Solution of equations, Newton iteration and dichotomy

Newton Iteration

#include <iostream>#include<string.h>#include<math.h>using namespacestd;floatFfloatx) {    return(Pow (x,3)-5*pow (x,2)+ -*x+ the);}floatF1 (floatx) {    return(3*pow (x,2)-5*x+ -);}intMain () {//x*x*x-5*x*x+16*x+80;    floatx=1, X1,y1,y2; CIN>>x;  Do{x1=x; Y1=f (x); Y2=F1 (x1); X=x1-y1/Y2; } while(Fabs (X-X1) >=0.000001); cout<<x1<<Endl; System ("Pause"); return 0;}

If you want to calculate the root under 3, then the equation is x*x-3=0;

Fabs the absolute value of floating-point numbers

The equation is solved by the dichotomy method:

#include <iostream>#include<math.h>using namespacestd;DoubleFDoublex) {    returnPow (x,x)-Ten;}intMain () {DoubleA=2, b=3, limit=0.00001;  while((B-A) >limit) {        if(f (a+b)/2) *f (b) <0) {a= (a+b)/2; }        Else //Same numberb= (A+B)/2; } cout<<a<<" "<<b<<Endl; System ("Pause"); return 0;}

x^x=10;

Solution of equations, Newton iteration and dichotomy