Celebratio Mathematica

Thomas Lig­gett’s last pa­per was un­pub­lished at the time of his death in 2020. The text that fol­lows is an ed­ited draft of that pa­per, pre­pared by Wen­pin Tang, who also provided the an­nota­tions throughout. Note: Strong Rayleigh ran­dom vari­ables are also called Pois­son bi­no­mi­al ran­dom vari­ables. Some res­ults in this pa­per are presen­ted in Sec­tion 4 of the sur­vey art­icle of W. Tang and F. Tang (“The Pois­son bi­no­mi­al dis­tri­bu­tion — old & new” (2019), arX­iv:1908.10024), which provides fur­ther ref­er­ences.

Sup­pose that $$\begin{array}{}\text{(1)}& f(u)=\sum _{i=0}^n{c}_i{u}^i\end{array}$$ is a poly­no­mi­al with pos­it­ive coef­fi­cients $c_i>0$.1 Then:

1. $f$ is said to be strong Rayleigh (SR) if all of its roots are real (and hence neg­at­ive). This is equi­val­ent to say­ing that $f$ can be factored in­to poly­no­mi­als of de­gree 1 with pos­it­ive coef­fi­cients.
2. $f$ is said to be Hur­witz (H) if all of its roots have neg­at­ive real part. Since $$(u-z)(u-\overline{z})=u^2-2u\mathrm{Re}(z)+|z{|}^2,$$ this is equi­val­ent to say­ing that $f$ can be factored in­to poly­no­mi­als of de­gree at most 2 with pos­it­ive coef­fi­cients.

In this pa­per we con­sider the fol­low­ing ques­tion: When can $f$ be factored in­to poly­no­mi­als of de­gree at most $j$ with pos­it­ive coef­fi­cients? We will call this prop­erty $P_j$.

One mo­tiv­a­tion for this is the fol­low­ing: Ex­cept for the nor­mal­iz­a­tion $f(1)=1$, the $f$ in $\text{(1)}$ is the prob­ab­il­ity gen­er­at­ing func­tion (pgf) $E{u}^X$ of a ran­dom vari­able $X$ with dis­tri­bu­tion $c_i=P(X=i)$. If it2 has prop­erty $P_j$, then there are in­de­pend­ent ran­dom vari­ables $X_i$ with val­ues in $\{0,1,\dots ,j\}$ so that $X$ and $\sum _i{X}_i$ have the same dis­tri­bu­tion. This prop­erty was ex­ploited in [◊] in the case $j=1$ in or­der to ob­tain vari­ous prob­ab­il­ist­ic lim­it the­or­ems.

An­oth­er mo­tiv­a­tion comes from [◊]. In our con­sid­er­a­tion of mul­tivari­ate cent­ral lim­it the­or­ems for SR vec­tors, we raised the fol­low­ing ques­tion: if $X$ has an SR pgf and $j,k\ge 1$, how well can one ap­prox­im­ate $\frac{j}{k}X$ by an SR ran­dom vari­able? We were able to solve this prob­lem for $j=1$, show­ing that $\lfloor \frac{1}{k}X\rfloor$ is SR. It turns out that $\lfloor \frac{2}{k}X\rfloor$ is very far from be­ing SR, in the sense that the ima­gin­ary parts of some roots of the pgf are large. Nev­er­the­less, we will see that a com­bin­a­tion of the res­ults in [◊] with the Hermite–Biehler the­or­em (see [◊] for ex­ample) im­plies that $\lfloor \frac{2}{k}X\rfloor$ is H. We then spec­u­late about pos­sible ex­ten­sions of this to $j\ge 3$. For the ap­plic­a­tions we have in mind, this prop­erty is gen­er­ally as use­ful as SR.

A suf­fi­cient con­di­tion for SR is that the coef­fi­cients of $f$ sat­is­fy $$\begin{array}{}\text{(2)}& c_i^2>4c_{i-1}{c}_{i+1},1\le i\le n-1;\end{array}$$ see [◊]. A suf­fi­cient con­di­tion for H if $n>5$ is that the coef­fi­cients of $f$ sat­is­fy $$\begin{array}{}\text{(3)}& c_i{c}_{i+1}\ge (2.147\dots )c_{i-1}{c}_{i+2},1\le i\le n-2;\end{array}$$ see [◊].

