,0,0,1/3,0, 0,0,1,0,0,1, 0,0,0,0,1/3,0]) 
       
show(A) 
       
A^37*vector([0,0,1,0,0,0]).n() 
       
(2110.51078377014, 556.579350102467, 2110.51078377014, 556.579350102467,
2111.51078377014, 556.579350102467)
(2110.51078377014, 556.579350102467, 2110.51078377014, 556.579350102467, 2111.51078377014, 556.579350102467)
9672./2552. 
       
3.78996865203762
3.78996865203762
2110./556. 
       
3.79496402877698
3.79496402877698
A.eigenvalues() 
       
[1, -1, 0, 0, -0.2637626158259734?, 1.263762615825974?]
[1, -1, 0, 0, -0.2637626158259734?, 1.263762615825974?]
A^44*vector([7,3,6,2,5,1]).n() 
       
(170541.275545315, 44981.7909239450, 170539.275545315, 44981.7909239450,
170540.275545315, 44981.7909239450)
(170541.275545315, 44981.7909239450, 170539.275545315, 44981.7909239450, 170540.275545315, 44981.7909239450)

If we assume that the model is correct, then we'd end up with about $l_n=ce^{an}$ numbers on level $n$.

Since $l_{40}=6049$ and $l_{60} = 651407$, 

exp(log(651407/6049.)/20) 
       
1.26359658012361
1.26359658012361

This appears to be very close to the largest eigenvalue of the matrix above; which appears to be

the root of $-1 - 3x +3x^2$ 

 

solve(-1+3*x+3*x^2,x) 
       
[x == -1/6*sqrt(21) - 1/2, x == 1/6*sqrt(21) - 1/2]
[x == -1/6*sqrt(21) - 1/2, x == 1/6*sqrt(21) - 1/2]
x=sqrt(21.)/6 + 1/2 print x 
       
1.26376261582597
1.26376261582597
-1+3*x+3*x^2 
       
-3.05311331771918e-16
-3.05311331771918e-16