Newton–Raphson Iteration

x_{n + 1} = x_{n} - \frac{f ( x _{n} )}{f ^{'} ( x _{n} )}

Newton’s method finds a root of f(x)=0 by following the tangent. One handbook iteration is enough on the exam if x0 is close and f' is given or easy to form. Stop when |x_{n+1}−x_n| meets the requested tolerance.

$x_{n}$: Current guess
$f (x)$: Residual function
$f^{'} (x)$: Derivative

Step-by-step solved example

f(x)=x²−5, x0=2. Perform one Newton iteration.

Newton follows the tangent from xn to the next intercept xn+1.

1. Evaluate
f(2)=4−5=−1, f'(x)=2x so f'(2)=4.
$f (2) = - 1, f^{'} (2) = 4$
2. Update
x1=2−(−1)/4=2.25.
$x_{1} = 2.25$

Answer: x1 = 2.25

10 practice questions

0/10 correct

1.Newton update is

2.f(x)=x²−9, x0=4: x1 =

3.If f'(x_n)=0 Newton

4.Quadratic convergence means errors

5.f(x)=e^x−2, x0=0. x1 =

6.Secant method replaces f' with

7.A good x0 is

8.f(x)=x³−x−1, x0=1: f(1) and f'(1) are

9.Tolerance is usually checked on

10.Compared with bisection, Newton is