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This is a problem about the practicalities of removing singularities in multivariable complex functions.

In trying to derive the generating function (in two variables) for a certain problem in combinatorics, we have obtained an expression of the form $S(x,u) = N(x,u)/D(x,u)$ where $$ D(x,u) = (1 + u - u^2) x^2 + (3 u^3 - 2 u) x + u^2 (1 - u - u^2) $$ and $$ N(x,u) = P(x,u) + S(x) Q(x,u) + T(x) R(x,u) $$ Here $P$, $Q$ and $R$ are low degree polynomials in $x$ and $u$, and $S$ and $T$ are known algebraic functions (having convergent power series expansions in a neighbourhood of 0).

The theorem on singularity removal that we would hope to apply is that if $N/D$ is bounded on a neighbourhood of $(0,0)$ excluding the set where $D(x,u) = 0$, then the singularity which that set represents is actually removable, and there is an analytic $F(x,u) = N(x,u)/D(x,u)$ on a neighbourhood of $(0,0)$ wherever the right hand side is defined.

If it matters (probably not), note that setting $D(x,u) = 0$ and solving for $x$ actually gives $x$ as (one of two) rational functions of $u$ (since the discriminant is $5 u^6$). We can also verify (of course or this would be a very silly question indeed) that $N(x,u) = 0$ whenever $D(x,u) = 0$.

What's tripping us up is the practical issues of establishing the boundedness criteria. Can anyone make suggestions about how to do that?

For more specific details, the problem concerns the example discussed in section 4.1 of Generating permutations with restricted containers (pages 9 to 11) which, as it stands, is missing the final piece that an answer to this question would provide.

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    $\begingroup$ Considering you can verify $D(x,u)=0$ implies $N(x,u)=0$, did you check that actually $D\mid N$ (which is indeed true if Nullstellensatz applies here)? $\endgroup$
    – Fan Zheng
    Commented Nov 1, 2015 at 21:42
  • $\begingroup$ @FanZheng, since $N(x,u)$ is not a polynomial, but an analytic function, does the Nullstellensatz apply here? $\endgroup$
    – Jay Pantone
    Commented Nov 2, 2015 at 5:05
  • $\begingroup$ Or you can use Weierstrass division theorem to write $N=f(u)x+g(u)\pmod D$, where $f$ and $g$ are convergent power series in $u$ Then substituting two solutions for $x$ in terms of $u$ to conclude that $f=g=0$. $\endgroup$
    – Fan Zheng
    Commented Nov 3, 2015 at 4:01

1 Answer 1

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I have a solution inspired by Fan Zheng's comment. There is a generalization of Study's lemma [1, $\S$6.13] (which is itself a special case of the Nullstellensatz), that says if $D(x,u)$ is irreducible and if $D(x,u)=0$ implies $N(x,u)=0$ then $D(x,u)$ divides $N(x,u)$.

The zero condition ended up being not easy to check. Let $x_1(u)$ and $x_2(u)$ be two components of the curve on which the denominator is zero. We have $S(x)$ and $T(x)$ in terms of their minimal polynomials (of degree 4 -- explicit expressions are attainable, but quite messy). When substituting $x = x_1(u)$ or $x = x_2(u)$ into the numerator, Maple refuses to simplify to zero.

To get around this, I had to use resultant tricks to find a minimal polynomial for the entire numerator. (In theory, Groëbner bases can be used to do this, but I gave up on the computation after a few hours.) Once this expression was in hand, Maple was happy to verify that the numerator is zero after the substitutions were made.

[1] G. Fischer. Plane algebraic curves, volume 15 of Student Mathematical Library. American Mathematical Society, Providence, RI, 2001. Translated from the 1994 German original by Leslie Kay.

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