Algorithm

Newton Raphson method Steps (Rule)
Step-1:	Find points `a` and `b` such that `a < b` and `f(a) * f(b) < 0`.
Step-2:	Take the interval `[a, b]` and find next value `x_0 = (a+b)/2`
Step-3:	Find `f(x_0)` and `f'(x_0)` `x_1 = x_0 - f(x_0)/(f'(x_0))`
Step-4:	If `f(x_1) = 0` then `x_1` is an exact root, else `x_0 = x_1`
Step-5:	Repeat steps 3 and 4 until `f(x_i) = 0` or `|f(x_i)| <= "Accuracy"`

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Example-1
1. Find a root of an equation `f(x)=x^3-x-1` using Newton Raphson method

Solution:
Here `x^3-x-1=0`

Let `f(x) = x^3-x-1`

`:. f'(x) = 3x^2-1`

Here

`x`	0	1	2
`f(x)`	-1	-1	5

Here `f(1) = -1 < 0 and f(2) = 5 > 0`

`:.` Root lies between `1` and `2`

`x_0 = (1 + 2)/2 = 1.5`


`1^(st)` iteration :

`f(x_0) = f(1.5) = 0.875`

`f'(x_0) = f'(1.5) = 5.75`

`x_1 = x_0 - f(x_0)/(f'(x_0))`

`x_1 = 1.5 - (0.875)/(5.75)`

`x_1 = 1.34783`


`2^(nd)` iteration :

`f(x_1) = f(1.34783) = 0.10068`

`f'(x_1) = f'(1.34783) = 4.44991`

`x_2 = x_1 - f(x_1)/(f'(x_1))`

`x_2 = 1.34783 - (0.10068)/(4.44991)`

`x_2 = 1.3252`


`3^(rd)` iteration :

`f(x_2) = f(1.3252) = 0.00206`

`f'(x_2) = f'(1.3252) = 4.26847`

`x_3 = x_2 - f(x_2)/(f'(x_2))`

`x_3 = 1.3252 - (0.00206)/(4.26847)`

`x_3 = 1.32472`


`4^(th)` iteration :

`f(x_3) = f(1.32472) = 0`

`f'(x_3) = f'(1.32472) = 4.26463`

`x_4 = x_3 - f(x_3)/(f'(x_3))`

`x_4 = 1.32472 - (0)/(4.26463)`

`x_4 = 1.32472`


Approximate root of the equation `x^3-x-1=0` using Newton Raphson method is `1.32472`

`n`	`x_0`	`f(x_0)`	`f'(x_0)`	`x_1`	Update
1	1.5	0.875	5.75	1.34783	`x_0 = x_1`
2	1.34783	0.10068	4.44991	1.3252	`x_0 = x_1`
3	1.3252	0.00206	4.26847	1.32472	`x_0 = x_1`
4	1.32472	0	4.26463	1.32472	`x_0 = x_1`

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