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
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
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 |
| 1 | 1.5 | 0.875 | 5.75 | 9 | 1.3273 | `x_0 = x_1` |
| 2 | 1.3273 | 0.0108 | 4.2848 | 7.9635 | 1.3247 | `x_0 = x_1` |
| 3 | 1.3247 | 0 | 4.2646 | 7.9483 | 1.3247 | `x_0 = x_1` |
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
