Alex Townsend Cornell [email protected]
Vanni Noferini Sujit Rao
ALGEBRAIC-GEOMETRIC METHODS FOR MULTIDIMENSIONAL POLYNOMIAL ROOTFINDING
Joint work with:
MULTIDIMENSIONAL ROOTFINDING
• Polynomials are of maximal degree (degree in each variable)
• Number of isolated solutions (Bernstein’s Theorem)
n n
d!nd
Rootfinding problem: Find all the solutions to:
Assumptions (the algebraically boring situation):•Zero-dimensional polynomial system with well-conditioned roots•Finite and simple solutions (Jacobian is invertible at roots)
0
B@p1(x1, . . . , xd)
...pd(x1, . . . , xd)
1
CA = 0, (x1, . . . , xd) 2 ⌦ ⇢ Cd
TWO DIFFICULT BUT FEASIBLE REGIMES
Robotics (inverse kinematics):
Witness sets convert this to a zero dimensional solution set [Sommese & Wampler, 2005].
Small n, large d Large n, small d
Rootfinders: Resultants, Homotopy, Möller-Stetter, Gröbner bases, Newton's method, etc.
−1 −0.5 0 0.5 1−1
−0.8
−0.6
−0.4
−0.2
0
0.2
0.4
0.6
0.8
1
[Belyaev, 2015]
Random plane waves:
NUMERICAL STABILITY
Goal: A stable numerical algorithm for polynomial rootfinding.
Absolute condition number for rootfinding: In 1D:
x
⇤is a root, [J(x
⇤)]
jk
=
@pj
@xk(x
⇤)
A stable numerical method should compute a root with errorO(kJ(x⇤
)
�1k2✏), ✏ = unit roundo↵
(x⇤) =1
|p0(x⇤)|(x⇤) = kJ(x⇤)�1k2
[Higham, 2002]
INHERIT ROBUSTNESS FROM EIGENSOLVER
(�, A) =||v||2||w||2
|v⇤w|Av = �v w = left eigenvector
Algebraic-geometric
method
For a simple eigenvalue:
[Kressner, Peláez, & Moro, 09]
UNIVARIATE POLYNOMIAL ROOTFINDING
p(x) = a0 + a1x+ · · ·+ anxn
For example,
is the characteristic polynomial forthe companion matrix:
C =
2
6664
0 � a0an
1 � a1an
. . ....
1 �an�1
an
3
7775
Leads to a numerically stable algorithm for rootfinding on the unit circle. [Van Dooren & Dewilde, 83]
ROOTFINDING IN ONE VARIABLE
DEMO
PARTIAL SURVEY OF GLOBAL ROOTFINDERS
Rootfinding problem: Find all the solutions to:0
B@p1(x1, . . . , xd)
...pd(x1, . . . , xd)
1
CA = 0, (x1, . . . , xd) 2 ⌦ ⇢ Cd
TWO IDEAS FROM ALGEBRAIC GEOMETRY
1) Multidimensional resultants
2) Möller-Stetter matrices
[Noferini & T., 2015]
[Noferini, Rao, & T., 2018]
Michael Möller Hans Stetter
LINEAR EXAMPLE: CRAMER’S RULE
Cramer’s rule as an eigenproblem: A1v = �x1(A(:, 1)eT1 )v
HIDDEN-VARIABLE RESULTANTS FOR BIVARIATE POLYNOMIALS
p(x, y) = q(x, y) = 0
p(x, y) =nX
j,k=0
ajkxjy
kq(x, y) =
nX
j,k=0
bjkxjy
k
Rootfinding problem:
where
q[x](y) =nX
k=0
bk(x)yk
p[x](y) =nX
k=0
ak(x)yk
“Hide” one of the variables:
,
Bézout resultant:
The matrix R is singular, where
p(s)q(t)� p(t)q(s)
s� t=
n�1X
j,k=0
Rjksjtk
p(y) and q(y) have a common root
RESULTANTS WITH MATRIX POLYNOMIALS
In algebraic geometry they form the resultant: R(x) = det(R[x])
We avoid the determinant: [Nakatsukasa, Noferini, T., 2015]
CAYLEY RESULTANT MATRIX (TRYING TO AVOID THE EYE-WATERING TENSOR MANIPULATIONS)
0
B@p1(x1, . . . , xd)
...pd(x1, . . . , xd)
1
CA = 0, (x1, . . . , xd) 2 ⌦ ⇢ Cd
In higher dimensions, one can still use multidimensional resultants:
R(x1)v = 0Still, obtain a poly. eig. problem:
DEVASTATING EXAMPLE FOR CAYLEY RESULTANTS
Devastating example for absolute (and relative) conditioning: 0
B@p1(x1, . . . , xd)
.
