Find a root of an equation `f(x)=root(3)(48)` using Fixed Point Iteration method
Solution:
`x=root(3)(48)`
`:.x^3=48`
`:.x^3-48=0`
Here
| `x` | 0 | 1 | 2 | 3 | 4 |
|---|
| `f(x)` | 48 | 47 | 40 | 21 | -16 |
|---|
`:.` Root lies between `3` and `4`
`x_0 = (3 + 4)/2 = 3.5`
`:.48-x^3=0`
Adding `10x` in both the sides, we get
`10x=48-x^3+10x`
`:.x=(48-x^3+10x)/10`
`:.phi(x)=(48-x^3+10x)/10`
`x_1 = phi(x_0) = phi(-0.5) = 4.3125`
`x_2 = phi(x_1) = phi(4.3125) = 1.09226`
`x_3 = phi(x_2) = phi(1.09226) = 5.76195`
`x_4 = phi(x_3) = phi(5.76195) = -8.56777`
`x_5 = phi(x_4) = phi(-8.56777) = 59.12531`
`x_6 = phi(x_5) = phi(59.12531) = -20605.1162`
The method lead us away from the solution, so it is divergent.
| `n` | `x_0` | `x_1=phi(x_0)` | Update | Difference
`|x_1-x_0|` |
| 2 | -0.5 | 4.3125 | `x_0 = x_1` | 4.8125 |
| 3 | 4.3125 | 1.09226 | `x_0 = x_1` | 3.22024 |
| 4 | 1.09226 | 5.76195 | `x_0 = x_1` | 4.66969 |
| 5 | 5.76195 | -8.56777 | `x_0 = x_1` | 14.32972 |
| 6 | -8.56777 | 59.12531 | `x_0 = x_1` | 67.69308 |
| 7 | 59.12531 | -20605.1162 | `x_0 = x_1` | 20664.24151 |
This material is intended as a summary. Use your textbook for detail explanation.
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