Is it possible, for an arbitrary polynomial in one variable with integer coefficients, to determine the roots of the polynomial in the Complex Field to arbitrary accuracy? When I was looking into this, I found some papers on homotopy continuation that seem to solve this problem (for the Real solutions at least), is that correct? Or are there restrictions on whether homotopy continuation will work? Does the solution region need to be bounded?

1$\begingroup$ The answer is "yes" and modern computer algebra systems have already done this for you. I confess I don't know howbut you don't make it clear whether you want to know how or you just want to know the answer. If you have a particular polynomial in mind, fire up the free maths package pari, set the precision to 1000 with \p 1000, and then use the polroots command. $\endgroup$ – Kevin Buzzard Mar 31 '10 at 20:53

14$\begingroup$ The fact that it's implemented doesn't mean it is solved! CASes have all sorts of routines which 'solve' undecidable problems... because all undecidable problems have (often large) subclasses which are semidecidable. It turns out that, for this problem, there is a complete algorithm which is guaranteed to terminate and find all roots. As far as I know, none of the CASes actually implement that (it's much too slow), instead they all implement algorithms which might fail (but with extremely low probability). $\endgroup$ – Jacques Carette Mar 31 '10 at 21:29

2$\begingroup$ Dror, NewtonRaphson is not guaranteed to converge. Even if it does, it finds only ONE solution. The question asked for ALL solutions. $\endgroup$ – user1855 Apr 1 '10 at 2:17

4$\begingroup$ Well, from a theoretical perspective, this follows from the decidability of the theory of the real numbers as an ordered field, as proved by Tarski. I agree of course that if you want an efficient algorithm, that's a separate question. $\endgroup$ – Pete L. Clark Apr 1 '10 at 7:16

9$\begingroup$ @Dror: dividing out by the root you found is known as 'deflation' as is amazingly badly behaved numerically. After you've deflated out about 10 roots, what you're left with is usually a total mess (from an error analysis point of view) and in practice the 'roots' you get after deflation are useless. $\endgroup$ – Jacques Carette Apr 29 '10 at 1:11
This argument is problematic; see Andrej Bauer's comment below.
Sure. I have no idea what an efficient algorithm looks like, but since you only asked whether it's possible I'll offer a terrible one.
Lemma: Let $f(z) = z^n + a_{n1} z^{n1} + \cdots + a_0$ be a complex polynomial and let $R = \max(1, a_{n1} + \cdots + a_0)$. Then all the roots of $f$ lie in the circle of radius $R$ centered at the origin.
Proof. If $z > R$, then $z^n > R z^{n1} \ge a_{n1} z^{n1} + \cdots + a_0$, so by the triangle inequality no such $z$ is a root.
Now subdivide the disk of radius $R$ into, say, a mesh of squares of side length $\varepsilon > 0$ and evaluate the polynomial at all the lattice points of the mesh. As the mesh size tends to zero you'll find points that approximate the zeroes to arbitrary accuracy.
There are also lots of specialized algorithms for finding roots of polynomials at the Wikipedia article.

5$\begingroup$ To speed up your "Now subdivide..." argument you'd compute some Lipschitz bounds for the polynomial's derivative on your subdivisions and apply Kantorovich's theorem (using Newton's method to find the roots). In practice this is very fast. $\endgroup$ – Ryan Budney Mar 31 '10 at 23:04

1

8$\begingroup$ The proposed answer does not work. If the task is to list all the zeroes then the algorithm must decide which squares of side length $\epsilon$ contain zeroes and which do not. How is it supposed to do that? Just because an approximate value at a point in the mesh is close to zero does not mean there is an actual zero there. What you are proposing is to compute a sequence of nested compact sets (finite unions of squares) whose intersection is the set of zeroes. But the trouble is that some of the squares may "disappear" after a while, so it's hard to tell where the zeroes actuall are. $\endgroup$ – Andrej Bauer Apr 29 '10 at 11:59

