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
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` |
This material is intended as a summary. Use your textbook for detail explanation.
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