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

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

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

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

`x_0 = 1.5`


`1^(st)` iteration :

`f(x_0)=f(1.5)=1.5^3-1.5-1=0.875`

`f(x_0+f(x_0))=f(1.5+0.875)=10.0215`

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

`x_1=1.5-((0.875)^2)/(10.0215-0.875)`

`x_1=1.4163`


`2^(nd)` iteration :

`f(x_1)=f(1.4163)=1.4163^3-1.4163-1=0.4246`

`f(x_1+f(x_1))=f(1.4163+0.4246)=3.398`

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

`x_2=1.4163-((0.4246)^2)/(3.398-0.4246)`

`x_2=1.3557`


`3^(rd)` iteration :

`f(x_2)=f(1.3557)=1.3557^3-1.3557-1=0.1357`

`f(x_2+f(x_2))=f(1.3557+0.1357)=0.8259`

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

`x_3=1.3557-((0.1357)^2)/(0.8259-0.1357)`

`x_3=1.3289`


`4^(th)` iteration :

`f(x_3)=f(1.3289)=1.3289^3-1.3289-1=0.0181`

`f(x_3+f(x_3))=f(1.3289+0.0181)=0.0973`

`x_4=x_3-(f(x_3)^2)/(f(x_3+f(x_3))-f(x_3))`

`x_4=1.3289-((0.0181)^2)/(0.0973-0.0181)`

`x_4=1.3248`


`5^(th)` iteration :

`f(x_4)=f(1.3248)=1.3248^3-1.3248-1=0.0004`

`f(x_4+f(x_4))=f(1.3248+0.0004)=0.0019`

`x_5=x_4-(f(x_4)^2)/(f(x_4+f(x_4))-f(x_4))`

`x_5=1.3248-((0.0004)^2)/(0.0019-0.0004)`

`x_5=1.3247`


Approximate root of the equation `x^3-x-1=0` using Steffensen's method is `1.3247` (After 5 iterations)

`n`	`x_0`	`f(x_0)`	`f(x_0+f(x_0))`	`x_1`	Update
1	1.5	0.875	10.0215	1.4163	`x_0 = x_1`
2	1.4163	0.4246	3.398	1.3557	`x_0 = x_1`
3	1.3557	0.1357	0.8259	1.3289	`x_0 = x_1`
4	1.3289	0.0181	0.0973	1.3248	`x_0 = x_1`
5	1.3248	0.0004	0.0019	1.3247	`x_0 = x_1`