2$\begingroup$ @Andrej Bauer: this is true, but maybe it can be fixed as follows. First, reduce to the case in which the roots are simple (by dividing by the gcd with the derivative). So now we know how many roots we are looking for. Fix a precision and start with Qiaochu's strategy. As you decrease the size of $\epsilon$, the sets (not nec connected) of squares will start zooming in on the set of zeros, leaving out the "faraway" regions where there is no zero. When you found an $\epsilon$ such that, within your precision, you only have $\deg(f)$ "basins" for your candidate roots, you stop. $\endgroup$ – damiano Apr 29 '10 at 16:07

1$\begingroup$ @Pace: the original problem specified integer coefficients. But I understand your concern. $\endgroup$ – Qiaochu Yuan Apr 29 '10 at 19:47
Edit  Big News: Below I give an algorithm and show that it works, in a specified sense. I acknowledge that it does not work perfectly, in a sense. We have fix for that  a bit of precalculation and it works perfectly!
The algorithm below produces $a_1,\dots,a_k$ such that every $a_j$ is withing $\epsilon$ of a zero of $p$ and every zero of $p$ is within $\epsilon$ of some $a_j$. It's not perfect, the problem being that it can happen that several $a_j$ cluster around the same zero; in a sense we've located all the zeroes within $\epsilon$, but we don't know which $a_j$ are near more than one zero. If we knew that $z_iz_j\ge 5\epsilon$ for every two zeroes $z_i\ne z_j$ we could fix that. (Because then $D(a_j,\epsilon)$ contains exactly one zero $z_j$ of $p$, and if $a_ia_j<2\epsilon$ then $z_i=z_j$. So $(a_1,\dots,a_k)$ gives us $(z_1,\dots,z_n)$: We know there is a zero $z_1\in D(a_1,\epsilon)$. Call $a_1$ the approximation to $z_1$. Throw away the $a_i$ with $a_ia_1<2\epsilon$ and continue...)
So we can fix that:
Lemma 1. Given $p$ with no multiple roots we can calculate $\epsilon>0$ such that any two roots of $p$ are separated by at least $\epsilon$.
First
Lemma 2. Given $p$ with no multiple roots we can calculate $\delta>0$ such that $p(z)=0$ implies $p'(z)\ge\delta$.
Proof: The Euclidean algorithm gives us $r_1,r_2\in\Bbb Q[z]$ such that $$1=r_1(z)p(z)+r_2(z)p'(z).$$Calculate $R$ so $p(z)=0$ implies $z\le R$, then calculate $\delta$ so that $z\le R$ implies $r_2(z)\le1/\delta$.
Proof of Lemma 2: Say all the zeroes of $p$ are contained in $z\le R$, and say $p''(z)\le c$ in $z\le R$. Let $\delta$ be as in Lemma 1, and assume $p(a)=0$. Then Taylor's theorem shows that $$p(z)\ge\deltaza\frac12 cza^2\quad(z\le R),$$so if $0<za<2\delta/c$ then $p(z)>0$.
So you begin the algorithm by first replacing $p$ by $p/gcd(p,p')$ to ensure there are no multiple zeroes and then decreasing $\epsilon$ if necessary to ensure all the zeroes are separated by $3\epsilon$.
Right now this post sort of starts in the middle  I wanted to put the big news at the top (Lemma 1 is sommething that's been bugging me). It starts here:
Start: In fact the algorithm in the accepted answer does work, at least with a little elaboration regarding how we decide to reject or accept a given square during the subdivision.
Edit: For some reason I thought the accepted answer used a binary search, as below. Looking again I see it doesn't, it starts with a grid of squares of size $\epsilon$. So never mind what I said about the accepted answer. It seems likely that the binary search below is much more efficient, since a square of size much larger than $\epsilon$ will often be rejected in one step. Also there's a proof below that the binary search works.
Regarding efficiency: I made a little Python implementation (see implementation below) and searched for the roots of $z^2+1$ with various $\epsilon$. The results appeared "instantaneously". Didn't bother figuring out how to actually time it, but I did count the number of iterations, i.e. the number of squares "tested". I got this:
eps = 0.1: count = 516
eps = 0.01: count = 868
eps = 0.001: count = 1156
eps = 0.0001: count = 1428
eps = 1e05: count = 1812
eps = 1e06: count = 2100
eps = 1e07: count = 2388
eps = 1e08: count = 2772
