Find `root(3)(48)` using Newton Raphson method
Solution:
Let `x=48^(1/3)`
`:.x^3=48`
`:.x^3-48=0`
i.e. `f(x)=x^3-48`
`d/(dx)(x^3-48)=3x^2`
`d/(dx)(x^3-48)`
`=d/(dx)(x^3)-d/(dx)(48)`
`=3x^2-0`
`=3x^2`
`:. f'(x) = 3x^2`
Here
| `x` | 0 | 1 | 2 | 3 | 4 |
|---|
| `f(x)` | -48 | -47 | -40 | -21 | 16 |
|---|
Here `f(3) = -21 < 0 and f(4) = 16 > 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)=3.5^3-48=-5.125`
`f'(x_0)=f'(3.5)=3*3.5^2=36.75`
`x_1 = x_0 - f(x_0)/(f'(x_0))`
`x_1=3.5 - (-5.125)/(36.75)`
`x_1=3.6395`
`2^(nd)` iteration :
`f(x_1)=f(3.6395)=3.6395^3-48=0.2069`
`f'(x_1)=f'(3.6395)=3*3.6395^2=39.7369`
`x_2 = x_1 - f(x_1)/(f'(x_1))`
`x_2=3.6395 - (0.2069)/(39.7369)`
`x_2=3.6342`
`3^(rd)` iteration :
`f(x_2)=f(3.6342)=3.6342^3-48=0.0003`
`f'(x_2)=f'(3.6342)=3*3.6342^2=39.6233`
`x_3 = x_2 - f(x_2)/(f'(x_2))`
`x_3=3.6342 - (0.0003)/(39.6233)`
`x_3=3.6342`
Approximate root of the equation `x^3-48=0` using Newton Raphson method is `3.6342` (After 3 iterations)
| `n` | `x_0` | `f(x_0)` | `f'(x_0)` | `x_1` | Update |
| 1 | 3.5 | -5.125 | 36.75 | 3.6395 | `x_0 = x_1` |
| 2 | 3.6395 | 0.2069 | 39.7369 | 3.6342 | `x_0 = x_1` |
| 3 | 3.6342 | 0.0003 | 39.6233 | 3.6342 | `x_0 = x_1` |
This material is intended as a summary. Use your textbook for detail explanation.
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