THE COMPLEX ZEROS OF RANDOM SUMS

Transcription

1 THE COMPLEX ZEROS OF RANDOM SUMS ROBERT J VANDERBEI This paper is dedicated to the memory of Larry Shepp ABSTRACT Mark Kac gave an explicit formula for the expectation of the number, ν n (Ω, of zeros of a random polynomial, P n (z = η j z j, in any measurable subset Ω of the reals Here, η 0,, η n are independent standard normal random variables In fact, for each n > 1, he obtained an explicit intensity function g n for which Eν n (Ω = g n (xdx Inspired by that result, Larry Shepp and I found an explicit formula for the expected number of zeros in any measurable subset Ω of the complex plane IC Namely, we showed that Eν n (Ω = h n (x, ydxdy + g n (xdx, Ω where h n is an explicit intensity function We also studied the asymptotics of h n showing that for large n its mass lies close to, and is uniformly distributed around, the unit circle Following the same general methodology, in this paper we derive analogous explicit formulas for a wider class of random sums, P n (z = η j f j (z, Ω where the functions f j are given analytic functions that are real-valued on the real line We consider some special cases including Weyl polynomials, Taylor polynomials, and truncated Fourier series Ω IR 1 INTRODUCTION The problem of characterizing the distribution of the roots of random polynomials has a long history In 1943, Kac [12] studied the real roots of random polynomials with iid normal coefficients He obtained an explicit formula for the density function for the distribution of the real roots 1991 Mathematics Subject Classification Primary 30C15, Secondary 30B20, 26C10, 60B99 Research supported by ONR through grant N

2 2 ROBERT J VANDERBEI Following the initial work of Kac, a large body of research on zeros of random polynomials has appeared see [2] for a fairly complete account of the early work in this area including an extensive list of references Most of this early work focused on the real zeros; [5], [7] and [24] being a few notable exceptions The paper of Edelman and Kostlan [4] gives a very elegent geometric treatment of the problem In more recent years, the work has branched off in a number of directions For example, in 1995, Larry Shepp and I derived an explicit formula for the distribution of the roots in the complex plane (see [20] when the coefficients are assumed to be iid normal random variables A short time later, Ibragimov and Zeitouni [11] took a different approach and were able to rederive our results and also find limiting distributions as the degree n tends to infinity under more general distributional assumptions At about the same time, it was pointed out that understanding deeper statistical properties of the random roots, such as k-point correlations among the roots, was both interesting mathematically and had important implications in physics (see, eg, [18] and [6] A number of papers have appeared that attempt to prove certain specific properties under increasingly general distributional assumptions For example, in 2002, Dembo et al [3] derived a formula for the probability that none of the roots are real (assuming n is even, of course in the case when the coefficients of the polynomial are iid but not necessarily normal Another property that has been actively studied in recent years is the fact that as n gets large the complex roots tend to distribute themselves close to and uniformly about the unit circle in the complex planes see, eg, the papers by Shiffman and Zelditch [21], Hughes and Nikeghbali [8], and Ibragimov and Zaporozhets [10] Also, Li and Wei [15] have considered harmonic polynomials polynomials in the complex variable z and it s conjugate z Recently, Tao and Vu [23], drawing on the close connection with random matrix theory, derived asymptotic formulas for the correlation functions of the roots of random polynomials They specifically address the question of how many zeros are real The results summarized above mostly establish certain properties of the roots under very general distributional assumptions The price paid for that generality is that most results only hold asymptotically as n In contrast, this paper introduces a modest generalization to the core assumptions underlying the results in [20] and we show that analogous explicit formulas can still be derived for any value of n Specifically, instead of considering polynomials, n η jz j, we generalize the basis functions z j to be any set of analytic functions, f j (z, that are real on the real line So, to that end, we let P n (z = η j f j (z, z IC, where n is a fixed integer, the η j s are independent identically distributed N(0, 1 random variables, and the functions f j are given analytic functions that are real-valued on the real line We derive