These suf­fi­cient con­di­tions are quite re­strict­ive. For ex­ample, if $f(u)=(u+1{)}^n$, so that $c_i=(\genfrac{}{}{0ex}{}{n}{i})$, then $$\frac{c_i^2}{c_{i-1}{c}_{i+1}}=\frac{i+1}{i}\frac{n-i+1}{n-i}.$$ There­fore $$\underset{i\to \mathrm{\infty }}{lim}\underset{n\to \mathrm{\infty }}{lim}\frac{c_i^2}{c_{i-1}{c}_{i+1}}=1,$$ so $\text{(2)}$ fails for large $n$.

The Hermite–Biehler the­or­em gives an NASC3 for H:

By ana­logy with this res­ult, we con­sider the fol­low­ing: Giv­en a poly­no­mi­al $f$ as in $\text{(1)}$ with pos­it­ive coef­fi­cients, write $$\begin{array}{}\text{(4)}& f(u)=\sum _{i=0}^{j-1}{u}^i{h}_i(u^j).\end{array}$$ We will say that $f$ has prop­erty $Q_j$ if the roots of $h_0,h_1,\dots ,h_{j-1}$ are neg­at­ive and in­ter­lace, with the largest be­ing a root of $h_0$, the second largest be­ing a root of $h_1$, etc.

Note that $Q_1=P_1$. The Hermite–Biehler the­or­em is the state­ment that $Q_2=P_2$. However, $Q_3\ne P_3$ as the fol­low­ing ex­ample shows. Let $$f(u)=u^5+u^4+u^3+2u^2+\frac{3}{2}u+\frac{1}{3}.$$ By defin­i­tion $\text{(4)}$, we have $$h_0(u)=u+\frac{1}{3},h_1(u)=u+\frac{3}{2},h_2(u)=u+2.$$ The roots of $f$ are $z_1$, ${\overline{z}}_1$, $z_2$, ${\overline{z}}_2$, $w$, where the (roun­ded) val­ues are $$z_1=-.725+.100i,z_2=.435+1.137i,w=-.420.$$ Then $$(u-w)(u-z_2)(u-{\overline{z}}_2)=.623+1.116u-.449u^2+u^3,$$ so $f$ is not $P_3$. However, the roots of $h_0,h_1,h_2$ are $-\frac{1}{3}$, $-\frac{3}{2}$, $-2$ re­spect­ively, which do in­ter­lace cor­rectly. We are not aware of a use­ful cri­terion for $P_j$ for $j\ge 3$.

In the fol­low­ing sec­tion, we prove that $\lfloor \frac{2}{k}X\rfloor$ is $P_2$5 if $X$ is $P_1$, and spec­u­late about the likely situ­ation for $\lfloor \frac{3}{k}X\rfloor$. In par­tic­u­lar, we con­jec­ture that $\lfloor \frac{3}{4}X\rfloor$ is $P_3$. $\lfloor \frac{3}{5}X\rfloor$ is not $P_3$, but we con­jec­ture that it is nearly so, in that the pgf can be factored in­to poly­no­mi­als of de­gree at most 3 with at most one of the factors hav­ing a neg­at­ive coef­fi­cient. We will call this situ­ation al­most $P_3$. More gen­er­ally, it may be the case that $\lfloor \frac{3}{k}X\rfloor$ is $P_3$ if $k\equiv 1(\mathrm{mod}3)$ and al­most $P_3$ if $k\equiv 2(\mathrm{mod}3)$.

Sup­pose the roots of $h_0,h_1,h_2$ are neg­at­ive and in­ter­lace. Let these roots be de­noted by $$t_m<t_{m-1}<\cdots <t_0<0.$$ Then $$f(u)=h_0(u^3)+u{h}_1(u^3)+u^2{h}_2(u^3),$$ where $h_0$ has roots $t_0,t_3,\dots ,$ and $h_1$ has roots $t_1,t_4,\dots ,$ and $h_2$ has roots $t_2,t_5,\dots .$ The de­gree of $f$ is $m+3$.