eps = 1e09: count = 3060
eps = 1e10: count = 3348
Note that's obsolete. With the improved version of reject($Q$) below we get count = 1108 for $\epsilon = 10^{10}$.
The dependence of the running time on $\epsilon$ appears to be much better than I expected  it certainly totally annihilates the $c/\epsilon^2$ in the accepted answer.
In hindsight this makes sense. Say we run the algorithm with $\epsilon=\epsilon_0$ and then again with $\epsilon=\epsilon_0/10$. The definition of reject($Q$) below is independent of $\epsilon$, so every square rejected the first time will also be rejected the second time; the only extra work on the second run is in subdividing the squares that were accepted the first time, and the number of accepted squares is very small. (With $z^2+1$ there were eight accepted squares for each $\epsilon$, which is what I expected, or at least hoped.) Hmm, by that argument it should be something like $\log(1/\epsilon)$, which seems roughly consistent with the experiment above.
It "works" in this sense: Given $\epsilon>0$ we can find finitely many points $a_1,\dots a_k$ such that every $a_j$ is within $\epsilon$ of a zero and every zero is within $\epsilon$ of some $a_j$.
Which is to say we can approximate the zero set in the Hausdorff metric. That seems like it should be good enough. It doesn't work perfectly, in that it could happen that two zeroes are close to the same $a_j$. It's clear that multiple roots are going to cause problems with this sort of search; seems clear that two roots close together are also going to make things harder. We can eliminate multiple roots but there's not much we can do about two roots close together...
It was said that the accepted algorithm doesn't work because $p(z)$ small doesn't imply that $z$ is close to a zero of $p$. But it does imply that under some conditions, for a suitable value of "small":
Lemma 3. Suppose $p$ is entire, $p(a)\ne0$ and $p'(a)\ne 0$. Let $r=2p(a)/p'(a)$, and choose $\delta_2>0$ such that $p''(z)\le \delta_2$ for $za\le r$. If $\frac12p(a)\delta_2<p'(a)^2$ then $p$ has exactly one zero in $za<r$.
Proof: Let $g(z)=p(a)+p'(a)(za)$ and note that $g$ has exactly one zero in $za<r$. If $za=r$ then $g(z)\ge rp'(a)p(a)=p(a)$, while Taylor's theorem shows that $p(z)g(z)\le\frac12 r^2\delta_2$. The hypothesis implies that $\frac12 r^2\delta_2<p(a)$, so Rouche's theorem says that $p$ has the same number of zeroes as $g$ in $za<r$.
Now for the algorithm: Let $\epsilon>0$. Replacing $p$ by $p/gcd(p,p')$, we may assume that every zero of $p$ is simple.
Given the coefficients of $p$ we can certainly find a (closed) square $Q_0$ that contains all the zeroes of $p$, and then we can find $\delta_1$ and $\delta_2$ such that $p'(z)\le\delta_1$ and $p''(z)\le\delta_2$ whenever $d(z,Q_0)\le\epsilon$.
We construct a tree of squares, dividing $Q_0$ into four equal subsquares and then dividing each of them into four subsquares, etc. For each square $Q$ obtained this way we perform two tests:
def reject($Q$):
Say $a$ is the center of $Q$ and let $r$ be the distance from $a$ to a corner of $Q$. If $p(a)>r\delta_1$ we declare $Q$ to be a terminal node and perform no further subdivision; we will say below that $Q$ was "rejected". Note that if $Q$ is rejected it's clear that $Q$ cannot contain a zero of $p$.
Edit: That's stupid, using a global upper bound on $p'$ for an upper bound in $Q$. In fact $\min(\delta_1,p'(a)+r\delta_2)$ is an upper bound for $p'$ in $Q$. So change the above: Reject $Q$ if $p(a)>r\min(\delta_1,p'(a)+r\delta_2)$. (This made a huge improvement in a search for the zeroes of $\sin(z)$ in $z<4$, since $\cos$ is usually much smaller than its global maximum.)
In fact, we get down to small squares very quickly, and once $Q$ is small it will rarely if ever happen that $\min(\delta_1,p'(a)+r\delta_2)=\delta_1$. So why not just forget $\delta_1$ and simplify things by saying we reject $Q$ if $p(a)>r(p'(a)+r\delta_2)$? That might work, but it messes up the proof below that the algorithm terminates.
def accept($Q$):