3 ZEROS OF RANDOM POLYNOMIALS 3 an explicit formula for the expected number of zeros in any measurable subset Ω of the complex plane IC The formula will be expressed in terms of the following functions: A 0 (z = f j (z 2, B 0 (z = f j (z 2, and A 1 (z = A 2 (z = zf j (zf j(z, B 1 (z = z 2 f j(z 2, B 2 (z = (11 D 0 (z = B 0 (z 2 A 0 (z 2 and, lastly, zf j (zf j(z, zf j(z 2, E 1 (z = A 2 (za 0 (z A 1 (z 2 Notes: (1 As usual, an overbar denotes complex conjugation and primes denote differentiation with respect to z (2 The function E 1 will only be needed in places where the argument of the square root is a positive real At such places, the square root is assumed to be a positive real (3 Throughout the paper, we follow the usual convention of denoting the real and imaginary parts of a complex variable z by x and y, respectively, ie, z = x + iy Let ν n (Ω denote the (random number of zeros of P n in a set Ω in the complex plane Our first theorem asserts that throughout most of the plane this random variable has a density with respect to Lebesgue measure on the plane: Theorem 11 For each measurable set Ω {z IC D 0 (z 0}, (12 Eν n (Ω = h n (x, ydxdy, where Ω h n (z = B 2D0 2 B 0 ( B A (A 0 B 1 A 1 + A 0 B 1 A 1 π z 2 D0 3 It is easy to see that the density function h n is real valued It is less obvious that it is nonnegative We leave this sanity check to the reader As we see from the above theorem, places where D 0 vanishes are special and must be studied separately The real axis is one such place: Theorem 12 On the real line, the function D 0 vanishes For each measurable set Ω IR, (13 Eν n (Ω = g n (xdx, Ω

4 4 ROBERT J VANDERBEI where g n (x = E 1(x π x B 0 In the case where the f j (z s are just powers of z, our results reduce to those given in [20] The proof here parallels the analogous proof given in [20] but there are a few differences, the most important one being the fact that, in general, the function B 1 is not real-valued like it was in [20] It seems that this explicit formula has not been derived before and, as shown in later sections, there are interesting new examples that can now be solved While the definition of h n in Theorem 11 looks rather complicated, it is nevertheless amenable both to computation and, with some choices of the functions f j, it is amenable to asymptotic analysis as well In the following section, we derive the explicit formulas given above for the intensity functions h n and g n Next, in Section 3, we look at some specific examples and finally, in Section 4, we offer some speculation and suggest future research directions 2 THE INTENSITY FUNCTIONS h n AND g n This section is devoted to the proof of Theorems 11 and 12 The proofs follow the same general approach layed out in [20] But, for completeness, we give them in their entirety here We begin with the following proposition Proposition 21 For each region Ω IC whose boundary intersects the set {z D 0 (z = 0} at most only finitely many times, (21 Eν n (Ω = 1 1 2πi z F (zdz, where (22 F = B 1D 0 + B 0 B 1 Ā0A 1 B 0 D 0 + B 2 0 Ā0A 0 Ω Proof The argument principle (see, eg, [1], p 151 gives an explicit formula for the random variable ν n (Ω, namely (23 ν n (Ω = 1 P n(z 2πi P n (z dz Ω

5 ZEROS OF RANDOM POLYNOMIALS 5 Taking expectations in (23 and then interchanging expectation and contour integration (the justification of which is tedious but doable, we get (24 Eν n (Ω = 1 E P n(z 2πi P n (z dz To facilitate computations, it is advantagous to multiply and divide by z (multiplying inside the expectation and dividing outside The following Lemma shows that, away from the set {z D 0 (z = 0}, the function (25 F (z = E zp n(z P n (z simplifies to the expression given in (22 and, since we ve assumed that Ω intersects this set at only finitely many points, this finishes the proof Lemma 22 Let F denote the function defined by (25 For z {z D 0 (z = 0}, Ω F = B 1D 0 + B 0 B 1 Ā0A 1 B 0 D 0 + B 2 0 Ā0A 0 Proof Note that P n (z and zp n(z are complex Gaussian random variables It is convenient to work with their real and imaginary parts, P n (z = ξ 1 + iξ 2, zp n(z = ξ 3 + iξ 4, which are just linear combinations of the original standard normal random variables: ξ 1 = a j η j, ξ 2 = b j η j, ξ 3 = c j η j, ξ 4 = The coefficients in these linear combinations are given by (26 d j η j a j = Re(f j (z = f j(z + f j (z, 2 b j = Im(f j (z = f j(z f j (z, 2i c j = Re(zf j(z = zf j(z + zf j (z, 2 d j = Im(zf j(z = zf j(z zf j (z 2i

