Random harmonic polynomials

# On the zeros of random harmonic polynomials: the naive model

Andrew Thomack
###### Abstract.

A complex harmonic polynomial is the sum of a complex polynomial and a conjugated complex polynomial, of degrees and respectively. Li and Wei present a formula for the expected number of zeros of a random harmonic polynomial in . In this paper we prove that if is a fixed number, this expectation is asymptotically as , and if we find a lower and upper bound of order for this expectation for sufficiently large .

## 1. Introduction

A harmonic polynomial is a function of the form where and are analytic polynomials so that is harmonic over  [9]. We will refer to as a harmonic polynomial of degree and where has degree and has degree , and will use the convention . Since is not analytic, the Fundamental Theorem of Algebra does not tell us the number of zeros of . As with the number of real roots of real polynomials, there is a wide range of possibilities for the number of zeros of , which we will denote as or just . Even if the number cannot be completely determined by , some restrictions can yet be made on . Wilmhurst proved is bounded above by  [11] if certain degenerate cases of are excluded (occuring with probability ), while the lower bound is found to be by the generalized argument principle. Both bounds are sharp, though for , the upper bound has been improved (see  [4]) to , and it has been conjectured (see  [11],  [5], and  [3]) that there is a general improvement that is linear in for each fixed .

Because of the range of values could take, one might ask for the average value of given a random harmonic polynomial, . A useful formula for determining the exact number of zeros of random real polynomials was provided by Kac  [2] while Dunnage provided a similar formula for trigonometric polynomials  [1]. Shepp and Vanderbei go on to extend Kac’s result to determining the number of zeros of complex polynomials for a given complex domain  [10].

The Kac formula and its generalizations have been used by Li and Wei to give an explicit formula for for a fixed and when the coefficients are independent Gaussian random variables. In the case when the variances of the coefficients are scaled by a binomial factor, they also determined the asymptotics of their formula as to be

 ENn,m∼{π4n3/2,m=nn,m=αn+o(n), α∈[0,1)

An altered definition for the random coefficients, referred to as the “truncated model”, was studied by Lerario and Lundberg who found different asymptotics for their model, for , , only dependent on  [6].

Although Li and Wei focused mainly on harmonic polynomials of the form

 hn,m(z)=n∑j=0ajzj+m∑j=0bj¯¯¯zj

where and are independent Gaussians, with and , they also considered what I will refer to as the “naive” model of coefficients of random polynomials, namely so that the coefficients are i.i.d. They proved the following  [7].

###### Theorem 1.

The expected number of zeros of

 hn,m(z)=n∑j=0ajzj+m∑j=0bj¯¯¯zj

on an open domain , denoted by , is given by

 EN(T)=1π∫T1|z|2r21+r22−2r212r23√(r1+r2)2−4r212dσ(z)

where denotes the Lebesgue measure on the complex plane and

 r12=(n∑j=1j|z|2j)(m∑j=1j|z|2j),r3=n∑j=0|z|2j, r1=r3n∑j=0j2|z|2j−(n∑j=0j|z|2j)2,r2=r3m∑j=0j2|z|2j−(m∑j=0j|z|2j)2.

Also in the publication by the same authors,

Numerical analysis suggests that the asymptotic of above expectation for is for fixed , but a rigorous analytic asymptotic hasn’t been found.

We will prove this analytically, that is,

###### Theorem 2.

for fixed .

We will also explore the case when . We conjecture for some constant , because we were able to find both an upper bound and lower bound of this order.

###### Theorem 3.

When , there exists such that for sufficiently large

 c1≤1nlognEN(C)≤c2.

Acknowledgment. I wish to thank Erik Lundberg and Antonio Lerario for leading me in the direction of this problem and discussing it with me at length.

## 2. Preliminary Results

The main tool in proving Theorem 2 will be the generalization of a common theorem in Real Analysis, the General Lebegue Dominated Convergence Theorem; see  [8] for proof.

###### Theorem 4 (General Lebegue Dominated Convergence Theorem).

Let be a sequence of measurable functions on that converges pointwise almost everywhere on to . Suppose there is a sequence of nonnegative measurable functions on that converges pointwise almost everywhere on to and dominates on in the sense that on for all . If

 limn→∞∫Egn=∫Eg<∞,  then  limn→∞∫Efn=∫Ef.

