Home > Numerical methods calculators > Halley's method example

7. Halley'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

6. Muller 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
Halley'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)`, `f'(x_0)` and `f''(x_0)`
`x_1=x_0-(2*f(x_0)*f'(x_0))/(2*f'(x_0)^2-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 Halley's method

Solution:
Here `x^3-x-1=0`

Let `f(x) = x^3-x-1`

`d/(dx)(x^3-x-1)=3x^2-1`


`d/(dx)(x^3-x-1)`

`=d/(dx)(x^3)-d/(dx)(x)-d/(dx)(1)`

`=3x^2-1-0`

`=3x^2-1`


`d/(dx)(3x^2-1)=6x`


`d/(dx)(3x^2-1)`

`=d/(dx)(3x^2)-d/(dx)(1)`

`=6x-0`

`=6x`


`:. f'(x) = 3x^2-1`

`:. f ''(x) = 6x`

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'(1.5)=3*1.5^2-1=5.75`

`f''(x_0)=f'(1.5)=6*1.5=9`

`x_1=x_0-(2*f(x_0)*f'(x_0))/(2*f'(x_0)^2-f(x_0)*f''(x_0))`

`x_1=1.5-(2xx0.875xx5.75)/(2xx(5.75)^2-(5.75)xx(9))`

`x_1=1.3273`


`2^(nd)` iteration :

`f(x_1)=f(1.3273)=1.3273^3-1.3273-1=0.0108`

`f'(x_1)=f'(1.3273)=3*1.3273^2-1=4.2848`

`f''(x_1)=f'(1.3273)=6*1.3273=7.9635`

`x_2=x_1-(2*f(x_1)*f'(x_1))/(2*f'(x_1)^2-f(x_1)*f''(x_1))`

`x_2=1.3273-(2xx0.0108xx4.2848)/(2xx(4.2848)^2-(4.2848)xx(7.9635))`

`x_2=1.3247`


`3^(rd)` iteration :

`f(x_2)=f(1.3247)=1.3247^3-1.3247-1=0`

`f'(x_2)=f'(1.3247)=3*1.3247^2-1=4.2646`

`f''(x_2)=f'(1.3247)=6*1.3247=7.9483`

`x_3=x_2-(2*f(x_2)*f'(x_2))/(2*f'(x_2)^2-f(x_2)*f''(x_2))`

`x_3=1.3247-(2xx0xx4.2646)/(2xx(4.2646)^2-(4.2646)xx(7.9483))`

`x_3=1.3247`


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

`n``x_0``f(x_0)``f'(x_0)``f''(x_0)``x_1`Update
11.50.8755.7591.3273`x_0 = x_1`
21.32730.01084.28487.96351.3247`x_0 = x_1`
31.324704.26467.94831.3247`x_0 = x_1`





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6. Muller method
(Previous method)
2. Example-2 `f(x)=2x^3-2x-5`
(Next example)





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