6 6 ROBERT J VANDERBEI Put ξ = [ ξ 1 ξ 2 ξ 3 ξ 4 ] T The covariance among these four Gaussian random variables is easy to compute: a T a a T b a T c a T d (27 Cov(ξ = Eξξ T = b T a b T b b T c b T d c T a c T b c T c c T d d T a d T b d T c d T d We now represent these four correlated Gaussian random variables in terms of four independent standard normals To this end, we seek a lower triangular matrix L = [ l ij ] such that the vector ξ is equal in distribution to Lζ, where ζ = [ ζ 1 ζ 2 ζ 3 ζ 4 ] T is a vector of four independent standard normal random variables The following simple calculation shows that L is the Cholesky factor for the covariance matrix: (28 Cov(ξ = Eξξ T = ELζζ T L T = LL T Now, since ξ D = Lζ and L is lower triangular (the symbol D = denotes equality in distribution, we get that zp n(z P n (z = ξ 3 + iξ 4 ξ 1 + iξ 2 D = (l 31 + il 41 ζ 1 + (l 32 + il 42 ζ 2 + (l 33 + il 43 ζ 3 + il 44 ζ 4 (l 11 + il 21 ζ 1 + il 22 ζ 2 Hence, exploiting the independence of the ζ i s, we see that (29 F (z = E zp n(z P n (z = Eαζ 1 + βζ 2 γζ 1 + δζ 2, where α = l 31 + il 41 β = l 32 + il 42 γ = l 11 + il 21 δ = il 22 Splitting up the numerator in (29 and exploiting the exchangeability of ζ 1 and ζ 2, we can rewrite the expectation as follows: F (z = α δ f(γ/δ + β γ f(δ/γ, where f is a complex-valued function defined on IC \ IR by ζ 1 f(w = E wζ 1 + ζ 2 The expectation appearing in the definition of f can be explicitly computed Indeed, f(w = 1 2π = 1 2π 2π 0 2π 0 0 ρ cos θ wρ cos θ + ρ sin θ e ρ2 /2 ρdρdθ dθ w + tan θ

7 ZEROS OF RANDOM POLYNOMIALS 7 and this last integral can be computed explicitly giving us 1, Im(w > 0, w + i f(w = 1, Im(w < 0 w i Recalling the definition of δ and γ, we see that γ δ = l 21 i l 11 l 22 l 22 In general, l 11 and l 22 are just nonnegative However, it is not hard to show that they are both strictly positive whenever z has a nonzero imaginary part Hence, γ/δ lies in the lower half-plane, δ/γ lies in the upper half-plane, and (210 F (z = α 1 γ δ i + β 1 γ δ δ + i γ = iα + β iγ + δ = l 32 l 41 + i(l 31 + l 42 l 21 + i(l 11 + l 22 At this point, we need explicit formulas for the elements of the Cholesky factor L From (27 and (28, we see that a T a = l 2 11 b T a = l 21 l 11 b T b = l l 2 22 c T a = l 31 l 11 c T b = l 31 l 21 + l 32 l 22 d T a = l 41 l 11 d T b = l 41 l 21 + l 42 l 22 Solving these equations in succession, we get l 11 = at a at a l 21 = bt a at a l 31 = ct a at a l 41 = dt a at a l 22 = (at a(b T b (b T a 2 at ar l 32 = (at a(c T b (c T a(b T a at ar l 42 = (at a(d T b (d T a(b T a at ar where R = (a T a(b T b (b T a 2

