How to Recognize Zero

Daniel Richardson

Department of Mathematics University of Bath

email dsrmathsbathacuk

January

Abstract

An elementary p oint is a p oint in complex n space which is an isolated

non singular solution of n equations in n variables each equation b eing

either or the form p where p is a p olynomial in Qx x or of

1 n

x

i

the form x e An elementary numb er is the p olynomial image

j

of an elementary p oint In this article a semi algorithm is given to decide

whether or not a given elementary numb er is zero It is proved that this

semi algorithm is an algorithm i e that it always terminates unless it

is given a problem containing a counter example to Schanuels conjecture

Intro duction

In computing a lazy sequence is a nite initial segment of a sequence together

with a pro cess which generates more elements of the sequence if desired See

for example Paulson

By analogy with this we may say that a lazy complex or real numb er is

a b ounded precision oating p oint complex or real numb er together with a

pro cess which could b e used to increase the precision to any desired extent

We will say that two such lazy numb ers are equal if rep eated application of

their pro cesses results in sequences which converge to the same ordinary numb er

Of course there are other reasonable denitions of equality for lazy numb ers

The one given here might b e called standard equality It will b e assumed in

the following that real and complex numb ers are given in lazy form rather than

as completed innities of some kind

The following fundamental question immediately presents itself

For which natural subsets of the real and complex numbers can we do exact

computations

The computations of interest include the eld op erations a test for equality

among complex numb ers and determination of the sign of a real numb er

It is clear that we can patch our approximations and pro cesses together in

order to eect addition subtraction and multiplication If two numb ers are

unequal we will eventually b e able to recognize this by calculating them to

sucient precision However there may b e problems recognizing equality

If we could recognize zero we could recognize equality since we can do

subtraction If we can recognize zero we can also avoid mistakes with division

by zero so we can do division Also if we can recognize zero we can order real

numb ers eectively

So the central part of the ab ove problem reduces to In which natural subsets

of the real and complex numb ers can we recognize zero

In the following section a denition is given for a subset of the complex

numb ers which is called elementary This set will b e denoted by E

E is algebraically closed and is also closed under application of elementary

x

functions such as e sinx cosx Also isolated solutions of systems of

equations involving elementary functions and p olynomials with co ecients in E

have co ordinates which are in E

The main result b elow is that we can recognize zero among the elementary

numb ers unless we are given a problem which contains a counterexample to

Schanuels conjecture The Schanuel conjecture is explained b elow

Let Q b e the rational numb ers

If B is a set of complex numb ers and z is complex we will say that z is

algebraically dep endent on B if there is a p olynomial

d

pt a t a

0 d

in QB t with a d and pz

0

If S is a set of complex numb ers a transcendence basis for S is a subset B

so that no numb er in B is algebraically dep endent on the rest of B and so that

every numb er in S is algebraically dep endent on B

The transcendence rank of a set S of complex numb ers is the cardinality of

a transcendence basis B for S It can b e shown that all transcendence bases

for S have the same cardinality

Schanuels conjecture If z z are complex numb ers which are linearly

1 n

z z

n 1

g has transcendence rank at least e indep endent over Q then fz z e

1 n

n

It is generally b elieved that this conjecture is true but that it would b e

extremely hard to prove

The history of the zero recognition problem is somewhat confused by the

fact that many p eople do not recognize it as a problem at all

In the algebraic case the nature of the problem dep ends up on what we

decide to accept as the denition of a complex algebraic numb er

In general our way of understanding the algebraic numb ers has b een in

uenced by the historical struggle to separate out the abstract algebra from

interpretation in the complex numb ers From this p oint of view it has b een as

sumed that algebra should avoid oating p oint approximations So it has b een

considered that the right way to do an algebraic computation is to put all the

numb ers involved into an algebraic numb er eld an abstract ob ject in which

there is a canonical form See Frohlich and Shepherdson Note that

we do not have a useful canonical form for the whole set of algebraic numb ers

but only for the numb ers in each particular nitely generated algebraic numb er

eld

If we are ultimately interested in oating p oint numb ers it is not clear that

it is sensible to construct an enclosing algebraic numb er eld in order to do

one computation But in any case this option disapp ears when we work with

elementary numb ers There is at present no suciently develop ed theory of

elementary numb er eld We do not for example know which abstract elds

with exp onentiation can b e emb edded into the complex numb ers The ideal of

separation b etween algebra and geometric interpretation do es not seem to work

very well in this case

The rst go o d result ab out recognition of zero among non algebraic numb ers

is due to Caviness Caviness shows in essence that if a weak version

of the Schanuel conjecture is true it is p ossible to dene a canonical form

and thus to solve the zero recognition problem in the set of numb ers which

are obtained by starting with the rationals and i and closing under addition

subtraction multiplication and exp onentiation Of course this set is probably

not algebraically closed

More recently Wilkie and Macintyre have proved that if the Schanuel con

x

jecture is true then the theory of R e is decidable where R is the ordered

eld of the reals

In particular the Wilkie and Macintyre result solves the zero recognition

problem for the necessarily elementary numb ers in the minimal mo del for this

x

theory Their metho ds can b e extended also to the theory of R e sin x

[01]

