3. Example-3 `f(x)=9x^4+12x^3+13x^2+12x+4` and `p0=-0.5`
Find a root of polynomial using Birge-Vieta method `f(x)=9x^4+12x^3+13x^2+12x+4` and `p0=-0.5`
Solution: `9x^4+12x^3+13x^2+12x+4=0`
In this problem the coefficients are `a_0=9,a_1=12,a_2=13,a_3=12,a_4=4`
Let the initial approximation to p be `p_0=-0.5`
`1^(st)` iteration :
`p_0` `-0.5` | `a_0` `9` | `a_1` `12` | `a_2` `13` | `a_3` `12` | `a_4` `4` | | | `p_0*b_0` `-4.5` | `p_0*b_1` `-3.75` | `p_0*b_2` `-4.625` | `p_0*b_3` `-3.6875` | | `b_0` `9` | `b_1=a_1+p_0*b_0` `7.5` | `b_2=a_2+p_0*b_1` `9.25` | `b_3=a_3+p_0*b_2` `7.375` | `b_4=a_4+p_0*b_3` `0.3125` | | | `p_0*c_0` `-4.5` | `p_0*c_1` `-1.5` | `p_0*c_2` `-3.875` | | | `c_0` `9` | `c_1=b_1+p_0*c_0` `3` | `c_2=b_2+p_0*c_1` `7.75` | `c_3=b_3+p_0*c_2` `3.5` | |
`p_1=p_0-b_4/c_3=-0.5-(0.3125)/(3.5)=-0.5893`
`2^(nd)` iteration :
`p_1` `-0.5893` | `a_0` `9` | `a_1` `12` | `a_2` `13` | `a_3` `12` | `a_4` `4` | | | `p_1*b_0` `-5.3036` | `p_1*b_1` `-3.9461` | `p_1*b_2` `-5.3353` | `p_1*b_3` `-3.9274` | | `b_0` `9` | `b_1=a_1+p_1*b_0` `6.6964` | `b_2=a_2+p_1*b_1` `9.0539` | `b_3=a_3+p_1*b_2` `6.6647` | `b_4=a_4+p_1*b_3` `0.0726` | | | `p_1*c_0` `-5.3036` | `p_1*c_1` `-0.8208` | `p_1*c_2` `-4.8516` | | | `c_0` `9` | `c_1=b_1+p_1*c_0` `1.3929` | `c_2=b_2+p_1*c_1` `8.2331` | `c_3=b_3+p_1*c_2` `1.813` | |
`p_2=p_1-b_4/c_3=-0.5893-(0.0726)/(1.813)=-0.6293`
`3^(rd)` iteration :
`p_2` `-0.6293` | `a_0` `9` | `a_1` `12` | `a_2` `13` | `a_3` `12` | `a_4` `4` | | | `p_2*b_0` `-5.664` | `p_2*b_1` `-3.9875` | `p_2*b_2` `-5.6719` | `p_2*b_3` `-3.9825` | | `b_0` `9` | `b_1=a_1+p_2*b_0` `6.336` | `b_2=a_2+p_2*b_1` `9.0125` | `b_3=a_3+p_2*b_2` `6.3281` | `b_4=a_4+p_2*b_3` `0.0175` | | | `p_2*c_0` `-5.664` | `p_2*c_1` `-0.4229` | `p_2*c_2` `-5.4057` | | | `c_0` `9` | `c_1=b_1+p_2*c_0` `0.672` | `c_2=b_2+p_2*c_1` `8.5896` | `c_3=b_3+p_2*c_2` `0.9224` | |
`p_3=p_2-b_4/c_3=-0.6293-(0.0175)/(0.9224)=-0.6483`
`4^(th)` iteration :
