# NEWTON RAPHSON METHOD C/C++ with code

## Newton Raphson method c/C++ with code and algorithm

Newton-Raphson method is used to find simple real roots of a polynomial equation.

It has the fastest rate of convergence. The method is quite sensitive to the starting value.

It may also diverge if the *first derivative* i.e.** f'(x) **of the function is near zero during the iterative cycle.

**see more**

**Algorithm for Newton Raphson method c/c++:**

**Read****x0**, e, n, N where**x0**is the initial guess of the root, e the allowed error, n the order of the polynomial and N the total number of iterations.**for i=0 to n in steps of 1 do Read bi end for.****for i=0 to n-1 in steps of 1 do Read bi end for.****P=an****bn-1=an****S=bn-1****for k=1 to N in steps of 1 do****for i=1 to n-1 in steps of 1 do****b**_{n-(i+1)}=a_{n-i}+**x0**b_{n-i}**S=b**_{n-(i+1)}+**x0**S

endfor**P=a****0**+ b**0**x0**x1**=**x0**-(P/S)**if |****x1**–**x0**/**x1**| ≤ e goto step 18

else**x0**=**x1**

endif

endfor**write “root not found in N iterations”****write S, P,****x1**,**x0****stop****write “root found in k iterations”****x0**=**x1****write****x0**, S, P**stop**

#### Some observations about Newton Raphson method c:

- Newton’s method is useful in cases of large values of f'(x) e.e.
- when the graph of f(x) while crossing the x-axis in nearly vertical.

For if f'(x) is small in the vicinity of the root, then by*h=-f(x)/f'(x),* - h will be large and the computation of the root is slow or may not be possible.
- Thus this method is not suitable in those cases where the graph of f(x) is nearly horizontal while crossing the x-axis.
- Newton’s method is applicable to both algebraic and transcendental equations.
- Newton’s method is useful when x0 is chosen sufficiently close to the root.
- Newton’s Method has second order convergence.

#include<iostream.h> #include<math.h> float fun(float x)//we are finding the root of x^4-x-10 { return x*x*x*x-x-10; } float diff(float x) differential of x^4-x-10; { return 4*x*x*x-1; } int main() { int itr,maxitr; float h,x0,x1,aerr; cout<<"Enter x0,allowed error, maximum iterations"<<endl; cin>>x0>>aerr>>maxitr; for(itr=1;itr<=maxitr;itr++) { h=fun(x0)/diff(x0); x1=x0-h; if(fabs(h)<aerr) { cout<<"after "<<itr<<" root = "<<x1<<endl; return 0; } x0=x1; } cout<<"Iterations not sufficient, Solution does not converge"<<endl; return 1; }

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