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ExaktAI: AI Mathematics with Validation and Human-in-the-Loop

A note on what I've been working on for the past while. Some of you may have seen the announcement on LinkedIn yesterday; this is for the home audience.

The question I've been chasing is the one that's underneath the Physics package, the dsolve / pdsolve formal methods and heuristics, the advanced Mathematical Functions and FunctionAdvisor, and most of what I've written for Maple over the years. How can mathematicians and physicists speed up significantly their work using Computer Algebra Systems (CAS) and at the same time trust the result a computer hands back? The new chapter is what happens when AI sits between the human and the CAS, and the answer to that, in my view, turns out to be a much harder problem than the AI hype suggests.

Why? Because AI is increasingly the driver of computational mathematics in research, engineering, and education. And the unsolved problem isn't whether AI can do mathematics. It can. The problem is that an incorrect AI result arrives with the same confidence as a correct one.

On 100 challenging problems of undergraduate mathematics we tested, six independent state-of-the-art AIs returned mathematically equivalent answers on only 21% of them, and even within a single AI, repeated runs disagreed with themselves on 3% to 57% of the problems (details). The gap this validation crosses, between probabilistic inference and certified mathematical computation, is epistemological, not technological. It won't close with more training data. It needs validation across multiple AIs and multiple CAS, with no single engine having the final word.

ExaktAI aims to address that gap. It guides AI through mathematical computation, validates each step against Maple and Mathematica, and automatically generates and opens a corresponding CAS document where the validation can be audited and reproduced for every step, and where one can continue working on the problem. The goal: AI-mathematics that is validated, with the human in the loop.

ExaktAI is now well developed (TRL 6: System prototype demonstration in a simulated environment, on the ISED / Innovative Solutions Canada TRL scale). At the end an image. A Beta is scheduled for late summer / fall 2026; details at exaktai.ai.

In summary: ExaktAI is my present, and if you work on AI for mathematics and computer algebra, or the validation problem for AI, I'd love to hear your perspective.

Edgardo S. Cheb-Terrab
ExaktAI
Research Fellow Emeritus at Maplesoft.

May 27

8 7

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dharr

9192

Characterizing polynomial roots with RegularChains and related routines

Recently @salim-barzani asked a question about a paper that involved analysing the different types of roots of polynomials. The appendix in that paper gave the example of the roots of using the analysis in Lu et al, "A complete discrimination system for polynomials", Science in China (Ser. E), 39 (1996) 628-646. The analysis uses the discriminant sequence and extensions. Maple provides this through RegularChains:-ParametricSystemTools:-DiscrminantSequence. For example for this polynomial we find there is a real root of multiplicity 2 and a complex conjugate pair when where the are the th entries in the discriminant sequence .

The problem with these conditions is that they are in terms of the and not directly in terms of the parameters. One can derive these conditions and then solve them to find the conditions on the parameters, but Maple has various routines in the RegularChains, RootFinding:-Parametric and SolveTools packages that directly find conditions on parameters to find when there are specified numbers of real or complex roots for polynomial systems. So this post is my attempt to use these tools to find the conditions on the parameters of the above polynomial that give various types of roots. One immediate difficulty is that generally these routines count distinct roots irrespective of multiplicity, and so some indirect analysis is required. There are several different types of commands and analyses that could be used, and my choices here are more to do with my learning experience than an optimum analysis.

The first conclusion is that it is possible, although RealComprehensiveTriangularize did not work as I expected when asking for zero real roots (see cases (a) and (b)) (bug?). Assuming it had worked, RealComprehensiveTriangularize could cover all the cases here, though that will not be true for higher-degree polynomials with more parameters. There doesn't seem to be an obvious systematic way of doing this analysis, which is a downside. Another downside is the large number of subcase conditions, which look as if they could be combined into fewer subcases. CellDecomposition works well for cases without multiplicity.

Main worksheet [not all is displayed below]:

Download RootAnalysis4.mw

Consider the following polynomial in with the three real parameters . We would like to know the conditions on these parameters that lead to different numbers of real and complex-conjugate pairs of roots of different multiplicities.

Consider first how many cases there are. We can set this up as a combinatorial problem in the combstruct package.

For a degree 4 polynomial there are 9 different cases to consider. Here means a (non-real) complex-conjugate pair of roots and means a real root; the exponents indicate the multiplicities.

These are (in order)
(a) A duplicate pair of complex-conjugate roots

(b) Two distinct pairs of complex-conjugate roots

(d) A real root of multiplicity 4

(e) Two distinct real roots of multiplicity 1 and a complex-conjugate pair

(f) Two distinct real roots each of multiplicity 2

(g) A real root of multiplicity 3 and a real root of multiplicity 1

(h) Three distinct real roots, of multiplicities 2, 1, and 1

(i) Four distinct real roots of multiplicity 1

Declare the variables first and parameters last. is the number of parameters. Use the suggested order.

vp := SuggestVariableOrder([p = 0], [x]); R := PolynomialRing(vp); np := nops(`minus`({vp[]}, {x}))

Define derivative polynomials. when there is a root of multiplicity 2 or more; when there is a root of multiplicity 3 or more and when there is a root of multiplicity 4.

Discriminant is zero if and only if there are repeated roots.