where sin x means sinx restricted to the interval and dened to

[01]

b e outside this interval In this theory all the real elementary numb ers are

denable The real and imaginary part of complex elementary numb ers are real

elementary So the metho ds of Wilkie and Macintyre can b e used to show that

the zero recognition problem can b e solved for the elementary numb ers

The work rep orted in this article is the result of a long indep endent devel

opment however the intention of which is ultimately to develop algorithms to

solve problems in the real and complex numb ers See Richardson

The techniques used here ie Wus metho d and

the LLL algorithm have their origins in computer algebra rather than in mo del

theory

Elementary p oints and numb ers

Denition An exponential system in variables x x is S E where

1 n r k

S p p is a list of r polynomials in Qx x and E w

r 1 r 1 n k 1

z z z

i k 1

with fw w z z g is a list of k terms w e w e e

1 k 1 k i k

fx x g

1 n

Let C b e the complex numb ers

In all the following we will use S E to denote an exp onential system

r k

as describ ed ab ove We will use J S E to denote the r k by n matrix of

r k

partial derivatives f x where f f S and f f E

i j 1 r r r +1 r +k k

n

Denition If is a point in C we wil l say that S E is non singular at

r k

if r k n and if the matrix J S E is non singular at

r k

n

Denition A point in C is elementary if there is an exponential system

S E with r k n so that S E and so that S E is non

r k r k r k

singular at

Denition A complex number c is elementary if there is an elementary point

and a polynomial q in Qx x so that c q

1 n

At each stage in the following we will assume that we have approximated

pr n

some elementary numb ers to within some tolerance We will use

pr n for the numb er of decimal places which are currently assumed to b e known

n

If x x is in C dene dx x to b e the maximum of the absolute

1 n 1 n

values of the real and imaginary parts of x x ie

1 n

dx x M axj Rex j j I mx j j Rex j j I mx j

1 n 1 1 n n

n

For in C let N f d g

2n

N can b e visualized as a co ordinate aligned b ox in R around We

will use N to denote the b oundary of N

It is assumed that we have an interval arithmetic pro cedure for p olynomials

x x

n 1

with the following prop erty if p is an expression for e in x x e

1 n



such a p olynomial and is an n tuple of pr n precision complex oats then the



interval arithmetic pro cedure applied to p over N gives a pair of intervals

I I with rational endp oints so that the b ox which is the pro duct of them

r i



in the complex plane is guaranteed to contain the image of N under p

Furthermore the pro cedure is such that the lengths of I and I tend to zero as

r i

pr n increases See Alefeld and Herzb erger



We will say p in N if the intervals I I pro duced by the

pr n r i

interval arithmetic pro cedure do not b oth contain zero ie if the complex is

not in the b ox which is the pro duct of the intervals



We will say possibl ep N in the complementary case in which the

intervals I I pro duced by the interval arithmetic pro cedure do b oth contain

r i

zero



Of course p in N dep ends on a particular expression for p

pr n

and not just on p as a function since we do not for example assume that our

interval arithmetic pro cedure ob eys the distributive law



p in N implies

pr n



X N pX but is much stronger As an imp ortant sp ecial

case of this if p is the unsimplied determinant of J S E with r k n

r k



and if p in N then the system of equations

pr n

S E

r k



can have at most one solution in N See Ab erth for a discussion

of this

How is an elementary numb er given

We will assume in all the following that an elementary numb er c E is given

to us in the following way

We are given an exp onential system S E in n variables x x with

r k 1 n

r k n

 pr n 

We are given a neighb ourho o d N with and is an

n tuple of precision pr n complex oats Let J b e the determinant of



J S E We have J in N

r k pr n



We are given a pro of that there is a p oint in the interior of N so

that S E

r k

The numb er c is dened as q where q is a p olynomial in Qx x

1 n



and is the elementary solution of S E in N

r k

We need to say some more ab out what kind of pro ofs might b e acceptable for



item ab ove Supp ose there is no solution of S E in N In this

r k



case the top ological degree of S E over N is dened and can b e cal

r k

culated by any one of a numb er of algorithms One algorithm to calculate such

a degree is in Ab erth This calculation will use oating p oint arithmetic

but at a precision higher than pr n and will also verify that there is no solution

on the b oundary of the b ox Because of item ab ove the only p ossible values

for the degree are or and there is a solution in the neighb ourho o d if

and only if the degree is non zero



Item also implies that any solution in N is necessarily non singular



and that there is at most one solution of S E in N

r k

If pr n is suciently large we can use Newtons metho d to get a sequence of

increasingly go o d approximations to



0

1

J S E S E

i+1 i r k r k i

Not only is convergence of this guaranteed for suciently large pr n but



there are also standard tests to verify convergence in N provided again

that pr n is suciently large One such is given by the Kantoravitch theorem in

Rabinowitz Others are given in Alefeld and Herzb erger and in

Richardson These tests for Newton convergence can also b e used instead

of computation of top ological degree to verify the existence of a solution in the

