Решение нелинейного уравнения

double f(int n, double c, double b, double d){
   double a = 0;
   for (int k = 0; k < n; k++){
      a += b * sqrt(c * pow(a,2) - pow(d,2))   
   }
   return a;
}

float f(float x){
  float a=0.0;
  for (int k=1; k<=n; k++){a+=...}
}
 
 
x=0; x1=f(x);dx=fabs(x1-x);i=0;
while (dx>epsilon && (i<100)){
x=x1;
x1=f(x);
dx=fabs(x1-x);
i++;
}
if (i==100)cout<<"FAIL";
else cout<<x1;

#include <cmath>
#include <iostream>
#include <vector>
#include <iomanip>
#include <algorithm>
using namespace std;
 
double f(int n,double a,double b[],double c[],double d)
 {
 double sum=0;
 for(int k=0;k<n;k++) sum+=  b[k]*sqrt(c[k]*a*a-d*d);
 return a-sum;
 }
//--------------------------------------------------
double dfda(int n,double a,double b[],double c[],double d)
 {
 double sum=0;
 for(int k=0;k<n;k++) sum+= b[k]*c[k]*a/sqrt(c[k]*a*a-d*d);
 return 1-sum;
 }
//--------------------------------------------------
int _tmain(int argc, _TCHAR* argv[])
{
int n=0;
cout<<"n= ";
cin>>n;
double *b= new double[n];
double *c= new double[n];
for(int k=0;k<n;k++)
 {
 cout<<"b"<<k+1<<"= "; cin>>b[k];
 cout<<"c"<<k+1<<"= "; cin>>c[k];
 }
double d;
cout<<"d= "; cin>>d;
double e=10e-15;
//cout<<"e= "; cin>>e;
double minc= *min_element(&c[0],&c[n]);
 
double a=sqrt(d*d/minc)+e;
//cout<<"a= "; cin>>a;
double fa=f(n,a,b,c,d);
const int count=100;
for(int i=0;i<count && fabs(fa)> e ;i++)
 {
 cout<<"f("<<a<<")= "<<setprecision(10)<<f(n,a,b,c,d)<<endl;
 cout<<"df("<<a<<")= "<<setprecision(10)<<dfda(n,a,b,c,d)<<endl;
 
 a= a - f(n,a,b,c,d)/dfda(n,a,b,c,d);
 fa= f(n,a,b,c,d);
 
 cout<<"a["<<i<<"]= "<<setprecision(10)<<a<<"  f("<<a<<")="
                     <<setprecision(10)<<fa<<endl;
 }
cout<<"result: a= "<<a<<endl;
 
delete[] b,c;
system("pause");
return 0;
}
//---------------------------------------------------------------------------