Реализация метода простой итерации, релаксации и метод Ньютона, страница 3

w

итераций

выход

погрешность

1

300

0.013403

0.0031079

1

300

0.0043695

0.0041238

0.9

48

7.1526e-07

0.0046067

0.9

19

7.1526e-07

0.0046155

0.8

24

7.1526e-07

0.0046067

0.8

10

9.5367e-07

0.0046154

0.7

13

4.7684e-07

0.0046069

0.7

6

9.5367e-07

0.0046154

0.6

8

4.7684e-07

0.0046067

0.6

4

4.7684e-07

0.0046154

0.5

3

2.8578e-10

0.0046068

0.5

2

4.7684e-07

0.0046155

0.4

8

4.7684e-07

0.0046069

0.4

4

7.1526e-07

0.0046155

0.3

14

9.3751e-10

0.0046068

0.3

6

4.7684e-07

0.0046155

0.2

23

9.8302e-07

0.0046069

0.2

10

7.5395e-07

0.0046155

0.1

48

7.5395e-07

0.0046069

0.1

18

7.1526e-07

0.0046155

x=1.2

точное

2.0736

h_x=0.2,

h_t=0.5

t=1.5

2.053012

t=2

2.052973

h_x=0.1,

h_t=0.5

t=1.5

2.068448

t=2

2.068435

h_x=0.1,

h_t=0.25

t=1.5

2.0684366

t=2

2.0684356

h_x=0.05,

h_t=0.25

t=1.5

2.0723095

t=2

2.0723095

h_x=0.05,

h_t=0.1

t=1.5

2.0723099

t=2

2.0723099

x=1.2

точное

2.0736

h_x=0.2,

h_t=0.5

t=1.5

2.05297303

t=2

2.05297374

h_x=0.1,

h_t=0.5

t=1.5

2.0684359

t=2

2.0684359

h_x=0.1,

h_t=0.25

t=1.5

2.0684359

t=2

2.0684359

h_x=0.05,

h_t=0.25

t=1.5

2.0723097

t=2

2.0723095

h_x=0.05,

h_t=0.1

t=1.5

2.0723097

t=2

2.0723097

x=1.2

точное

2.0736

h_x=0.2,

h_t=0.5

t=1.5

2.039104

t=2

2.039104

h_x=0.1,

h_t=0.5

t=1.5

2.064891

t=2

2.064891

h_x=0.1,

h_t=0.25

t=1.5

2.064891

t=2

2.064891

h_x=0.05,

h_t=0.25

t=1.5

2.071417

t=2

2.071417

h_x=0.05,

h_t=0.1

t=1.5

2.071417

t=2

2.071417