neighb ourho o d

In terms of giving the pro ofs required in item ab ove it is not clear whether

we should prefer top ological degree computation or a Newton convergence test

or some other metho d All we are claiming here is that there exist feasible

standard metho ds for giving such pro ofs All these metho ds dep end up on the

fact that the numb er of equations is the same as the numb er of unknowns and



that we have a guarantee for the non singularity of the Jacobian in N

Once we have a pro of that there is a unique solution of S E in

r k



N we can use either Newtons metho d or recursive sub division and in



terval arithmetic to increase pr n reduce and improve our approximation

This metho d of giving the numb er c is consistent with the lazy philosophy stated

earlier

Note that the pro of is part of the presentation of an elementary numb er If

we are not given enough information for example if the precision is not high

enough to carry out either of the typ es of verication mentioned ab ove then we

have not b een given a correct sp ecication of an elementary numb er

A consequence of Schanuels conjecture

We will say that a list of p olynomials S p p is indep endent at in

r 1 r

n

C if the r by n matrix of partial derivatives p x has rank r at ie if

i j

the gradients of p p are linearly indep endent at

1 r

If there are r indep endent p olynomials S in Qx x at then by

r 1 n

the implicit function theorem the equation S implicitly denes r of the

r

co ordinate variables as algebraic functions of the other variables in some neigh

b ourho o d of In this case the transcendence rank of the co ordinates of can

b e at most n r

We have the following obvious consequence of Schanuels conjecture

Prop osition Assuming Schanuels conjecture Suppose S E

r k

z z

1 k

and r k n and E w e w e and S is independent at

k 1 k r

Then at z z are linearly dependent over Q

1 k

Pro of If z z were linearly indep endent over Q at then by Schanuels

1 k

conjecture fz z w w g would have transcendence rank at least k at

1 k 1 k

Thus fx x g would have transcendence rank at least k In this case there

1 n

could exist at most n k indep endent p olynomials in Qx x which are

1 n

also zero at Since r n k we have a contradiction

2

As will b e seen later this implies that any identity among elementary num

b ers is either an algebraic consequence of their denitions or is explained by

a linear relationship b etween numb ers which app ear in the denitions as argu

ments of the exp onential function So in order to detect the presence of zero

we need to b e able to detect linear relationships over Q among elementary num

b ers and we also need a systematic way to make algebraic deductions The rst

problem is solved b elow with the help of the LLL algorithm and the second is

solved subsequently using a variation of Wus metho d

Finding rational linear relationships among el

ementary numb ers

n

Supp ose we are given an elementary p oint in C as describ ed ab ove



The p oint is the unique solution of S E in N and E

r k k

z z

k 1

w e w e

k 1

We supp ose that z z and w w are co ordinate values of These

1 k 1 k



values are xed although we only currently know them approximately as



with error no more than Using the values in and the error b ound we get

a zb ox containing z z and a wb ox containing w w

1 k 1 k

P

We wish to nd if p ossible a non trivial sum n z with n n integral

i i 1 k

so that the sum is indistinguishable from zero according to the current precision

ie so that

P



possibl e n z N

i i

A sum of this sort will b e called a candidate sum

P

Dene the height of a sum n z to b e M axj n j j n j

i i 1 k

We prefer candidates with small height Of course if one of the z is itself

i

indistinguishable from zero we can immediately nd a candidate sum of height

one We assume in the following therefore that for all z

i

z

pr n i

This means that the zb ox containing the p ossible z values is a pro duct of

intervals none of which straddles zero Let m b e the minimum of the absolute

values of the co ordinates of the zb ox

We will also supp ose in the following that pr n is suciently large so that

pr n4k pr n4k

Condition pr n k and m

P

Dene z z to b e the minimum of j n z j for non trivial sums of

N 1 k i i

height N

Let M b e an upp er b ound for the mean of j z j j z j

1 k

k

There are N sums of height N and all such sums have absolute

value no more than N k M

k 2

Around each such p oint put a circle of radius N k M N At least

two of these small circles overlap since the sum of the areas of the small circles

