Rational function is defining as a polynomial with real coefficients over polynomial with real coefficents, how to find the removeable or infinite discontinuity of any rational function without the factoring of the polynomial since it is very troublesome?
How to find the removeable or infinite discontinuity of any rational function without the factoring of the polynomial?
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0@Peter and Victor: I might $p$ost it later if I find time. For now, you can search around on that scheme, called the *quotient-difference* (QD) algorithm. – 2012-05-03
1 Answers
Factoring is not needed, only much more efficient gcd computations. Suppose that your rational function $\rm\:F/G\:$ reduces to $\rm\:f/g\:$ in lowest terms, after cancelling out $\rm\:d = gcd(F,G).\:$ Then any root $\rm\:r\:$ of $\rm\:g\:$ is non-removable, since it cannot also be a root of $\rm\:f,\:$ else $\rm\:x-r\:$ divides both $\rm\:f,g\:$ contra $\rm\:gcd(f,g) = 1.\:$ So roots of $\rm\:d = gcd(F,G)\:$ not also roots of $\rm\:g\:$ are removed singularities, since $\rm\:F/G\:$ is singular at $\rm\:r\:$ by $\rm\:d(r)=0\:\Rightarrow\:F(r) = 0 = G(r),\:$ but $\rm\:f/g\:$ is not singular at $\rm\:r\:$ since $\rm\:g(r)\ne 0.\:$ These are precisely the roots whose numerator multiplicity is $\ge$ denominator multiplicity, so they are removed from the denominator when cancelling out $\rm\:gcd(F,G).$