该程序用于使用二分法求解方程给出错误“函数显示返回值”。
在这种方法中，我们给定一个函数f(x)，我们近似 2 个根a，b
对于函数，使得f(a).f(b)<0.
然后我们找到另一个点
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;
我们对给定的迭代次数重复这些步骤
#include<stdio.h>
#include<math.h>
#define e 0.000001/*e is the prescribed accuracy*/
main()
{
double g1,g2,g,v,v1,v2,prev;
int found=0,converged=0,i=0;
double f(double);
printf("We apply Bisection method to find a real root of the non-linear equation f(x) = 0, where f(x) = x^(2.7182818)-3cosx+1n");
while(found==0)/*This loop will continue until a range is found in between which a real root lies*/
{
printf("nEnter the first guess : ");
scanf("%lf",&g1);
v1=f(g1);
printf("nEnter the second guess : ");
scanf("%lf",&g2);
v2=f(g2);
if(v1*v2>0)
{
found=0;
g1++;
printf("nRoot does not lie in [%lf,%lf].n",g1-1,g2);
printf("nn..Enter two new guesses..nn");/*Previous two guesses are inappropriate*/
}
else
found=1;
}
printf("nThere is a real root which lies in [%lf,%lf].n",g1,g2);
while(converged==0)/*This loop will continue until a real root is found*/
{
printf("nnIteration = %dnn",i);
printf("a[%d](-ve)tb[%d](+ve)tbbx[%d]=(a[%d]+b[%d])/2tf(x[%d])n",i,i,i+1,i,i,i+1);
printf("%lft",g1);
printf("%lft",g2);
g=(g1+g2)/2;
v=f(g);
printf("%lft",g);
printf("t%lf",v);
if(v<0)
g1=g;
else
g2=g;
if(fabs(prev-v)<e)
converged=1;
else
prev=v;
i=i+1;
}
printf("nnThe approximate value of the root is : %lfn",g);
}
/*This function returns values of f(x)*/
double f(double x)
{
return pow(2.7182818,x)-3*cos(x)+1;
}