We be­gin with the case $m=1$, so $$h_0(y)=a(y-t_0),h_1(y)=b(y-t_1),h_2(y)=c,$$ with $a,b,c>0$. Then $$\begin{array}{}\text{(5)}& f(u)=-a{t}_0-b{t}_{1}u+c{u}^2+a{u}^3+b{u}^4.\end{array}$$ If we write $$f(u)=(u+d)(c_3{u}^3+c_2{u}^2+c_{1}u+c_0)$$ and solve, we get $$c_3=b,c_2=a-bd,c_1=c-ad+b{d}^2,c_0=-b{t}_1-dc+a{d}^2-b{d}^3,$$ where $f(-d)=0$. Now we seek the con­di­tion un­der which $f\in P_3$. For the coef­fi­cients to be non­neg­at­ive, we need $$b\ge 0,a-bd\ge 0,c-ad+b{d}^2\ge 0,-b{t}_1-dc+a{d}^2-b{d}^3\ge 0.$$ Writ­ing $$f(-d)=d(b{d}^3-a{d}^2+cd+b{t}_1)-a{t}_0=d^2(b{d}^2-ad+c)-(-bd{t}_1+a{t}_0),$$ we see that these in­equal­it­ies are equi­val­ent to $$a-bd\ge 0,-b{t}_{1}d+a{t}_0\ge 0.$$

A ne­ces­sary and suf­fi­cient con­di­tion for the ex­ist­ence of a $d>0$ so that6 $$f(-d)=0,\frac{a{t}_0}{b{t}_1}\le d\le \frac{a}{b},$$ is that $f$ take on op­pos­ite signs (in­clud­ing 0) at two points in the in­ter­val $$(-\frac{a{t}_0}{b{t}_1},-\frac{a}{b}).$$ To find a use­ful suf­fi­cient con­di­tion, write $$b^{3}f(-\gamma \frac{a}{b})=a^{2}bc{\gamma }^2-a{b}^3(-\gamma t_1+t_0)-{\gamma }^3(1-\gamma )a^4.$$ If $\gamma =1$, this is $$ab[ac-b^2(-t_1+t_0)].$$ This is $\ge 0$ if and only if $-t_1+t_0\le ac/b^2$. It is $\le 0$ if $\gamma t_1\le t_0$ and $bc\le \gamma (1-\gamma )a^2$. There­fore, a suf­fi­cient con­di­tion for the ex­ist­ence of the re­quired $d$ is that there ex­ist a $\gamma \in [0,1]$ so that $$\begin{array}{}\text{(6)}& t_0-t_1\le \frac{ac}{b^2},\gamma t_1\le t_0,\text{and}bc\le \gamma (1-\gamma )a^2.\end{array}$$

When is $Q_3$ sat­is­fied? If $n=3$, $P_3$ and $Q_3$ are both auto­mat­ic. If $n=4$, $Q_3$ is equi­val­ent to $c_1{c}_3\ge c_0{c}_4$.7 If $f$ is $P_3$ but not $P_2$, then $$\begin{array}{rl}f(u)& =(a_0+a_{1}u)(b_0+b_{1}u+b_2{u}^2+b_3{u}^3),\\ c_1{c}_3-c_0{c}_4& =a_1^2{b}_0{b}_2+a_0{a}_1{b}_1{b}_2+a_0^2{b}_1{b}_3,\end{array}$$ so $f$ is $Q_3$. However, $P_2$ does not im­ply $Q_3$. For ex­ample, $f(u)=(1+u^2{)}^2$ is $P_2$ but not $Q_3$, since $c_1{c}_3-c_0{c}_4=-1$. More gen­er­ally, if $$f(u)=(a_0+a_{1}u+a_2{u}^2)(b_0+b_{1}u+b_2{u}^2),$$ then $$c_1{c}_3-c_0{c}_4=(a_0{b}_1+a_1{b}_0)(a_1{b}_2+a_2{b}_1)-a_0{a}_2{b}_0{b}_2.$$ This is non­neg­at­ive if $a_0{a}_2\le 4a_1^2$, $b_0{b}_2\le 4b_1^2$.8

If $n=5$, $Q_3$ is equi­val­ent to $c_1{c}_3\ge c_0{c}_4$ and $c_2{c}_4\ge c_1{c}_5$.9 If $$f(u)=(a_0+a_{1}u+a_2{u}^2)(\frac{25}{2}+\frac{1}{2}{u}^2+u^3),$$ then $$c_1{c}_3-c_0{c}_4=\frac{25}{2}(a_1^2-a_0{a}_2).$$ In this case, $f$ is $P_3$ but not $P_2$, but it may not be $Q_3$.

We be­gin by show­ing that in gen­er­al, $X$ be­ing SR does not im­ply that $\lfloor \frac{2}{3}X\rfloor$ is SR, and in fact that this is very far from be­ing the case.

