Northeastern Illinois University What are roots of the Wilkinson polynomial? Zhonggang Zeng Northeastern Illinois University May 12, 2001

Can you solve (x-1.0 )100 = 0 Can you solve x100-100 x99 +4950 x98 - 161700 x97+3921225x96 - ... - 100 x +1 = 0

The Wilkinson polynomial p(x) = (x-1)(x-2)...(x-20) = x20 - 210 x19 + 20615 x18 + ... Wilkinson wrote in 1984: Speaking for myself I regard it as the most traumatic experience in my career as a numerical analyst.

Classical textook methods for multiple roots Newton’s iteration xj+1 = xj - f(xj)/f’(xj), j=0,1,2,... converges locally to a multiple root of f(x) with a linear rate. The modified Newton’s iteration xj+1 = xj - mf(xj)/f’(xj), j=0,1,2,... converges locally to a m-fold root of f(x) with a quadratic rate. Newton’s iteration applied to g(x) = f(x)/f’(x) converges locally and quadratically to a root of f(x) regardliss of its multiplicity. None of them work!

Example: f(x) = (x-2)7(x-3)(x-4) in expanded form. Modified Newton’s iteration with m = 7 intended for root x = 2: x1 = 1.9981 x2 = 1.7481 x3 = 1.9892 x4 = 0.4726 x5 = 1.8029 x6 = 1.9931 x7 = 4.2681 x8 = 3.3476 ... ...

How do we justify the answer?

Conclusion: the problem is “bad” The forward error: 5 The backward error: 5 x 10-10 Conclusion: the problem is “bad”

Who is asking a wrong question? What is the wrong question? If the answer is highly sensitive to perturbations, you have probably asked the wrong question. Maxims about numerical mathematics, computers, science and life, L. N. Trefethen. SIAM News A: “Customer” B: Numerical analyst Who is asking a wrong question? A: The polynomial B: The computing subject What is the wrong question?

The question we used to ask: Given a polynomial p(x) = xn + a1 xn-1+...+an-1 x + an find ( z1, ..., zn ) such that p(x) = ( x - z1 )( x - z2 ) ... ( x - zn ) Right - or - Wrong ?

Kahan’s pejorative manifolds All n-polynomials having certain multiplicity structure form a pejorative manifold xn + a1 xn-1+...+an-1 x + an <=> (a1 , ..., an-1 , an ) Example: ( x-t )2 = x2 + (-2t)x + t2 Pejorative manifold: a1= -2t a2= t2

Pejorative manifolds of 3-polynomials The edge: a1 = -3s a2 = 3s2 a3 = -s3 The wings: a1= -s-2t a2=2st+t2 a3= -st2 General form of pejorative manifolds u = G(z)

Ill-condition is caused by solving polynomial W. Kahan, Conserving confluence curbs ill-condition, 1972 1. Ill-condition occurs when a polynomial is near a pejorative manifold. 2. A small “drift” by a polynomial on that pejorative manifold does not cause large forward error to the multiple roots, except 3. If a multiple root is sensitive to small perturbation on the pejorative manifold, then the polynomial is near a pejorative submanifold of higher multiplicity. Ill-condition is caused by solving polynomial equations on a wrong manifold

Given a polynomial p(x) = xn + a1 xn-1+...+an-1 x + an / / / / / / / / / / / / / / / / / / Find ( z1, ..., zn ) such that p(x) = ( x - z1 )( x - z2 ) ... ( x - zn ) The wrong question: because you are asking for simple roots! Find ( z1, ..., zm ) such that p(x) = ( x - z1 ) s1( x - z2 )s2 ... ( x - zm )sm s1+...+ sm = n, m < n The right question: do it on the pejorative manifold!

For ill-conditioned polynomial p(x)= xn + a1 xn-1+...+an-1 x + an ~ a = (a1 , ..., an-1 , an ) The objective: find u*=G(z*) that is nearest to p(x)~a

I.e. An over determined polynomial system Let ( x - z1 ) s1( x - z2 )s2 ... ( x - zm )sm = xn + g1 ( z1, ..., zm ) xn-1+...+gn-1 ( z1, ..., zm ) x + gn ( z1, ..., zm ) Then, p(x) = ( x - z1 ) s1( x - z2 )s2 ... ( x - zm )sm <==> g1 ( z1, ..., zm ) =a1 g2( z1, ..., zm ) =a2 ... ... ... gn ( z1, ..., zm ) =an I.e. An over determined polynomial system G(z) = a

zi+1=zi - J(zi )+[ G(zi )-a ], i=0,1,2 ... The Gauss-Newton iteration zi+1=zi - J(zi )+[ G(zi )-a ], i=0,1,2 ...

zi+1=zi - J(zi )+[ G(zi )-a ], i=0,1,2 ... Theorem: If z=(z1, ..., zm) with z1, ..., zm distinct, then the Jacobian J(z) of G(z) is of full rank. Theorem: Let u*=G(z*) be nearest to p(x)~a, if 1. z*=(z*1, ..., z*m) with z*1, ..., z*m distinct; 2. z0 is sufficiently close to z*; 3. a is sufficiently close to u* then the Gauss-Newton iteration converges with a linear rate. Further assume that a = u* , then the convergence is quadratic.

The edge: u1 = -3s u2 = 3s2 u3 = -s3 The wings: u1= -s-2t u2=2st+t2 u3= -st2

Example: p(x) = ( x- 0.5)18( x-1.0 )10( x-1.5 )16 The Gauss-Newton iteration: x1 x2 x3 ---- ---- ---- 0.45 1.05 1.55 0.51 0.86 1.57 0.5002 0.9983 1.5007 0.4999996 0.999997 1.500002 0.499999997 1.00000001 1.499999993

What are the roots of the Wilkinson polynomial? (x-1)(x-2)...(x-19)(x-20) ~ (x-z1 )(x-z2 )(x-z3 )2(x-z4 )3(x-z5 )4(x-z6 )4(x-z7 )3(x-z8 )2 Where roots multiplicity 1.00031227 + 1.98468140 + 3.36763082 + + 5.99316993 + + + 9.29701289 + + + + 13.85522338 + + + + 16.66437572 + + + 19.84916622 + + These roots are not sensitive to perturbation

Conclusion Ill-condition is caused by a wrong “identity”. Multiple roots are stable and can be computed with high accuracy, if they are calculated on a proper pejorative manifold. As a related work, isolated multiple roots/eigenvalues can be computed as simple, stable zeros of an extended polynomial system with high accuracy.