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Bairstow's Method

A procedure for finding the quadratic factors for the Complex Conjugate Roots of a Polynomial $P(x)$ with Real Coefficients.


\begin{displaymath}[x-(a+ib)][x-(a-ib)] = x^2+2ax+(a^2+b^2) \equiv x^2+Bx+C.
\end{displaymath} (1)

Now write the original Polynomial as
\begin{displaymath}
P(x)=(x^2+Bx+C)Q(x)+Rx+S
\end{displaymath} (2)


\begin{displaymath}
R(B+\delta B,C+\delta C)\approx R(B,C)+{\partial R\over\partial B}\,dB+{\partial R\over\partial C}\,dC
\end{displaymath} (3)


\begin{displaymath}
S(B+\delta B,C+\delta C)\approx S(B,C)+{\partial S\over\partial B}\,dB+{\partial S\over\partial C}\,dC
\end{displaymath} (4)


\begin{displaymath}
{\partial P\over\partial C} = 0 = (x^2+Bx+C){\partial Q\over...
...Q(x)+
{\partial R\over\partial C}+{\partial S\over\partial C}
\end{displaymath} (5)


\begin{displaymath}
-Q(x) = (x^2+Bx+C){\partial Q\over\partial C}+{\partial R\over\partial C}+{\partial S\over\partial C}
\end{displaymath} (6)


\begin{displaymath}
{\partial P\over\partial B} = 0 = (x^2+Bx+C){\partial Q\over...
...Q(x)+
{\partial R\over\partial B}+{\partial S\over\partial B}
\end{displaymath} (7)


\begin{displaymath}
-xQ(x) = (x^2+Bx+C){\partial Q\over\partial B}+{\partial R\over\partial B}+{\partial S\over\partial B}.
\end{displaymath} (8)

Now use the 2-D Newton's Method to find the simultaneous solutions.


References

Press, W. H.; Flannery, B. P.; Teukolsky, S. A.; and Vetterling, W. T. Numerical Recipes in C: The Art of Scientific Computing. Cambridge, England: Cambridge University Press, pp. 277 and 283-284, 1989.




© 1996-9 Eric W. Weisstein
1999-05-26