Proof.  Up to a con­stant mul­tiple, the pgf of $\lfloor \frac{2}{3}X\rfloor$ is $$f(u)=\sum _{i=0}^{3n}(\genfrac{}{}{0ex}{}{3n}{i})u^{\lfloor 2i/3\rfloor }=\sum _{i\ge 0}(\genfrac{}{}{0ex}{}{3n+1}{3i+1})u^{2i}+\sum _{i\ge 0}(\genfrac{}{}{0ex}{}{3n}{3i+2})u^{2i+1},$$ which has de­gree $d=2n$. Let $z_1,z_2,\dots$ be the com­plex roots with pos­it­ive ima­gin­ary part and $w_1,w_2,\dots$ be the real roots. Then $$f(u)=\prod _j(u^2-2u\mathrm{Re}(z_j)+|z_j{|}^2)\prod _j(u-w_j).$$ Both of these ex­pres­sions have $u^d$ coef­fi­cients equal to 1. Identi­fy­ing the coef­fi­cients of $u^{d-1}$ gives $$\begin{array}{}\text{(7)}& -2\sum _j\mathrm{Re}(z_j)-\sum _j{w}_j=3n.\end{array}$$ Identi­fy­ing the coef­fi­cients of $u^{d-2}$ gives $$4\sum _{i<j}\mathrm{Re}(z_i)\mathrm{Re}(z_j)+\sum _j|z_j{|}^2+2\sum _j\mathrm{Re}(z_j)\sum _i{w}_i+\sum _{i<j}{w}_i{w}_j=(\genfrac{}{}{0ex}{}{3n+1}{3n-2}).$$ Com­bin­ing this with $\text{(7)}$ leads to $$\sum _i[\mathrm{Im}(z_i){]}^2-\sum _i[\mathrm{Re}(z_i){]}^2-\frac{1}{2}\sum _i{w}_i^2=\frac{1}{2}n(9n^2-9n-1).$$ There­fore, since there are at most $n$ pairs of com­plex con­jug­ate roots, $$nmax[\mathrm{Im}(z_i){]}^2\ge \sum _i[\mathrm{Im}(z_i){]}^2\ge \frac{1}{2}n(9n^2-9n-1).◻$$

Here is the res­ult from [◊] that we used to show that $\lfloor \frac{1}{k}X\rfloor$ is SR if $X$ is SR. For a poly­no­mi­al $f$ of de­gree $n$, write $$\begin{array}{}\text{(8)}& f(x)=\sum _{i=0}^{k-1}{x}^i{g}_i(x^k),\end{array}$$ where $g_i$ is a poly­no­mi­al of de­gree $\lfloor \frac{n-i}{k}\rfloor$.

This state­ment is il­lus­trated in the fol­low­ing ar­ray, in the case $k=3$: $$\left(\begin{array}{cccccccccc}& \cdots & s_6& s_5& s_4& s_3& s_2& s_1& s_0& 0\\ \\ g_0& \cdots & 0& +& +& 0& -& -& 0& +\\ g_1& \cdots & +& +& 0& -& -& 0& +& +\\ g_2& \cdots & +& 0& -& -& 0& +& +& +\end{array}\right).$$ Note that each row is peri­od­ic of peri­od 6, and each row is ob­tained from the pre­vi­ous row via a shift. The proof in [◊] is by in­duc­tion on the de­gree of $f$. The pgf of $\lfloor \frac{1}{k}X\rfloor$ is $\sum _i{g}_i$, which al­tern­ates signs at the points $s_{mk}$. There­fore, it has a root in each of the in­ter­vals $(s_{(m+1)k},s_{mk})$, thus show­ing that $\lfloor \frac{1}{k}X\rfloor$ is SR $(=P_1)$.

Proof.  It suf­fices to prove the res­ult for $k$ odd. If $k$ is odd, the pgf of $\lfloor \frac{2}{k}X\rfloor$ is $$\begin{array}{}\text{(9)}& \sum _{i=0}^{(k-1)/2}{g}_i(u^2)+u\sum _{i=(k+1)/2}^{k-1}{g}_i(u^2).\end{array}$$ By the Hermite–Biehler the­or­em, all roots of $\text{(9)}$ have neg­at­ive real parts if and only if all the roots of $$\sum _{i=0}^{(k-1)/2}{g}_i(-u^2)\text{and}\sum _{i=(k+1)/2}^{k-1}{g}_i(-u^2)$$ are neg­at­ive and in­ter­lace, with the largest of these be­ing a root of $$\sum _{i=0}^{(k-1)/2}{g}_i(-u^2).$$