8 8 ROBERT J VANDERBEI Substituting these expressions into (210 and simplifying, we see that (211 F (z = dt a + ic T a i ( a T a( d T b + ic T b ( d T a + ic T ab T a /R b T a + ia T a + ir Recalling the definitions of a j, b j, c j, and d j given in (26, it is easy to check that the following identities hold: a T a = 1 4 (A 0 + 2B 0 + Ā0, b T a = i 4 (A 0 Ā0, b T b = 1 4 (A 0 2B 0 + Ā0, c T a = 1 4 (A 1 + B 1 + B 1 + Ā1, c T b = i 4 (A 1 B 1 + B 1 Ā1, d T a = i 4 (A 1 + B 1 B 1 Ā1, d T b = 1 4 (A 1 B 1 B 1 + Ā1 Plugging these expressions into (211 and simplifying, we get that (212 F (z = A 1 + B 1 + (A 0 B 1 + B 0 B 1 A 1 B 0 Ā0A 1 /D 0 A 0 + B 0 + D 0, where D 0 is as given in (11 It turns out that further simplification occurs if we make the denominator real by the usual technique of multiplying and dividing by its complex conjugate We leave out the algebraic details except to mention that a factor of A 0 + 2B 0 + Ā0 cancels out from the numerator and denominator leaving us with (213 F (z = B 1D 0 + B 0 B 1 Ā0A 1, D 0 (B 0 + D 0 or, expanding out D 2 0, (214 F (z = B 1D 0 + B 0 B 1 Ā0A 1 B 0 D 0 + B 2 0 Ā0A 0 Lemma 23 On the real axis, F has a jump discontinuity Indeed, for each a IR, and lim F = B 1(a i E 1 (a z a: Im(z>0 B 0 (a lim F = B 1(a + i E 1 (a z a: Im(z<0 B 0 (a Proof Consider a point a on the real axis On the reals, A k = B k, for k = 0, 1, and so D 0 = 0 Hence, the right-hand side in (213 is an indeterminate form To analyze the limiting behavior of

9 ZEROS OF RANDOM POLYNOMIALS 9 F near the real axis, we first divide the numerator and denominator by D 0 : B 1 + B 0B 1 Ā0A 1 D (215 F = 0 B 0 + D 0 Now, only the ratio in the numerator is indeterminate To study it, we start by expressing things in terms of the f j functions: B 0 B 1 Ā0A 1 = j,k = 2i j,k ( zf j (zf k(z f j (zf k (z f j (zf k (z ( zf j (zf k(zim f j (zf k (z and D 2 0 = B 2 0 A 0 2 = j,k = 2i j,k ( f j (zf k (z f j (zf k (z f j (zf k (z ( f j (zf k (zim f j (zf k (z Next, we write the first few terms of the Taylor series expansion of the f j s about the point z = a, substitute the expansions into the formulas above and then drop high order terms to derive the first few terms of the Taylor expansions for B 0 B 1 Ā0A 1 and B 2 0 A 0 2 For the first expression, we only need to go to linear terms to get B 0 B 1 Ā0A 1 = 2i af j (af k(a ( f j(af k (a f j (af k(a y + o(z a j,k ( A1 (a 2 = 2i A 0(aA 2 (a y + o(z a a a (as usual, we use y to denote the imaginary part of z For the second expression, we need to go to quadratic terms The result is ( f j (a 2 f k (a 2 f j (af j(af k (af k(a y 2 + o((z a 2 D0 2 = 4 j,k ( ( ( 2 = 4 f j(a 2 f j (a 2 f j(af j (a y 2 + o((z a 2 = 4 A 2(aA 0 (a A 1 (a 2 a 2 y 2 + o((z a 2 Hence, we see that (216 B 0 B 1 Ā0A 1 D 0 = i E 1 (a sgn(a sgn(y + o(z a

