Find a root of an equation `f(x)=2x^3-2x-5` using Halley's method
Solution:
Here `2x^3-2x-5=0`
Let `f(x) = 2x^3-2x-5`
`d/(dx)(2x^3-2x-5)=6x^2-2`
`d/(dx)(2x^3-2x-5)`
`=d/(dx)(2x^3)-d/(dx)(2x)-d/(dx)(5)`
`=6x^2-2-0`
`=6x^2-2`
`d/(dx)(6x^2-2)=12x`
`d/(dx)(6x^2-2)`
`=d/(dx)(6x^2)-d/(dx)(2)`
`=12x-0`
`=12x`
`:. f'(x) = 6x^2-2`
`:. f ''(x) = 12x`
Here
Here `f(1) = -5 < 0 and f(2) = 7 > 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)=2*1.5^3-2*1.5-5=-1.25`
`f'(x_0)=f'(1.5)=6*1.5^2-2=11.5`
`f''(x_0)=f'(1.5)=12*1.5=18`
`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-(2xx-1.25xx11.5)/(2xx(11.5)^2-(11.5)xx(18))`
`x_1=1.6002`
`2^(nd)` iteration :
`f(x_1)=f(1.6002)=2*1.6002^3-2*1.6002-5=-0.0057`
`f'(x_1)=f'(1.6002)=6*1.6002^2-2=13.3633`
`f''(x_1)=f'(1.6002)=12*1.6002=19.2021`
`x_2=x_1-(2*f(x_1)*f'(x_1))/(2*f'(x_1)^2-f(x_1)*f''(x_1))`
`x_2=1.6002-(2xx-0.0057xx13.3633)/(2xx(13.3633)^2-(13.3633)xx(19.2021))`
`x_2=1.6006`
`3^(rd)` iteration :
`f(x_2)=f(1.6006)=2*1.6006^3-2*1.6006-5=0`
`f'(x_2)=f'(1.6006)=6*1.6006^2-2=13.3715`
`f''(x_2)=f'(1.6006)=12*1.6006=19.2072`
`x_3=x_2-(2*f(x_2)*f'(x_2))/(2*f'(x_2)^2-f(x_2)*f''(x_2))`
`x_3=1.6006-(2xx0xx13.3715)/(2xx(13.3715)^2-(13.3715)xx(19.2072))`
`x_3=1.6006`
Approximate root of the equation `2x^3-2x-5=0` using Halleys method is `1.6006` (After 3 iterations)
| `n` | `x_0` | `f(x_0)` | `f'(x_0)` | `f''(x_0)` | `x_1` | Update |
| 1 | 1.5 | -1.25 | 11.5 | 18 | 1.6002 | `x_0 = x_1` |
| 2 | 1.6002 | -0.0057 | 13.3633 | 19.2021 | 1.6006 | `x_0 = x_1` |
| 3 | 1.6006 | 0 | 13.3715 | 19.2072 | 1.6006 | `x_0 = x_1` |
This material is intended as a summary. Use your textbook for detail explanation.
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