Say $a$ is the center of $Q$, and let $r=2p(a)/p'(a)\in[0,\infty]$. If $diam(Q)<\epsilon$, $\frac12p(a)\delta_2<p'(a)^2$ and $r<\epsilon$ we declare $Q$ to be a terminal node and perform no further subdivison; we will say below that $Q$ was "accepted". Note that if $Q$ is accepted then $p$ has exactly one zero in $za<r$, by the lemma. (Unless $p(a)=0$, in which case $r=0$; in any case $p$ has a zero in $za\le r$.)
Note it can and probably will happen that reject($Q$)=accept($Q$)=True. I didn't realize until I ran some code that this means it's important to test reject($Q$) first. Doing it the other way around increases the length of the list $a_1,\dots,a_k$ returned, with more $a_j$ clustering around each zero.
If the algorithm terminates we've partitioned $Q_0$ into finitely many squares, each of which was accepted or rejected, and we're done.
Why we're done, assuming the algorithm terminates: Say $a_1,\dots,a_k$ are the centers of the accepted squares. For every $j$, $p$ has exactly one root in $za_j\le r_j$; since $r_j<\epsilon$ this says every $a_j$ is within $\epsilon$ of a root. Conversely, since the rejected squares do not contain any roots, every root lies in some accepted square $Q$; since $diam(Q)<\epsilon$ this shows that every root is within $\epsilon$ of some $a_j$.
The algorithm terminates.
Proof: By Konig's lemmma it's enough to show that our tree has no infinite branches. Say $Q_1,Q_2,\dots$ is a sequence of squares such that $Q_{k+1}$ is one of thhe four quarters of $Q_k$; we need to show that some $Q_k$ is accepted or rejected.
Say $a_k$ is the center of $Q_k$. Say $a_k\to z$.
Suppose $p(z)\ne0$, and let $r_k$ be the distance from $a_k$ to a corner of $Q_k$. Then there certainly exists $k$ such that $p(a_k)>r_k\delta_1$, since $r_k\to0$ but $p(a_k)\not\to0$. So $Q_k$ is rejected.
Suppose $p(z)=0$. Then $p'(z)\ne0$. So $p(a_k)\to0$ and $p'(a_k)\not\to0$, and if you look at the conditions in the definition of "accepted" this makes it clear that some $Q_k$ was accepted.
Implementation: I may as well include the code I mentioned above, in case anyone wants to try it. Note that this is not exactly what's described above: search(Q_0,f,df,eps,d1,d2) works for any holmorphic function $f$, returning all the zeroes in $Q_0$. In particular it doesn't try to calculate a $Q_0$ that contains all the zeroes, since $f$ is not a polynnomial; also it doesn't calculate $\delta_1$ and $\delta_2$, you have to work out appropriate upper bounds d1 and d2 for $f'$ and $f''$ yourself. Also it's simply not right, in that it uses floatingpoint calculations to test whether $p(a)=0$. Not intended as a production version, just a quick and filthy test of whether the thing actually works. It does, much better than I expected... (If you get a SyntaxError or IndentError I screwed up adding four spaces to the start of each line.)
Anyway:
from math import sqrt
class square():
def __init__(Q, a, r):
"""If a= x+iy and r>0 then square(a,r) is supposed to represent the
square [xr,x+r] x [yr,y+r]."""
Q.a = a
Q.r = float(r)
def quarters(Q):
vs = [0.5+0.5j, 0.50.5j, 0.5+0.5j, 0.50.5j]
return [square(Q.a + v*Q.r, Q.r/2)for v in vs]
def search(Q_0,f,df,eps,d1,d2):
"""f should be analytic in the epsnbd of Q_0. NOTE proof that the algorithm
terminates requires that all zeroes of f are simple. df=f', d1 d2 upper bounds
for f' and f'' in epsnbd of Q_0. Returns list approximating all zeroes in
Q_0, in the sense that everything in returned list is within eps of a zero
and every zero is within eps of something in list"""
s = sqrt(2)
def reject(Q):
#return abs(f(Q.a)) > s*d1*Q.r
#No! There's a better upper bound for the derivative in Q
#if the derivative at a happens to be small:
return abs(f(Q.a)) > s*min(d1,abs(df(Q.a))+d2*s*Q.r)*Q.r
def accept(Q):
z = abs(f(Q.a))
dz = abs(df(Q.a))
return (2*s*Q.r<eps) and (2*z<eps*abs(dz)) and (z*d2<2*dz*dz)
todo = [Q_0]
accepted = []
while len(todo)>0:
newtodo = []
for Q in todo:
for q in Q.quarters():
if not reject(q):
if accept(q):
accepted.append(q)
else:
newtodo.append(q)
todo = newtodo
return [Q.a for Q in accepted]

