Newton-Raphson Method:

Unlike the earlier methods, this method requires only one appropriate starting point, $x_{0}$ as an initial assumption of the root of the function $f(x)=0$. At $(x_{0},f(x_{0}))$ a tangent to $f(x)=0$ is drawn. Equation of this tangent is given by

$$y=f'(x_{0})(x-x_{0})+f(x_{0})$$

The point of intersection, say $x_{1}$, of this tangent with x-asis (y = 0) is taken to be the next approximation to the root of f(x) = 0. So on substituting y = 0 in the tangent equation we get

$$x_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0})}$$

If $\vert f(x_{1})\vert < \epsilon=10^{-6}$ (say) we have got an acceptable approximate root of $f(x)=0$, otherwise we replace $x_{0}$ by $x_{1}$, and draw a tangent to $f(x)=0$ at $(x_{1},f(x_{1}))$ and consider its intersection, say $x_{2}$, with x-axis as an improved approximation to the root of f(x)=0. If $\vert f(x_{2})\vert>\epsilon$, we iterate the above process till the convergence criteria is satisfied. This geometrical description of the method may be clearly visualized in the figure below:

The various steps involved in calculating the root of $f(x)=0$ by Newton Raphson Method are described compactly in the algorithm below.

Algorithm:
Given a continuously differentiable function $f(x)$ and an initial approximation $x_{0}$ to the root of $f(x)=0$ the sets involved in calculating an approximation $\xi$ to the root of $f(x)=0$ s.t. $\vert f(\xi)\vert<\tilde{\epsilon}$ are:
(1) Calculate $f(x_{0}),f'(x_{0})$ and set $\epsilon=\tilde{\epsilon}$
(2) For n = 0,1,2... until convergence criteria is satisfied do:
Calculate

$$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$$

Remark (1)=1in=1     This method converges faster than the earlier methods. In fact the method converges at a quadratic rate. We will prove this later.

Remark (2)=1in =1    This method can be derived directly by the Taylor expansion f(x) in the neighborhood of the root $\xi$ of $f(x)=0$. The starting approximation $x_{0}$ to $\xi$ is to be properly chosen so that the first order Taylor series approximation of $f(x_{0}+h)$ in the neighbour of $x_{0}$ leads to $x_{1}$, an improved approximation to $\xi$. i.e

$$f(x_{0}+h)=f(x_{0})+hf^{'}(x_{0})+\frac{h^{2}}{2}f''(x_{0})+.....=0$$

$\because \qquad h\,\, << \,\, 1$, neglecting $h^2$ and its higher powers, we get

$$f(x_{0})+hf'(x_{0})=0$$

i.e.

$$h=-\frac{f(x_{0})}{f'(x_{0})},\qquad \because h=x_{1}-x_{0}$$

$$x_{1}=x_{0}-\frac{f(x_{0})}{f'(x_{0})}$$

Now the successive approximations $x_{2}\quad x_{3}...x_{n}...$ etc may be the iterative formula:

$$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$$

Remark(3) =1in =1    One may also derive the above iteration formulation starting with the iteration formula for the secant method. In a way this may help one to visualize Newton-Raphson method as an improvement over the secant method. So, let us consider the iteration formula for the secant method i.e.

$$x_{n+1}=\frac{f(x_{n})x_{n-1}-f(x_{n-1})x_{n}}{f(x_{n})-f(x_{n-1})}$$

Add and subtract $f(x_{n})x_n$ to the numeration on the r.h.s. to get

$$x_{n+1}=x_{n}-\frac{f(x_{n})(x_{n}-x_{n-1})}{f(x_{n})-f(x_{n-1})}$$

$$x_{n+1}=x_{n}-\frac{f(x_{n})}{[f(x_{n})-f(x_{n-1})]/[x_{n}-x_{n-1}]}$$

Clearly, $${\frac{f(x_{n})-f(x_{n-1})}{(x_{n}-x_{n-1})}}$$ is the slope of the secant to the curve $f(x)=0$ through the points $(x_{n-1},f(x_{n-1}))$, $(x_{n},f(x_{n}))$. This also represents slope of the tangent to f(x)=0 parallel to the secant intersecting x-axis between $x_{n-1}$ and $x_{n}$. If $f(x)$ is differentiable one may as well approximate two slope by $f'(x_{n})$ and thus arrive at the iteration formula.

$$x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}$$

=0in =1
Example:
Solve $2x^3-2.5x-5=0$ for the root in [1,2] by Newton Raphson method.

Solution:
Given

$$f(x)=2x^3-2.5x-5=0$$

$$f'(x)=8x^2-2.5$$

Take $x_0=2$ and $\epsilon =0$

$$\therefore \qquad f(x_0)=6\qquad ; \qquad f'(x_0)=29.5$$

$$\therefore \qquad x_1=x_0 -\frac{ f(x_0)}{f'(x_0)}=2-\frac{6.0}{29.5}=1.709302187$$

$$\vert f(x_1)\vert=0.8910911679 > 10^{-6}\qquad$$

$\therefore$ repeat the process.
Results are tabulated below:

	Newton Raphson Method
Iteration no.	$x_n$	$x_{n+1}$	$\vert f(x_{n+1})\vert$
0	2.0000000000	1.7209302187	0.8910911679
1	1.7209302187	1.6625729799	0.0347661413
2	1.6625729799	1.6601046324	0.0000604780
3	1.6601046324	1.6601003408	0.0000002435

Example:
Solve $5 \sin^{2}x-8\cos^{5}x=0$ in [0.5,1.5] for the root by Newton-Raphson method.

Solution: Given

$$f(x)=5\sin^{2}x-8cos^{5}x$$

$$f'(x)=10 \sin x \cos x + 40 cos^{4}x \sin x$$

Say, $x_{0}=0.5, \quad \epsilon = 10^{-6}$
The results are tabulated below:

	Newton Raphson Method
Iteration no.	$x_n$	$x_{n+1}$	$\vert f(x_{n+1})\vert$
0	0.5000000000	0.6934901476	-0.1086351126
1	0.6934901476	0.7013291121	-0.0005313741
2	0.7013291121	0.7013678551	-0.0000003363