/*
g1(x) = 10 / (x^3-1)
g2(x) = (x + 10)^1/4
g3(x) = ((x+10)^1/2) / x
*/
#include<conio.h>
#include<stdio.h>
#include<math.h>
#define f(x) (10 / (x*x*x-1))
#define e 0.0001
#define MAX 30
int main(void)
{
float x0,x1,f1 ;
int i=1;
clrscr();
printf("Using Fixed Point Method / Iterative Method to find root of x4-x-10 = 0 \n");
printf("Enter value for x0\n");
printf("x0: ");
scanf("%f",&x0);
printf("steps\tx0\tx1\n");
while (i<=MAX)
{
x1=f(x0);
printf("%d\t%.4f\t%.4f\n",i,x0,x1);
if(fabs((x1-x0)/x1)<=e)
{
printf("The root is %.4f",x1);
exit(0);
}
else
{
x0=x1;
i++;
}
}
getch();
}
輸出畫面
Using Fixed Point Method / Iterative Method to find root of x4-x-10 = 0
Enter value for x0
x0: -5
steps x0 x1
1 -5.0000 -0.0794
2 -0.0794 -9.9950
3 -9.9950 -0.0100
4 -0.0100 -10.0000
5 -10.0000 -0.0100
6 -0.0100 -10.0000
7 -10.0000 -0.0100
8 -0.0100 -10.0000
9 -10.0000 -0.0100
10 -0.0100 -10.0000
11 -10.0000 -0.0100
12 -0.0100 -10.0000
13 -10.0000 -0.0100
14 -0.0100 -10.0000
15 -10.0000 -0.0100
16 -0.0100 -10.0000
17 -10.0000 -0.0100
18 -0.0100 -10.0000
19 -10.0000 -0.0100
20 -0.0100 -10.0000
21 -10.0000 -0.0100
22 -0.0100 -10.0000
23 -10.0000 -0.0100
24 -0.0100 -10.0000
25 -10.0000 -0.0100
26 -0.0100 -10.0000
27 -10.0000 -0.0100
28 -0.0100 -10.0000
29 -10.0000 -0.0100
30 -0.0100 -10.0000
無法收斂
/*
g1(x) = 10 / (x^3-1)
g2(x) = (x + 10)^1/4
g3(x) = ((x+10)^1/2) / x
*/
#include<conio.h>
#include<stdio.h>
#include<math.h>
#define f(x) pow(x+10,0.25)
#define e 0.0001
#define MAX 30
Using Fixed Point Method / Iterative Method to find root of x4-x-10 = 0
Enter value for x0
x0: -5
steps x0 x1
1 -5.0000 1.4953
2 1.4953 1.8413
3 1.8413 1.8550
4 1.8550 1.8556
5 1.8556 1.8556
The root is 1.8556
另一個程式
#include <stdio.h>
#include <math.h>
double f(double x)
{
return x*x*x*x-x-10; //change equation for each problem
}
double g(double x)
{
//return 10 / (x*x*x-1); //change equation for each problem
//return pow ( (x+10), 0.25); //change equation for each problem
return (1/x)*pow((x+10), 0.5); //change equation for each problem
/*
g1(x) = 10 / (x^3-1)
g2(x) = (x + 10)^1/4
g3(x) = ((x+10)^1/2) / x
*/
}
int main()
{
double p, p0, Tol;
int i=1;
int No;
printf("Enter approximate p: ");
scanf ("%lf", &p0);
printf("Desired Tolerance: ");
scanf ("%lf", &Tol);
printf("Maximum Iterations: ");
scanf ("%d", &No);
printf("\n\n");
while (i<=No)
{
p = g(p0);
if((fabs(p-p0))<Tol)
{
//printf("%lf", &p);
break;
}
printf("Iteration %d: Current value = %lf , the root value = %lf\n", i, p , f(p));
i++; //i=i+1
p0=p;
if (i>No)
{
printf("Method Failed after %d", No);
printf(" iterations");
}
}
}
-5
0.0001
20
Enter approximate p: Desired Tolerance: Maximum Iterations:
Iteration 1: Current value = -0.079365 , the root value = -9.920595
Iteration 2: Current value = -9.995003 , the root value = 9980.023728
Iteration 3: Current value = -0.010005 , the root value = -9.989995
Iteration 4: Current value = -9.999990 , the root value = 9999.959930
Iteration 5: Current value = -0.009990 , the root value = -9.990010
Iteration 6: Current value = -9.999990 , the root value = 9999.960110
Iteration 7: Current value = -0.009990 , the root value = -9.990010
Iteration 8: Current value = -9.999990 , the root value = 9999.960110
Iteration 9: Current value = -0.009990 , the root value = -9.990010
Iteration 10: Current value = -9.999990 , the root value = 9999.960110
Iteration 11: Current value = -0.009990 , the root value = -9.990010
Iteration 12: Current value = -9.999990 , the root value = 9999.960110
Iteration 13: Current value = -0.009990 , the root value = -9.990010
Iteration 14: Current value = -9.999990 , the root value = 9999.960110
Iteration 15: Current value = -0.009990 , the root value = -9.990010
Iteration 16: Current value = -9.999990 , the root value = 9999.960110
Iteration 17: Current value = -0.009990 , the root value = -9.990010
Iteration 18: Current value = -9.999990 , the root value = 9999.960110
Iteration 19: Current value = -0.009990 , the root value = -9.990010
Iteration 20: Current value = -9.999990 , the root value = 9999.960110
Method Failed after 20 iterations
-5
0.0001
20
Enter approximate p: Desired Tolerance: Maximum Iterations:
Iteration 1: Current value = 1.495349 , the root value = -6.495349
Iteration 2: Current value = 1.841325 , the root value = -0.345977
Iteration 3: Current value = 1.855026 , the root value = -0.013701
Iteration 4: Current value = 1.855563 , the root value = -0.000536
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