C中的这段代码（附在帖子中）使用Newton - Raphson method来查找特定区间内多项式的根。
对于像x^3 + x^2 + x + 1这样的多项式，此代码非常适用，但运算符对于像x^3 - 6*x^2 + 11*x - 6这样的多项式来说是无限的。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，则它将无限期地运行。
如果有人找到解决方案，请告诉我。我在代码中写了评论来引导读者，但是如果有人发现难以理解，可以在评论中提问，我会解释它。
#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;
}
答案 0 :(得分：8)
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，但您并未考虑到这一点。
当您打印出公式时，您 帐户，打印order-i：
printf(" %fx^%d ",coeff[i],order-i);
所以你需要在func中做同样的事情：
fc=fc+(double)(*p)*pow(x,(double)(order-i));
和derv：
dv=dv+(double)(*q)*(pow(x,(double)((order-i)-1)))*(double)(order-i);
它为x^3 + x^2 + x + 1这样的多项式工作的原因是因为在这个例子中所有的系数是相同的，所以如果你向前或向后读数组，它就不会有所作为。< / p>
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。
调试类似这样的东西的好方法（当然除了使用调试器之外）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。
答案 1 :(得分：0)
Yehh，我终于在C中编写程序来计算用户在用户输入的区间内输入的多项式的根，它给出了输出 - 根所进行的根的多重性。使用的算法是首先我们将间隔划分为度数+ 1个等间隔的数量，然后在每个间隔中应用Newton-Raphson方法来计算最大根数。当用户给出无效输入时，还包括几个检查。对于某些多项式，它可能会产生不希望的输出
#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("\tYou entered an alphabet. Kindly, retry with valid entry\n");
else
printf("\tYou entered a character. Kindly, retry with valid entry\n");
return 0;
}
char signum(float x)
{
if (x>=0) return '+';
else return '-';
}
double func(double *p,int degree,double x) //calculates the value of the function required in the formmula in main
{
double fc=0.0;
int i;
for(i=0;i<=degree;i++)
{
fc=fc+(*p)*pow(x,degree-i);
p++;
}
return fc;
}
int fact(int n)
{
if (n>0) return n*fact(n-1);
else return 1;
}
double D(int k,int a,double x) //calculates the derivative of the function required in the formmula in main
{
if(x!=0 && a>=k) return fact(a)*pow(x,a-k)/fact(a-k);
else return 0;
}
double derv(double *q,int degree,double x,int order) //calculates the derivative of the function required in the formmula in main
{
double dv=0.0;
int i;
for(i=0;i<=degree;i++)
{
dv=dv+(*q)*D(order,degree-i,x);
q++;
}
return dv;
}
int main()
{
double coeff[1000],t,x1=0.0,roots[100];
int degree=-1,count=0,i,j=0,k,l=0,flag=1;
char ch;
float a,b;
coeff[0]=0;
printf("\t**This Programme helps you find roots of a polynomial using a manipulated version of Newton-Raphson Method in a real interval [a,b]**\n");
while (degree<0)
{
printf("\tEnter a NON-NEGATIVE INTEGRAL value for the DEGREE of the polynomial: ");
while(scanf("%d",°ree)!=1)
{
printf("\tInvalid Entry. Kindly, retry with valid entry: ");
while(getchar()!='\n'){}
}
}
int w[degree];
printf("\tEnter the coefficients...\n");
while(coeff[0]==0 && degree!=0)
{
printf("\t-->for x^%d (Remember that this coefficient cannot be zero since degree is %d): ", degree,degree);
while(scanf("%lf",&coeff[0])!=1)
{
check(coeff[0]);
}
}
if (degree==0)
{
printf("\t-->for x^%d: ",0);
while(scanf("%lf",&coeff[0])!=1)
{
check(coeff[0]);
}
}
for(i=1;i<=degree;i++)
{
printf("\t-->for x^%d: ",degree-i);
while(scanf("%lf",&coeff[i])!=1)
{
check(coeff[i]);
}
}
while(getchar()!='\n'){} //this clears the buffer accumulated upto pressing enter
printf("\tThe input polynomial is:\n");
printf("\t");
for(i=0;i<=degree;i++)
{
printf("%c %.2fx^%d ",signum(coeff[i]),fabs(coeff[i]),degree-i);
}
printf("\n");
//while(getchar()!='\n'){};
/* fflush(stdout);
fflush(stdin);*/
printf("\tFor the interval domain [a,b]; enter 'a' followed by 'b': ");
scanf("%f %f",&a,&b);
while(getchar()!='\n'){}
printf("\tSo, you entered [%f,%f].\n",a,b);
//while(getchar()!='\n'){}
fflush(stdout);
fflush(stdin);
double d=(b-a)/(degree+1), val[degree+2],x=a;
for (k = 0; k<=degree+1; k++)
{
val[k]= a+d*k;
// printf("%lf\t", val[k]);
}
while(l<(degree+2))
{
if(derv(&coeff[0],degree,x,1)<=0.00001)
{
if(func(&coeff[0],degree,x)<=0.00001)
{
w[j]=2;
roots[j]=x;
//printf("Axk before %f\n", x);
j++;
// printf("jC is %d\n", j);
l++;
x=val[l]; // 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
//printf("Axk after %f\n", x);
t=0.0;
x1=0.0;
count=(-1);
}
else
{
t=0.0;
x1=0.0;
count=(-1);
l++;
//printf("Bxk before%f\n", x);
x=val[l];
// printf("Bxk after%f\n", x);
}
}
else
{
t=x1;
x1=(x-(func(&coeff[0],degree,x)/derv(&coeff[0],degree,x,1))); //this is the newton's formula
// printf("jB is %d\n", j);
//printf("Cxk before%f\n", x);
x=x1;
//printf("f %f\td %f\tCxk after%f\tc %i\n",func(&coeff[0],degree,x), derv(&coeff[0],degree,x,1), x,count);
if(count>500)
{
if(j>degree)
break;
printf("\tPolynomial started to diverge. So, no roots can be found\n");
l++;
//printf("Dxk before %f\n", x);
x=val[l];
//printf("Dxk after %f\n", x);
t=0.0;
x1=0.0;
count=(-1);
}
if((fabs(t - x1))<=0.00001 && count!=0)
{
w[j]=1;
if(j>degree)
break;
roots[j]=x;
j++;
l++;
// printf("jC is %d\n", j);
x=val[l]; // 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
//printf("Exk after%f\n", x);
t=0.0;
x1=0.0;
count=(-1);
flag=0;
if(derv(&coeff[0],degree,x,1)<=0.00001) w[j]=2;
else w[j]=1;
}
count++;
}
}
if(flag==0)
{
printf("\tThe roots are = \t");
// int p=0;
for(j=0;j<degree;j++)
{
if(j==0 && roots[0]>=a && roots[0]<=b)
{
printf(" %d %f\n", w[0], roots[0]);
//p++;
}
if(fabs(roots[j]-roots[j-1])>0.001 && j!=0 && roots[j]>=a && roots[j]<=b)
{
printf(" %d %f\n", w[j], roots[j]);
//p++;
}
}
}
else
printf("\tNo roots found in the interval you entered.\n");
return 0;
}