F(x)=X2-6X+8
G(X)=X=(X2+8)/6;
We have to initiate a value then put it on G(X) and get value of G(X) and then put this on F(X).it continues until it F(X)=0 or we get tolerable fraction into F(X) to output or finishing maximum iteration step.
///Bismillahir Rahmanir Rahim
///Author: Tanvir Ahmmad
///CSE,Islamic University,Bangladesh
#include<iostream>
#include<math.h>
using namespace std;
double func(double num)
{
return ((num*num)-(6*num)+8);
}
double value(double num)
{
return ((8+(num*num))/6);
}
int main()
{
double guess,tol,val,fund,ans;
int it_time,i;
cout<<"Enter initial guess: ";
cin>>guess;
cout<<"Enter tolerable error: ";
cin>>tol;
cout<<"Enter maximum iteration: ";
cin>>it_time;
for(i=1; i<=it_time; i++)
{
val=value(guess);
if(abs(func(val))>=tol)
{
ans=val;
fund=func(val);
cout<<"Iteration-"<<i<<": x = "<<val<<" and f(x) = "<<fund<<endl;
guess=val;
}
else break;
}
if(i==(it_time+1)) cout<<"Using "<<it_time<<" step root is : "<<ans<<endl;
else cout<<"Root is : "<<ans<<endl;
return 0;
}
Just count the first three & last three number ///Bismillahir Rahmanir Rahim ///Author: Tanvir Ahmmad ///CSE,Islamic University,Bangladesh #include< iostream > #include< cstdio > #include< algorithm > #include< string > #include< cstring > #include< sstream > #include< cmath > #include< cstring > #include< vector > #include< queue > #include< map > #include< set > #include< stack > #include< vector > #include< iterator > #include < functional > ///sort(arr,arr+n,greater<int>()) for decrement of array /*every external angle sum=360 degree angle find using polygon hand(n) ((n-2)*180)/n*/ ///Floor[Log(b) N] + 1 = the number of digits when any number is repres...
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