`p_3` `-0.6483` | `a_0` `9` | `a_1` `12` | `a_2` `13` | `a_3` `12` | `a_4` `4` | | | `p_3*b_0` `-5.8349` | `p_3*b_1` `-3.997` | `p_3*b_2` `-5.8368` | `p_3*b_3` `-3.9957` | | `b_0` `9` | `b_1=a_1+p_3*b_0` `6.1651` | `b_2=a_2+p_3*b_1` `9.003` | `b_3=a_3+p_3*b_2` `6.1632` | `b_4=a_4+p_3*b_3` `0.0043` | | | `p_3*c_0` `-5.8349` | `p_3*c_1` `-0.2141` | `p_3*c_2` `-5.698` | | | `c_0` `9` | `c_1=b_1+p_3*c_0` `0.3303` | `c_2=b_2+p_3*c_1` `8.7889` | `c_3=b_3+p_3*c_2` `0.4652` | |
`p_4=p_3-b_4/c_3=-0.6483-(0.0043)/(0.4652)=-0.6576`
`5^(th)` iteration :
`p_4` `-0.6576` | `a_0` `9` | `a_1` `12` | `a_2` `13` | `a_3` `12` | `a_4` `4` | | | `p_4*b_0` `-5.9181` | `p_4*b_1` `-3.9993` | `p_4*b_2` `-5.9186` | `p_4*b_3` `-3.9989` | | `b_0` `9` | `b_1=a_1+p_4*b_0` `6.0819` | `b_2=a_2+p_4*b_1` `9.0007` | `b_3=a_3+p_4*b_2` `6.0814` | `b_4=a_4+p_4*b_3` `0.0011` | | | `p_4*c_0` `-5.9181` | `p_4*c_1` `-0.1077` | `p_4*c_2` `-5.8478` | | | `c_0` `9` | `c_1=b_1+p_4*c_0` `0.1637` | `c_2=b_2+p_4*c_1` `8.8931` | `c_3=b_3+p_4*c_2` `0.2336` | |
`p_5=p_4-b_4/c_3=-0.6576-(0.0011)/(0.2336)=-0.6621`
`6^(th)` iteration :
`p_5` `-0.6621` | `a_0` `9` | `a_1` `12` | `a_2` `13` | `a_3` `12` | `a_4` `4` | | | `p_5*b_0` `-5.9592` | `p_5*b_1` `-3.9998` | `p_5*b_2` `-5.9594` | `p_5*b_3` `-3.9997` | | `b_0` `9` | `b_1=a_1+p_5*b_0` `6.0408` | `b_2=a_2+p_5*b_1` `9.0002` | `b_3=a_3+p_5*b_2` `6.0406` | `b_4=a_4+p_5*b_3` `0.0003` | | | `p_5*c_0` `-5.9592` | `p_5*c_1` `-0.054` | `p_5*c_2` `-5.9236` | | | `c_0` `9` | `c_1=b_1+p_5*c_0` `0.0815` | `c_2=b_2+p_5*c_1` `8.9462` | `c_3=b_3+p_5*c_2` `0.117` | |
`p_6=p_5-b_4/c_3=-0.6621-(0.0003)/(0.117)=-0.6644`
`7^(th)` iteration :
`p_6` `-0.6644` | `a_0` `9` | `a_1` `12` | `a_2` `13` | `a_3` `12` | `a_4` `4` | | | `p_6*b_0` `-5.9797` | `p_6*b_1` `-4` | `p_6*b_2` `-5.9797` | `p_6*b_3` `-3.9999` | | `b_0` `9` | `b_1=a_1+p_6*b_0` `6.0203` | `b_2=a_2+p_6*b_1` `9` | `b_3=a_3+p_6*b_2` `6.0203` | `b_4=a_4+p_6*b_3` `0.0001` | | | `p_6*c_0` `-5.9797` | `p_6*c_1` `-0.027` | `p_6*c_2` `-5.9617` | | | `c_0` `9` | `c_1=b_1+p_6*c_0` `0.0407` | `c_2=b_2+p_6*c_1` `8.973` | `c_3=b_3+p_6*c_2` `0.0586` | |
`p_7=p_6-b_4/c_3=-0.6644-(0.0001)/(0.0586)=-0.6655`
`8^(th)` iteration :
