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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?

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    You may please want to clarify what an infinite discontinuity is. Does it mean that the function has a vertical asymptote at that point?2012-05-02
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    @KannappanSampath It must be that case.2012-05-02
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    Look at $P(x)/Q(x)$. There is discontinuity at $x=a$ precisely if $x-a$ is a factor of $Q(x)$, so to know the discontinuities, you will have to know the real $a$ such that $Q(a)=0$. This would give at least a partial factorization of $Q(x)$. To identify the removables can be easier, since the factor $x-a$ would have to occur in $Q(x)$ no more often than it does in $P(x)$. You can identify common factors by finding the greatest common divisor of $P(x)$ and $Q(x)$, which can be done by a variant of the Euclidean Algorithm.2012-05-02
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    You can (approximately!) find the poles of a rational function without factoring, but the method I know requires a series expansion of your rational function, and I don't think that is any easier to do...2012-05-02
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    @KannappanSampath Ok. You can delete the comment.2012-05-02
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    @J.M. Easier or not, it might be interesting!2012-05-02
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    @AndréNicolas - How do you use Euclidean Algorithm to find their common factor?2012-05-02
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    @J.M. - It would be interesting if you post your series expansion answer. Thanks in advance for posting it2012-05-02
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    @Victor: If $A(x)$, $B(x)$ are polynomials, say with degree of $B(x)$ less than or equal to the degree of $A(x)$, divide $A(x)$ by $B(x)$. We get say $A(x)=B(x)Q(x)+R(x)$. Then $\gcd(A,B)=\gcd(B,R)$. Continue.2012-05-02
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    @Peter and Victor: I might post it later if I find time. For now, you can search around on that scheme, called the *quotient-difference* (QD) algorithm.2012-05-03

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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 $\rm\:gcd(F,G).$