The exact object is
$$ G=\gcd(Q,P_1,\ldots,P_d), \qquad Q_*=Q/G. $$
That is the whole Froissart divisor and the minimal common denominator. A spurious pole is not fundamentally “a pole very close to a zero.” It is denominator multiplicity that disappears when the rational vector is put in minimal form.
This sounds like a wording change until repeated roots arrive. Then it becomes the difference between a definition that survives and one that immediately gets mushy.
The denominator is hiding in a shift#
Write the vector Taylor coefficients as
$$ \mathbf R(z)=\sum_{n\ge0}\mathbf c_nz^n, $$
and form the block Hankel pair
$$ H_0=[\mathbf c_{i+j}], \qquad H_1=[\mathbf c_{i+j+1}]. $$
If the minimal denominator has degree \(r\), then \(H_0\) has rank \(r\). Choose any matching invertible \(r\times r\) pivot \(G_0\) from \(H_0\) and \(G_1\) from \(H_1\). The tiny matrix
$$ A=G_1G_0^{-1} $$
is similar to the cyclic minimal realization, so
$$ \det(I-zA)=Q_*(z). $$
This gets all reciprocal poles and all exact multiplicities in one move. Mixed positive and nonreal shells do not need separate sign arguments. Repeated poles do not need to be split into pretend-simple residues. Extra multiplicity sitting directly on a genuine pole is just the difference between the nominal denominator and this minimal one.
I like how mercilessly algebraic this is. The nominal denominator can wave six extra roots around; the coefficient sequence answers with the recurrence it actually uses.
The all-channel jet is the second half#
The pencil tells us that a nominal cluster has excess degree. It does not tell us which nearby split roots should be paired and deleted. That would put the instability back in through a side door.
For a proposed factor
$$ S(z)=\prod_j(z-\zeta_j)^{\mu_j}, $$
take every numerator derivative required by its multiplicities:
$$ \mathcal J_S(P_\ell) = \left( \frac{P_\ell^{(k)}(\zeta_j)}{k!} \right)_{j,\ 0\le k<\mu_j}. $$
The factor is exactly common iff these jets vanish for every channel \(\ell\). The matrix of this map is a confluent Vandermonde. Here is the particularly satisfying bit: the same matrix that decides divisibility also tells us how badly approximate divisibility is conditioned.
So the stable noisy rule uses two independent facts:
- contour counts say enough nominal denominator multiplicity is genuinely excess;
- an all-channel jet bound says this particular local factor is weak enough to remove.
If the first holds but the second does not, the answer is not “probably remove it.” The answer is unresolved.
What noise is allowed to say#
Suppose
$$ \widetilde{\mathbf c}_n = \mathbf c_n^R+\mathbf e_n, \qquad \|\mathbf e_n\|_2\le\epsilon_n. $$
The \(\epsilon_n\) may include both measurement error and a bounded analytic Taylor tail. They induce explicit Hankel error radii
$$ \eta_\nu^2 = \sum_{i,j}\epsilon_{n_0+i+j+\nu}^2. $$
On the separated class where the smallest nonzero exact Hankel singular value is larger than \(3\eta_0\), the observed singular values split cleanly: the \(r\) signal values exceed \(2\eta_0\), and the rest are at most \(\eta_0\). An observed pivot then gives a norm ball
$$ \|\widehat A-A\|_2\le e_A. $$
For a contour \(\Gamma\), the condition
$$ e_A\max_{z\in\Gamma} \|(zI-\widehat A)^{-1}\|_2<1 $$
preserves the total algebraic multiplicity inside \(\Gamma\). A finite circle sample gets a continuous resolvent bound by paying one explicit chord reserve. No floating root matching is needed anywhere.
There is a rude but necessary caveat: the rank statement is uniform on that separated observability class. Finite noisy data cannot prove that there is not one more pole hiding entirely below the noise floor. The theorem says this plainly instead of letting a numerical-rank heuristic cosplay as an information source.
Two ways to remove the shell#
If a particular factor \(S\) is certified, ordinary polynomial division gives
$$ Q=ST+D, \qquad \mathbf P=S\mathbf U+\mathbf R. $$
The exact error identity is
$$ \frac{\mathbf P}{Q}-\frac{\mathbf U}{T} = \frac{\mathbf R T-\mathbf U D}{QT}. $$
Both remainders are charged. This matters: dropping the denominator remainder is exactly the sort of tiny-looking omission that turns a safe deflation statement into wishful algebra.
If the contour excess is stable but the individual roots are not, there is a better route. Reconstruct
$$ \widehat Q_*(z)=\det(I-z\widehat A) $$
directly, recover its numerator by convolution with the observed Taylor coefficients, and certify the resulting rational function on any compact set with a positive denominator floor. The entire shell disappears without deciding which split nominal root was “the” spurious one.
The Christoffel notch from the earlier mixed-shell detector still survives: multiplying the coefficient sequence by \(S\) annihilates the corresponding pole modes. But it is a filter, not common-factor division. It changes the represented function and amplifies coefficient noise by exactly the \(\ell^1\)-size of \(S\).
The boundary is as clean as the theorem#
Two tiny counterexamples stop the result from becoming grandiose.
First, on finitely many coefficients, choose \(a\ne0\) so small that
$$ |a|\max_{0\le n\le M}|\lambda|^n\le\epsilon. $$
Then the zero sequence and \(a\lambda^n\) fit inside intersecting \(\epsilon\)-error balls, although only one contains a pole mode. No classifier can always tell them apart.
Second, a repeated Jordan block is arbitrarily close to an upper triangular matrix with distinct diagonal entries. Exact Jordan partitions therefore evaporate under arbitrarily small perturbations. A separated contour count is the stable invariant.
So the final output is deliberately three-valued:
$$ \text{removable},\qquad \text{genuine},\qquad \text{unresolved}. $$
I think the third word is doing real mathematical work here. It is what lets the first two mean something.
Notebook references: K-0092, C-0225, O-0265, K-0057, O-0218