`p_7` `-0.6655` | `a_0` `9` | `a_1` `12` | `a_2` `13` | `a_3` `12` | `a_4` `4` | | | `p_7*b_0` `-5.9898` | `p_7*b_1` `-4` | `p_7*b_2` `-5.9898` | `p_7*b_3` `-4` | | `b_0` `9` | `b_1=a_1+p_7*b_0` `6.0102` | `b_2=a_2+p_7*b_1` `9` | `b_3=a_3+p_7*b_2` `6.0102` | `b_4=a_4+p_7*b_3` `0` | | | `p_7*c_0` `-5.9898` | `p_7*c_1` `-0.0135` | `p_7*c_2` `-5.9808` | | | `c_0` `9` | `c_1=b_1+p_7*c_0` `0.0203` | `c_2=b_2+p_7*c_1` `8.9865` | `c_3=b_3+p_7*c_2` `0.0293` | |
`p_8=p_7-b_4/c_3=-0.6655-(0)/(0.0293)=-0.6661`
`9^(th)` iteration :
`p_8` `-0.6661` | `a_0` `9` | `a_1` `12` | `a_2` `13` | `a_3` `12` | `a_4` `4` | | | `p_8*b_0` `-5.9949` | `p_8*b_1` `-4` | `p_8*b_2` `-5.9949` | `p_8*b_3` `-4` | | `b_0` `9` | `b_1=a_1+p_8*b_0` `6.0051` | `b_2=a_2+p_8*b_1` `9` | `b_3=a_3+p_8*b_2` `6.0051` | `b_4=a_4+p_8*b_3` `0` | | | `p_8*c_0` `-5.9949` | `p_8*c_1` `-0.0068` | `p_8*c_2` `-5.9904` | | | `c_0` `9` | `c_1=b_1+p_8*c_0` `0.0102` | `c_2=b_2+p_8*c_1` `8.9932` | `c_3=b_3+p_8*c_2` `0.0147` | |
`p_9=p_8-b_4/c_3=-0.6661-(0)/(0.0147)=-0.6664`
`10^(th)` iteration :
`p_9` `-0.6664` | `a_0` `9` | `a_1` `12` | `a_2` `13` | `a_3` `12` | `a_4` `4` | | | `p_9*b_0` `-5.9975` | `p_9*b_1` `-4` | `p_9*b_2` `-5.9975` | `p_9*b_3` `-4` | | `b_0` `9` | `b_1=a_1+p_9*b_0` `6.0025` | `b_2=a_2+p_9*b_1` `9` | `b_3=a_3+p_9*b_2` `6.0025` | `b_4=a_4+p_9*b_3` `0` | | | `p_9*c_0` `-5.9975` | `p_9*c_1` `-0.0034` | `p_9*c_2` `-5.9952` | | | `c_0` `9` | `c_1=b_1+p_9*c_0` `0.0051` | `c_2=b_2+p_9*c_1` `8.9966` | `c_3=b_3+p_9*c_2` `0.0073` | |
`p_10=p_9-b_4/c_3=-0.6664-(0)/(0.0073)=-0.6665`
`11^(th)` iteration :
`p_10` `-0.6665` | `a_0` `9` | `a_1` `12` | `a_2` `13` | `a_3` `12` | `a_4` `4` | | | `p_10*b_0` `-5.9987` | `p_10*b_1` `-4` | `p_10*b_2` `-5.9987` | `p_10*b_3` `-4` | | `b_0` `9` | `b_1=a_1+p_10*b_0` `6.0013` | `b_2=a_2+p_10*b_1` `9` | `b_3=a_3+p_10*b_2` `6.0013` | `b_4=a_4+p_10*b_3` `0` | | | `p_10*c_0` `-5.9987` | `p_10*c_1` `-0.0017` | `p_10*c_2` `-5.9976` | | | `c_0` `9` | `c_1=b_1+p_10*c_0` `0.0025` | `c_2=b_2+p_10*c_1` `8.9983` | `c_3=b_3+p_10*c_2` `0.0037` | |
`p_11=p_10-b_4/c_3=-0.6665-(0)/(0.0037)=-0.6666`
Approximate root = -0.6666
This material is intended as a summary. Use your textbook for detail explanation. Any bug, improvement, feedback then
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