Algorithm
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Halley's method Steps (Rule)
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Step-1:
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Find points a and b such that a < b and f(a) * f(b) < 0.
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Step-2:
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Take the interval [a, b] and
find next value x_0 = (a+b)/2
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Step-3:
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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))
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Step-4:
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If f(x_1) = 0 then x_1 is an exact root,
else x_0 = x_1
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Step-5:
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Repeat steps 2 to 4 until f(x_i) = 0 or |f(x_i)| <= "Accuracy"
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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.
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