2

is more than N k M

So there must b e two such sums with dierent co ecients which dier by

k 2

no more than N k M N If we subtract these two sums we get a

sum of height N We have

k 2

z z N k M N

2N 1 k

k 2

So if N is even z z N k M N and in general

N 1 k

k 2

z z N k M N

N 1 k

If z z are all real we get a b etter b ound

1 k

k

z z N k M N

N 1 k

We also would like to have a lower b ound for z z but of course this

N 1 k

dep ends on z z It is not clear to me how to estimate this lower b ound

1 k

even when z z are chosen at random

1 k

Op en Question Suppose z z are chosen at random independently

1 k

with uniform distribution in What is the expected value and variance of

z z

N 1 k

There is some numerical evidence for the following

Conjecture For almost al l z z and for al l suciently large N

1 k

k +1

mN z z

N 1 k

In view of this conjecture we will say that a non trivial sum of height no

P

2k

more than N is surprisingly smal l if j n z j mN

i i

We would like to calculate pr n and N from the denition of in such a way

that

X X

possibl e n z N n z

i i i i

P

if n z has height N In other words we would like some sort of gap

i i

theorem for elementary numb ers This is not available at present So we im

p ose the following somewhat arbitrary condition derived from the concept of a

surprisingly small sum mentioned ab ove

pr n 12k

Condition N is the integer ceiling of m

In order to nd a candidate sum we can use the LLL algorithm as follows

Let I b e the k by k identity matrix Let Re b e the column vector obtained by

k k

pr n2    

transp osing Rez Rez where z z are the precision pr n

1 1

k k

oating p oint approximations to z z Similarly let I m b e the column

1 k k

pr n2  

vector obtained by transp osing I mz I mz Let V b e the k

k k +2

1

k

by k matrix of precision pr n oating p oint numb ers whose rst k columns

are the same as I and whose last two columns are Re and I m resp ectively

k k k

V I Re I m

k k +2 k k k

k +2

Supp ose the rows of V are v v Form a lattice LV in R with

k k +2 1 k

basis v v Now use the LLL algorithm to nd a reduced basis v v

1 k 1 k

for LV See Lenstra

P

Supp ose v is n n Then n z is a p ossible candidate sum

1 1 k 1 2 i i

We accept it as a candidate if it passes the tests ie if its height is b ounded by

N and if it is not distinguishable from zero using the current precision If one

of these tests fails then our attempt to nd a candidate has also failed How

ever the following lemma shows that if z z actually are linearly dep endent

1 k

over Q then our metho d applied on an increasing sequence of precisions will

eventually nd a correct relationship

Lemma Suppose that at z z are linearly dependent over Q If pr n

1 k



is suciently large and v n n is the rst vector in the reduced

1 k 1 2

1

basis found by the LLL algorithm as described above then

P

n z at

i i

Pro of

The idea of the pro of is that as pr n is increased false candidates get pushed

out of contention eventually the only p ossible initial vector in a reduced basis

is a true candidate if there is any such

We rely on the following result ab out reduced bases

If v min is a minimal length non zero vector in the lattice LV then

 2 k 2

j v j j v min j

1

See Lenstra

Supp ose z z are linearly dep endent over Q Then there is an upp er

1 k

b ound B valid for all pr n on the length of v min a minimal length non zero

vector in the lattice In other words no matter how much pr n is increased there

will b e a non zero vector in the lattice of length no more than B It follows that