(d) A real root of multiplicity 4

This is perhaps the simplest case and can be done using RealComprehensiveTriangularize. Specifying means use the last 3 variables in the PolynomialRing as parameters. The argument 1 means we want the cases where there is one distinct real root. We specify that all of are zero so that the 1 real root is the common root of these polynomials, i.e., is a root on multiplicity 4. (I find the cadcell output a little easier to use, but it is not critical.)

PiecewiseTools:-Is, "Wrong kind of parameters in piecewise"

This means that the conditions on the parameters to get a single real root of multiplicity 4 are

and that the polynomial to solve to find this root under these conditions is

which we can check:

	(g) A real root of multiplicity 3 and a real root of multiplicity 1

	(f) Two distinct real roots each of multiplicity 2

	(i) Four distinct real roots of multiplicity 1

	(h) Three distinct real roots, of multiplicities 2, 1, and 1

	(e) Two distinct real roots of multiplicity 1 and a complex-conjugate pair

	(c) A real root of multiplicity 2 and a pair of complex-conjugate roots

	(b) Two distinct pairs of complex-conjugate roots

(a) A duplicate pair of complex-conjugate roots

Here we want multiplicity 2 but no real roots. The discriminant is expected to be zero, but in most cases is not.

rct := RealComprehensiveTriangularize({p = 0, p2 = 0, p3 <> 0, p4 <> 0}, np, R, 0, output = cadcell); nops(rct[2]); Display(rct[2][1 .. 4], R)

Consider one of the cases with an equality for e[0], suggesting a zero discriminant.
Find a sample point satisfying the conditions, and see what the roots are like

j := 37; cell := rct[2][j][1]; conds := Info(cell, R); pts := Info(SamplePoints(cell, R), R)[1]; `~`[is](eval(conds, pts))

[0 < e[2], e[1] = 0, e[0] = (1/4)*e[2]^2]

We find two complex roots of multiplicity 2, which are complex conjugates, as expected

Check that discriminant is zero.

But, for example, the first cell does not have the discriminant zero, and does not give a correct result.

j := 1; cell := rct[2][j][1]; conds := Info(cell, R); pts := Info(SamplePoints(cell, R), R)[1]; `~`[is](eval(conds, pts))

[e[2] < 0, e[1] < -(2/9)*(-6*e[2]^3)^(1/2), e[0] < RootOf(256*_Z^3-128*e[2]^2*_Z^2+(16*e[2]^4+144*e[1]^2*e[2])*_Z-4*e[1]^2*e[2]^3-27*e[1]^4, index = real[1])]

We find a complex conjugate pair and two distinct real roots, which is not expected for this case.

The discriminant is indeed nonzero.

We did not yet find the conditions for this case. We can ask for the conditions for different numbers of complex roots (complex in this context includes real).

We are interested in the conditions for exactly two distinct roots, which is found from the second constructible_set. There are two subcases.

PIECEWISE([12*e[0]+e[2]^2 = 0, ``], [27*e[1]^2+8*e[2]^3 = 0, ``], [e[2] <> 0, ``]), PIECEWISE([4*e[0]-e[2]^2 = 0, ``], [e[1] = 0, ``], [e[2] <> 0, ``])

Consider the second case

solve(case2[1], {e[0], e[1]}); q1 := eval(p, %); rts2 := [solve(%, x)]

${e[0] = (1/4)*e[2]^2, e[1] = 0}$

[(1/2)*(-2*e[2])^(1/2), -(1/2)*(-2*e[2])^(1/2), (1/2)*(-2*e[2])^(1/2), -(1/2)*(-2*e[2])^(1/2)]

We saw this before when e[2]<0 as case (f) with real roots of multiplicity 2. Now, for e[2]>0 we indeed have two duplicate pairs of complex conjugate roots

Consider the first case

ans1 := {solve(case1[1], {e[0], e[1]}, explicit)}; q11 := eval(p, ans1[1]); rts11 := [solve(%, x, explicit)]

${{e[0] = -(1/12)*e[2]^2, e[1] = -(2/9)*(-6*e[2])^(1/2)*e[2]}, {e[0] = -(1/12)*e[2]^2, e[1] = (2/9)*(-6*e[2])^(1/2)*e[2]}}$

x^4+x^2*e[2]-(2/9)*x*(-6*e[2])^(1/2)*e[2]-(1/12)*e[2]^2

[(1/6)*(-6*e[2])^(1/2), (1/6)*(-6*e[2])^(1/2), (1/6)*(-6*e[2])^(1/2), -(1/2)*(-6*e[2])^(1/2)]

The rts11 subcase polynomial q11 must have real coefficients and therefore only applies for e[2]<0. The roots are real with multiplicity 3 and multiplicity 1 and this case is just case (g) above. The rts12 subcase polynomial q12 (below) also requires e[2]<0 and corresponds to case (g), but with signs reversed.

x^4+x^2*e[2]+(2/9)*x*(-6*e[2])^(1/2)*e[2]-(1/12)*e[2]^2

[(1/2)*(-6*e[2])^(1/2), -(1/6)*(-6*e[2])^(1/2), -(1/6)*(-6*e[2])^(1/2), -(1/6)*(-6*e[2])^(1/2)]

Therefore the rts2 case is the solution for case (a); the cell37 example was a special case of this. In fact if we have a duplicate pair of complex-conjugate roots, the polynomial must be a pefect square, as we see it is