牛顿的方法程序(在C中)循环无限运行

Question

C中的这段代码(附在帖子中)使用Newton-Raphson方法在特定区间中找到多项式的根.

对于某些多项式,这段代码可以完美地x^3 + x^2 + x + 1运行,但运算符对于某些多项式来说是无限的x^3 - 6*x^2 + 11*x - 6.也就是说,此代码适用于输入间隔中具有一个或零根的多项式,但如果存在多个根,则它将无限期运行.

如果有人找到解决方案,请告诉我.我在代码中写了一些评论来引导读者,但如果有人发现很难理解,可以在评论中提问,我会解释一下.

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

int check(float num)                          //just a function to check for  the correct input
{
    char c;
    scanf("%c",&c);

    if(isalpha((int)c))
        printf("you entered an alphabet\n");

    else
        printf("you entered a character, please retry\n");

    return 0;
}

float func(float *p,int order,double x)                     //calculates the value of the function required in the formmula in main
{
    double fc=0.0;
    int i;

    for(i=0;i<=order;i++)
    {
        fc=fc+(double)(*p)*pow(x,(double)i);
        p++;
    }

    return fc;
}

float derv(float *q,int order,double x)               //calculates the derivative of the function required in the formmula in main
{     
    double dv=0.0,i;

    for(i=1;i<=order;i++)
    {
        dv=dv+(double)(*q)*(pow(x,(double)(i-1)))*(double)i;
        q++;
    }

    return dv;
}


int main()
{
    float coeff[1000];
    int order,count=0,i,j=0;
    char ch;
    float a,b;
    double val[5];

    printf("roots of polynomial using newton and bisection method\n");
    printf("enter the order of the equation\n");

    while(scanf("%d",&order)!=1)
    {
        printf("invalid input.please retry\n");
        while(getchar()!='\n'){}          
    }     

    printf("enter the cofficients\n");

    for(i=0;i<=order;i++)
    {
        printf("for x^%d  :",order-i);
        printf("\n");

        while(scanf("%f",&coeff[i])!=1)
        {
            check(coeff[i]);
        }   
    }

    while(getchar()!='\n'){}                                 //this clears the buffer accumulated upto pressing enter

    printf("the polynomial you entered is :\n");

    for(i=0;i<=order;i++)
    {
        printf(" %fx^%d ",coeff[i],order-i);
    }

    printf("\n");

    //while(getchar()!='\n'){};

    /* fflush(stdout);
    fflush(stdin);*/

    printf("plese enter the interval domain [a,b]\n");
    printf("enter a and b:\n");
    scanf("%f %f",&a,&b);

    while(getchar()!='\n'){}

    printf("the entered interval is [%f,%f]",a,b);

    fflush(stdout);
    fflush(stdin);

    //this array is used to choose a different value of x to apply newton's formula recurcively in an interval to scan it roperly for 3 roots

    val[0]=(double)b;       
    val[1]=(double)a;
    val[2]=(double)((a+b)/2);

    double t,x=val[0],x1=0.0,roots[10];

    while(1)
    {

        t=x1;
        x1=(x-(func(&coeff[0],order,x)/derv(&coeff[0],order,x)));           //this is the newton's formula

        x=x1;

        if((fabs(t - x1))<=0.0001 && count!=0)
        {
            roots[j]=x;
            j++;
            x=val[j];   // every time a root is encountered this stores the root in roots array and runs the loop again with different value of x to find other roots
            t=0.0;
            x1=0.0;
            count=(-1);

            if(j==3)
                break;
        }

        count++;
    }

    printf("the roots are = \n");

    int p=0;

    for(j=0;j<3;j++)
    {
        if(j==0 && roots[j]>=a && roots[j]<=b)
        {
            printf("  %f  ",roots[j]);
            p++;
        }

        if(fabs(roots[j]-roots[j-1])>0.5 && j!=0 && roots[j]>=a && roots[j]<=b)
        {
            printf(" %f  ",roots[j]);
            p++;
        }
    }

    if(p==0)
        printf("Sorry,no roots are there in this interval \n");

    return 0;
}

Answer 1

您没有正确计算函数或导数,因为您以相反的顺序存储系数,但您没有考虑到这一点.

当您打印出公式时,您会通过打印来解释它order-i:

printf(" %fx^%d ",coeff[i],order-i);

所以你需要做同样的事情func:

fc=fc+(double)(*p)*pow(x,(double)(order-i));

并且derv:

dv=dv+(double)(*q)*(pow(x,(double)((order-i)-1)))*(double)(order-i);

它对多项式工作的原因x^3 + x^2 + x + 1是因为在那个例子中所有的系数是相同的,所以如果你向前或向后读取数组,它将没有什么区别.

此外,正如Johnathon Leffler在评论中所提到的,您可能需要考虑该方法无法收敛的其他原因.您可以为循环设置最大迭代次数,如果超过最大值则突破.

调试这样的东西(当然除了使用调试器)之外的一个好方法是添加一些额外的printf语句来显示正在计算的值.您可以通过在Google搜索中输入等式来检查输出,它将提供该功能的交互式图表.