$\begingroup$ Could I ask you to please copy over whatever parts of the Math.SE answer you think are relevant for answering the question here? It's just a bit awkward flitting back and forth between two pages, in case someone would like to give feedback here. Thanks (and a belated welcome to MO)! $\endgroup$ – Todd Trimble♦ Aug 9 '18 at 20:36

1$\begingroup$ @ToddTrimble Instead I posted a new and much improved version. Hope that's good enough, hehe. (Finally saw how to define "accept" and "reject" to give a finite tree...) $\endgroup$ – David C. Ullrich Aug 10 '18 at 17:41
The wikipedia article http://en.wikipedia.org/wiki/Rootfinding_algorithm gives links to many different methods for finding roots of polynomials. (Start at the section entitled "Finding roots of polynomials".) Many of the methods are incomparable, in the sense that they work faster or slower than others depending on the specific polynomial.
Homotopy continuation method is good for finding all COMPLEX solutions to arbitrary accuracy, and it is implemented in the Numerical Algebraic Geometry package in Macaulay 2, for example. The method is more general. It can solve a system of polynomial equations in many variables. In fact, it is a more difficult problem to find all REAL solutions WITHOUT finding all complex solutions.
From what I understand, the solution region does not need to be bounded for homotopy continuation to work. You can also "projectify" your problem if necessary, so that you don't have to worry about homotopy paths going off to infinity. Some methods assume that the solutions are all simple, but there're ways to work around it. One is the method of "deflation".
For univariate polynomials you should look at "An Efficient Algorithm for the Complex Roots Problem" by Andy Neff and John Reif http://citeseerx.ist.psu.edu/viewdoc/download;jsessionid=5E9156BAF80D8D6AEDCA2F42C11AB4B2?doi=10.1.1.33.3353&rep=rep1&type=pdf
I give a course on matrix analysis, and I like to mention the following thing. It is equivalent to finding the roots of polynomials, or to finding the eigenvalues of matrices, because to a polynomial $P$, you can associate its companion matrix $C_P$, and to a square matrix $M$, you can associate its characteristic polynomial $\chi_M$.
Now, because the calculation of $\chi_M$ is costly, and direct solving methods are illconditionned, it is usually a bad idea to use polynomial solving in order to compute eigenvalues.
Instead, the efficient idea is to apply the QR method to the companion matrix $C_P$ when you want to calculate the roots of $P$. This way has several advantages:
 the method does converge. Proven.
 it is fast. Actually, each iteration produces a Hessenberg matrix (only one nonzero subdiagonal) and this ensures that the cost of an iteration is only $O(n^2)$, instead of $O(n^3)$.
 it is stable. Because an iteration acts as a unitary conjugation. Therefore the roundoff errors only add up, but are not amplified.
I agree that because of the roundoff errors, this produces on a computer approximations of limited (but outstanding) precision. What you can do is to stop after a few hundred iterations, and then apply Newton's method or something similar, starting from one of the approximate eigenevalues.

