Find a root of an equation `f(x)=2x^3-2x-5` using Newton Raphson method
Solution:
Here `2x^3-2x-5=0`
Let `f(x) = 2x^3-2x-5`
`:. f'(x) = 6x^2-2`
Here
Here `f(1) = -5 < 0 and f(2) = 7 > 0`
`:.` Root lies between `1` and `2`
`x_0 = (1 + 2)/2 = 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`
`x_1 = x_0 - f(x_0)/(f'(x_0))`
`x_1 = 1.5 - (-1.25)/(11.5)`
`x_1 = 1.6087`
`2^(nd)` iteration :
`f(x_1)=f(1.6087)=2*1.6087^(3)-2*1.6087-5=0.1089`
`f'(x_1)=f'(1.6087)=6*1.6087^(2)-2=13.52741`
`x_2 = x_1 - f(x_1)/(f'(x_1))`
`x_2 = 1.6087 - (0.1089)/(13.52741)`
`x_2 = 1.60065`
`3^(rd)` iteration :
`f(x_2)=f(1.60065)=2*1.60065^(3)-2*1.60065-5=0.00062`
`f'(x_2)=f'(1.60065)=6*1.60065^(2)-2=13.37239`
`x_3 = x_2 - f(x_2)/(f'(x_2))`
`x_3 = 1.60065 - (0.00062)/(13.37239)`
`x_3 = 1.6006`
`4^(th)` iteration :
`f(x_3)=f(1.6006)=2*1.6006^(3)-2*1.6006-5=0`
`f'(x_3)=f'(1.6006)=6*1.6006^(2)-2=13.37149`
`x_4 = x_3 - f(x_3)/(f'(x_3))`
`x_4 = 1.6006 - (0)/(13.37149)`
`x_4 = 1.6006`
Approximate root of the equation `2x^3-2x-5=0` using Newton Raphson method is `1.6006`
| `n` | `x_0` | `f(x_0)` | `f'(x_0)` | `x_1` | Update |
| 1 | 1.5 | -1.25 | 11.5 | 1.6087 | `x_0 = x_1` |
| 2 | 1.6087 | 0.1089 | 13.52741 | 1.60065 | `x_0 = x_1` |
| 3 | 1.60065 | 0.00062 | 13.37239 | 1.6006 | `x_0 = x_1` |
| 4 | 1.6006 | 0 | 13.37149 | 1.6006 | `x_0 = x_1` |
This material is intended as a summary. Use your textbook for detail explanation.
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