10 10 ROBERT J VANDERBEI Combining (215 and (216, we get the desired limits expressing the jump discontinuity on the real axis Proof of Theorem 11 Without loss of generality, it suffices to consider regions Ω that are either regions that do not intersect the real axis or small rectangles centered on the real axis We begin by considering a region Ω that does not intersect the real axis Applying Stokes theorem to the expression for Eν n (Ω given in Proposition 21, we see that Eν n (Ω = 1 π Ω 1 z F (z, zdxdy z Note that we are now writing F (z, z to emphasize the fact that F depends on both z and z Letting the dagger symbol stand for the derivative with respect to z, we see from Lemma 22 that F { (217 = (B 0 D 0 + B0 2 z Ā0A 0 (B 1D 0 + B 1 D 0 + B 0B 1 + B 0 B 1 Ā 0A 1 } (B 1 D 0 + B 0 B 1 Ā0A 1 (B 0D 0 + B 0 D 0 + 2B 0 B 0 Ā 0A 0 /(B 0 D 0 + B 2 0 A Recall that we have assumed that the functions f j are analytic and are real-valued on the real line Hence, they have the property that f j (z = f j ( z Their derivatives also have this property Exploiting these facts, it is easy to check that (218 B 0 = B 1 z, Recalling that D 0 = B 2 0 A 0 2, we get that Ā 0 = 2Ā1 z, B 1 = B 2 z (219 D 0 = B B 0 1 A 0 Ā 1 zd 0 As explained in [20], substituting these formulas for the derivatives into the expression given above for F/ z followed by tedious algebraic simplifications eventually leads to the fact that (πz 1 F (z, z/ z equals the expression given for h n in the statement of the theorem Proof of Theorem 12 Now consider a narrow rectangle that straddles an interval of the positive real axis: Ω = [a, b] [ ε, ε] where 0 a < b Writing the contour integral for Eν n (Ω given by Proposition 21 and letting ε tend to 0, we see that Eν n ((a, b = 1 2πi b a F (x F (x+ dx, x where ν n ((a, b denotes the number of zeros in the interval (a, b of the real axis and F (x = lim F (z and F (x+ = lim F (z z x:im(z<0 z x:im(z>0

11 ZEROS OF RANDOM POLYNOMIALS 11 FIGURE 1 Random degree 10 polynomial: η 0 + η 1 z + η 2 z η 10 z 10 In this figure and the following ones, the left-hand plot is a grey-scale image of the intensity functions h n and g n (the latter being concentrated on the x-axis The right-hand plot shows 200,000 roots from randomly generated polynomials Note that, for the left-hand plots, the grey-scales for h n and g n are scaled separately and in such a way that both use the full range from white to black Hence, from Lemma 23, we see that g n (x = 1 F (x F (x+ 2πi x = E 1(x πxb 0 On the negative half of the real axis the analysis is the same except that x in the denominator needs to be negated This completes the proof 3 EXAMPLES In this section, we consider some examples The simplest example corresponds to the f j simply being the power functions: f j (z = z j As this case was studied carefully in [20], other that showing a particular example (n = 10 in Figure 1, we refer the reader to that previous paper for more information about this example

12 12 ROBERT J VANDERBEI Each figure in this section shows two plots On the left is a grey-scale plot of the intensity functions g n and h n On the right is a plot of hundreds of thousands of zeros obtained by generating random sums and explicitly finding their zeros The intensity plots appearing on the left were produced by partitioning the given square domain into a 440 by 440 grid of pixels and computing the intensity function in the center of each pixel The grey-scale was computed by assigning black to the pixel with the smallest value and white to the pixel with the largest value and then linearly interpolating all values in between This grey-scale computation was performed separately for h n and for g n (which appears only on the x-axis and so no conclusions should be drawn comparing the intensity shown on the x-axis with that shown off from it The applet used to produce these figures can be found at rvdb/java/roots/rootshtml Of course, the intensity function g n is one-dimensional and therefore it would be natural (and more informative to make separate plots of values of g n verses x, but such plots appear in many places (see, eg, [13] and so it seemed unnecessary to produce them here 31 Weyl Polynomials Sums in which the f j s are given by f j (z = zj j! are called Weyl polynomials (also sometimes called flat polynomials Figure 3 shows the empirical distribution for the case where n = 10 For this case, the limiting forms of the various functions defining the densities are easy to compute: lim 0(z n = e z2 lim B 0 (z n = e z 2 lim 1(z n = z 2 e z2 lim B 1 (z n = z 2 e z 2 lim 2(z n = z 2 (z 2 + 1e z2 lim B 2 (z n = z 2 ( z 2 + 1e z 2 lim n D 0(z = e 2(x2 +y 2 e 4(x2 y 2 lim n E 1 (z = z e z2 The random Weyl polynomials are interesting because in the limit as n, the distribution of the real roots becomes uniform over the real line: Theorem 31 If f j (z = z j /z! for all j, then lim g n(x = 1 n π Proof Follows trivially from Theorem 12 and the formulas above

