Rational functions with nonnegative powerseries coefficients.

Ira M. Gessel

Department of MathematicsBrandeis University

AMS Special Session on Special Functions andCombinatorics

in honor of Dennis Stanton’s 65th birthday

Joint Mathematics MeetingSan Diego, CA

Thursday, January 11, 2018

Rational functions withnonnegative power series coefficients

Consider the following three rational functions:

A(x , y , z) =1

1− 2(x + y + z) + 3(xy + xz + yz)

B(x , y , z) =1

1− x − y − z + 4xyz

C(x , y , z) =1

1− x − y − xz − yz + 4xyz

=1

1− (1 + z)(x + y) + 4xyz

All three are known to have nonnegative coefficients (Szego,Kaluza 1933).

Rational functions withnonnegative power series coefficients

Consider the following three rational functions:

A(x , y , z) =1

1− 2(x + y + z) + 3(xy + xz + yz)

B(x , y , z) =1

1− x − y − z + 4xyz

C(x , y , z) =1

1− x − y − xz − yz + 4xyz

=1

1− (1 + z)(x + y) + 4xyz

All three are known to have nonnegative coefficients (Szego,Kaluza 1933).

How can we prove that they have nonnegative coefficients?

The nonnegativity of A is implied by that of B and thenonnegative of B is implied by that of C:

A(x , y , z) =1

(1− x)(1− y)(1− z)B(

x1− x

,y

1− y,

z1− z

)and

B(x , y , z) =1

1− zC(

x , y ,z

1− z

).

So it suffices to show that the coefficients of C are nonnegative.

But first, there is another way to show that the coefficients ofB(x , y , z) are nonnegative (due to Richard Askey):

How can we prove that they have nonnegative coefficients?

The nonnegativity of A is implied by that of B and thenonnegative of B is implied by that of C:

A(x , y , z) =1

(1− x)(1− y)(1− z)B(

x1− x

,y

1− y,

z1− z

)and

B(x , y , z) =1

1− zC(

x , y ,z

1− z

).

So it suffices to show that the coefficients of C are nonnegative.

But first, there is another way to show that the coefficients ofB(x , y , z) are nonnegative (due to Richard Askey):

How can we prove that they have nonnegative coefficients?

The nonnegativity of A is implied by that of B and thenonnegative of B is implied by that of C:

A(x , y , z) =1

(1− x)(1− y)(1− z)B(

x1− x

,y

1− y,

z1− z

)and

B(x , y , z) =1

1− zC(

x , y ,z

1− z

).

So it suffices to show that the coefficients of C are nonnegative.

But first, there is another way to show that the coefficients ofB(x , y , z) are nonnegative (due to Richard Askey):

How can we prove that they have nonnegative coefficients?

The nonnegativity of A is implied by that of B and thenonnegative of B is implied by that of C:

A(x , y , z) =1

(1− x)(1− y)(1− z)B(

x1− x

,y

1− y,

z1− z

)and

B(x , y , z) =1

1− zC(

x , y ,z

1− z

).

So it suffices to show that the coefficients of C are nonnegative.

But first, there is another way to show that the coefficients ofB(x , y , z) are nonnegative (due to Richard Askey):

Askey’s proof

The coefficients of

D(x , y , z) =

√1− 4xy

1− x − y − z + 4xyz

are nonnegative.

Therefore so are the coefficients of

B(x , y , z) =1√

1− 4xyD(x , y , z).

D(x , y , z) is nonnegative because there is an explicit formulafor its coefficients:

D(x , y , z) =∑a,b,c

S(a,b, c)xaybzc ,

where

S(a,b, c) =

(a− b + c)! (a + b − c − 1)!(a− b − c − 1)!a!b! c!

, if a > b + c

S(b,a, c), if b > a + c

(a− b + c)! (b + c − a)!(c − a− b)!a!b! c!

, if a + b ≤ c

0, otherwise

This follows by equating coefficients in

(1− x − y − z + 4xyz)D(x , y , z) =√

1− 4xy

or from a hypergeometric identity:

We have√1− 4xy

1− x − y − z + 4xyz=∞∑

c=0

(1− 4xy)c+ 12

(1− x − y)c+1 zc .

Using the Pfaff-Saalschütz theorem we obtain

(1− 4xy)γ+12

(1− x − y)γ+1 =∑a,b

(−γ)a+b(γ + 1)a−b

a!b! (−γ)a−bxayb;

we need to take a limit as γ → c.

Here (γ)n = γ(γ + 1) · · · (γ + n − 1), (γ)−n = (−1)n/(1− γ)n.

This follows by equating coefficients in

(1− x − y − z + 4xyz)D(x , y , z) =√

1− 4xy

or from a hypergeometric identity: We have√1− 4xy

1− x − y − z + 4xyz=∞∑

c=0

(1− 4xy)c+ 12

(1− x − y)c+1 zc .

