Perfidious Polynomials and Elusive RootsZhonggang Zeng

Northeastern Illinois University

Nov. 2, 2001 at Northern Illinois University

What are the “best” functions?

1. Linear functions2. Polynomials3. Exponential functions4. Trigonometric functions... ...

Weierstrass Theorem: Any continuous function can be approximated to any accuracy by a polynomial of sufficiently high degree.

Why by a

polynomial ?

Polynomial root-finding

xn + a1 xn-1 + a2 xn-2 + ... + an-1x + a0 = 0

has played a key role in the history of mathematics

- Rational/irrational numbers, complex numbers- Fundamental Theorem of Algebra- Abel and Galois Theorem... ...

Can you solve (x-1.0 )100 = 0 Can you solve

x100-100 x99 +4950 x98 - 161700 x97+3921225x96 - ... - 100 x +1 = 0

1 01 1 1 1 ... ... 1 1

0 1 1

A = X X-1

Eigenvalues of

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 = 1.9981x2 = 1.7481x3 = 1.9892x4 = 0.4726x5 = 1.8029x6 = 1.9931x7 = 4.2681x8 = 3.3476 ... ...

Myths on polynomial roots:

- multiple roots are ill-conditioned, or even intractable

- extension of machine precision is necessary to calculate multiple roots

- there is an “attainable precision” for multiple roots:

machine precision attainable precision = ----------------------------- multiplicity

Example: for 100-fold root, to get 5 digits right

500 digits in machine precision 5 digits precision = -----------------------------------------

100 in multiplicity

How do we justify the answer?

The condition number

[Forward error] ~ [Condition number] [Backward error]

A large condition number <=> The problem is sensitive or, ill-conditioned

From computational method

From problem

The backward error: 5 x 10-10

The forward error: 5

Conclusion: the problem is “bad”

-- method is good!

-- Ouch! Who’s responsible?

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

Who is asking a wrong question?

What is the wrong question?

A: “Customer”

B: Numerical analyst

A: The polynomial

B: The computing objective

The question we used to askThe question we used to ask:(Fundamental Theorem of Algebra)

Given a polynomial

p(x) = xn + a1 xn-1+...+an-1 x + an

find z = ( z1, ..., zn ) such that

p(x) = ( x - z1 )( x - z2 ) ... ( x - zn )

Right - or - Wrong ?

Kahan’s pejorative manifolds

xn + a1 xn-1+...+an-1 x + an <=> (a1 , ..., an-1 , an )

All n-polynomials having certain multiplicity structure form a pejorative manifold

Example: ( x-t )2 = x2 + (-2t) x + t2

Pejorative manifold: a1= -2t a2= t2

Pejorative manifolds of 3-polynomials

( x - s )( x - t )2 = x3 + (-s-2t) x2 + (2st+t2) x + (-st2)

( x - s )3 = x3 + (-3s) x2 + (3s2) x + (-s3)

Pejorative manifold of multiplicity structure [1,2]

a1= -s-2ta2= 2st+t2

a3= -st2

Pejorative manifold ofmultiplicity structure [ 3 ]

a1 = -3sa2 = 3s2

a3 = -s3

Pejorative manifolds of 3-polynomials

The wings: a1= -s-2t a2= 2st+t2

a3= -st2

The edge: a1 = -3s a2 = 3s2

a3 = -s3

General form ofpejorative manifolds

u = G(z)

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 manifolddoes not cause large forward error to the multiple roots, except

3. If a multiple root is sensitive to small perturbation on thepejorative manifold, then the polynomial is near a pejorativesubmanifold of higher multiplicity.

Ill-condition is caused by solving polynomialequations on a wrong manifold

Pejorative manifolds of 3-polynomials

The wings: a1= -s-2t a2= 2st+t2

a3= -st2

The edge: a1 = -3s a2 = 3s2

a3 = -s3

Given a polynomial p(x) = xn + a1 xn-1+...+an-1 x + an

The wrong question: Find ( z1, ..., zn ) such thatp(x) = ( x - z1 )( x - z2 ) ... ( x - zn )

because you are asking for simple roots!