If $k=3$, one can see from the above ar­ray that $$g_0(\cdot )+g_1(\cdot )\text{ has a root in each of the intervals }\cdots (s_7,s_6),(s_4,s_3),(s_1,s_0)$$ and $$g_2(\cdot )\text{ has a root in each of the intervals }\cdots (s_9,s_7),(s_6,s_4),(s_3,s_1),$$ since the val­ues of the func­tion at the two en­d­points of each in­ter­val have op­pos­ite signs. There­fore, the neg­at­ive roots in­ter­lace. For lar­ger odd $k$, the ar­gu­ment is sim­il­ar. Now $$\sum _{i=0}^{(k-1)/2}{g}_i(\cdot )\text{ has a root in each of the intervals }\cdots (s_{(k-1)/2+2k},s_{2k}),(s_{(k-1)/2+k},s_k),(s_{(k-1)/2},s_0)$$ and $$\sum _{i=(k+1)/2}^{k-1}{g}_i(\cdot )\text{ has a root in each of the intervals }\cdots (s_{3k-1},s_{(k+1)/2+2k}),(s_{2k-1},s_{(k+1)/2+k}),(s_{k-1},s_{(k+1)/2}).$$ It fol­lows that $\lfloor \frac{2}{k}X\rfloor$ is H $(=P_2)$.  ◻

The situ­ation for the case $\lfloor \frac{3}{k}X\rfloor$, where $k$ is not a mul­tiple of 3, is sim­il­ar. Its pgf is $$h_0(u)+u{h}_1(u)+u^2{h}_2(u),$$ where $$h_0(u)=\sum _{i=0}^{\lfloor k/3\rfloor }{g}_i(u),h_1(u)=\sum _{\lceil k/3\rceil }^{\lfloor 2k/3\rfloor }{g}_i(u),h_2(u)=\sum _{i=\lceil 2k/3\rceil }^{k-1}{g}_i(u).$$ Then $$\begin{array}{rl}& h_0\text{ has a root in each of the intervals }(s_{\lfloor k/3\rfloor +ik},s_{ik})\text{ for }i\ge 0,\\ & h_1\text{ has a root in each of the intervals }(s_{\lfloor 2k/3\rfloor +ik},s_{\lceil k/3\rceil +ik})\text{ for }i\ge 0,\\ & h_2\text{ has a root in each of the intervals }(s_{k-1+ik},s_{\lceil 2k/3\rceil +ik})\text{ for }i\ge 0.\end{array}$$ In a few cases, the in­ter­val is of the form $(a,a)$, which is in­ter­preted as the singleton $\{a\}$. So, the roots of $h_0,h_1,h_2$ are neg­at­ive and in­ter­lace. Un­for­tu­nately, as was poin­ted out earli­er, this prop­erty does not im­ply $P_3$.

The case of $\lfloor \frac{3}{k}X\rfloor$ is more chal­len­ging, since for each root with pos­it­ive real part, one must find a neg­at­ive root that com­pensates for it. It is not simply a mat­ter of prov­ing some prop­erty for each root, as was the case for $P_1$ and $P_2$. All we can of­fer at this point is spec­u­la­tion based on com­pu­ta­tions of roots in spe­cial cases. If $f$ is $P_3$ but not $P_2$, $f$ will have com­plex roots of pos­it­ive real part. For a neg­at­ive real root $w$ and a con­jug­ate pair of roots $z,\overline{z}$ with pos­it­ive real part, let $$⟨z,w⟩=(u-w)(u-z)(u-\overline{z}).$$ For this cu­bic poly­no­mi­al to have pos­it­ive coef­fi­cients, it is ne­ces­sary and suf­fi­cient that $$z+\overline{z}\le -w\le \frac{z\overline{z}}{z+\overline{z}}.$$

Here is what seems to be the case. Let $w_1,w_2,\dots$ be the real roots of $f$ ordered in in­creas­ing or­der of their ab­so­lute val­ues, and $z_1,z_2,\dots$ be the roots with pos­it­ive real and ima­gin­ary parts, ordered in in­creas­ing or­der of the real parts. As­sume that $X$ is SR and takes val­ues $0,1,\dots ,n$.

