2019年3月27日星期三

習題1-7 (b) H'n(x)的計算　

Ｈn(x)=

到google " online differential equation of first order" 搜尋
進入
https://www.symbolab.com/solver/linear-first-order-differential-equation-calculator

輸入資料

\left(x-1\right)+\left(x-1\right)^2\cdot \left(-0.307\right)+\left(x-2\right)\left(x-1\right)^2\cdot \left(0.114\right)\:+\left(-0.0323\right)\left(x-2\right)^2\left(x-1\right)^2+\left(x-1\right)^2\left(x-2\right)^2\left(x-3\right)\left(0.0091\right)+\left(x-1\right)^2\left(x-2\right)^2\left(x-3\right)^2\left(-0.0091\right)+\left(x-1\right)^2\left(x-2\right)^2\left(x-3\right)^2\left(x-4\right)\left(0.0004\right)

計算結果為
0.0004x^7-0.0155x^6+0.1607x^5-0.7924x^4+2.2079x^3-3.7649x^2+4.3624x-2.1586

=0.0004x^7-0.0155x^6+0.1607x^5-0.7924x^4+2.2079x^3-3.7649x^2+4.3624x-2.1586

2019年3月26日星期二

C語言　例題 EX2-9 定點回路法求非線性方程式 f(x)=1/5^x - x = 0

/* ex2-9.c is used for solving nonlinear equation
 * based on Fixed-Point Algorithm g(x)=x with initial
 * approximation P0.
 */
#include <stdio.h>
#include <math.h>
#define MAX  50
#define TOL 0.0001
#define  g(x)   (1/pow(5,x))
void main()
{
   int i=1;
   double x0,x;
   x0=0.45;
   while(i<=MAX)
   {
      x=g(x0);
      printf("%-2d  %10.7lf\n",i-1,x0);
      if(fabs(x-x0) < TOL)
      {
         printf("The Root=%10.7lf  x-x0=%10.7lf\n",x,fabs(x-x0));
         exit(0);
      }
      i++;
      x0=x;
   }
   printf("Fixed-point failed after %d iteration.\n",i);
   return;
}

輸入畫面 : 無

輸出畫面

0    0.4500000
1    0.4846894
2    0.4583705
3    0.4782035
4    0.4631803
5    0.4745160
6    0.4659374
7    0.4724151
8    0.4675155
9    0.4712167
10   0.4684181
11   0.4705327
12   0.4689340
13   0.4701421
14   0.4692289
15   0.4699191
16   0.4693974
17   0.4697917
18   0.4694936
19   0.4697189
20   0.4695486
21   0.4696773
The Root= 0.4695801  x-x0= 0.0000973

2-10 定點迴路法 C語言求根

2-10 定點迴路法 C語言求根
Fixed-Point Iteration. Root finding.

https://www.codeproject.com/Questions/723378/Fixed-Point-Iteration-Root-finding

Algorithm:

INPUT initial approximation p0; tolerance TOL; maximum number of iterations N0.
OUTPUT approximate solution p or message of failure.

Step 1 Set i=1.
Step 2 While i <= N0 do Steps 3-6.
Step 3 Set p=g(p0). (Compute pi.)
Step 4 If |p-p0|<tol mode="hold"> OUTPUT (p); (The procedure was successful.)
STOP.
Step 5 Set i=i+1.
Step 6 Set p0=p. (Update p0.)
Step 7 OUTPUT ('The method failed after N0 iterations, N0=', N0);
(The procedure was unsuccessful.)
STOP.

程式

#include <stdio.h>
#include <math.h>

double f(double x)
{
return x*x*x*x-3*x*x-3;  //change equation for each problem
}

double g(double x)
{
return pow(3*x*x+3,.25);
}

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);
    scanf ("%lf %lf %d",&p0,&Tol, &No);

    while (i<=No)
    {
        p = g(p0);

        if((fabs(p-p0))<Tol)
        {
            //printf("%lf", &p);
            break;
        }
        printf("Iteration %d: Current value = %lf\n", i, p);

        i++;  //i=i+1
        p0=p;

        if (i>No)
        {
        printf("Method Failed after %d", No);
        printf(" iterations");
        }

    }

}

輸入資料

10.0 0.001 50

輸出結果

Iteration 1: Current value = 4.172157
Iteration 2: Current value = 2.725997
Iteration 3: Current value = 2.242595
Iteration 4: Current value = 2.062271
Iteration 5: Current value = 1.992422
Iteration 6: Current value = 1.965006
Iteration 7: Current value = 1.954192
Iteration 8: Current value = 1.949919
Iteration 9: Current value = 1.948229

C語言例題2-7已知方程式 e^x + x^-2 + 2 cosx -6 正割法(secant method)

C語言例題2-7已知方程式 e^x + x^-2 + 2 cosx -6 正割法(secant method)

