Demak bu shart bajarildi deb hisoblash mumkun, bu esa yuqoridagi hisoblashlar to’g’ri ekanligini isbotlaydi.
Shuning bilan yuqoridagi sistemasinig yechimini birinchi yaqinlashuvchisi topildi. Bundan so’ng, berilgan sistemasining yechimining keyingi yaqinlashuvchilarida ko’rsatilgan tartib bilan aniqlanadi. Olingan natijalar quyidagi jadvalga joylashtirilgan
K
|
|
|
|
|
|
|
0
|
0.000000
|
0.000000
|
0.000000
|
0.000000
|
0.000000
|
9.646378
|
1
|
0.655795
|
0.340081
|
0.453441
|
0.662151
|
0.524424
|
4.283544
|
2
|
0.829120
|
0.203306
|
0.180018
|
0.885773
|
0.521819
|
0.026856
|
3
|
0.797559
|
0.217627
|
0.170817
|
0.921591
|
0.511841
|
0.000649
|
4
|
0.798707
|
0.224296
|
0.172697
|
0.920801
|
0.510045
|
0.000017
|
5
|
0.798379
|
0.224531
|
0.172231
|
0.920644
|
0.510904
|
0.000002
|
6
|
0.798379
|
0.224531
|
0.172232
|
0.920644
|
0.510903
|
0.000000
|
K
|
|
|
|
|
|
|
0
|
0.000000
|
0.000000
|
0.000000
|
0.000000
|
0.000000
|
0.000000
|
1
|
0.655795
|
0.340081
|
0.453441
|
0.662151
|
0.524424
|
0.061357
|
2
|
0.173325
|
0.136677
|
0.273422
|
0.223621
|
0.002604
|
0.009699
|
3
|
0.031561
|
0.014320
|
0.003936
|
0.035817
|
0.009977
|
0.023949
|
4
|
0.001147
|
0.006669
|
0.003384
|
0.000789
|
0.001796
|
0.021915
|
5
|
0.000328
|
0.000234
|
0.000465
|
0.000156
|
0.000858
|
-0.000038
|
6
|
0.000000
|
0.000001
|
0.000002
|
0.000000
|
0.000001
|
0.000011
|
K
|
|
|
|
|
|
|
0
|
0.83
|
3.21
|
4.289
|
6.25
|
4.96
|
0.000000
|
1
|
0.672009
|
0.795986
|
1.521095
|
0.924002
|
0.175626
|
0.198044
|
2
|
0.105898
|
0.049845
|
0.004092
|
0.107192
|
0.031715
|
0.313910
|
3
|
0.006292
|
0.020912
|
0.010876
|
0.005445
|
0.005147
|
0.302529
|
4
|
0.001373
|
0.000341
|
0.001519
|
0.000541
|
0.003489
|
0.257363
|
5
|
0.000000
|
0.000001
|
0.000002
|
0.0000000
|
0.0000001
|
0.105842
|
6
|
0.000000
|
0.000000
|
0.000000
|
0.000000
|
0.000000
|
0.246918
|
Bu topilgan jadvaldagi natijalardan kelib chiqqan natija y (2.15) sistemasining Qo’shma gradeintlar usulining yangi varyanti bilan topilgan yechimini keltiramiz
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