/ / / / / / / / / / / / / / / / / // / / / / / / / / / / / / / / / / /

The right question:Find distinct z1, ..., zm such thatp(x) = ( x - z1 1 x - z2 )2 ... ( x - zm )m

m = n, m < n

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

Let ( x - z1 1 x - z2 )2 ... ( x - zm )m =

xn + g1 ( z1, ..., zm ) xn-1+...+gn-1 ( z1, ..., zm ) x + gn ( z1, ..., zm )

Then, p(x) = ( x - z1 1 x - z2 )2 ... ( x - zm )m <==>

g1 ( z1, ..., zm ) =a1

g2( z1, ..., zm ) =a2

... ... ...

gn ( z1, ..., zm ) =an

I.e. An over determined polynomial system

G(z) = a

(m<n)n

m

tangent plane P0 :

u = G(z0)+J(z

0)(z- z0)

initial iterate

u0 =

G(z

0 )

pejorative root

u* =

G(z

* )

The polynomiala

Project to tangent plane

u 1 = G(z 0

)+J(z 0)(z 1

- z 0)

~

new iterate

u1 =

G(z

1 )

Pejora

tive m

anifo

ld

u = G

( z )

Solve G( z ) = a for nonlinear least squares solution z=z*

Solve G(z0)+J(z0)( z - z0 ) = a for linear least squares solution z = z1

G(z0)+J(z0)( z - z0 ) = aJ(z0)( z - z0 ) = - [G(z0) - a ] z1 = z0 - [J(z0)+] [G(z0) - a]

Theorem: Let u*=G(z*) be nearest to p(x)~a, if1. 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 iteration converges with a linear rate.

Further assume that a = u* , then the convergence is quadratic.

Theorem: If z=(z1, ..., zm) with z1, ..., zm distinct, then the Jacobian J(z) of G(z) is of full rank.

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

Calculation of G(z) and J(z) ( x - z1 1 x - z2 )2 ... ( x - zm )m =

xn + g1 ( z1, ..., zm ) xn-1+...+gn-1 ( z1, ..., zm ) x + gn ( z1, ..., zm )

Explicit polynomail formula gj ( z1, ..., zm ) isneither necessary nor desirable.

G(z) and every column of J(z) are calculated viasimple numerical procedures, as key componentsof the algorithm.

The conventional condition number

Let f(x*) = 0

f(xg(xx(0) = x*

f’(xx’g(xg’(xx’/ / / / / / / / / / / / -----x*

-----x*

x’(0) = - g(x*)/ f’(x*)

xx* - g(x*)/ f’(x*)h.o.t.

| xx* | < | 1 / f’(x*)| .| g(x*) | + h.o.t.

Condition number | 1 / f’(x*)| = infinity at multiple a root

The “pejorative” condition number

u = G(y)

v = G(z)

|| u - v ||2 = backward error|| y - z ||2 = forward error

u - v = G(y) - G(z) = J(z) (y - z) + h.o.t.

|| u - v ||2 = || J(z) (y - z) ||2 > y - z ||2

y - z ||2 < (1/) u - v ||2

1/ is the pejorative condition numberwhere is the smallest singular value of J(z) .

Example 1 (x-1.0 )100 = 0

To make the problem interesting:round the coefficients to 5 digits

Step iterates ... ... ... ... 43 1.007 44 1.001 45 1.0001 46 1.0000003 47 .999999998

conventional condition: infinity pejorative condition: 0.0017

Is it ill-conditioned?

Example 2 (x-0.9)18(x-1.0)10(x-1.1)16 = 0 Step z1 z2 z3

--------------------------------------------------------------------0 .92 .95 1.121 .87 1.05 1.102 .92 .95 1.113 .88 1.01 1.104 .90 .97 1.125 .901 .992 1.1016 .89993 .9998 1.10027 .9000003 .999998 1.10000078 .899999999997 .999999999991 1.1000000000099 .900000000000006 .99999999999997 1.10000000000001

forward error: 6 x 10-15

backward error: 8 x 10-16

Pejorative condition: 58

Even clustered multiple roots are pejoratively well conditioned

Multiplicity pejorative roots backward error pejorative condition structure------------------------------------------------------------------------------------------------------------- [1,1,...1] erratic .0000000000000006 1390704851032436. [18,10,16] (.9000, 1.0000, 1.1000) .0000000000000008 58.2 [17,11,16] (.8980, .9934, 1.1006) .0000005 51.9 [14,16,14] (.8890, .9892, 1.1090) .000006 29.0 [10,24,10] (.8711, .9906, 1.1316) .00001 25.8 [ 2,40, 2] (.7393, .9917, 1.3282) .0001 23.6 [ 1,43] (.6559, 1.0053 ) .004 12.9 [44] ( .9925 ) .04 .0058

If you are on a wrong manifold ...