利用正割法(secant method) 採用誤差0.00001 找出f(x)=0的根

程式

// C Program to find root of an equations using secant method 
#include <stdio.h>
#include <math.h>


// function takes value of x and returns f(x) 
float f(float x) 
{ 
    // we are taking equation as  e^x + x^-2 + 2 cosx -6
    float f =  exp(x) + (1/pow(x,2)) + 2*cos(x) - 6.0;
    return f; 
} 
  
void secant(float x1, float x2, float E) 
{ 
    float n = 0, xm, x0, c; 
    if (f(x1) * f(x2) < 0)
    { 
        do { 
            // calculate the intermediate value 
            x0 = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1)); 
  
            // check if x0 is root of equation or not 
            c = f(x1) * f(x0); 
  
            // update the value of interval 
            x1 = x2; 
            x2 = x0; 
  
            // update number of iteration 
            n++; 
  
            // if x0 is the root of equation then break the loop 
            if (c == 0) 
                break; 
            xm = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));
            printf("xm=%0.5lf\n", xm); 
        
            
        } while (fabs(xm - x0) >= E); // repeat the loop 
                                // until the convergence 
  
        printf("Root of the given equation=%0.5lf", x0); 
        printf("\nNo. of iterations = %0.0lf", n); 
    
        
    } else
        
        printf("Can not find a root in the given inteval"); 
} 
  
// Driver code 
int main() 
{ 
    // initializing the values 
    float x1 = 1.8, x2 = 2.0, E = 0.0001; 
    secant(x1, x2, E); 
    return 0; 
}

輸出畫面

xm=1.82444                                                                                                                                       
xm=1.82498                                                                                                                                       
xm=1.82498                                                                                                                                       
Root of the given equation=1.82498                                                                                                               
No. of iterations = 3                                                                                                                            
                                                                                                                                                 
...Program finished with exit code 0                                                                                                             
Press ENTER to exit console.

C語言正割法 Program to find root of an equations using secant method

2-7 正割法 Program to find root of an equations using secant method

源自於

https://www.geeksforgeeks.org/program-to-find-root-of-an-equations-using-secant-method/

Examples :

Input : equation = x3 + x - 1 
        x1 = 0, x2 = 1, E = 0.0001
Output : Root of the given equation = 0.682326
         No. of iteration=5

Algorithm

Initialize: x1, x2, E, n // E = convergence indicator calculate f(x1),f(x2) if(f(x1) * f(x2) = E); //repeat the loop until the convergence print 'x0' //value of the root print 'n' //number of iteration } else print "can not found a root in the given interval"

程式:

// C Program to find root of an equations using secant method 
#include <stdio.h>

// function takes value of x and returns f(x) 
float f(float x) 
{ 
    // we are taking equation as x^3+x-1 
    float f = pow(x, 3) + x - 1; 
    return f; 
} 
  
void secant(float x1, float x2, float E) 
{ 
    float n = 0, xm, x0, c; 
    if (f(x1) * f(x2) < 0)
    { 
        do { 
            // calculate the intermediate value 
            x0 = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1)); 
  
            // check if x0 is root of equation or not 
            c = f(x1) * f(x0); 
  
            // update the value of interval 
            x1 = x2; 
            x2 = x0; 
  
            // update number of iteration 
            n++; 
  
            // if x0 is the root of equation then break the loop 
            if (c == 0) 
                break; 
            xm = (x1 * f(x2) - x2 * f(x1)) / (f(x2) - f(x1));
            printf("xm=%0.5lf\n", xm); 
        
            
        } while (fabs(xm - x0) >= E); // repeat the loop 
                                // until the convergence 
  
        printf("Root of the given equation=%0.5lf", x0); 
        printf("\nNo. of iterations = %0.0lf", n); 
    
        
    } else
        
        printf("Can not find a root in the given inteval"); 
} 
  
// Driver code 
int main() 
{ 
    // initializing the values 
    float x1 = 0, x2 = 1, E = 0.0001; 
    secant(x1, x2, E); 
    return 0; 
}

輸出結果

xm=0.69005                                                                                                                                     
xm=0.68202                                                                                                                                     
xm=0.68233                                                                                                                                     
xm=0.68233                                                                                                                                     
Root of the given equation=0.68233                                                                                                             
No. of iterations = 5