Let $f$ be the pgf of $\lfloor \frac{3}{4}X\rfloor$. The de­gree of $f$ is $d=\lfloor \frac{3}{4}n\rfloor$. Then $f$ ap­pears to be $P_3$ in gen­er­al. The num­ber of $z_i$’s and the num­ber of $w_i$’s are close to $\frac{n}{4}$. This has been checked in a num­ber of cases, in­clud­ing $X\sim B(n,p)$ for $n=54$ and 75, and $p=\frac{1}{100}$, $\frac{1}{2}$ and $\frac{100}{101}$.10 If $X\sim B(n,\frac{1}{2})$, the pgf of $\lfloor \frac{3}{4}X\rfloor$ is $P_2$ for $n\le 9$. If $10\le n\le 30$, it is $P_3$ and can be factored in­to $\lfloor \frac{n}{4}\rfloor$ ir­re­du­cible cu­bics, with an ex­tra lin­ear factor if $d\equiv 1\mathrm{mod}3$ and an ex­tra ir­re­du­cible quad­rat­ic factor if $d\equiv 2mod3$. Here ir­re­du­cible means that it can­not be fur­ther factored in­to poly­no­mi­als with pos­it­ive coef­fi­cients. The ap­pro­pri­ate pair­ing of real and com­plex roots is $w_i$ with $z_i$. This pic­ture con­tin­ues to hold for many $X$’s that are sums of in­de­pend­ent Bernoulli ran­dom vari­ables with ran­domly chosen para­met­ers.

Here is some more de­tail when $X$ is $B(n,\frac{1}{2})$ and $f$ is the pgf of $\lfloor \frac{3}{4}X\rfloor$, omit­ting mul­ti­plic­at­ive con­stants: $$\begin{array}{rl}& n=2:f(u)=u+3\text{ is }{P}_1,\\ & n=3:f(u)=u^2+3u+4\text{ is }{P}_2,\\ & n=4:f(u)=u^3+4u^2+6u+5=(u+a)(u^2+bu+c)\text{ is }{P}_2,\\ & \phantom{n=4:}f(-a)=0,b=4-a,c=\frac{5}{a},f(0)=5,f(-4)=-19,\end{array}$$ so $f(-a)=0$ for some $0<a<4$. There­fore $f$ is $P_2$. It is not $P_1$ since the dis­crim­in­ant of the quad­rat­ic factor is $<0$ for $0<a<4$. The ac­tu­al val­ues are $a=2.35$, $b=1.65$, $c=2.12$. $$\begin{array}{rl}& n=5:f(u)=u^3+\frac{5}{3}{u}^2+\frac{5}{3}u+1=(u+a)(u^2+bu+c)\text{ is }{P}_2,\\ & \phantom{n=5:}f(-a)=0,c=\frac{1}{a},b=\frac{5}{3}-a,f(0)=1,f(-\frac{5}{3})=-\frac{16}{9},\end{array}$$ so $f(-a)=0$ for some $0<a<\frac{5}{3}$. Again, $f$ is $P_2$ but not $P_1$. The ac­tu­al val­ues are $a=1$, $b=\frac{2}{3}$, $c=1$. $$\begin{array}{rl}& n=6:f(u)=u^4+21u^3+20u^2+15u+7=(u+a)(u+b)(u^2+cu+d)\text{ is }{P}_2,\\ & \phantom{n=6:}f(-a)=f(-b)=0,d=\frac{7}{ab},c=21-a-b.\end{array}$$ Com­put­ing $f(x+iy)$ with $y\ne 0$ and set­ting it $=0$ leads to $$f(x)-\frac{1}{2}{f}^{\prime \prime }(x)y^2+y^4=0,f^{\prime }(x)-\frac{1}{6}{y}^2{f}^{\prime \prime \prime }(x)=0.$$ Solv­ing the second equa­tion for $y^2$ and us­ing that in the first equa­tion leads to $$f(x)[f^{\prime \prime \prime }(x){]}^2-3f^{\prime \prime }(x)f^{\prime }(x)f^{\prime \prime \prime }(x)+36[f^{\prime }(x){]}^2=0.$$ Ex­pand­ing this yields a poly­no­mi­al in $x$, all of whose coef­fi­cients are neg­at­ive, so $x<0$. It fol­lows that $f$ is $P_2$. The ac­tu­al val­ues are $a=20.04,b=.66,c=.30,d=.53$.

