Find a root of an equation `f(x)=x^3+2x^2+x-1` using Steffensen's method

Solution:
Here `x^3+2x^2+x-1=0`

Let `f(x) = x^3+2x^2+x-1`

Here

`x`	0	1
`f(x)`	-1	3

Here `f(0) = -1 < 0 and f(1) = 3 > 0`

`:.` Root lies between `0` and `1`

`x_0 = (0 + 1)/2 = 0.5`

`x_0 = 0.5`


`1^(st)` iteration :

`f(x_0)=f(0.5)=0.5^3+2*0.5^2+0.5-1=0.125`

`f(x_0+f(x_0))=f(0.5+0.125)=0.6504`

`x_1=x_0-(f(x_0)^2)/(f(x_0+f(x_0))-f(x_0))`

`x_1=0.5-((0.125)^2)/(0.6504-0.125)`

`x_1=0.4703`


`2^(nd)` iteration :

`f(x_1)=f(0.4703)=0.4703^3+2*0.4703^2+0.4703-1=0.0165`

`f(x_1+f(x_1))=f(0.4703+0.0165)=0.0761`

`x_2=x_1-(f(x_1)^2)/(f(x_1+f(x_1))-f(x_1))`

`x_2=0.4703-((0.0165)^2)/(0.0761-0.0165)`

`x_2=0.4657`


`3^(rd)` iteration :

`f(x_2)=f(0.4657)=0.4657^3+2*0.4657^2+0.4657-1=0.0003`

`f(x_2+f(x_2))=f(0.4657+0.0003)=0.0015`

`x_3=x_2-(f(x_2)^2)/(f(x_2+f(x_2))-f(x_2))`

`x_3=0.4657-((0.0003)^2)/(0.0015-0.0003)`

`x_3=0.4656`


Approximate root of the equation `x^3+2x^2+x-1=0` using Steffensen's method is `0.4656` (After 3 iterations)

`n`	`x_0`	`f(x_0)`	`f(x_0+f(x_0))`	`x_1`	Update
1	0.5	0.125	0.6504	0.4703	`x_0 = x_1`
2	0.4703	0.0165	0.0761	0.4657	`x_0 = x_1`
3	0.4657	0.0003	0.0015	0.4656	`x_0 = x_1`

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