.
.
pd(x1, . . . , xd)
1
CA =
0
B@x
21.
.
.
x
2d
1
CA+ uQ
0
B@x1.
.
.
xd
1
CA , Q = orthogonal
Cond. no. of rootfinding problem: kJ(x⇤)�1k2 =1
u
kQ�1k2 =1
u
Cond. no. of eigenproblem: (x⇤, R(x1)) =
1
det(J(x⇤))=
1
u
d
J(x⇤) = uQ
x
⇤ = 0
[Noferini & T., 2015]
A devastating example exists for any reasonable basis, any linear combination of solution components, and symbolic calculation of resultant. Also, devastating for relative conditioning.
A NEGATIVE RESULT WITH (HOPEFULLY) A POSITIVE IMPACT
BivariatePolynomial (in MAPLE), chebfun2v/roots (in Chebfun), & Roots (in Mathematica). Numerically unstable bivariate rootfinders:
ROOTFINDING IN TWO DIMENSIONS
DEMO
TWO IDEAS FROM ALGEBRAIC GEOMETRY
1) Multidimensional resultants
2) Möller-Stetter matrices
[Noferini & T., 2015]
[Noferini, Rao, & T., 2018]
Michael Möller Hans Stetter
BACK TO THE COMPANION MATRIX
p(x) = 0, p(x) = a0 + a1x+ · · ·+ anxn
2
6664
0 � a0an
1 � a1an
. . ....
1 �an�1
an
3
7775
2
6664
b0b1...
bn�1
3
7775=
2
66664
� bn�1a0
an
b0 � bn�1a1
an
...
bn�2 � bn�1an�1
an
3
77775
Companion matrix as Euclidean division:q(x) = b0 + b1x+ . . .+ bn�1x
n�1
xq(x) = b0x+ b1x2 + . . .+ bn�1x
n
Euclidean division: xq(x) = c(x)p(x) + r(x), deg(r) < n
c(x) = bn�1
an, r(x) = xq(x)� bn�1
anp(x)
C : Pn�1 ! Pn�1, C(q) = mod(xq, p)
MÖLLER-STETTER MATRICES
One can construct a matrix
0
B@p1(x1, . . . , xd)
...pd(x1, . . . , xd)
1
CA = 0Suppose that has N solutions.
M1 2 CN⇥N
[Lazard, 81], [Möller, 93], [Möller-Stetter, 95]whose eigs are the x1
component of the solutions.
M1 =
2
664
0 0 0 01 �u 0 �u2
0 �u 0 �u2
0 0 1 �u
3
775
For example: p(x, y) = x
2 + u(x+ y)q(x, y) = y
2 + u(x� y)
represents multivariable Euclidean division wrtM1 {1, x, y, xy}
u =p22
DEVASTATING EXAMPLE FOR MÖLLER-STETTER
Devastating example for absolute (and relative) conditioning: 0
B@p1(x1, . . . , xd)
.
.
.
pd(x1, . . . , xd)
1
CA =
0
B@x
21.
.
.
x
2d
1
CA+ uQ
0
B@x1.
.
.
xd
1
CA , Q = orthogonal
A devastating example exists for any reasonable basis, any linear combination of solution components, and symbolic calculation of the Möller-Stetter matrix.
[Noferini, Rao, & T., 2018]
Cond. no. of rootfinding problem: kJ(x⇤)�1k2 =1
u
kQ�1k2 =1
u
Cond. no. of eigenproblem:
J(x⇤) = uQ
x
⇤ = 0
(x⇤,M1) �
1
u
d
THANK YOU
Thanks to: Anthony Austin, Daniel Bates, John Boyd, Martin Lotz, Nick Higham, Tyler Jarvis, Gregorio Malajovich, Yuji Nakatsukasa, Bor Plestenjak, and Andrew Sommese, and Mike Stillman.
Is it possible to design a practical and stable multidimensional rootfinder based on eigensolvers?
Promising ongoing work by:
Bor Plestenjak Marc Van BarelTyler Jarvis
SYLVESTER RESULTANT MATRIX