Algorithm
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Newton Raphson method Steps (Rule)
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Step-1:
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Find points `a` and `b` such that `a < b` and `f(a) * f(b) < 0`.
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Step-2:
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Take the interval `[a, b]` and
find next value `x_0 = (a+b)/2`
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Step-3:
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Find `f(x_0)` and `f'(x_0)`
`x_1 = x_0 - f(x_0)/(f'(x_0))`
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Step-4:
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If `f(x_1) = 0` then `x_1` is an exact root,
else `x_0 = x_1`
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Step-5:
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Repeat steps 3 and 4 until `f(x_i) = 0` or `|f(x_i)| <= "Accuracy"`
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Example-1
1. Find a root of an equation `f(x)=x^3-x-1` using Newton Raphson method
Solution:
Here `x^3-x-1=0`
Let `f(x) = x^3-x-1`
`:. f'(x) = 3x^2-1`
Here
Here `f(1) = -1 < 0 and f(2) = 5 > 0`
`:.` Root lies between `1` and `2`
`x_0 = (1 + 2)/2 = 1.5`
`1^(st)` iteration :
`f(x_0) = f(1.5) = 0.875`
`f'(x_0) = f'(1.5) = 5.75`
`x_1 = x_0 - f(x_0)/(f'(x_0))`
`x_1 = 1.5 - (0.875)/(5.75)`
`x_1 = 1.34783`
`2^(nd)` iteration :
`f(x_1) = f(1.34783) = 0.10068`
`f'(x_1) = f'(1.34783) = 4.44991`
`x_2 = x_1 - f(x_1)/(f'(x_1))`
`x_2 = 1.34783 - (0.10068)/(4.44991)`
`x_2 = 1.3252`
`3^(rd)` iteration :
`f(x_2) = f(1.3252) = 0.00206`
`f'(x_2) = f'(1.3252) = 4.26847`
`x_3 = x_2 - f(x_2)/(f'(x_2))`
`x_3 = 1.3252 - (0.00206)/(4.26847)`
`x_3 = 1.32472`
`4^(th)` iteration :
`f(x_3) = f(1.32472) = 0`
`f'(x_3) = f'(1.32472) = 4.26463`
`x_4 = x_3 - f(x_3)/(f'(x_3))`
`x_4 = 1.32472 - (0)/(4.26463)`
`x_4 = 1.32472`
Approximate root of the equation `x^3-x-1=0` using Newton Raphson method is `1.32472`
| `n` | `x_0` | `f(x_0)` | `f'(x_0)` | `x_1` | Update |
| 1 | 1.5 | 0.875 | 5.75 | 1.34783 | `x_0 = x_1` |
| 2 | 1.34783 | 0.10068 | 4.44991 | 1.3252 | `x_0 = x_1` |
| 3 | 1.3252 | 0.00206 | 4.26847 | 1.32472 | `x_0 = x_1` |
| 4 | 1.32472 | 0 | 4.26463 | 1.32472 | `x_0 = x_1` |
This material is intended as a summary. Use your textbook for detail explanation.
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