Find a root of polynomial using Birge-Vieta method
f(x) = x^4-3x^3+3x^2-3x+2 and p0 = 0.5
Solution:
`x^4-3x^3+3x^2-3x+2=0`
In this problem the coefficients are `a_0=1,a_1=-3,a_2=3,a_3=-3,a_4=2`
Let the initial approximation to p be `p_0=0.5`
`1^(st)` Iteration
`p_0`
`0.5` `p_0=0.5` | `a_0`
`1` `a_0=1` | `a_1`
`-3` `a_1=-3` | `a_2`
`3` `a_2=3` | `a_3`
`-3` `a_3=-3` | `a_4`
`2` `a_4=2` |
| | `p_0*b_0`
`0.5` `0.5=0.5*1`
`p_0*b_0` | `p_0*b_1`
`-1.25` `-1.25=0.5*-2.5`
`p_0*b_1` | `p_0*b_2`
`0.88` `0.88=0.5*1.75`
`p_0*b_2` | `p_0*b_3`
`-1.06` `-1.06=0.5*-2.12`
`p_0*b_3` |
| `b_0`
`1` `b_0=1` | `b_1=a_1+p_0*b_0`
`-2.5` `-2.5=-3+0.5`
`b_1=a_1+p_0*b_0` | `b_2=a_2+p_0*b_1`
`1.75` `1.75=3-1.25`
`b_2=a_2+p_0*b_1` | `b_3=a_3+p_0*b_2`
`-2.12` `-2.12=-3+0.88`
`b_3=a_3+p_0*b_2` | `b_4=a_4+p_0*b_3`
`0.94` `0.94=2-1.06`
`b_4=a_4+p_0*b_3` |
| | `p_0*c_0`
`0.5` `0.5=0.5*1`
`p_0*c_0` | `p_0*c_1`
`-1` `-1=0.5*-2`
`p_0*c_1` | `p_0*c_2`
`0.38` `0.38=0.5*0.75`
`p_0*c_2` | |
| `c_0`
`1` `c_0=1` | `c_1=b_1+p_0*c_0`
`-2` `-2=-2.5+0.5`
`c_1=b_1+p_0*c_0` | `c_2=b_2+p_0*c_1`
`0.75` `0.75=1.75-1`
`c_2=b_2+p_0*c_1` | `c_3=b_3+p_0*c_2`
`-1.75` `-1.75=-2.12+0.38`
`c_3=b_3+p_0*c_2` | |
`p_1=p_0-b_4/c_3=0.5-(0.94)/(-1.75)=1.04`
`2^(nd)` Iteration
`p_1`
`1.04` `p_1=1.04` | `a_0`
`1` `a_0=1` | `a_1`
`-3` `a_1=-3` | `a_2`
`3` `a_2=3` | `a_3`
`-3` `a_3=-3` | `a_4`
`2` `a_4=2` |
| | `p_1*b_0`
`1.04` `1.04=1.04*1`
`p_1*b_0` | `p_1*b_1`
`-2.03` `-2.03=1.04*-1.96`
`p_1*b_1` | `p_1*b_2`
`1` `1=1.04*0.97`
`p_1*b_2` | `p_1*b_3`
`-2.07` `-2.07=1.04*-2`
`p_1*b_3` |
| `b_0`
`1` `b_0=1` | `b_1=a_1+p_1*b_0`
`-1.96` `-1.96=-3+1.04`
`b_1=a_1+p_1*b_0` | `b_2=a_2+p_1*b_1`
`0.97` `0.97=3-2.03`
`b_2=a_2+p_1*b_1` | `b_3=a_3+p_1*b_2`
`-2` `-2=-3+1`
`b_3=a_3+p_1*b_2` | `b_4=a_4+p_1*b_3`
`-0.07` `-0.07=2-2.07`
`b_4=a_4+p_1*b_3` |
| | `p_1*c_0`
`1.04` `1.04=1.04*1`
`p_1*c_0` | `p_1*c_1`
`-0.96` `-0.96=1.04*-0.93`
`p_1*c_1` | `p_1*c_2`
`0` `0=1.04*0`
`p_1*c_2` | |
| `c_0`
`1` `c_0=1` | `c_1=b_1+p_1*c_0`
`-0.93` `-0.93=-1.96+1.04`
`c_1=b_1+p_1*c_0` | `c_2=b_2+p_1*c_1`
`0` `0=0.97-0.96`
`c_2=b_2+p_1*c_1` | `c_3=b_3+p_1*c_2`
`-2` `-2=-2+0`
`c_3=b_3+p_1*c_2` | |
`p_2=p_1-b_4/c_3=1.04-(-0.07)/(-2)=1`
`3^(rd)` Iteration
`p_2`
`1` `p_2=1` | `a_0`
`1` `a_0=1` | `a_1`
`-3` `a_1=-3` | `a_2`
`3` `a_2=3` | `a_3`
`-3` `a_3=-3` | `a_4`
`2` `a_4=2` |
| | `p_2*b_0`
`1` `1=1*1`
`p_2*b_0` | `p_2*b_1`
`-2` `-2=1*-2`
`p_2*b_1` | `p_2*b_2`
`1` `1=1*1`
`p_2*b_2` | `p_2*b_3`
`-2` `-2=1*-2`
`p_2*b_3` |
| `b_0`
`1` `b_0=1` | `b_1=a_1+p_2*b_0`
`-2` `-2=-3+1`
`b_1=a_1+p_2*b_0` | `b_2=a_2+p_2*b_1`
`1` `1=3-2`
`b_2=a_2+p_2*b_1` | `b_3=a_3+p_2*b_2`
`-2` `-2=-3+1`
`b_3=a_3+p_2*b_2` | `b_4=a_4+p_2*b_3`
`0` `0=2-2`
`b_4=a_4+p_2*b_3` |
| | `p_2*c_0`
`1` `1=1*1`
`p_2*c_0` | `p_2*c_1`
`-1` `-1=1*-1`
`p_2*c_1` | `p_2*c_2`
`0` `0=1*0`
`p_2*c_2` | |
| `c_0`
`1` `c_0=1` | `c_1=b_1+p_2*c_0`
`-1` `-1=-2+1`
`c_1=b_1+p_2*c_0` | `c_2=b_2+p_2*c_1`
`0` `0=1-1`
`c_2=b_2+p_2*c_1` | `c_3=b_3+p_2*c_2`
`-2` `-2=-2+0`
`c_3=b_3+p_2*c_2` | |
`p_3=p_2-b_4/c_3=1-(0)/(-2)=1`
Approximate root = 1
This material is intended as a summary. Use your textbook for detail explanation.
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