13 ZEROS OF RANDOM POLYNOMIALS F IGURE 2 Random degree 10 Weyl polynomials: η0 + η1 z + η2 z 2! + + η10 z10! The empirical distribution on the right was generated using 500,000 random sums It is interesting to note that, in addition to the asymptotic uniformity of the distribution of the real roots, the complex roots are also much more uniformly distributed than was the case when we did not have the 1/ j! factors 32 Taylor Polynomials Another obvious set of polynomials to consider are the random Taylor polynomials; ie, those polynomials with fj (z = zj j! Figure 2 shows the n = 10 empirical distribution for these polynomials 33 Fourier Cosine Series Now let s consider a family of random sums that are not polynomials, namely, random (truncated Fourier cosine series: fj (z = cos(jz This case is interesting because these functions are real-valued not only on the real axis but on the imaginary axis as well: cos(iy = cosh(y Hence, D0 vanishes on both the real and the imaginary axes and, therefore, both axes have a density of zeros as can be seen in Figure 4

14 14 ROBERT J VANDERBEI 2 10 F IGURE 3 Random degree 10 Taylor polynomials: η0 + η1 z + η2 z2! + + η10 z10! The empirical distribution on the right was generated using 500,000 random sums F IGURE 4 Random sum of first 11 terms in a cosine Fourier series: η0 +η1 cos(z+ η2 cos(2z+ +η10 cos(10z The empirical distribution on the right was generated using 10,000 random sums

15 ZEROS OF RANDOM POLYNOMIALS 15 FIGURE 5 Random sum of first 3 terms in a sine/cosine Fourier series: η 0 + η 1 cos(z + η 2 cos(2z + + η 10 cos(10z The empirical distribution on the right was generated using 50,000 random sums 34 Fourier Sine/Cosine Series Finally, we consider random (truncated Fourier sine/cosine series: { cos( j z, j even, 2 f j (z = sin( j+1 z, j odd 2 The n = 2 case is shown in Figure 5 and the n = 10 case is shown in Figure 6 For this case, it is easy to compute the key functions Assuming that n is even and letting m = n/2, we get m A 0 (z = m + 1 B 0 (z = m sinh 2 (jy A 1 (z = 0 B 1 (z = 2zi j=1 m j cosh(jy sinh(jy m A 2 (z = z 2 m(m + 1(2m + 1/6 B 2 (z = 2 z 2 j 2 sinh 2 (jy From these explicit formulas, it is easy to check that the density function h n (z depends only on the imaginary part of z as in evident in Figures 5 and 6 It is also easy to check that the distribution on the real axis is uniform; ie, the density function g n (x is a constant: g n (x = 1 π m(2m + 1/6 j=1 j=1

16 16 ROBERT J VANDERBEI FIGURE 6 Random sum of first 11 terms in a sine/cosine Fourier series: η 0 + η 1 sin(z + η 2 cos(z + + η 9 sin(5z + η 10 cos(5z The empirical distribution on the right was generated using 26,000 random sums 4 FINAL COMMENTS AND SUGGESTED FUTURE RESEARCH The machinery developed in this paper can be applied in many situations that we have not covered For example, if the coefficients are assumed to be independent complex Gaussains (instead of real, then we can apply the same methods and we expect that the computations will be simpler In this case, the intensity function does not have mass concentrated on the real axis (ie, g n = 0 and the intensity function is rotationally invariant REFERENCES 1 LV Ahlfors, Complex analysis, McGraw-Hill, New York, AT Bharucha-Reid and M Sambandham, Random polynomials, Academic Press, Amir Dembo, Bjorn Poonen, Qi-Man Shao, and Ofer Zeitouni, Random polynomials having few or no real zeros, Journal of the American Mathematical Society 15 (2002, no 4, Alan Edelman and Eric Kostlan, How many zeros of a random polynomial are real?, Bulletin of the American Mathematical Society 32 (1995, no 1, P Erdös and P Turán, On the distribution of roots of polynomials, Ann Math 51 (1950, PJ Forrester and G Honner, Exact statistical properties of the zeros of complex random polynomials, Journal of Physics A: Mathematical and General 32 (1999, no 16, J Hammersley, The zeros of a random polynomial, Proc Third Berkeley Symp Math Stat Probability, vol 2, 1956, pp

