Algorithm
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Fixed Point Iteration method Steps (Rule)
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Step-1:
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First write the equation `x = phi(x)`
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Step-2:
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Find points `a` and `b` such that `a < b` and `f(a) * f(b) < 0`.
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Step-3:
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If `f(a)` is more closer to 0 then `f(b)`
then `x_0=a` else `x_0=b`
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Step-4:
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`x_1 = phi(x_0)`
`x_2 = phi(x_1)`
`x_3 = phi(x_2)`
...
Repeat until `|f(x_i)-f(x_(i-1))| ~~ 0`
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Example-1
1. Find a root of an equation `f(x)=x^3-x-1` using Fixed Point Iteration method
Solution:
Method-1
Let `f(x) = x^3-x-1`
Here `x^3-x-1=0`
`:.x^3=x+1`
`:.x=root (3)(x+1)`
`:.phi(x)=root (3)(x+1)`
Here
Here `f(1) = -1 < 0` and `f(2) = 5 > 0`
`:.` Root lies between `1` and `2`
`x_0 = (1 + 2)/2 = 1.5`
`x_1 = phi(x_0) = phi(1.5) = 1.35721`
`x_2 = phi(x_1) = phi(1.35721) = 1.33086`
`x_3 = phi(x_2) = phi(1.33086) = 1.32588`
`x_4 = phi(x_3) = phi(1.32588) = 1.32494`
`x_5 = phi(x_4) = phi(1.32494) = 1.32476`
Approximate root of the equation `x^3-x-1` using Iteration method is `1.32476` (After 5 iterations)
| `n` | `x_0` | `x_1=phi(x_0)` | Update | Difference
`|x_1-x_0|` |
| 2 | 1.5 | 1.35721 | `x_0 = x_1` | 0.14279 |
| 3 | 1.35721 | 1.33086 | `x_0 = x_1` | 0.02635 |
| 4 | 1.33086 | 1.32588 | `x_0 = x_1` | 0.00498 |
| 5 | 1.32588 | 1.32494 | `x_0 = x_1` | 0.00094 |
| 6 | 1.32494 | 1.32476 | `x_0 = x_1` | 0.00018 |
This material is intended as a summary. Use your textbook for detail explanation.
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