POLYNOMIALS
WITH
REAL ZEROS
Richard P. Stanley
Department of Mathematics
M.I.T. 2-375
Cambridge, MA 02139
rstan@math.mit.edu
http://www-math.mit.edu/~rstan
Transparencies available at:
http://www-math.mit.edu/~rstan/trans.html
1
Rolle's theorem. If f is ontinuous on [a; b℄ and dierentiable on (a; b),
and f (a) = f (b) = 0, then there exists
a < < b suh that f 0() = 0.
Corollary. If P (x) 2 R [x℄ and every zero of P (x) is real, then every
zero of P 0(x) is real.
2
Let P (x) =
an
xn+
n
n
2
+ 2 a2x + 1 a1x+a0 2 R [x℄:
Theorem (Newton). If all zeros of
P (x) are real, then
a2i ai
Proof. P
1
ai+1; 1 i n
n i
(
) Q(x) := xi
)Qi
(
But Q
i
(
1)
+1
1)
1)
1:
(x) has real zeros
P (n i
1)
(1=x) has real zeros
(x) has real zeros:
(x) = n2!
ai+1 + 2aix + ai
) ai ai
2
1
3
ai+1:
2
1
x
2
P
Let P (x) =
aixi have only nonpositive real zeros. Let
i = mode(P ) if ai = max aj :
(If ai = ai+1 = max aj , let mode(P ) =
i + 21 .)
Theorem (J. N. Darroh, 1964):
0
P (1)
P (1)
mode(P ) < 1:
4
Example.
Hermite polynomials:
bX
n= k
2
Hn(x) =
k =0
Hn(x) =
( 1) n! (2x)n 2k
k ! (n 2k )!
2
d
x
x
e
e Hn
dx
By indution, Hn
zeros. Sine
e
x2 H
n
2
1
1
(x) has n
1
(x)
1 real
(x) ! 0 as x ! 1;
it follow that Hn(x) has n real zeros
interlaed by the zeros of Hn 1(x).
5
400
200
6
-3
-2
-1
00
1
2
x
-200
-400
3
2.2
0
-10
7
-20
-30
-40
2.3
2.4
2.5
x
2.6
2.7
2.8
2.9
3
Example (Heilmann-Lieb, 1972). Let
G be a graph with ti i-sets of edges
with no vertex in ommon (mathing
P
of size i). Then i tixi has only real
zeros.
3x3 + 11x2 + 7x + 1
8
Let
n 2 R [x℄:
+
a
x
+
+
a
x
n
0
1
T (x) = a
Set ak = 0 for k < 0 or k > n. Dene
AT =
aj i i;j 1 ;
an innite Toeplitz
matrix.
Theorem (Aissen-Shoenberg-Whitney,
1952) TFAE:
Every
minor of AT is 0, i.e.,
AT is totally nonnegative.
Every zero of T (x) is real and 0.
Gives innitely many onditions, even
for ax2 + bx + .
9
Culture:
Edrei-Thoma generalization (onjetured by Shoenberg). Let
T (x) = 1 + a1x + 2 R [[x℄℄. As
before, let
AT =
aj i i;j 1 :
TFAE:
Every minorQof AT is nonnegative.
(1 + rix)
x
i
T (x) = e Q (1 s x) , where
; ri; sj
0;
j
X
10
j
ri +
X
sj < 1:
Q
(1 + rix)
x
i
T (x) = e Q
j (1
Note:
sj x)
AT easily seen to be t.n. for
1
T (x) = 1+ax; a 0; or T (x) =
; b 0:
1 bx
A; B t.n. ) AB t.n. (by Binet-Cauhy)
AT U = AT AU
n
x
ex = nlim
1+
!1
n
11
Connetion with S1 (Thoma, Ver-
shik, Kerov, et al). Let n ` n and
n
~ = normalized irred: harater of
Sn
n
Then limn!1 ~ exists if and only if
ri = lim ni=n
n!1
sj = lim (n)0j =n
n!1
exist.
12
An appliation of A-S-W:
Let P be a nite poset. Let i be the
number of i-element hains of P .
c0 = 1
c1 = 5
c2 = 5
c3= 1
Chain polynomial: CP (x) =
13
P
ixi
Theorem (Gasharov (essentially), Skandera) Let P have no indued 3 + 1.
Then CP (x) has only real zeros.
good
bad
Proof of Gasharov based on ombinatorial interpretation of minors of AC .
14
Speial ase: P is a unit interval
order or semiorder, i.e., a set of real
numbers with
P
u<v
R
,u<v
1:
2.6
1.2
1.7
1.3
.5
0
Same as no indued 3 + 1 or
15
2 + 2.
For any poset, dene the
eny matrix NP by
(
(NP )st =
0; if s < t
1; otherwise:
5
6
3
4
2
1
2
1
61
6
6
61
6
61
6
41
1
1
1
1
1
1
1
0
1
1
1
1
1
antiadja-
0
1
1
1
1
1
16
0
0
0
0
1
1
3
0
07
7
7
17
7
17
7
15
1
Fats.
det(I + xNP ) = CP (x)
P an be ordered so that NP is totally nonnegative , P is a semiorder.
(Gantmaher-Krein) Eigenvalues of
t.n. matries are real.
Corollary. If P is a semiorder, then
CP (x) has only real zeros.
17
Conjeture (S.-Stembridge) (implies
Gasharov-Skandera theorem) Let P be
a (3 + 1)-avoiding poset. Dene
XP =
0
X
f :P !P
skt)f (s)6=f (t)
Y
t2P
1
xf (t)A ;
the \hromati symmetri funtion" of
the inomparability graph of P . Then
XP is an e-positive symmetri funtion.
Above onjeture, in the speial ase
of semiorders, follows from:
Conjeture (Stembridge) Monomial
immanants of Jaobi-Trudi matries are
s-positive.
