Найти корень многочлена методом простой итерации

#include <cmath>
#include <iomanip>
#include <iostream>
using namespace std;
 
double f(double x)
{
return 1.78*pow(x,5.0) + 3,2*pow(x, 4.0) - 5*pow(x,3.0)-9.7*pow(x,2.0)+x-21;
}
 
double g(double x)
{
return x + 0.5*f(x);//xk = g(xk-1) где g(x) = x + b*f(x) где b - произвольное число
}
 
int main()
{
double x;
double eps;
cout<<"Enter initial root value : ";cin>>x;
cout<<"Enter error of calculation : ";cin>>eps;
for(double iter = 1; eps < fabs(f(x)); iter = iter + 1)
{
//system("cls");
 
cout<<"Iteration : "<<setprecision(0)<<iter<<endl;
cout<<"x = "<<x <<endl;
cout<<"g(x) = "<<g(x)<<endl;
cout<<"f(x) = "<<f(x)<<endl;
x = g(x);
}
system("pause");
return 0;
}

#include <cmath>
#include <iomanip>
#include <iostream>
using namespace std;
 
double f(double x)
{
return 1.78*pow(x,5.0) + 3,2*pow(x, 4.0) - 5*pow(x,3.0)-9.7*pow(x,2.0)+x-21;
}
 
double g(double x)
{
return x + 0.5*f(x);//xk = g(xk-1) где g(x) = x + b*f(x) где b - произвольное число
}
 
int main()
{
double x0, x1;
double eps;
cout<<"Enter initial root value : ";cin>>x0;
x1 = g(x0);
cout<<"Enter error of calculation : ";cin>>eps;
for(int iter = 1; eps < fabs(f(x1) - f(x0)); iter++)
{
//system("cls");
 
cout<<"Iteration : "<<iter<<endl;
cout<<"x = "<<x1 <<endl;
cout<<"g(x) = "<<g(x1)<<endl;
cout<<"f(x) = "<<f(x1)<<endl;
x0 = x1;
x1 = g(x1);
}
system("pause");
return 0;
}

eps < fabs(f(x1) - f(x0))

#include <cmath>
#include <iomanip>
#include <iostream>
using namespace std;
 
double f(double x)
{
return 1.78*pow(x,5.0) + 3,2*pow(x, 4.0) - 5*pow(x,3.0)-9.7*pow(x,2.0)+x-21;
}
 
double g(double x)
{
return x -(f(x)/(8.9*pow(x,4)+12.8*pow(x,3)-15*pow(x,2)-19.4*x+1));
}
 
int main()
{
double x;
//x=1.92245;
double eps;
cout<<"Enter initial root value : ";cin>>x;
cout<<"Enter error of calculation : ";cin>>eps;
for(double iter = 1; eps < fabs(f(x)); iter = iter + 1)
{
//system("cls");
cout<<"Iteration : "<<setprecision(0)<<iter<<endl;
cout<<"x = "<<x <<endl;
cout<<"g(x) = "<<g(x)<<endl;
cout<<"f(x) = "<<f(x)<<endl;
x = g(x);
}
system("pause");
return 0;
}

#include <stdio.h>
#include <conio.h>
#include <math.h>
//Âû÷èñëåíèå çíà÷åíèÿ ôóíêöèè F(x)=0
double f(double x)
{   
        return 1.78*x*x*x*x*x + 3.2*x*x*x*x - 5*x*x*x-9.7*x*x+x-21;
}
 
 
//Âû÷èñëåíèå çíà÷åíèÿ ïåðâîé ïðîèçâîäíîé ôóíêöèè F'(x)=0
double f1(double x)
{     
        return 8.9*x*x*x*x+12.8*x*x*x-15*x*x-19.4*x+1;
}
 
//Âû÷èñëåíèå çíà÷åíèÿ âòîðîé ïðîèçâîäíîé ôóíêöèè F''(x)=0
double f2(double x)
{ 
        return 35.6*x*x*x+38.4*x*x-30*x-19.4;
}
 
//Âû÷èñëåíèå ýêâèâàëåíòíîé ôóíêöèè äëÿ ìåòîäà èòåðàöèé
double fe(double x,double lambda)
{
    return x-lambda*f(x);
}
 
//Âû÷èñëåíèå ïåðâîé ïðîèçâîäíîé ýêâèâàëåíòíîé ôóíêöèè äëÿ ìåòîäà èòåðàöèé
double fe1(double x,double lambda)
{
    return 1-lambda*f1(x);
}
 
//Ïîèñê êîðíÿ ìåòîäîì Íüþòîíà
int newton (double a,double b,double e, double *x)
{
int n=0;
double x0=(b+a)/2.0,fx,last;
//Ïðîâåðêà ñõîäèìîñòè
if(fabs(f(a)*f2(a))>=f1(a)*f1(a)||fabs(f(b)*f2(b))>=f1(b)*f1(b))return -1;
fx=x0;
do
  {
  last=fx;
  fx=last-f(last)/f1(last);
  n++;
  }
  while(fabs(fx-last)>e);
*x=fx;
return n;
}
//Ïîèñê êîðíÿ ìåòîäîì äèõîòîìèè
int deleniya(double a,double b,double e, double *x)
{
int n=0;
//Ïðîâåðêà ñõîäèìîñòè
if(f(a)*f(b)>=0) return -1;
while(b-a>e)
  {
  if(f(a)*f((a+b)/2)>0)
    a=(a+b)/2;
  else
    b=(a+b)/2;
  n++;
  }
  *x=(a+b)/2.0;
return n;
}
//Ïîèñê êîðíÿ ìåòîäîì èòåðàöèè
int iterac(double a, double b, double e, double *x,double lambda)
{
int n=0;
double x0=(b+a)/2.0,fx,last;
//Ïðîâåðêà ñõîäèìîñòè
if(fabs(fe1(a,lambda))>=1||fabs(fe1(b,lambda))>=1) return -1;
fx=x0;
do
  {
  last=fx;
  fx=fe(last,lambda);
  n++;
  }
  while(fabs(fx-last)>e);
*x=fx;
return n;
}
 
int main()
{
    int it;     //êîë-âî òî÷åê è èòåðàöèé
    double e,x;
    printf("e = ");
    scanf("%lf",&e);    //ââîä ïîãðåøíîñòè âû÷èñëåíèé
    printf("Metod Newtona.\n");
    it=newton(1.0,2.0,e,&x);
    if(it==-1) printf("Metod ne shoditsya.\n");
    else
    {
        printf("x=%lf\t%d iteraciy\t\n",x,it);
    }
 
    printf("Metod dihotomii.\n");
    it=deleniya(1.0,2.0,e,&x);
    if(it==-1) printf("Metod ne shoditsya.\n");
    else
    {
        printf("x=%lf\t%d iteraciy\t\n",x,it);
    }
 
    printf("Metod iteracii.\n");
    it=iterac(1.0,2.0,e,&x,1/f1(1.5));
    if(it==-1) printf("Metod ne shoditsya.\n");
    else
    {
        printf("x=%lf\t%d iteraciy\t\n",x,it);
    }
 
    getch();        //ïàóçà
    return 0;
}