Find a root of an equation `f(x)=x^3+2x^2+x-1` using Newton Raphson method
Solution:
Here `x^3+2x^2+x-1=0`
Let `f(x) = x^3+2x^2+x-1`
`d/(dx)(x^3+2x^2+x-1)=3x^2+4x+1`
`d/(dx)(x^3+2x^2+x-1)`
`=d/(dx)(x^3)+d/(dx)(2x^2)+d/(dx)(x)-d/(dx)(1)`
`=3x^2+4x+1-0`
`=3x^2+4x+1`
`:. f'(x) = 3x^2+4x+1`
Here
Here `f(0) = -1 < 0 and f(1) = 3 > 0`
`:.` Root lies between `0` and `1`
`x_0 = (0 + 1)/2 = 0.5`
`x_0 = 0.5`
`1^(st)` iteration :
`f(x_0)=f(0.5)=0.5^3+2*0.5^2+0.5-1=0.125`
`f'(x_0)=f'(0.5)=3*0.5^2+4*0.5+1=3.75`
`x_1 = x_0 - f(x_0)/(f'(x_0))`
`x_1=0.5 - (0.125)/(3.75)`
`x_1=0.4667`
`2^(nd)` iteration :
`f(x_1)=f(0.4667)=0.4667^3+2*0.4667^2+0.4667-1=0.0039`
`f'(x_1)=f'(0.4667)=3*0.4667^2+4*0.4667+1=3.52`
`x_2 = x_1 - f(x_1)/(f'(x_1))`
`x_2=0.4667 - (0.0039)/(3.52)`
`x_2=0.4656`
`3^(rd)` iteration :
`f(x_2)=f(0.4656)=0.4656^3+2*0.4656^2+0.4656-1=0`
`f'(x_2)=f'(0.4656)=3*0.4656^2+4*0.4656+1=3.5126`
`x_3 = x_2 - f(x_2)/(f'(x_2))`
`x_3=0.4656 - (0)/(3.5126)`
`x_3=0.4656`
Approximate root of the equation `x^3+2x^2+x-1=0` using Newton Raphson method is `0.4656` (After 3 iterations)
| `n` | `x_0` | `f(x_0)` | `f'(x_0)` | `x_1` | Update |
| 1 | 0.5 | 0.125 | 3.75 | 0.4667 | `x_0 = x_1` |
| 2 | 0.4667 | 0.0039 | 3.52 | 0.4656 | `x_0 = x_1` |
| 3 | 0.4656 | 0 | 3.5126 | 0.4656 | `x_0 = x_1` |
This material is intended as a summary. Use your textbook for detail explanation.
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