C語言例題2-7 已知方程式 e^x + x^-2 + 2 cosx -6 利用正割法找出f(x)=0的根

C語言例題2-6 已知方程式 e^x + x^-2 + 2 cosx -6 利用正割法找出f(x)=0的根

C語言例題2-7 已知方程式 e^x + x^-2 + 2 cosx -6 利用正割法找出f(x)=0的根

/* ex2-7.c Secant Method is similar to Newton-Raphson
* Method used for find solutions to f(x)=0 given
* initial approximations x0 and x1.
*/
#include <stdio.h>
#include <math.h>
#define MAX 50
#define PI 3.14159
#define TOL 0.00001
#define f(x) (exp(x)+1/pow(2,x)+2*cos(x)-6)
void main()
{
int i=2;
double x0,x1,x,q0,q1;
x0=1.8;
x1=2.0;
q0=f(x0);
q1=f(x1);
printf("i xi f(x)\n");
printf("%-2d %10.6lf %10.6lf\n",0,x0,q0);
printf("%-2d %10.6lf %10.6lf\n",1,x1,q1);
while(i<=MAX)
{
x=x1-q1*(x1-x0)/(q1-q0);
printf("%-2d %10.6lf %10.6lf\n",i,x,f(x));
if(fabs(x-x1) < TOL)
{
printf("The Root=%10.6lf f(%10.6lf)=%10.6lf\n",x,x,f(x));
exit(0);
}
else
{
i++;
x0=x1;
q0=q1;
x1=x;
q1=f(x);
}
}
if(i>MAX)
printf("Secant Method faileds!!!\n");
return;
}

輸出畫面

i xi f(x)
0 1.800000 -0.117582
1 2.000000 0.806762
2 1.825441 -0.016116
3 1.828860 -0.002147
4 1.829385 0.000007
5 1.829384 -0.000000

The Root= 1.829384 f( 1.829384)= -0.000000

2019年3月22日星期五

Bisection Method – Algorithm, Flowchart and Code in C

源自於

https://justcode.me/uncategorized/bisection-method-algorithm-flowchart-code-c/

The bisection methodis one of the simplest and most reliable of iterative methods for the solution of nonlinear equations. This method, also known as binary chopping or half-interval method, relies on the fact that if f(x) is real and continuous in the interval a < x < b, and f(a) and f(b) are of opposite signs, that is,

f(a)*f(b) < 0

then there is at least one real root in the interval between a and b.

Bisection method is a closed bracket method and requires two initial guesses. It is the simplest method with slow but steady rate of convergence.

Features of Bisection Method:

Type – closed bracket
No. of initial guesses – 2
Convergence – linear
Rate of convergence – slow but steady
Accuracy – good
Programming effort – easy
Approach – middle point

Bisection Method Algorithm and Flowchart:

Algorithm:

Start

Decide initial values for x1 and x2 and stopping criterion, E.
Compute f1 = f(x1) and f2 = f(x2).
If f1 * f2>0, x1 and x2 do not bracket any root and go to step 7;
Otherwise continue.
Compute x0 = (x1+x2)/2 and compute f0 = f(x0)
If f1*f0 < 0 then
set x2 = x0
else
set x1 = x0
set f1 = f0
If absolute value of (x2 – x1)/x2 is less than error E, then
root = (x1 + x2)/2
write the value of root
go to step 7
else
go to step 4
Stop.

Flow Chart

Above given Algorithm and Flowchart of Bisection Methods Root computation is a simple and easier way of understanding how the bracketing system works, algorithm and flowchart may not follow same procedure, yet they give the same outputs.

Tabular Example of Bisection Method Numerical Computation

Consider f(x) = x3 + 3x – 5, where [x1 = 1, x2 = 2] and E = 0.001.

i	x1	x0	x2	f(x1)	f(x0)	f(x2)
1	1	1.5	2	–1	2.875	9
2	1	1.25	1.5	–1	0.703125	2.875
3	1	1.125	1.25	–1	–0.201171875	0.703125
4	1.125	1.1875	1.25	–0.201171875	0.237060546875	0.703125
5	1.125	1.15625	1.1875	–0.201171875	0.014556884765625	0.237060546875
6	1.125	1.140625	1.15625	–0.201171875	–0.0941429138183594	0.014556884765625
7	1.140625	1.1484375	1.15625	–0.0941429138183594	–0.0400032997131348	0.014556884765625
8	1.1484375	1.15234375	1.15625	–0.0400032997131348	–0.0127759575843811	0.014556884765625
9	1.15234375	1.154296875	1.15625	–0.0127759575843811	0.000877253711223602	0.014556884765625

Note: Bisection method guarantees the convergence of a function f(x) if it is continuous on the interval [a,b] (denoted by x1 and x2 in the above algorithm. For this, f(a) and f(b) should be of opposite nature i.e. opposite signs. The slow convergence in bisection method is due to the fact that the absolute error is halved at each step. Due to this the method undergoes linear convergence, which is comparatively slower than the Newton-Raphson, secant or false-position method.

Bisection Method – Code in C Programming

METHOD 1:

This program in C is used to demonstrate bisection method. Bisection method is one of the many root finding methods.
In this method we are given a function f(x) and we approximate 2 roots a and b for the function such that f(a).f(b)<0
Then we find another point
c=(a+b)/2
if f(c)==0
then root=c;
else
if f(a).f(c)<0
b=c;
if f(b).f(c)<0
a=c;
and we repeat these steps for the given number of iterations.