k

the rst vector in a reduced basis must have length B no matter how large

k k

pr n is Since the initial part of V is I there are only B vectors

k k +2 k

k

in the lattice which could p ossibly have length B The initial vector in

a reduced basis must b e one of these nitely many p ossibilities But if pr n is

k

suciently large all of these will have size bigger than B except for those

which corresp ond to linear combinations of z z which are actually zero at

1 k

2

In fact it is also true that if there are d indep endent linear relationships

b etween z z at and if pr n is suciently large the rst d vectors in the

1 k

reduced basis will give us d correct relationships So we could take more than

one candidate from the reduced basis This computationally useful p ossibility

is ignored in the following in order to simplify the exp osition

P

k

Supp ose we have found a candidate sum n z Renumb er z z if

i i 1 k

1

necessary so that n is non zero but is minimal in absolute value among the

k

non zero co ecients We now intend to use this to replace

z z

k 1

E w e w e

k k 1

by

z z

k1 1

E w e w e

k 1 k 1 1

together with the pair of algebraic conditions

P

k 1

n z n z

i i k k

1

Q

k 1

n n

i k

w w

i

1 k

We assume pr n is large enough to satisfy conditions and given earlier

As b efore we also assume that there is a solution of S E in N

r k

Let Z Z b e any p oint in the zb ox and let W W b e any p oint

1 k 1 k



in the wb ox These are obtained by varying the numb ers in by no more

than

P Q

k k

n

Z Z

i

k1 1

and W e Lemma If W e n Z and W

k 1 1 i i

i

1 1

Z

k

then W e

k

Pro of

The problem is to exclude the p ossibility that W Z are on the wrong

k k

branch of the algebraic solution

We have

P

k 1

n Z and n Z

i i k k

1

Q

k 1

n n

n Z

k i

k k

W e W

i

k 1

So

Z

k

W e

k

where is an n th ro ot of unity We need to show that

k

If then

j j N

There are values z and w among the co ordinates of so that

k k

j Z z j

k k

j W w j

k k

z

k

w and e

k

z

k

j j jj e

z z

k k

j e j e

z Z Z z

k k k k

j e e e j e

z Z

k k

e e j j W w j

k k

z

k

j j e

z

k

j we have j j and thus N which is imp ossible If j e

z

k

j we get On the other hand if j e

m j j and thus

pr n4k

N m which is also imp ossible b ecause m

Z

k

Thus and W e

k

2

Wu stratication

n

A subset S of C is a d dimensional manifold if for every p oint in S there is

d

a numb er so that S B is dieomorphic to an op en ball in C where

B is the op en ball of radius around

A stratication of a set is a decomp osition of it into nitely many manifolds

The manifolds in a stratication are called strata

An algorithm is given b elow which uses Wus metho d for stratifying sets

dened by p olynomial equalities and inequalities

Characteristic sets

The following denitions are taken from Wu derived in some cases from the

ideas of JF Ritt Their purp ose is to dene a metho d of putting a system of

p olynomial equations into triangular form

Supp ose we are dealing with p olynomials in Qx x and we order the

1 n

variables by imp ortance

x x x

1 2 n

The leading variable of a p olynomial is the variable most imp ortant in the

ordering among those which o ccur in the p olynomial We will write l v p for

the leading variable of p

We assume here unless otherwise stated that p olynomials are written in

normal form as p olynomials in their leading variable with co ecients which

are p olynomials also in normal form in less imp ortant variables So if y is the

leading variable of a p olynomial p p would b e in the form

n

C y C

n 0

where n is called the degree of p and C assumed to b e non zero is called

n

the leading coecient of p Of course the leading co ecient may itself b e a

p olynomial in variables b elow y in the ordering

We will write l cp for the leading co ecient of p

If p and q are p olynomials q is not a constant and y is the leading variable

of q we will say that p is reduced with respect to q if the degree of y in p is less

than the degree of y in q It may happ en that p is reduced with resp ect to q

although the leading variable of p is more imp ortant than the leading variable

of q

For p olynomials p and q we will say p q if the leading variable of p is less

imp ortant than the leading variable of q or if the leading variables are the same

and the degree of p is less than the degree of q If b oth the leading variables

and the degrees of p and q are the same we will say p q

Let S p p b e a list of p olynomials We will say that S is an

r 1 r r

ascending set if for each i r the leading variable of p is less imp ortant than

i

the leading variable of p and if for all j i p is reduced with resp ect to

i+1 i

p

j

The next step is to put an order on ascending sets If S p p and

r 1 r

S q q are ascending sets we will say S S if for some k p q

s 1 s r s 1 1

and and p q and p q or if s r and p q and and

k k k +1 k +1 1 1

p q

s s

Note that if we add a new p olynomial to the end of an ascending set the

result if it is an ascending set is lower in the ordering than the original

The ascending sets are well ordered by the ordering describ ed ab ove This

means that any descending sequence of ascending sets must b e nite This

is useful to prove termination of algorithms Any pro cess which pro duces a

descending sequence of ascending sets must eventually terminate If we have a

pro cess which pro duces a tree in which the no des are lab elled with ascending

sets and if the ascending sets on each branch are descending and if no no de has

more than nitely many children then the tree itself must b e nite

Supp ose p and q are p olynomials and the leading variable of p is y The

usual pseudor emainder p q is dened as follows Supp ose p has degree n in

y and q has degree m in y In case m n we let pseudor emainder p q

q Otherwise pseudor emainder p q is the remainder after dividing p into

mn+1

l cp q considering this as a p olynomial in y with co ecients in the other

variables

Supp ose S p p is an ascending set and q is a p olynomial Dene

r 1 r

RemS q which is called the Wu remainder of q with resp ect to S recursively

r r

in terms of the usual pseudoremainder by

Remp q pseudor emainder p q

1 1

in the case r and for r

Remp p q Remp p pseudor emainder p q

1 j +1 1 j j +1

It follows from this that RemS q is reduced with resp ect to every p oly

r

nomial in S and satises an equation of the form

r

Y X

n

i

q d p RemS q I

i i r

i

ir

where each I is the leading co ecient of p and n n are some natural

i i 1 r

numb ers and the d are some p olynomials

i

If S is ascending and RemS q q we will say that q is reduced with

r r

resp ect to S If S p p then q is reduced with resp ect to S if for

r r 1 r r

each i q is reduced with resp ect to p ie if the degree of the leading variable

i

of p in q is less than it is in p

i i

Note that if S p p is ascending then for each i the leading

r 1 r

co ecient of p is reduced with resp ect to S and also the partial derivative of

i r

p with resp ect to its leading variable is reduced with resp ect to S

i r

Denition If S is a set of polynomials in Qx x we wil l say that an

1 n

ascending set A is a characteristic set for S if

A is contained in the ideal generated by S

If q is any polynomial in S then RemA q

If S is a set of p olynomials the notation S means the conjunction of

the conditions p for all p in S

Note that if A is characteristic for S then S implies A and also

A implies S provided that I A where I A is the pro duct of the

leading co ecients of A

If ascending set A is contained in the ideal generated by S but is not

1

characteristic then there must b e a p olynomial q in S so that RemA q

1

In this case supp ose RemA q r and A p p We can now use r

1 1 1 k

to construct an ascending set A which is b elow A in the ordering but which

2 1

is also contained in the ideal generated by S If r is a non zero constant or

if the leading variable of r is less than or equal to all the leading variables of