17 ZEROS OF RANDOM POLYNOMIALS 17 8 Christopher P Hughes and A Nikeghbali, The zeros of random polynomials cluster uniformly near the unit circle, Compositio Mathematica 144 (2008, no 03, IA Ibragimov and NB Maslova, On the expected number of real zeros of random polynomials i coefficients with zero means, Teor Veroyatn Ee Primen 16 (1971, Ildar Ibragimov and Dmitry Zaporozhets, On distribution of zeros of random polynomials in complex plane, Prokhorov and contemporary probability theory, Springer, 2013, pp Ildar Ibragimov and Ofer Zeitouni, On roots of random polynomials, Transactions of the American Mathematical Society 349 (1997, no 6, M Kac, On the average number of real roots of a random algebraic equation, Bull Amer Math Soc 49 (1943, ,938 13, Probability and related topics in physical sciences, Interscience, London, SV Konyagin, On minimum of modulus of random trigonometric oolynomials with coefficients ±1, Matem zametki 56 (1994, , In Russian 15 Wenbo Li and Ang Wei, On the expected number of zeros of a random harmonic polynomial, Proceedings of the American Mathematical Society 137 (2009, no 1, BF Logan and LA Shepp, Real zeros of random polynomials, Proc London Math Soc Third Series 18 (1968, , Real zeros of random polynomials ii, Proc London Math Soc Third Series 18 (1968, Tomaz Prosen, Exact statistics of complex zeros for gaussian random polynomials with real coefficients, Journal of Physics A: Mathematical and General 29 (1996, no 15, SO Rice, Mathematical theory of random noise, Bell System Tech J 25 (1945, LA Shepp and RJ Vanderbei, The complex zeros of random polynomials, Transactions of the AMS 347 (1995, no 11, Bernard Shiffman and Steve Zelditch, Equilibrium distribution of zeros of random polynomials, International Mathematics Research Notices 2003 (2003, no 1, DC Stevens, The average and variance of the number of real zeros of random functions, PhD Dissertation, New York University, New York, Terence Tao and Van Vu, Local universality of zeroes of random polynomials, International Mathematics Research Notices (2014, DI Šparo and MG Šur, On the distribution of roots of random polynomials, Vestn Mosk Univ, Ser 1: Mat, Mekh (1962, Appendix: Algebraic Simplification of the Formula for F/ z Substituting the derivatives given in (218 and (219 into the formula (217 for F/ z, we get that the denominator simplifies to ( F denom z = B 0 D 0 + B 2 0 Ā0A 0 = B 0 D 0 + D 2 0 = (B 0 + D 0 2 D 2 0

