Find a root of an equation `f(x)=2x^3-2x-5` using Steffensen's method

Solution:
Here `2x^3-2x-5=0`

Let `f(x) = 2x^3-2x-5`

Here

`x`	0	1	2
`f(x)`	-5	-5	7

Here `f(1) = -5 < 0 and f(2) = 7 > 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)=2*1.5^3-2*1.5-5=-1.25`

`f(x_0+f(x_0))=f(1.5+-1.25)=-5.4688`

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

`x_1=1.5-((-1.25)^2)/(-5.4688--1.25)`

`x_1=1.8704`


`2^(nd)` iteration :

`f(x_1)=f(1.8704)=2*1.8704^3-2*1.8704-5=4.3454`

`f(x_1+f(x_1))=f(1.8704+4.3454)=462.8797`

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

`x_2=1.8704-((4.3454)^2)/(462.8797-4.3454)`

`x_2=1.8292`


`3^(rd)` iteration :

`f(x_2)=f(1.8292)=2*1.8292^3-2*1.8292-5=3.5823`

`f(x_2+f(x_2))=f(1.8292+3.5823)=301.1224`

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

`x_3=1.8292-((3.5823)^2)/(301.1224-3.5823)`

`x_3=1.7861`


`4^(th)` iteration :

`f(x_3)=f(1.7861)=2*1.7861^3-2*1.7861-5=2.823`

`f(x_3+f(x_3))=f(1.7861+2.823)=181.6021`

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

`x_4=1.7861-((2.823)^2)/(181.6021-2.823)`

`x_4=1.7415`


`5^(th)` iteration :

`f(x_4)=f(1.7415)=2*1.7415^3-2*1.7415-5=2.0801`

`f(x_4+f(x_4))=f(1.7415+2.0801)=98.978`

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

`x_5=1.7415-((2.0801)^2)/(98.978-2.0801)`

`x_5=1.6968`


`6^(th)` iteration :

`f(x_5)=f(1.6968)=2*1.6968^3-2*1.6968-5=1.3775`

`f(x_5+f(x_5))=f(1.6968+1.3775)=46.9661`

`x_6=x_5-(f(x_5)^2)/(f(x_5+f(x_5))-f(x_5))`

`x_6=1.6968-((1.3775)^2)/(46.9661-1.3775)`

`x_6=1.6552`


`7^(th)` iteration :

`f(x_6)=f(1.6552)=2*1.6552^3-2*1.6552-5=0.7592`

`f(x_6+f(x_6))=f(1.6552+0.7592)=18.32`

`x_7=x_6-(f(x_6)^2)/(f(x_6+f(x_6))-f(x_6))`

`x_7=1.6552-((0.7592)^2)/(18.32-0.7592)`

`x_7=1.6224`


`8^(th)` iteration :

`f(x_7)=f(1.6224)=2*1.6224^3-2*1.6224-5=0.2959`

`f(x_7+f(x_7))=f(1.6224+0.2959)=5.282`

`x_8=x_7-(f(x_7)^2)/(f(x_7+f(x_7))-f(x_7))`

`x_8=1.6224-((0.2959)^2)/(5.282-0.2959)`

`x_8=1.6048`


`9^(th)` iteration :

`f(x_8)=f(1.6048)=2*1.6048^3-2*1.6048-5=0.0567`

`f(x_8+f(x_8))=f(1.6048+0.0567)=0.8503`

`x_9=x_8-(f(x_8)^2)/(f(x_8+f(x_8))-f(x_8))`

`x_9=1.6048-((0.0567)^2)/(0.8503-0.0567)`

`x_9=1.6008`


`10^(th)` iteration :

`f(x_9)=f(1.6008)=2*1.6008^3-2*1.6008-5=0.0024`

`f(x_9+f(x_9))=f(1.6008+0.0024)=0.0344`

`x_10=x_9-(f(x_9)^2)/(f(x_9+f(x_9))-f(x_9))`

`x_10=1.6008-((0.0024)^2)/(0.0344-0.0024)`

`x_10=1.6006`


`11^(th)` iteration :

`f(x_10)=f(1.6006)=2*1.6006^3-2*1.6006-5=0`

`f(x_10+f(x_10))=f(1.6006+0)=0.0001`

`x_11=x_10-(f(x_10)^2)/(f(x_10+f(x_10))-f(x_10))`

`x_11=1.6006-((0)^2)/(0.0001-0)`

`x_11=1.6006`


Approximate root of the equation `2x^3-2x-5=0` using Steffensen's method is `1.6006` (After 11 iterations)

`n`	`x_0`	`f(x_0)`	`f(x_0+f(x_0))`	`x_1`	Update
1	1.5	-1.25	-5.4688	1.8704	`x_0 = x_1`
2	1.8704	4.3454	462.8797	1.8292	`x_0 = x_1`
3	1.8292	3.5823	301.1224	1.7861	`x_0 = x_1`
4	1.7861	2.823	181.6021	1.7415	`x_0 = x_1`
5	1.7415	2.0801	98.978	1.6968	`x_0 = x_1`
6	1.6968	1.3775	46.9661	1.6552	`x_0 = x_1`
7	1.6552	0.7592	18.32	1.6224	`x_0 = x_1`
8	1.6224	0.2959	5.282	1.6048	`x_0 = x_1`
9	1.6048	0.0567	0.8503	1.6008	`x_0 = x_1`
10	1.6008	0.0024	0.0344	1.6006	`x_0 = x_1`
11	1.6006	0	0.0001	1.6006	`x_0 = x_1`

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