A then we can take A r Otherwise pick j maximal so that the leading

1 2

variable of p is less than the leading variable of r Then let A p p r

j 2 1 j

Wus Characteristic set algorithm

We can given any nite set S of p olynomials nd a characteristic set A for

S

First pick any ascending set A which is a subset of S and among ascending

1

subsets of S is minimal in the ascending set ordering A can be constructed

1

recursively starting with the smal lest polynomial in S Cal l such an A a basic

1

set for S Note that A may not be minimal in the ideal generated by S

1

Then nd Wu remainders of members of S with respect to A If al l the

1

Wu remainders are then A is characteristic If not al l Wu remainders are

1

zero use a non zero remainder to construct an ascending set A as explained

2

above so that A is in the ideal generated by S but A has lower order than A

2 2 1

Continue this process until a characteristic set is obtained

The ascending sets generated in this way are decreasing in order and the

ordering on ascending sets is well founded so the pro cess eventually terminates

with a characteristic set

Wu stratication algorithm

Let A p p b e an ascending set Let I A b e the pro duct of the

r 1 r r

leading co ecients of p p Recall that the leading co ecient of p may

1 r i

b e a p olynomial in variables b elow the leading variable of p Let D A b e

i r

the pro duct of the partial derivatives of p p with resp ect to their leading

1 r

variables ie

Q

D A p l v p

r i i

Note that there are r distinct leading variables in A and that D A is

r r

the Jacobian determinant of the matrix of partial derivatives of p p with

1 r

resp ect to these leading variables By the implicit function theorem if D A

r

is not zero but A the leading variables are lo cally dened implicitly as

r

functions of the other variables by A

r

Thus the condition A I A D A denes either the empty

r r r

n

set or an n r complex dimensional manifold in C

We can now use the Wu characteristic set algorithm as a to ol to construct

a stratication for the zero set of any nite set of p olynomials S The strat

ication for the zero set of S will b e presented as a nite tree lab elled with

conditions with each dening either the empty set or a manifold

1 k i

n

in C The zero set of S will b e the union of the manifolds dened by

1 k

The tree will b e called W u tr eeS and it will b e dened recursively

Wu stratication algorithm

Suppose given nite set S of polynomials Let A p p be the char

r 1 r

acteristic set for S obtained by the characteristic set algorithm

If one of p p is a non zero constant S is inconsistent and the

1 r

W u tr eeS wil l be a single node label led with the impossible condition

Otherwise dene to be A I A D A Dene W u

r r r

tr eeS to be the tree which has root label led with and which has subtrees

T T where for i r

1 r 1 r

T W u tr eeS A l cp

i r i

W u tr eeS A p l v p

i r i i

This construction terminates b ecause the characteristic sets obtained are

descending in order on each branch Note that even a basic set for S A l cp

r i

or for S A p l v p must b e b elow A in the ordering This is b ecause each

r i i r

p in A is reduced with resp ect to p p and thus RemA l cp

i r 1 i1 r i

0 0 0

l cp Similarly if p p l v p then RemA p p If p has

i i i r i

i i i

0

degree one in its leading variable then p is l cp If p has degree more than

i i

i

0

one in its leading variable then p p p p is an ascending set which is

1 2 i1

i

b elow A in the ordering

r

Warning We would hop e that in a stratication tree the b oundary of a

set dened at a no de would b e the set dened by the subtree b elow the no de

This do es happ en in nice cases with this algorithm But sometimes it do es

not happ en In fact the set dened at a no de may have lower dimension than

the sets dened at subtrees of the no de In exceptional cases a no de may b e

lab elled with a condition which denes the empty set but it may have subtrees

which dene non empty sets

2

Jetender Kang at Bath has implemented this algorithm using the Axiom

computer algebra system See Kang

It is not known what the average computational complexity of this algorithm

is In practice it seems to b ehave like the zero structure decomp osition given

by Wu

The Wu stratication describ ed ab ove together with the use of the LLL

algorithm describ ed previously is sucient to solve our problem However it

seems that we do not need the full strength of the Wu stratication and that it

would b e computationally useful to dene a weaker lo cal version This is done

b elow

Approximate lo cal Wu stratication

If we are mainly interested in a stratication for a set in a certain b ounded

n

region of C we can prune the Wu tree accordingly and thus improve the

computational b ehavior of the Wu stratication This improvement may b e

n

esp ecially dramatic if we have a p oint in C and we are only interested in the

limiting case of a very small neighb ourho o d around In this case we can pick

a small neighb ourho o d of and show using interval arithmetic that certain

semi algebraic sets do not intersect with the neighb ourho o d if that happ ens we

