Algorithm
|
Steffensen's method Steps (Rule)
|
|
Step-1:
|
Find points `a` and `b` such that `a < b` and `f(a) * f(b) < 0`.
|
|
Step-2:
|
Take the interval `[a, b]` and
find next value `x_0 = (a+b)/2`
|
|
Step-3:
|
Find `f(x_0)` and `f(x_0+f(x_0))`
`x_1=x_0-(f(x_0)^2)/(f(x_0+f(x_0))-f(x_0))`
|
|
Step-4:
|
If `f(x_1) = 0` then `x_1` is an exact root,
else `x_0 = x_1`
|
|
Step-5:
|
Repeat steps 2 to 4 until `f(x_i) = 0` or `|f(x_i)| <= "Accuracy"`
|
Example-1
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
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` |
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
