1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
|
//Example 2.25
//Mullers Method
//Page no. 43
clc;clear;close;
deff('y=f(x)','y=cos(x)-x*exp(x)')
zi=[-1;0;1];
s=["i","z2","z0","z1","f2","f0","f1","a0","a1","a2","zr+","zr-"]
li(1)=(zi(3,1)-zi(2,1))/(zi(2,1)-zi(1,1))
hi(1)=zi(3,1)-zi(2,1);
for i=2:7
for j=1:3
fz(j,i-1)=f(zi(j,i-1))
end
di(i-1)=1+li(i-1)
gi(i-1)=fz(1,i-1)*li(i-1)^2-fz(2,i-1)*di(i-1)^2+fz(3,i-1)*(li(i-1)+di(i-1))
D1(i-1)=gi(i-1)+sqrt(gi(i-1)^2-4*fz(3,i-1)*di(i-1)*li(i-1)*(fz(1,i-1)*li(i-1)-fz(2,i-1)*di(i-1)+fz(3,i-1)))
D2(i-1)=gi(i-1)-sqrt(gi(i-1)^2-4*fz(3,i-1)*di(i-1)*li(i-1)*(fz(1,i-1)*li(i-1)-fz(2,i-1)*di(i-1)+fz(3,i-1)))
if abs(D1(i-1))>abs(D2(i-1)) then
li(i)=-2*fz(3,i-1)*di(i-1)/D1(i-1)
else
li(i)=-2*fz(3,i-1)*di(i-1)/D2(i-1)
end
hi(i)=li(i)*hi(i-1);
z(i-1)=zi(3,i-1)+hi(i)
for j=1:2
zi(j,i)=zi(j+1,i-1)
end
zi(3,i)=z(i-1)
end
for i=1:12
if i==1 then
printf(s(i))
for j=1:6
printf('\t\t\t%i',j-1)
end
elseif i<=4
printf('\n %s',s(i))
for j=1:6
printf('\t\t%.10f',zi(i-1,j))
end
elseif i<=7
printf('\n %s',s(i))
for j=1:6
printf('\t\t%.10f',fz(i-4,j))
end
elseif i<=8
printf('\n %s',s(i))
for j=1:6
printf('\t\t%.10f',li(j))
end
elseif i<=9
printf('\n %s',s(i))
for j=1:6
printf('\t\t%.10f',di(j))
end
elseif i<=10
printf('\n %s',s(i))
for j=1:6
printf('\t\t%.10f',gi(j))
end
elseif i<=11
printf('\n %s',s(i))
for j=1:6
printf('\t\t%.10f',z(j))
end
elseif i<=12
printf('\n %s',s(i))
for j=1:6
printf('\t\t%.10f',zi(j))
end
end
end
printf('\n\nAt the end of the %i iteration, the root of the equation is %.10f',j-2,z(j))
|