Thus in section 3 we search for the appropriate sequence in order to reach the result.

To maintain the clarity of the proof of the theorems, let us discuss a few of the inequalities used.

###### Lemma 5.

If is a positive real number and , , and , then , for .

That is a direct application of Cauchy-Schwarz, and the lemma follows.

###### Lemma 6.

Let and be non-negative functions over and let . Then and is integrable over if and are.

###### Proof.

We first notice that the arithmetic mean of and is while the geometric mean is . The arithmetic and geometric mean inequality says that , which we relax for the purposes of this lemma. Since is a positive measurable function, integrability follows. ∎

## 3. Proof of Theorem 2

Within the context of Theorem 1, Theorem 2 can be restated as follows

 (1) limn→∞1nπ∫C1|z|2r21+r22−2r212r23√(r1+r2)2−4r212dσ(z)=1.

Beginning with the integral on the left side of (1), using polar coordinates and then a substitution of gives us

 limn→∞∫∞−n1n(n+t)r21+r22−2r212r23√(r1+r2)2−4r212dtwhere
 r3=n∑j=0(1+tn)j+m∑j=0(1+tn)jr12=(n∑j=1j(1+tn)j)(m∑j=1j(1+tn)j)
 r1=r3n∑j=1j2(1+tn)j−(n∑j=1j(1+tn)j)2
 r2=r3m∑j=1j2(1+tn)j−(m∑j=1j(1+tn)j)2.

We use to be the function yielding 1 for values greater than and 0 otherwise and, for a fixed , define

 fn(t)=χn(t)r21+r22−2r212n(n+t)r23√(r1+r2)2−4r212.

We define the following functions of to simplify notation:

 ak=k∑j=0(1+tn)j,bk=k∑j=1j(1+tn)j,ck=k∑j=1j2(1+tn)j.

It follows from Lemma 5 and that

 r1r2=r3((an+am)cncm−cnb2m−cmb2n)+r212≥r212.

Using this, we can bound

 fn(t)≤√r21+r22−2r212n(n+t)r23 =√(r1−r2)2+2(r1r2−r212)n(n+t)r23 ≤r1−r2n(n+t)r23+√2√r1r2−r212n(n+t)r23.

Then since

 r1r2−r212n(n+t)r23=cmn2(n+t)(an+am)⋅(an+am)cn−b2n−b2mcncm(an+am)2

and , by Lemma 6,

 fn(t)≤χn(t)[r1−r2n(n+t)r23+2cmn2(1+tn)(an+am)+(an+am)cn−b2n−bnbmn(n+t)(an+am)2].

We can see when and when . When , we have , so we see that

 fn(t)≤χn(t)⎡⎢ ⎢⎣(an+am)(2cn−cm)−2b2n+b2m−bnbmn(n+t)(an+am)2+⎧⎪ ⎪⎨⎪ ⎪⎩2m3n2(1+(1+tn)2)t<02m3n2(1+tn)n−m+1t≥0⎤⎥ ⎥⎦.

The above bound for we will call .

Now the integrability of is important if we wish to use Theorem 4. We first note that and . From this we see

 ddt[2bn−bman+am]=(an+am)(2cn−cm)−2b2n−bnbm+b2m(n+t)(an+am)2.

Furthermore,

and

 ddt⎡⎢ ⎢⎣−2m3n(n−m)(1+tn)n−m⎤⎥ ⎥⎦=2m3n2(1+tn)n−m+1.

Then has an antiderivative when or ,

 ∫gn=χnn⎡⎢ ⎢ ⎢ ⎢ ⎢ ⎢⎣2bn−bman+am+2m3⎧⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪ ⎪ ⎪⎩arctan(1+tn)t<0−1(n−m)(1+tn)n−mt>0⎤⎥ ⎥ ⎥ ⎥ ⎥ ⎥⎦.

It follows that and clearly the limit as of this integral is .

Next we compute the point-wise limit of as , which we will call . Because when , then for , . Then

 bk=n+tt2((kt−n)(1+tn)k+n), and ck=n+tt3((k2t2−n(2k−1)t+2n2)(1+tn)k−2n2−nt).