18 18 ROBERT J VANDERBEI and the numerator of the formula becomes ( F num = 1 z z (B 0D 0 + B0 2 Ā0A 0 ( B 2 D 0 + B 1 B 0 B1 A 0 Ā 1 + D B 1 B 1 + B 0 B 2 2Ā1A z ( (B 1D 0 + B 0 B 1 Ā0A B 0 B1 A 0 Ā 1 1 B 1 D 0 + B 0 + 2B 0 B1 2Ā1A 0 D 0 The first step to simplifying the numerator is to replace B0 2 Ā0A 0 in the first term with D 0 (like we did in the denominator and factor out a 1/D 0 to get ( F num z = 1 ( (B 0 + D 0 D 0 B2 D0 2 + B 1 B 0 B1 B 1 A 0 Ā 1 + zd B 1 B 1 D 0 + B 0 B 2 D 0 2Ā1A 1 D ((B 0 + D 0 B 1 zd Ā0A 1 ( B1 D0 2 + B 0 B 0 B1 B 1 A 0 Ā 1 + 2B 0 B1 D 0 2Ā1A 0 D 0 0 Next, we bundle together the terms that have a B 0 + D 0 factor: ( F 1 num = (B 0 + D 0 ( B 2 D0 3 + B 1 B 0 B1 D 0 B 1 A 0 Ā 1 D 0 + z zd B 1 B 1 D0 2 + B 0 B 2 D0 2 2Ā1A 1 D0 2 0 B 1 2 D0 2 B 0 B 0 B B1A 2 0 Ā 1 2B 0 B 1 2 D 0 + 2Ā1A 0 B 1 D ( Ā 0 A 1 B1 D0 2 + B 0 B 0 B1 B 1 A 0 Ā 1 + 2B 0 B1 D 0 2Ā1A 0 D 0 zd 0 Now, there are several places where we can find B 0 + D 0 factors For example, the big factor containing eleven terms can be rewritten as follows: B 2 D B 1 B 0 B1 D 0 B 1 A 0 Ā 1 D 0 + B 1 B 1 D B 0 B 2 D 2 0 2Ā1A 1 D 2 0 B 1 2 D 2 0 B 0 B 0 B B 2 1A 0 Ā 1 2B 0 B 1 2 D 0 + 2Ā1A 0 B 1 D 0 = (B 0 + D 0 (B 2 D 2 0 B 0 B A 0 Ā 1 B 1 2 A 1 2 D 2 0 We also look for B 0 + D 0 factors in the five-term factor: B 1 D B 0 B 0 B1 B 1 A 0 Ā 1 + 2B 0 B1 D 0 2Ā1A 0 D 0 = (B 0 + D 0 ( (B 0 + D 0 B 1 A 0 Ā 1 A0 Ā 1 D 0 Substituting these expressions into the formula for the numerator, we get ( F zd 0 num = (B 0 + D 0 ( (B 0 + D 0 (B 2 D0 2 B 0 B A 0 Ā 1 B 1 2 A 1 2 D0 2 z +Ā0A 1 ( (B0 + D 0 ( (B 0 + D 0 B 1 A 0 Ā 1 A0 Ā 1 D 0

19 ZEROS OF RANDOM POLYNOMIALS 19 Rearranging the terms, we see that ( F zd 0 num = (B 0 + D 0 2 (B 2 D0 2 B 0 B A 0 Ā 1 B 1 + z Ā0A 1 B1 (B 0 + D 0 ( 2 A 1 2 D0 2 + A 0 2 A 1 2 A 0 2 A 1 2 D 0 Here s the tricky part replace D 2 0 in the second row with B 2 0 A 0 2 and the second and third row simplify nicely: (B 0 + D 0 ( 2 A 1 2 D A 0 2 A A 0 2 A 1 2 D 0 = (B 0 + D 0 ( 2 A 1 2 B 2 0 A 0 2 A A 0 2 A 1 2 D 0 = 2 A 1 2 B 3 0 A 0 2 A 1 2 B A 1 2 B 2 0D 0 = A 1 2 B 0 ( 2B 2 0 A B 0 D 0 = A 1 2 B 0 (B 0 + D 0 2 Substituting this expression into our formula for the numerator, we now have ( F zd 0 num = (B 0 + D 0 2 (B 2 D0 2 B 0 ( A B A 0 Ā 1 B 1 + z Ā0A 1 B1 Finally, we get a simple formula for F/ z: F z = B 2D0 2 B 0 ( A B A 0 Ā 1 B 1 + Ā0A 1 B1 zd 0 3 ROBERT J VANDERBEI, DEPT OF OPERATIONS RESEARCH AND FINANCIAL ENGINEERING, PRINCETON UNIVERSITY, PRINCETON, NJ address: rvdb@princetonedu