The first case that is $P_3$ is $n=10$. Now $$\begin{array}{rl}f(u)& =11+45u+120u^2+462u^3+210u^4+120u^5+55u^6+u^7\\ & =(u+53)(u^3+.24u^2+.10u+.03)(u^3+1.96u^2+3.23u+7.69).\end{array}$$ Set­ting $f(x+iy)=0$ for $y\ne 0$ gives $$\sum _{j=0}^3\frac{(-1{)}^j{f}^{(2j)}(x)y^{2j}}{(2j)!}=0,\sum _{j=0}^3\frac{(-1{)}^j{f}^{(2j+1)}(x)y^{2j}}{(2j+1)!}=0.$$

Con­sider small $n$ with $X$ gen­er­al. Use $e_j$ for the ele­ment­ary sym­met­ric func­tions.

For $n=3$, $$f(u)=u^2{e}_0+u{e}_1+(e_2+e_3)$$ is $P_2$.

For $n=4$, $$f(u)=u^3{e}_0+u^2{e}_1+u{e}_2+(e_3+e_4)$$ is $P_3$, but may not be $P_2$.

For $n=5$, $$f(u)=u^3(e_0+e_1)+u^2{e}_2+u{e}_3+(e_4+e_5)$$ is $P_3$, but may not be $P_2$.

For $n=6$, $$f(u)=u^4{e}_0+u^3(e_1+e_2)+u^2{e}_3+u{e}_4+(e_5+e_6).$$ Set­ting this equal to $$(u+w)(u^3+a{u}^2+bu+c)$$ with $w\ge 0$ gives $f(-w)=0$ and $$\begin{array}{rl}{w}^{3}a& =w^2{e}_3-w{e}_4+e_5+e_6=w^3(e_1+e_2-e_{0}w),\\ w^{2}b& =w{e}_4-e_5-e_6,\\ wc& =e_5+e_6.\end{array}$$ So that $a,b,c\ge 0$, we need $$\frac{e_5+e_6}{e_4}\le w\le \frac{e_1+e_2}{e_0}.$$ Com­pute $$\begin{array}{c}{e}_0^{2}f(-\frac{e_1+e_2}{e_0})=e_1(e_1{e}_3-e_0{e}_4)+e_2(2e_1{e}_3-e_0{e}_4)+e_2^2{e}_3+e_0^2(e_5+e_6)\ge 0,\\ 16e_0^{3}f(-\frac{e_1+e_2}{2e_0})=-e_1^4-4e_1^2(e_1^2-e_0{e}_3)-8e_0^2(e_1{e}_4-2e_0{e}_5),\\ -2e_1{e}_2(3e_1{e}_2-4e_0{e}_3)-8e_0^2(e_2{e}_4-2e_0{e}_6)-4e_2^2(e_1{e}_2-4e_0{e}_3)-e_2^4\le 0.\end{array}$$ Since $$\frac{e_5+e_6}{e_4}\le \frac{e_1+e_2}{2e_0},$$ it fol­lows that $f$ has a root $-w$ which makes $a,b,c\ge 0$. The in­equal­it­ies above fol­low from the New­ton in­equal­it­ies $$\frac{1}{6}\frac{e_1}{e_0}\ge \frac{2}{5}\frac{e_2}{e_1}\ge \frac{3}{4}\frac{e_3}{e_2}\ge \frac{4}{3}\frac{e_4}{e_3}\ge \frac{5}{2}\frac{e_5}{e_4}\ge \frac{6}{1}\frac{e_6}{e_5}.$$ It fol­lows that $f$ is $P_3$.