Then, to compute , we can use the following limits,

 limn→∞ann+t=1t(et−1)limn→∞bnn(n+t)=1t2((t−1)et+1)
 limn→∞cnn2(n+t)=1t3((t2−2t+2)et−2).

It is clear that for a fixed and , the limits as increases to infinity of , and are all as well as the limits of and . Then we can calculate

 g(t)=2e2t−(t2+2)et+1t2(et−1)2=2ddt[(t−1)et+1t(et−1)].

Meanwhile, if is the point-wise limit of , then

 f(t)=e2t−(t2+2)et+1)t2(et−1)2=ddt[(t−1)et+1t(et−1)].

Then with a little computation we see and . Since

 limn→∞∫Rgn=∫Rg% then by Theorem ???,limn→∞∫Rfn=∫Rf=1

which proves Theorem 2.

## 4. Proof of Theorem 3

We again start with the formula provided by Theorem 1, divide by the appropriate function of and attempt to take the limit. This time we use the change of variables, . Then Theorem 3 is equivalent to

 (2) limn→∞1nlogn∫∞01wr21+r22−2r212r23√(r1+r2)2−4r212dw≤1

where , , and , again using the convention that

 ak=k∑j=0wj,bk=k∑j=1jwjandck=k∑j=1j2wj.

In the case where , then , , and . This allows us to simplify the integrand to

 √r21−r212wr23=√ancn(ancn−b2n)2wa2n

First we bound . By increasing all the coefficients of in from to , we can clearly see is bounded above by . Furthermore,

 cnn2≥n∑k=⌈n2⌉k2n2wk≥14n∑k=⌈n2⌉wk

and for ,

 an≤2n∑k=⌊n2⌋wk⟹cnn2an≥⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩18n even18(1−1/an+12)n odd≥116.

A better lower bound of can be found using algebraically complicated calculus but we simply require a lower bound greater than .

Next we inspect another factor of the square of the integrand. Similar to the case with fixed , we can write as a product of and the derivative of , and similarly . As before, we use the identity that . Then

 ancn−b2nwa2n =ddw[bnan]=ddw[wddw[logan]]=ddw[(n+1)wn+1wn+1−1−w(w−1)] =ddw[n+n+1wn+1−1−1w−1]=1(w−1)2(1−(n+1)2wna2n).

Here, can be bounded below by for all and above for for greater than some fixed . The lower bound of zero is clear. For the upper bound, we first note that it is less than for all . This can be seen by observing that and then by using the Cauchy Schwarz to see that

 (n+1)2=(n∑k=0w−k/2wk/2)2≤(n∑k=0w−k)(n∑k=0wk).

Then we see that

 ddw[wna2n]=wn−1a3n(nan−2bn)<0

because

 nan−2bn= n∑k=0(n−2k)wk = ⌊n/2⌋∑k=0(n−2k)wk−⌊(n−1)/2⌋∑k=0(n−2k)wn−k = ⌊(n−1)/2⌋∑k=0(n−2k)(wk−wn−k)

and . Thus is minimized at .

 1−(n+1)2wna2n≥1−(n+1)2(1+1n)nn2((1+1n)n+1−1)2

The limit of this lower bound as is and although this function seems to be decreasing as increases, we need not have a sharp lower bound and it will suffice that for large enough , we have , say, half of this limit as a lower bound. In summary, when

 (3) 116≤cnn2an≤1and12−e2(e−1)2≤1−(n+1)2wna2n≤1

moreover, the upper bounds apply for all .

We will proceed by bounding the integrals over the following three intervals,

 I=[0,1−1n),J=[1−1n,1+1n],andK=(1+1n,∞).

When ,

 0≤√ancn(ancn−b2n)2wa2n≤12√w(1−w)
 ⟹0≤∫I√ancn(ancn−b2n)2wa2n≤12(1+log(2+2√1−1n−1n))
 ⟹0≤limn→∞1logn∫I√ancn(ancn−b2n)2wa2n≤12.