$\begingroup$ Could you give a reference for the statement "the method does converge"? I spent some time looking through Watkins' "Fundamentals of matrix computations" which discusses the eigenvalue problem extensively, and was unable to find it. $\endgroup$ – Igor Belegradek Aug 13 '18 at 13:43

$\begingroup$ I talked to a local numerical algebra expert and according to him there is no no convergence theorem in the case of repeated eigenvalues. $\endgroup$ – Igor Belegradek Dec 15 '18 at 16:28

1$\begingroup$ @IgorBelegradek. True in general. But something remains. The obstacle to convergence is the fact that several eigenvalues have the same modulus. By choosing randomly a complex translation, all eigenvalues have distinct moduli, except for multiple eigenvalues. But then the QR method converges towards a blocktriangular form, where each diagonal block has a single eigenvalue, which is the mean value of its diagonal elements. Therefore we still have a method to approximate efficiently the eigenvalues of a given matrix (up to a translation by $\alpha I_n$). $\endgroup$ – Denis Serre Dec 15 '18 at 16:55
This can be done. Check this article by Hubbard, Schleicher, and Sutherland, entitled "How to find all roots of complex polynomials by Newton's method".
One of the semirecommended ones for finding roots in the complex plane is Laguerre's method, which for some reason is not included in the Wikipedia article on rootfinding.
http://en.wikipedia.org/wiki/Laguerre's_method
The reason I know of this is a colloquium lecture long ago by Steven Smale on the complexity of Newton's method, during which William Kahan stood up and held forth on why Newton's method was worthless and Laguerre's was much better.
I cannot tell whether you insist on finding all roots to high accuracy. One could perhaps divide out by $(x  r_k)^{n_k}$ each time a root $r_k$ with multiplicity $n_k$ is found, and search for roots for the new polynomial, using those results as seed values for finding accurate roots using the original polynomial.

1$\begingroup$ Never bet against Kahan. He has been shown right an astonishing number of times. [Same with David Parnas for software engineering issues]. $\endgroup$ – Jacques Carette Apr 29 '10 at 1:14
You have already seen McNamee's excellent bibliography on polynomial rootfinding methods?
Personally I have a preference for the "simultaneous iteration" methods (of which DurandKerner and EhrichAberth are two of the simplest and most wellknown); all you need to start from is a set of points equispaced around a circle in the complex plane (as to the radius of this circle, there are a number of suggestions in the literature; alternatively, formulas in Marden's "Geometry of Polynomials" might be of use here).

$\begingroup$ DurandKerner is quite a lovely method. As far as I know there isn't much known about its dynamics (a thread you know): math.stackexchange.com/questions/1661140/… I find the method bogs down on large degree polynomials  it seems to have a particularly hard time when a large number (hundreds) of roots cluster close together. But as far as conceptual purity and practical effectiveness for "everyday applications", it's amazing. $\endgroup$ – Ryan Budney Aug 13 '18 at 1:15

$\begingroup$ @Ryan, it has been a while since I last investigated this, but I do recall simultaneous iterations tending to bog down on root clusters unless suitably modified. I recall Dario Bini has a modification of Aberth's method to tackle clusters, which you might want to look at. $\endgroup$ – J. M. is not a mathematician Sep 24 '18 at 11:41
At least for real roots it can be completely solved by bracketing zeroes with Sturm sequences.
Although it's not specific to polynomials with integer coefficients, have a look at "Computing the Zeros of Analytic Functions".
A completely ineffective theoretical method goes as follows: Write $f(z)$ as $f_r(x,y)+if_i(x,y)$ where $f_r,f_i\in \mathbb R[x,y]$ are real polynomials in the real and complex part $x,y$ of $z=x+i y$. Compute a Groebner basis of the ideal $(f_r,f_i)$ with respect to an order which eliminates one of the variables in the first element of the basis and use real techniques (based on Sturm sequences) to compute, say, the real parts of all solutions. Use another element of the Groebner basis (or again real techniques) to compute the corresponding imaginary part and test for multiplicities (which can be avoided by computing first gcd$(f,f')$).
Completely useless (and equivalent) variation: Study the intersection giving the zeroes of $f$ of the two real curves determined by $f_r$ and by $f_i$.
Back in the days before elecronic computers, methods were known for this. I read about them in the nice book,
J. V. Uspensky, Theory of Equations (1948 ... 1963)
Look for:
Chapter VIII, "Approximate Evaluation of Roots" (real roots)
Appendix V, "Graeffe's Method" (complex roots)
All computations are done by hand, and you find as many places as you wish for the roots.