Algorithm

Secant method Steps (Rule)
Step-1:	Find points `x_0` and `x_1` such that `x_0 < x_1` and `f(x_0) * f(x_1) < 0`.
Step-2:	find next value using Formula-1 : `x_2=x_0-f(x_0)*(x_1-x_0)/(f(x_1)-f(x_0))` or Formula-2 : `x_2=(x_0*f(x_1)-x_1*f(x_0))/(f(x_1)-f(x_0))` or Formula-3 : `x_2=x_1-f(x_1)*(x_1-x_0)/(f(x_1)-f(x_0))` (Using any of the formula, you will get same x2 value)
Step-3:	If `f(x_2) = 0` then `x_2` is an exact root, else `x_0 = x_1` and `x_1 = x_2`
Step-4:	Repeat steps 2 & 3 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 Secant method

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

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

Here

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

`1^(st)` iteration :

`x_0 = 1` and `x_1 = 2`

`f(x_0) = f(1) = -1` and `f(x_1) = f(2) = 5`

`:. x_2 = x_0 - f(x_0) * (x_1 - x_0) / (f(x_1) - f(x_0))`

`x_2 = 1 - (-1) xx (2 - 1)/(5 - (-1))`

`x_2 = 1.16667`

`:. f(x_2) = f(1.16667) = -0.5787`


`2^(nd)` iteration :

`x_1 = 2` and `x_2 = 1.16667`

`f(x_1) = f(2) = 5` and `f(x_2) = f(1.16667) = -0.5787`

`:. x_3 = x_1 - f(x_1) * (x_2 - x_1) / (f(x_2) - f(x_1))`

`x_3 = 2 - 5 xx (1.16667 - 2)/(-0.5787 - 5)`

`x_3 = 1.25311`

`:. f(x_3) = f(1.25311) = -0.28536`


`3^(rd)` iteration :

`x_2 = 1.16667` and `x_3 = 1.25311`

`f(x_2) = f(1.16667) = -0.5787` and `f(x_3) = f(1.25311) = -0.28536`

`:. x_4 = x_2 - f(x_2) * (x_3 - x_2) / (f(x_3) - f(x_2))`

`x_4 = 1.16667 - (-0.5787) xx (1.25311 - 1.16667)/(-0.28536 - (-0.5787))`

`x_4 = 1.33721`

`:. f(x_4) = f(1.33721) = 0.05388`


`4^(th)` iteration :

`x_3 = 1.25311` and `x_4 = 1.33721`

`f(x_3) = f(1.25311) = -0.28536` and `f(x_4) = f(1.33721) = 0.05388`

`:. x_5 = x_3 - f(x_3) * (x_4 - x_3) / (f(x_4) - f(x_3))`

`x_5 = 1.25311 - (-0.28536) xx (1.33721 - 1.25311)/(0.05388 - (-0.28536))`

`x_5 = 1.32385`

`:. f(x_5) = f(1.32385) = -0.0037`


`5^(th)` iteration :

`x_4 = 1.33721` and `x_5 = 1.32385`

`f(x_4) = f(1.33721) = 0.05388` and `f(x_5) = f(1.32385) = -0.0037`

`:. x_6 = x_4 - f(x_4) * (x_5 - x_4) / (f(x_5) - f(x_4))`

`x_6 = 1.33721 - 0.05388 xx (1.32385 - 1.33721)/(-0.0037 - 0.05388)`

`x_6 = 1.32471`

`:. f(x_6) = f(1.32471) = -0.00004`


Approximate root of the equation `x^3-x-1=0` using Secant method is `1.32471`

`n`	`x_0`	`f(x_0)`	`x_1`	`f(x_1)`	`x_2`	`f(x_2)`	Update
1	1	-1	2	5	1.16667	-0.5787	`x_0 = x_1` `x_1 = x_2`
2	2	5	1.16667	-0.5787	1.25311	-0.28536	`x_0 = x_1` `x_1 = x_2`
3	1.16667	-0.5787	1.25311	-0.28536	1.33721	0.05388	`x_0 = x_1` `x_1 = x_2`
4	1.25311	-0.28536	1.33721	0.05388	1.32385	-0.0037	`x_0 = x_1` `x_1 = x_2`
5	1.33721	0.05388	1.32385	-0.0037	1.32471	-0.00004	`x_0 = x_1` `x_1 = x_2`

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