For $n=7$, $$f(u)=u^5{e}_0+u^4{e}_1+u^3(e_2+e_3)+u^2{e}_4+u{e}_5+(e_6+e_7).$$ Set this equal to $$(u^2+au+b)(u^3+c{u}^2+du+g).$$ Solv­ing for the coef­fi­cients gives $$\begin{array}{rl}c& =e_1-a,\\ d& =a^2-b-a{e}_1+e_2+e_3,\\ g& =-a^3+2ab+a^2{e}_1-b{e}_1-a(e_2+e_3)+e_4,\end{array}$$ with $$b^2-b[3a^2+(e_2+e_3)-2a{e}_1)]+a^4-a^3{e}_1+a^2(e_2+e_3)-a{e}_4+e_5=0$$ and $$(e_1-2a)b^2+b[a^3-a^2{e}_1+a(e_2+e_3]-e_4]+(e_6+e_7)=0.$$ Com­bin­ing these two equa­tions gives (and noth­ing is lost if $e_1\ne 2a$) $$b=\frac{2a^5-3e_1{a}^4+[e_1^2+2(e_2+e_3)]a^3-[e_1(e_2+e_3)+2e_4]a^2+(e_1{e}_4+2e_5)a-[(e_1{e}_5-(e_6+e_7)]}{5a^3-6e_1{a}^2+[2e_1^2+(e_2+e_3)]a-[e_1(e_2+e_3)-e_4]}.$$ Us­ing this value of $b$ in either of those equa­tions im­plies that $a$ must be a root of a de­gree-10 poly­no­mi­al $H(a)$. Also, let­ting $$D=5a^3-6e_1{a}^2+[2e_1^2+(e_2+e_3)]a-[e_1(e_2+e_3)-e_4],$$ we have $$\begin{array}{rl}Dd& =3a^5-8e_1{a}^4+[7e_1^2+4(e_2+e_3)]a^3-[2e_1^3+7e_1(e_2+e_3)-3e_4]a^2\\ & +[(e_2+e_3)(e_2+e_3+3e_1^2)-2(e_1{e}_4+e_5)]a\\ & -[e_1(e_2+e_3{)}^2-e_4(e_2+e_3)-e_1{e}_5+(e_6+e_7)]\end{array}$$ and $$\begin{array}{rl}Dg& =-a^6+3e_1{a}^5-[3e_1^2+2(e_2+e_3)]a^4+[e_1^3+4e_1(e_2+e_3)]a^3\\ & -[(e_2+e_3)(e_2+e_3+2e_1^2)+(e_1{e}_4-4e_5)]a^2+[e_1(e_2+e_3{)}^2+e_1^2{e}_4-4e_1{e}_5+2(e_6+e_7)]a\\ & -[e_1{e}_4(e_2+e_3)-e_4^2-e_1^2{e}_5+e_1(e_6+e_7)].\end{array}$$ Us­ing Poly­no­mi­alRe­duce, $$H(a)=[a^4-a^3{e}_1+a^2(e_2+e_3)-a{e}_4+e_5]Dg-\frac{1}{3}(e_6+e_7)[13Dd+K(a)],$$ where $$\begin{array}{rl}K(a)& =29e_1{a}^4-2[20e_1^2+17(e_2+e_3)]a^3+[14e_1^3+61(e_2+e_3)-24e_4]a^2\\ & -[27e_1^2(e_2+e_3)+10(e_2+e_3{)}^2-17e_1{e}_4-20e_5]a\\ & +10[e_1(e_2+e_3{)}^2-e_4(e_2+e_3)-e_1{e}_5+e_6+e_7].\end{array}$$ Now the New­ton in­equal­it­ies are $$\frac{1}{7}\frac{e_1}{e_0}\ge \frac{2}{6}\frac{e_2}{e_1}\ge \frac{3}{5}\frac{e_3}{e_2}\ge \frac{4}{4}\frac{e_4}{e_3}\ge \frac{5}{3}\frac{e_5}{e_4}\ge \frac{6}{2}\frac{e_6}{e_5}\ge \frac{7}{1}\frac{e_7}{e_6}.$$

Un­for­tu­nately, the pgf of $f$ $\lfloor \frac{3}{5}X\rfloor$ is not ne­ces­sar­ily $P_3$. Take $X$ to be $B(n,\frac{1}{2})$. For $n\le 16$, $f$ is $P_2$, and for $17\le n\le 20$ is $P_3$. For $n=21$, $d=12$, and there are 4 $w_i$’s and 4 $z_j$’s. For $f$ to be $P_3$, there would have to be a pair­ing of the $w_i$’s and $z_j$’s so that for each such pair, $⟨z_i,w_j⟩$ has pos­it­ive coef­fi­cients. But, no such pair­ing ex­ists, since $⟨w_1,z_i⟩$ has a neg­at­ive quad­rat­ic coef­fi­cient for $i=1,2,3,4$, so $f$ is not $P_3$. However, $⟨z_i,w_i⟩$ and $⟨z_{i-1},w_i⟩$ have pos­it­ive coef­fi­cients for $i=2,3,4$, so one can choose a fac­tor­iz­a­tion of $f$ so that only one factor has a neg­at­ive coef­fi­cient. Note that it is not $P_4$ either. It is $P_6$, since $$(u-w_1)(u-w_2)(u-z_1)(u-{\overline{z}}_1)(u-z_2)(u-{\overline{z}}_2)$$ has pos­it­ive coef­fi­cients. This may or may not be use­ful.