From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Distribution of roots Date: 19 Oct 1995 04:24:17 GMT In article , Nick Halloway wrote: >Some questions I have been wondering about roots: > >a) Are roots of polynomials over Q dense in C^n? That is for arbitrary > c_1, ..., c_n in C, and e > 0 in R, is there an irreducible polynomial > in Q[x] with roots z_1, ..., z_n, conj (z_1), ...., conj (z_n), with > |z_i - c_i| < e for i = 1, n? Yes. If P(z) = Prod(z-c_i), then P1(z) = P(z)*(conj P(z)) is a polynomial with real coefficients and roots c_i and conj(c_i). You can choose rational polynomials P2 as close as you like to this P1; as you do so, you will get as roots collections of n pairs of complex conjugates, which approach the c_i as P2 approaches P1 -- in particular, you can keep |z_i - c_i| < e/2. Likewise, all the other rational polynomials P3 in a neighborood of P2 have roots within e/2 of the roots of P2, and thus within e of the roots of P1. Now you need the Hilbert Irreducibility Theorem: that the rational polynomials which are irreducible over Q are dense in the collection of all rational polynomials. Thus there will be choices for P3 which meet your additional requirement of being irreducible. >b) Suppose that f is complex analytic over the whole complex plane. > Is there anything you can say about the distribution of its roots? > i.e. that the distance between them has a positive lower bound? > Or that the roots have no accumulation point? The distance between consecutive zeros of f(z) = sin(pi * z^2) is sqrt(n+1)-sqrt(n), which is roughly 1/(2 sqrt(n)) for large n. So, no, there is no positive lower bound on the distance between roots. (By the way, for non-polynomials, they're usually called zeros of f.) Let Z = { x : f(x)=0 }. Accumulation points lie in closures, but Z = f^(-1)(0) is already closed in C as f is continuous on all of C. Thus if the roots have an accumulation point, that accumulation point is also a root. (You need to consider e.g. f(z) = sin(1/z) to see why I bother with this paragraph.) But if f(z_0)=0 and f(z_n)=0 for a sequence of z_n approaching z_0, then the Taylor series of f at z_0 vanishes. If f is entire, then f must be identically zero. So, yes, you can rule out zero-sets with accumulation points. You ought to look at factorizations of entire functions. Roughly, an entire function f is a product f(z) = Prod(z - c_i) * exp(g(z)) for some other entire function g; then c_i are the zeros of f. Of course there's a lot to worry about when it comes to convergence and analyticity. As I recall there's a readable exposition of this topic towards the end of Conway's book on complex analysis. dave ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: Distribution of roots Date: 19 Oct 1995 17:28:16 GMT I just thought I'd add some of the details to this remark: In article <464jth$jl1@muir.math.niu.edu>, Dave Rusin wrote: >You ought to look at factorizations of entire functions. Roughly, an entire >function f is a product f(z) = Prod(z - c_i) * exp(g(z)) for >some other entire function g; then c_i are the zeros of f. Of course >there's a lot to worry about when it comes to convergence and analyticity. >As I recall there's a readable exposition of this topic towards the end >of Conway's book on complex analysis. It is in section VII.5 of that book. If {c_n} is a sequence of complex numbers with no accumulation points, (possibly with repitition, of course) then there are only finitely many c_n in any compact set; in particular, we will have lim c_n = \infty (in the Riemann sphere). You can make a function with the nonzero c_n as the roots, although you must do it carefully if you want a product to converge. Choose a sequence of integers p_n which makes the c_n grow fast enough in the sense that Sum (r/|c_n|)^(p_n+1) converges for all real r; for example, p_n = n-1 will do. Then the function h(z) = Prod E(p_n, z/c_n) converges to an entire function with roots precisely at the c_n, where E(0,z)=1-z and E(k,z)=(1-z)*exp(z+z^2/2+...+z^k/k) for k>0. (The E(k,z) converge pointwise to 1 for all z, which is why they are a better choice than the simple 1-z/c_n if the c_n grow slowly -- they allow the above product h to converge in the (complete) metric space of holomorphic entire functions.) Conversely, given a function f with roots c_n (counted according to multiplicity), you can make a function h as above with the same set of nonzero roots. Then (if f has a root of multiplicity n at zero) f/(z^n h) is an entire function with no zeros, so g(z)=log(f(z)/(z^n h(z))) is a well-defined entire function and we have a factorization f(z) = z^n * Prod E(p_n, z/c_n) * exp(g(z)) This is the Weierstrass Factorization Theorem. Note that the resulting factorization is _not_ unique; for example, for any _finite_ collection of the roots you may replace the 'E' factors above by (z-c_n). Naturally, there are extensions of this theorem, such as to other open domains in C. There are also applications, such as factorizations of the trigonometric functions, the gamma function, and so on. Oh, you can factor the zeta function too, but as an exercise you'll have to find its zeros first. :-) dave ============================================================================== From: russakov@jhunix.hcf.jhu.edu (Alexander Russakovskii) Newsgroups: sci.math Subject: Re: distribution of roots Date: 20 Oct 1995 11:22:01 GMT The best source is B.Ya.Levin, Distribution of zeros of entire functions. AMS, Providence, (several editions). ============================================================================== Date: Thu, 30 May 1996 19:32:51 -0700 (PDT) From: [Permission pending] To: rusin@math.niu.edu Subject: hilbert.irred From your file hilbert.irred: > Now you need the Hilbert Irreducibility Theorem: that the rational > polynomials which are irreducible over Q are dense in the collection of > all rational polynomials. You don't really need Hilbert Irreducibility here, just find an irreducible polynomial of the required degree over F_2, the field of two elements; its preimages under reduction mod 2 in Q[x] are irreducible in Q[x] by the Gauss lemma and are dense in Q[x]. [sig deleted] ==============================================================================