Find `sqrt(12)` using Newton Raphson method
Solution:
Let `x=sqrt(12)`
`:.x^2=12`
`:.x^2-12=0`
i.e. `f(x)=x^2-12`
`d/(dx)(x^2-12)=2x`
`d/(dx)(x^2-12)`
`=d/(dx)(x^2)-d/(dx)(12)`
`=2x-0`
`=2x`
`:. f'(x) = 2x`
Here
| `x` | 0 | 1 | 2 | 3 | 4 |
|---|
| `f(x)` | -12 | -11 | -8 | -3 | 4 |
|---|
Here `f(3) = -3 < 0 and f(4) = 4 > 0`
`:.` Root lies between `3` and `4`
`x_0 = (3 + 4)/2 = 3.5`
`x_0 = 3.5`
`1^(st)` iteration :
`f(x_0)=f(3.5)=3.5^2-12=0.25`
`f'(x_0)=f'(3.5)=2*3.5=7`
`x_1 = x_0 - f(x_0)/(f'(x_0))`
`x_1=3.5 - (0.25)/(7)`
`x_1=3.4643`
`2^(nd)` iteration :
`f(x_1)=f(3.4643)=3.4643^2-12=0.0013`
`f'(x_1)=f'(3.4643)=2*3.4643=6.9286`
`x_2 = x_1 - f(x_1)/(f'(x_1))`
`x_2=3.4643 - (0.0013)/(6.9286)`
`x_2=3.4641`
`3^(rd)` iteration :
`f(x_2)=f(3.4641)=3.4641^2-12=0`
`f'(x_2)=f'(3.4641)=2*3.4641=6.9282`
`x_3 = x_2 - f(x_2)/(f'(x_2))`
`x_3=3.4641 - (0)/(6.9282)`
`x_3=3.4641`
Approximate root of the equation `x^2-12=0` using Newton Raphson method is `3.4641` (After 3 iterations)
| `n` | `x_0` | `f(x_0)` | `f'(x_0)` | `x_1` | Update |
| 1 | 3.5 | 0.25 | 7 | 3.4643 | `x_0 = x_1` |
| 2 | 3.4643 | 0.0013 | 6.9286 | 3.4641 | `x_0 = x_1` |
| 3 | 3.4641 | 0 | 6.9282 | 3.4641 | `x_0 = x_1` |
This material is intended as a summary. Use your textbook for detail explanation.
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