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

#include"stdio.h"

#include"math.h"

double F(double x)

{

    return(pow(x,3)+3*x-5);//This return the value of the function

}

int main()

{

    printf("This program illustrates the bisection method in C\n");

    printf("x^3 + 3*x - 5 = 0\n");

    double x0,x1;

    printf("Enter the first approximation to the root\n");

    scanf("%lf",&x0);

    printf("Enter the second approximation to the root\n");

    scanf("%lf",&x1);

    int iter;

    printf("Enter the number of iterations you want to perform\n");

    scanf("%d",&iter);

    int ctr=1;

    double l1=x0;

    double l2=x1;

    double r,f1,f2,f3;

    //We check if the initail approximations are the root or not

    if(F(l1)==0)

    r=l1;

    else

    if(F(l2)==0)

    r=l2;

    else

    {

    while(ctr <= iter)

    {//this is an implementation of the algorithm mentioned above

        f1=F(l1);

        r=(l1+l2)/2.0;

        f2=F(r);

        f3=F(l2);

        if(f2==0)

        {

            r=f2;

            break;

        }

        printf("The root after %d iteration is %lf\n",ctr,r);

        if(f1*f2<0)

        l2=r;

        else

        if(f2*f3<0)

        l1=r;

        ctr++;

    }

    }

    printf("The approximation to the root is %lf",r);

    getch();

}

/*A sample run of the program was carried out and the results were found as:-

This program illustrates the bisection method in C

x^3 + 3*x - 5 = 0

Enter the first approximation to the root

1

Enter the second approximation to the root

2

Enter the number of iterations you want to perform

9

The root after 1 iteration is 1.500000

The root after 2 iteration is 1.250000

The root after 3 iteration is 1.125000

The root after 4 iteration is 1.187500

The root after 5 iteration is 1.156250

The root after 6 iteration is 1.146025

The root after 7 iteration is 1.148438

The root after 8 iteration is 1.152344

The root after 9 iteration is 1.154297

The root is 1.154297

*/

METHOD 2:

Source Code for Bisection Method
for equation f(x) = x^3 – 4*x – 9

Bisection Method - Method 2 for equation

C

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

#include "math.h"

#include "conio.h"

float fun (float x)

{

    return (x*x*x - 4*x - 9);

}

void bisection (float *x, float a, float b, int *itr)

/* this function performs and prints the result of one iteration */

{

    *x=(a+b)/2;

    ++(*itr);

    printf("Iteration no. %3d X = %7.5f\n", *itr, *x);

}

void main ()

{

    int itr = 0, maxmitr;

    float x, a, b, allerr, x1;

    printf("\nEnter the values of a, b, allowed error and maximum iterations:\n");

    scanf("%f %f %f %d", &a, &b, &allerr, &maxmitr);

    bisection (&x, a, b, &itr);

    do

    {

        if (fun(a)*fun(x) < 0)

            b=x;

        else

            a=x;

        bisection (&x1, a, b, &itr);

        if (fabs(x1-x)<allerr)

        {

            printf("After %d iterations, root = %6.4f\n", itr, x1);

            return 0;

        }

        x=x1;

    }

    while (itr<maxmitr);

    printf("The solution does not converge or iterations are not sufficient");

    return 1;

}

訂閱：文章 (Atom)

alex9ufo 聰明人求知心切

2019年3月27日星期三

習題1-7 (b) H'n(x)的計算

2019年3月26日星期二

C語言　例題 EX2-9 定點回路法求非線性方程式 f(x)=1/5^x - x = 0

2-10 定點迴路法 C語言求根

C語言例題2-7已知方程式 e^x + x^-2 + 2 cosx -6 正割法(secant method)

C語言正割法 Program to find root of an equations using secant method

2-7 正割法 Program to find root of an equations using secant method

C語言例題2-7 已知方程式 e^x + x^-2 + 2 cosx -6 利用正割法找出f(x)=0的根

2019年3月22日星期五

Bisection Method – Algorithm, Flowchart and Code in C

Bisection Method – Algorithm, Flowchart and Code in C

Bisection Method Algorithm and Flowchart:

Tabular Example of Bisection Method Numerical Computation

Bisection Method – Code in C Programming

METHOD 1:

METHOD 2:

2024產專班作業2

2019年3月27日 星期三

2019年3月26日 星期二

2-7 正割法 Program to find root of an equations using secant method

2019年3月22日 星期五

Bisection Method Algorithm and Flowchart:

Tabular Example of Bisection Method Numerical Computation

Bisection Method – Code in C Programming

METHOD 1:

METHOD 2:

Share this:

2019年3月27日星期三

2019年3月26日星期二

2019年3月22日星期五