18
Rephrasing of A-S-W theorem.
Let P (x) 2 R [x℄, P (0) = 1. Dene
FP (x) = P (x1)P (x2) ;
a symmetri formal series in x = (x1; x2; : : :).
TFAE:
Every zero of P (x) is real and < 0.
FP (x) is s-positive, i.e., a nonneg-
ative linear ombination of Shur funtions s.
FP (x) is e-positive, i.e., a nonnegative linear ombination of elementary symmetri funtions e.
19
Eulerian polynomial:
X
An(x) =
x w
w2Sn
des(
)+1
;
where
des(w) = #fi : w(i) > w(i + 1)g:
E.g., des(4175236) = 3.
Euler:
X
j 0
j nxj =
An(x)
:
n
+1
(1 x)
Theorem (Harper). An(x) has only
real zeros.
20
Example.
A5(x)
P (x) =
= 1+26x+66x2+26x3+x4
x
FP = 1 + 26s1 + (66s2 + 610s11)
+ (26s3 + 1690s21 + 14170s111) + = 1 + 26e1 + (544e2 + 66e11)
+ (12506e3 + 1638e21 + 26e111) + Problem.
(a) Let P (x) = An(x)=x.
Find a ombinatorial interpretation for
the oeÆients of the expansion of FP (x)
in terms of s's or e's, thereby showing
they are nonnegative.
(b) Generalize to other polynomials
P (x).
21
Let P be a partial ordering of 1; : : : ; n.
Let
LP = fw = w wn 2 Sn :
P
i < j ) w ( i) < w ( j )
(i.e., i preedes j in w)g:
1
1
WP (x) =
X
w2LP
Note. P
1
xdes(w):
= n-element antihain
n and WP (x) = An(x)=x.
LP = S
22
)
2
3
1
4
w des(w)
1423
1
4123
1
1432
2
4132
2
1243
1
WP (x) = 3x + 2x2 :
23
all zeros real!
Poset Conjeture (Neggers-S, . 1970)
For any poset P on 1; : : : ; n, all zeros
of WP (x) are real. (True for jP j 7
and naturally labelled P with jP j = 8.)
Let Q be a nite poset.
hain polynomial: CQ(x) =
X
x# ;
where ranges over all hains of Q.
Speial ase
(open). Let L be a
nite distributive lattie (a olletion of
sets losed under [ and \, ordered by
inlusion). Then all zeros of CL(x) are
real.
24
abcd
abc
abd
ab
bc
a
b
φ
CL(x) = (1 + 6x + 10x2 + 5x3)(1 + x)2
Also open:
All zeros of CL(x) are
real if L is a nite modular lattie.
25
Example. If A is a (real) symmetri
matrix, then every zero of det(I + xA)
is real.
Corollary. Let G be a graph. Let
ai(G) be the number of rooted spanP
ning forests with i edges. Then ai(G)xi
has only real zeros.
26
Proof. Dene the Laplaian matrix L(G), rows and olumns indexed
by vertex set V (G), by:
L(G)uv = #(edges between u and v ); u 6= v
L(G)uu = deg(u):
1
2
2
3
6
L(G) = 4 3
0
3
3
4
1
3
0
7
15
1
det(I + xL(G)) = 1 + 8x + 9x2
27
Matrix-Tree Theorem )
det(I + xL(G)) =
X
ai(G)xi:
2
Note. For unrooted spanning forests,
orresponding result is false. I.e, if fi
is the number of i-edge spanning forests
P
of G, then
fixi need not have only
P
real zeros. E.g., G = K3,
fixi =
3x2 + 3x + 1.
28
A-S-W gives innitely many inequalities for real zeros. Are there nitely
many inequalities?
Example. x
2
, b 4.
+ bx + : all zeros real
2
29
Let f (x) 2 R [x℄
have positive leading oeÆient. Apply
Eulidean algorithm to f (x) and f 0(x):
Sturm hains.
f (x) = q1(x)f 0(x) + r1(x)
f 0(x) = q2(x)r1(x) + r2(x)
rk
rk
(x) = qk (x)rk 1(x) + rk (x)
(x) = qk+1(x)rk (x)
1
2
Theorem. f (x) has only real zeros
, deg(ri) = deg(f ) i 1 and the
leading oeÆients of r (x); : : : ; rk (x)
have sign sequene
++
++ .
1
30
Theorem (soure?). Let
Y
V (y1; : : : ; yk) =
(yi yj );
i<j k
the Vandermonde produt. Let
1
f (x) =
n
Y
i=1
(x
i ) :
All zeros of f (x) are real if and only
if
Dk (f ) :=
2k
X
i1<<ik
V (i1 ; : : : ; ik )2 0;
n.
31
Dk (f ) =
X
i1<<ik
V (i1 ; : : : ; ik )2
Dk(f ) is a polynomial in the oeÆients of f
n 1 polynomial inequalities
Dn(f ) = dis(f )
Condition learly neessary
32
Example. f (x) = x
3
has real zeros
,
dis(f )
b2
33
+ bx2 + x + d
0
3:
Distribution of real zeros (M.
Ka, A. Edelman, et al.). Let the o-
eÆients of anxn + + a1x + a0 be
independent standard normals.
34
Density of expeted number of real
zeros at t 2 R :
s
1
n(t) =
1
(t2
1)2
(n + 1)t2n
:
2n+2
2
(t
1)
Hene zeros are onentrated near 1.
Expeted number of real zeros as n !
1:
2
2
En = log(n) + C +
+ O(1=n2);
n
where
C = 0:6257358072 :
Prob(all zeros real) = ompliated integral
35
Suppose the oeÆients ai are independent normals with mean 0 and varin
ane i . Now
En
p
= n:
36