do not need to consider branches of the tree corresp onding to such sets



As b efore we assume that we have a neighb ourho o d N of and

pr n 

and is an n tuple of complex oating p oint numb ers with precision

pr n

Let S b e a nite set of p olynomials We assume S and we are

lo oking for a stratication of the zero set of S near

In the construction of of W u tr eeS we can prune branches which



include conditions q where q in N

pr n

We wish however to prune more radically We will therefore adopt the

following eager annihilation rule



possibl eq N

q

This will only apply to p olynomials which app ear in conditions in W u

tr eeS

A somewhat alarming disadvantage of this rule is that it may give incorrect

results if the precision is not high enough It has the advantage however that it

collapses the Wutree into one no de lab elled with some condition S We

hope that S will correctly describ e the zero set of S near The rule will

b e used in a context in which incorrect results will eventually b e recognized as

incorrect and this will force increase of the precision which as will b e shown

will eventually imply that any results of the rule are correct

Approximate lo cal Wu stratication algorithm



Suppose given nite set S of polynomials and neighbourhood N as ex

plained above Find characteristic set A p p for S If for i r

r 1 r

we have

l cp

pr n i

and

p l v p

pr n i i



in N then let S be A I A D A

r r r

Otherwise let S be the union of S and fp p g and fl cp i r

1 r i



possibl el cp N g and f p l v p i r possibl e p l v p

i i i i i



N g

Then dene S S

2

This algorithm given a nite set S of p olynomials pro duces a condition

S of the form

A I A D A

s s s

which either denes a manifold or the empty set and which has the prop er

ties

RemA q for all q in S

s



I A and D A in N

pr n s pr n s

If therefore A we have found a manifold which includes and

s

which is included in the zero set of S which is what we want In this case we

will say that S is correct

However the algorithm may pro duce A with A In this case we

s s

will say that the result is incorrect

Supp ose the set S of p olynomials is xed

Lemma If the precision pr n is suciently large the result S of the

approximate local Wu stratication algorithm is correct

Pro of The ordering on ascending sets is well founded The characteristic

set algorithm given previously denes an eective map from nite sets of p oly

nomials S to their characteristic sets chS which of course are ascending If

the lemma were false there would b e a nite set S of p olynomials so that

The lemma is false for S

If S is any set of p olynomials with chS chS in the ascending set

ordering then the lemma is true for S

Let chS A p p We assume that S at So it must

r 1 r

happ en that A If I A and D A then for pr n

r r r

suciently large we will b e able to prove this and the result of the algorithm

will b e

A I A D A

r r r

which is correct

Next supp ose that some of the leading co ecients of A or some of the

r

leading partial derivatives are zero at In this case we can increase the

precision until

l cp l cp

i pr n i

p l v p p l v p

i i pr n i i



in N for i r

Form a sew set S by adding the terms l cp or p l cp which do

i i i

app ear to vanish at to S using this higher precision The new set S which

is formed with this precision is zero at and has characteristic set which is lower

in the ordering than the characteristic set of S But S was a counterexample

with minimal order characteristic set So the result for S and also for S will

b e correct for suciently large pr n We have a contradiction and therefore the

lemma is true

2

Solution of the zero recognition problem among

elementary numb ers

Assume as b efore that an elementary numb er c is dened as q and is an



elementary p oint satisfying the dening condition N and S E

r k

The pro cess describ ed b elow will either show that c or will show that

c is zero as a consequence of the algebraic and numerical information which

we already have or if this fails will pro ceed by nding go o d candidate linear

relationships and removing exp onential terms If it do es not prove p ossible

to verify the correctness of the candidates it may b e necessary eventually to

backtrack ie to reject the candidates to increase the precision and start over

with the original problem replaceing the exp onential terms which have b een

removed

We assume that pr n is initially at some reasonably large value for example

Zero recognition pro cess

Let S b e the union of S and fq g Calculate A the characteristic set of

r s

S We will use d to denote the numb er of exp onential terms which we have

removed from the initial set E Set d initially so that S S and

k r +d r

E E

k d k

pr n

Use Newtons metho d on S E to decrease to b elow and

r +d k d



reset



If q in N then return c

pr n

Use approximate lo cal Wu stratication to nd S hop efully cor

rect near If the result is A with s n k d and if J

s pr n



in N where J is the determinant of J A E calculate the top o

s k d

logical degree of

A E

s k d



over N If the result is non zero return c In all other cases

continue

Apply the LLL algorithm to lo ok for non trivial integral linear combina

P

k d

tions n z with height N so that

i i

1

k d

X



n z N possibl e

i i

1

If such a candidate sum is found renumb er z z so that n

1 k d k d

and n is minimal in absolute value among the non zero co ecients

k d

Expand S by adding

P

k d

n z

i i

1

and

Q Q

n n

i i

w w

i i

(ik d^n 0) (ik d^n 0)

