Sei eine reelle Folge durch folgenden Algorithmus gegeben:
n = 10
b = 2
a = sqrt(b)
s = a^a
for i in range(1,n):
s = a^s
print(numerical_approx(s))
mit
1.76083955588003
1.84091086929101
1.89271269682851
1.92699970184710
1.95003477380582
1.96566488651732
1.97634175440970
1.98366839930382
1.98871177341395
Ich vermute Konvergenz gegen 2.0.
Wie kann ich das beweisen?
Vielen Dank im voraus.
bis n = 100
1.76083955588003
1.84091086929101
1.89271269682851
1.92699970184710
1.95003477380582
1.96566488651732
1.97634175440970
1.98366839930382
1.98871177341395
1.99219088294706
1.99459445071210
1.99625666626586
1.99740700114134
1.99820347750870
1.99875513308459
1.99913731011939
1.99940211832500
1.99958562293568
1.99971279632964
1.99980093549297
1.99986202375778
1.99990436444334
1.99993371158210
1.99995405289782
1.99996815214924
1.99997792487387
1.99998469874709
1.99998939400781
1.99999264849993
1.99999490433494
1.99999646795725
1.99999755177603
1.99999830302118
1.99999882374426
1.99999918468182
1.99999943486458
1.99999960827801
1.99999972847903
1.99999981179601
1.99999986954694
1.99999990957683
1.99999993732344
1.99999995655592
1.99999996988686
1.99999997912716
1.99999998553205
1.99999998997158
1.99999999304883
1.99999999518182
1.99999999666029
1.99999999768509
1.99999999839543
1.99999999888779
1.99999999922908
1.99999999946564
1.99999999962961
1.99999999974326
1.99999999982204
1.99999999987665
1.99999999991450
1.99999999994074
1.99999999995892
1.99999999997153
1.99999999998026
1.99999999998632
1.99999999999052
1.99999999999343
1.99999999999544
1.99999999999684
1.99999999999781
1.99999999999848
1.99999999999895
1.99999999999927
1.99999999999949
1.99999999999965
1.99999999999976
1.99999999999983
1.99999999999988
1.99999999999992
1.99999999999994
1.99999999999996
1.99999999999997
1.99999999999998
1.99999999999999
1.99999999999999
1.99999999999999
2.00000000000000
2.00000000000000
2.00000000000000
2.00000000000000
2.00000000000000
2.00000000000000
2.00000000000000
2.00000000000000
2.00000000000000
2.00000000000000
2.00000000000000
Es existieren b > 2, wo die Folge nicht konvergiert.
n = 10
b = 3
a = sqrt(b)
s = a^a
for i in range(1,n):
s = a^s
print(numerical_approx(s))
4.14695063102757
9.75661446705656
212.590605146743
5.19694516711664e50
+infinity