Find a root of an equation `f(x)=sqrt(12)` using Fixed Point Iteration method
Solution:
`x=sqrt(12)`
`:.x^2=12`
`:.x^2-12=0`
Here
`:.` Root lies between `3` and `4`
`x_0 = (3 + 4)/2 = 3.5`
`:.12-x^2=0`
Adding `10x` in both the sides, we get
`10x=12-x^2+10x`
`:.x=(12-x^2+10x)/10`
`:.phi(x)=(12-x^2+10x)/10`
`x_1 = phi(x_0) = phi(3.5) = 3.475`
`x_2 = phi(x_1) = phi(3.475) = 3.46744`
`x_3 = phi(x_2) = phi(3.46744) = 3.46513`
`x_4 = phi(x_3) = phi(3.46513) = 3.46442`
`x_5 = phi(x_4) = phi(3.46442) = 3.4642`
Approximate root of the equation `(12-x^2+10x)/10` using Iteration method is `3.4642` (After 5 iterations)
| `n` | `x_0` | `x_1=phi(x_0)` | Update | Difference
`|x_1-x_0|` |
| 2 | 3.5 | 3.475 | `x_0 = x_1` | 0.025 |
| 3 | 3.475 | 3.46744 | `x_0 = x_1` | 0.00756 |
| 4 | 3.46744 | 3.46513 | `x_0 = x_1` | 0.00231 |
| 5 | 3.46513 | 3.46442 | `x_0 = x_1` | 0.00071 |
| 6 | 3.46442 | 3.4642 | `x_0 = x_1` | 0.00022 |
This material is intended as a summary. Use your textbook for detail explanation.
Any bug, improvement, feedback then
