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8. Steffensen's method example ( Enter your problem )
  1. Algorithm & Example-1 `f(x)=x^3-x-1`
  2. Example-2 `f(x)=2x^3-2x-5`
  3. Example-3 `f(x)=x^3+2x^2+x-1`
Other related methods
  1. Bisection method
  2. False Position method (regula falsi method)
  3. Newton Raphson method
  4. Fixed Point Iteration method
  5. Secant method
  6. Muller method
  7. Halley's method
  8. Steffensen's method
  9. Ridder's method

7. Halley's method
(Previous method)
2. Example-2 `f(x)=2x^3-2x-5`
(Next example)

1. Algorithm & Example-1 `f(x)=x^3-x-1`





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
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`012
`f(x)`-1-15



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
11.50.87510.02151.4163`x_0 = x_1`
21.41630.42463.3981.3557`x_0 = x_1`
31.35570.13570.82591.3289`x_0 = x_1`
41.32890.01810.09731.3248`x_0 = x_1`
51.32480.00040.00191.3247`x_0 = x_1`





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7. Halley's method
(Previous method)
2. Example-2 `f(x)=2x^3-2x-5`
(Next example)





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