# Newton Raphson Method MATLAB code (Program) with simple Example

Introduction to Iterative methods: There are number of iterative methods like Jacobi method, Gauss–Seidel method that have been tried and used successfully in various problem situations. All these methods typically generate a sequence of estimates of the solution which is expected to converge to the true solution. Newton Raphson method is also one of the iterative method which is used to find the roots of given expression.

### Newton Raphson Method with MATLAB code:

If point x0 is close to the root α, then tangent line to the graph of f(x) at x0 is a good approximation the f(x) near α. So root of the tangent line, where line cuts the X-axis; x1 is the better approximation to α than x0 is.

Slope of the tangent =

Repeating process, we obtain a better approximation,

Continue in this way. If xn is the current estimate, then the next estimate xn+1 is given by

if x0 is sufficiently close to α, xn→α as n→∞.

### Limitations of Newton Raphson method:

- If initial guess is too far away from the required root, the process may converge to some other root.
- Division by zero may occur if f’(xi) is zero or very close to zero.
- A particular value in the iteration sequence may repeat, resulting in an infinite loop.

## Newton Raphson MATLAB program :

% Newton Raphson Method clear all close all clc % Change here for different functions f=@(x) cos(x)-3*x+1 %this is the derivative of the above function df=@(x) -sin(x)-3 % Change lower limit 'a' and upper limit 'b' a=0; b=1; x=a; for i=1:1:100 x1=x-(f(x)/df(x)); x=x1; end sol=x; fprintf('Approximate Root is %.15f',sol) a=0;b=1; x=a; er(5)=0; for i=1:1:5 x1=x-(f(x)/df(x)); x=x1; er(i)=x1-sol; end plot(er) xlabel('Number of iterations') ylabel('Error') title('Error Vs. Number of iterations')

**ANSWER :**

f =

@(x)cos(x)-3*x+1

df =

@(x)-sin(x)-3

Approximate Root is 0.6071016481031231

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Bruno Kreutz

March 18, 2016@ 12:53 pmI did not get very well what is going on in this section of our code, could ou please explain?

a=0;b=1;

x=a;

er(5)=0;

for i=1:1:5

x1=x-(f(x)/df(x));

x=x1;

er(i)=x1-sol;

end

Tks!

mayure5h

March 20, 2016@ 6:44 amMake sure that you have defined f(x), df(x) and sol initially. For example if your function is sin(x)+3x+1 then you need to define it as f= @x sin(x)+3*x+1 then your code becomes:

f=@(x) sin(x)+3*x+1df=@(x) cos(x)+3

a=0; b=1;

x=a;

for i=1:1:100

x1=x-(f(x)/df(x));

x=x1;

end

sol=x;fprintf(‘Approximate Root is %.15f’,sol)

a=0;b=1;

x=a;

er(5)=0;

for i=1:1:5

x1=x-(f(x)/df(x));

x=x1;

er(i)=x1-sol;

end

Mao

April 16, 2016@ 10:58 amcan we find the root of the polynomial in term of a

f[x]=(42-8a)x^3+(29-6a)x^2+(-42+7a)x+(7-a), where, a belongs to N (set of natural number).