Example 3 (x-.3-.6i)100 (x-.1-.7i) 200 (x - .7-.5i) 300 (x-.3-.4i) 400 =0

Scary enough? Round coefficients to 6 digits.

Z1 z2 z3 z4

.289 +.601i .100 +.702i .702 +.498i .301 +.399i

.309 +.602i .097 +.698i .698 +.499i .299 +.401i

.293 +.596i .101 +.7003i .7002 +.5005i .3007 +.4003i

.300005 +.600006i .099998 +.6999992i .69999992+.4999993i .2999992 +.3999992i

.3000002+.60000005i .09999995+.69999998i .69999997+.49999998i .29999997+.400000002i

Roots are correct up to 7 digits!

Pejorative condition: 0.58

Example 4: The Wilkinson polynomial

p(x) = (x-1)(x-2)...(x-20) = x20 - 210 x19 + 20615 x18 + ...

There are 605 manifolds in total. It is near some manifolds, but which ones?

Multiplicity backward error condition Estimated structure number error------------------------------------------------------------------------[1,1,1,1,1,1,1,1,1...,1] .000000000000003 550195997640164 1.6[1,1,1,1,2,2,2,4,2,2,2] .000000003 29269411 .09[1,1,1,2,3,4,5,3] .0000001 33563 .003[1,1,2,3,4,6,3] .000001 6546 .007[1,1,2,5,7,4] .000005 812 .004[1,2,5,7,5] .00004 198 .008[1,3,8,8] .0002 25 .005[2,8,10] .003 6 .02[5,15] .04 1 .04[20] .9 .2 .2

What are the roots of the Wilkinson polynomial?

Choose your poison!

If the estimated error is the sole criterion...

Roots multiplicity---------------------------- .9999012 * 2.0012667 * 2.9758883 * 4.3848008 ** 6.9548807 ***10.3585762 ****15.0276741 *****19.2719863 ***

The polynomial with these roots matches W. Polynomial for 6 digitsThe condition 33563 is manageable.

Summary of the simultaneous method:- Given a polynomial p(x) ~ a, decide the structure ( x - z1 1 x - z2 )2 ... ( x - zm )m

implicitly define the coefficient operator G(z) and its Jacobian J(z)

- Have an initial guess z0 to the “pejorative root”

- Start the nonlinear least squares iteration zi+1=zi - J(zi )+[ G(zi )-a ], i=0,1,2 ...

- If it converges verify (i) the backward error || G(z* )-a || (ii) the pejorative condition 1/

- Accept the pejorative root if both are tiny

Isolated multiple roots:

The polynomial p(x) has an m-fold root p(x) = ( x-z* )m . q(x-z*)

The clue:p(z*) = p’(z*) = p”(z*) = ... = p(m-1)(z*) = 0 p(m)(z*) =/= 0

F

t1t2

...tm-1

x

=

p(x) - t1 p(m)(x)

p’(x) - t2 p(m)(x)

... ... ...

p(m-1)( x) - tm-1 p(m)(x)

p(m-1)( x)

F

t1t2

...tm-1

x

=

p(x) - t1 p(m)(x)

p’(x) - t2 p(m)(x)

... ... ...

p(m-1)( x) - tm-1 p(m)(x)

p(m-1)( x)

DF

t1t2

...tm-1

x

= diag(-p(m)(x), -p(m)(x),..., -p(m)(x), p(m)(x) )

z* is an m-fold root of p <==> (0,0,..., z* ) is a simple zero of F

Theorem: Let p(x) = [( x-z* )m . q(x-z*)] + [m-1(x-z* )m-1 + ... + 1(x-z* ) + 0] Then F has a simple zero

t2

...tm-1

z

t1

=

0...0

z*

0

+ [1/q(0)]

1

...m-2

m-1/m

0

z = z* + [1/(mq(0))] [m-1]

Conclusion

1. Ill-condition is cause by a wrong “identity”

4. To calculate multiple roots, one has to figure out the multiplicity structure (how?)

2. Multiple roots are pejoratively well conditioned, thereby tractable.

3. Extension of machine precision is not needed,a change in computing concept is.

Question: Can Jordan Canonical form be computed?