Let . Then in a similar way, for ,

 √α8√21√w(w−1)≤√ancn(ancn−b2n)2wa2n≤12√w(1−w)
 ⟹0.025≈√α8√2≤limn→∞1logn∫I√ancn(ancn−b2n)2wa2n≤12

Finally, for ,

 ancn−b2n4w2a2n≤cn−bn4w2an≤n2−n4w2≤n24(n−1)

so

 0≤limn→∞∫J√ancn−b2n2wanlogndw≤limn→∞n2√n−1logn⋅2n=0.

Thus,

 √α8√2≤limn→∞1nlogn∫∞0√ancn(ancn−b2n)2wa2ndw≤1.

## 5. Conclusion

We have determined the asymptotic order of growth of in two cases, namely when is fixed and when grows with . The outcome in the former case, where there are on average asymptotically zeros, is not completely unexpected. Indeed, one might expect that as increases, has a decreasing influence over the nature of when is fixed, so that may be heuristically treated as an analytic polynomial. Our result gives evidence to this conclusion since on average has zeros. On the other hand, when grows with we see a different behavior, implying that is quite distinguishable from an analytic polynomial of degree . Besides the additional factor in the total number of zeros, we also notice a disparity in the shape of the so-called “first intensity”, that is, in the integrand appearing in Theorem 1. In the case that is fixed the first intensity divided by converges to everywhere except along the unit circle. Moreover, after the change of variable, we find that the area to the left of under the curve is equal to the area to the right of , where corresponds to through the change of variables. This tells us that as increases, not only do the zeros concentrate on the unit circle, but they seem to do so in a symmetric manner (resembling the known symmetry of the first intensity of an analytic polynomial). In contrast, the first intensity when accumulates to the unit circle in an asymmetric manner when divided by , which can be seen when dividing the integrand by instead of and taking a limit. After the change of variables it is clear that this limit is for while it is for .

There are natural questions extending from this discussion. Though we know the order of growth of for to be , it is yet an open problem to prove the limit of the quotient exists and, if it does, to obtain the explicit value of the constant. One may also notice that we have only explored two cases for the behavior of . An obvious extension would be to investigate the case where , as was done in  [7] and  [6].

## References

• [1] J. E. A. Dunnage, The number of real zeros of a random trigonometric polynomial, Proceedings of the London Mathematical Society s3-16 (1966), 53–84.
• [2] M. Kac, On the average number of real roots of a random algebraic equation, Bulletin of the American Mathematical Society 49 (1943), no. 4, 314–320.
• [3] D. Khavinson, S-Y. Lee, and A. Saez, Zeros of harmonic polynomials, critical lemniscates and caustics, preprint at arXiv:1508.04439.
• [4] D. Khavinson and G. Swiatek, On the maximal number of zeros of certain harmonic polynomials, Proceedings of the American Mathematical Society 131 (2003), 409–414.
• [5] S-Y. Lee, A. Lerario, and E. Lundberg, Remarks on Wilmshurst’s theorem, Indiana University Mathematics Journal 64 (2015), no. 4, 1153–1167.
• [6] A. Lerario and E. Lundberg, On the zeros of random harmonic polynomials: the truncated model, J. Math. Analysis Appl. 438 (2016), 1041–1054.
• [7] W. V. Li and A. Wei, On the expected number of zeros of random harmonic polynomials, Proceedings of the American Mathematical Society 137 (2009), 195–204.
• [8] H. L. Royden and P. M. Fitzpatrick, Real analysis, fourth ed., Pearson Education, Inc., Boston, MA, 2010.
• [9] Terry Sheil-Small, Complex polynomials, Cambridge University Press, 2002.
• [10] L. Shepp and R. Vanderbei, The complex zeros of random polynomials, Transactions of the American Mathematical Society 347 (1995), no. 11, 4365–4384.
• [11] A. S. Wilmshurst, The valence of harmonic polynomials, Proceedings of the American Mathematical Society 126 (1998), no. 7, 2077–2081.
You are adding the first comment!
How to quickly get a good reply:
• Give credit where it’s due by listing out the positive aspects of a paper before getting into which changes should be made.
• Be specific in your critique, and provide supporting evidence with appropriate references to substantiate general statements.
• Your comment should inspire ideas to flow and help the author improves the paper.

The better we are at sharing our knowledge with each other, the faster we move forward.
The feedback must be of minimum 40 characters and the title a minimum of 5 characters