i i

z z

kd 1

g Go back to step w e Set d d Set E to fw e

kd k d 1

with new S E

k d

If no candidate sum is found double pr n reset d and backtrack to

step with S set to b e the union of S and fq g and A as initially

r s

calculated

Theorem If the zero recognition process given above terminates it does so

z z

k 1

e with the correct answer If the process does not terminate then z z e

1 k

contains a counterexample to Schanuels conjecture

Pro of

For the rst part of the theorem we observe that the pro cess can only termi

nate at step or step For correctness at step we rely on the supp osed

correctness of our interval arithmetic pro cedure

Supp ose we get termination at step This means that we have done

approximate lo cal Wu stratication near and found S to b e A

s

I A D A S implies S and S includes the original

s s

S which was used to dene the elementary p oint The dening condition for

r

was

S E

r k



in N

E may not b e currently equal to E but by lemma

k d k

S E E

k d k

 

in N Thus S E implies S E in N

k d r k



The inequalities of S are always true in N So

A E S E

s k d r k



in N

We have J where J is the Jacobian determinant of A E

pr n s k d



This implies that the top ological degree of A E over N is either plus

s k d

or minus one or zero and that the degree is non zero if and only if



N A E

s k d

This must b e the same as since satises a condition which is stronger

than the condition which we supp osed uniquely dened Since q is also in S

we must have

q

and thus the conclusion c is correct

In order to prove the second part of the theorem assume that the pro cess

do es not terminate Each lo op b etween step and step eliminates one

exp onential term so there can never b e more than k such successive lo ops

Thus pr n is doubles an unb ounded numb er of times in the non terminating

computation Since step never succeeds we must have q and thus

c

Supp ose that there are actually d indep endent linear integral relationships

among z z

1 k

If pr n is suciently large then after backtracking from step to step d

lo ops through steps and will by lemma pro duce d correct candidate

linear relationships At this p oint S and the remaining z z

1 k d

the arguments to the exp onential function are linearly indep endent over the

rationals at

After going into step we get S which is

A I A D A

s s s

Lemma implies that if the precision is suciently high these conditions

are true at

If s n k d we have a counterexample to Schanuels conjecture since

z z are linearly indep endent at

1 k d

It is not p ossible to have s n k d since A E implies

s k d

S E and the latter has non singular Jacobian

r k

It is claimed that unless z z is a counterexample to Schanuels conjec

1 k

ture it is also not p ossible to have J A E singular at For the sake of a

s k d

contradiction supp ose this did happ en



Note here that A E determines a single p oint in N

s k d

z z

kd 1

by the variables w w e Let J b e the result of replacing e

1 kd

which are equal to them if E Supp ose we found the Wu stratication

k d

of the solution set of A J S The p oint is in this solution

s r

set So would b e describ ed by some condition of the form

B I B D B

r r r

where RemB J and RemB q for all q in A

r r s

We can not have r s since the equations A are indep endent at

s

We can not have r s unless this is a counterexample to Schanuels conjecture

Supp ose r s B implies J A E is singular Since the A are

r s k d s

z

i

in E has a gradient in indep endent this must mean that some w e

k d i



N which is a linear combination of the gradients of B and the rest of E

r k d

on the solution set of B near Remove this dep endent term from E

r k d

to get E B E at and since there are less equations

r

k (d+1) k (d+1)

than variables there must b e a curve through on which B and E are

r

k (d+1)

identically zero E is also identically zero on this curve since it is zero at

k d

one p oint and the curve is orthogonal to the gradients of the terms in E

k d

A is identically on this curve since A is reduced to zero by B But this is

s s r

imp ossible since we supp osed that A E determined a single p oint in

s k d



N

We have decided that A E has non singular Jacobian at For pr n

s k d



suciently large we can prove J in N The top ological degree

pr n

metho d will then apply and will give termination

Lo oking back over the discussion we can see that the only case in which

termination can b e avoided is when z z w w is a counterexample

1 k 1 k

to Schanuels conjecture

2

It follows from the ab ove theorem that if Schanuels conjecture is true the

elementary numb ers are a computable eld Also the real elementary numb ers

are a computable real closed eld

Implementation

There is a zero recognition program written in Reduce It do es not use top o

logical degree to show existence of non singular solutions of n equations in n

unknowns but instead applies standard tests for convergence of Newton se

quences The program has only b een tried so far on ab out examples the

most interesting of which is

c atan atan

which is zero but which we might say is not obviously zero

So far on small problems of this typ e termination with the correct answer

has always b een obtained within a few minutes

The p ossibility of backtracking which is built in to the algorithm to ensure

correctness has not yet b een used That is it has not yet happ ened that a

false candidate linear relationship has b een pro duced by the LLL pro cess This

suggests the following dicult and interesting problem in applied numb er the

ory how can we set the thresholds in such a way that no false candidate is ever

produced From a theoretical p oint of view this problem lo oks to b e a little bit

harder than settling the Schanuel conjecture

Although the program has not failed yet on any small problems it is clear

that it is p ossible to create problems involving very large or very small numb ers

which will require such large precision that their solution will b e infeasible

There may b e some other computational diculties such as counterexam

ples or near counterexamples to Schanuels conjecture but so far none such

have app eared

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