Using the Pfaff-Saalschütz theorem we obtain

(1− 4xy)γ+12

(1− x − y)γ+1 =∑a,b

(−γ)a+b(γ + 1)a−b

a!b! (−γ)a−bxayb;

we need to take a limit as γ → c.

Here (γ)n = γ(γ + 1) · · · (γ + n − 1), (γ)−n = (−1)n/(1− γ)n.

The numbers S(a,b, c) are “super ballot numbers". For c = 0they reduce to the ballot numbers:

S(a,b,0) =a− ba + b

(a + b

a

), for a > b.

Also interesting is the special case of “super Catalan numbers":

T (m,n) = S(m + n,n,m − 1) =12

(2m)! (2n)!m!n! (m + n)!

.

In particular,T (1,n) = Cn =

(2n)!n! (n + 1)!

T (2,n) = 6(2n)!

n! (n + 2)!= 4Cn − Cn+1

Nice combinatorial interpretations of T (2,n) are known but notfor T (m,n) for m > 2.

The numbers S(a,b, c) are “super ballot numbers". For c = 0they reduce to the ballot numbers:

S(a,b,0) =a− ba + b

(a + b

a

), for a > b.

Also interesting is the special case of “super Catalan numbers":

T (m,n) = S(m + n,n,m − 1) =12

(2m)! (2n)!m!n! (m + n)!

.

In particular,T (1,n) = Cn =

(2n)!n! (n + 1)!

T (2,n) = 6(2n)!

n! (n + 2)!= 4Cn − Cn+1

Nice combinatorial interpretations of T (2,n) are known but notfor T (m,n) for m > 2.

A positive compositional inverse

There is an interesting consequence of the positivity of

A(x , y , z) =1

1− 2(x + y + z) + 3(xy + xz + yz).

Theorem. Let f = f (x) be the compositional inverse of thepower series x − (α+ β + γ)x2 + (αβ +αγ + βγ)x3. Then f haspositive coefficients.

Proof. We have f − (α+ β + γ)f 2 + (αβ + αγ + βγ)f 3 = x .Differentiating with respect to f and simplifying gives

f ′ =1

1− 2(α+ β + γ)f + 3(αβ + αγ + βγ)f 2

= A(αf , βf , γf ).

So the positivity of A(x , y , z) implies a recurrence for thecoefficients of f with positive coefficients.

Specializations

In the specialization α = β = γ = 1, f reduces to

1− 3√

1− 9x3

= x + 3x2 + 15x3 + 90x4 + 594x5 + 4158x6 + · · ·

a Catalan number generating function analogue with no knowncombinatorial interpretation.

Also α = 1, β = γ = 0 gives Catalan numbers andα = β = 1, γ = 0 gives the numbers 1

n+1

(3n+1n

).

Specializations

In the specialization α = β = γ = 1, f reduces to

1− 3√

1− 9x3

= x + 3x2 + 15x3 + 90x4 + 594x5 + 4158x6 + · · ·

a Catalan number generating function analogue with no knowncombinatorial interpretation.

Also α = 1, β = γ = 0 gives Catalan numbers andα = β = 1, γ = 0 gives the numbers 1

n+1

(3n+1n

).

Positivity of C(x , y , z)

Why are the coefficients of

C(x , y , z) =1

1− x − y − xz − yz + 4xyz

nonnegative?

They are essentially squares.

More generally, let

E(x , y , z;λ)

=1

1− (1− λ)x − λy − λxz − (1− λ)yz + xyz

=∑i,j,k

α(i , j , k ;λ)x iy jzk .

Note that C(x , y , z) = E(2x ,2y , z;1/2).

Positivity of C(x , y , z)

Why are the coefficients of

C(x , y , z) =1

1− x − y − xz − yz + 4xyz

nonnegative?

They are essentially squares.

More generally, let

E(x , y , z;λ)

=1

1− (1− λ)x − λy − λxz − (1− λ)yz + xyz

=∑i,j,k

α(i , j , k ;λ)x iy jzk .

Note that C(x , y , z) = E(2x ,2y , z;1/2).

Positivity of C(x , y , z)

Why are the coefficients of

C(x , y , z) =1

1− x − y − xz − yz + 4xyz

nonnegative?

They are essentially squares.

More generally, let

E(x , y , z;λ)

=1

1− (1− λ)x − λy − λxz − (1− λ)yz + xyz

=∑i,j,k

α(i , j , k ;λ)x iy jzk .

Note that C(x , y , z) = E(2x ,2y , z;1/2).

Ismail and Tamhankar (1979) showed, using MacMahon’smaster theorem, that

I If i + j < k then α(i , j , k ;λ) = 0.I If i + j ≥ k then

α(i , j , k ;λ) = λ2i+j−k (1− λ)k−i (i + j − k)! k !i! j!

×

[∑m

(im

)(j

k − i + m

)(1− λ−1)m

]2

which is clearly nonnegative for 0 < λ < 1.

A generalization of Ismail and Tamhankar’s result

Let A = (aij) be an m × n matrix. Let r = (r1, . . . , rn) ands = (s1, . . . , sm) be sequences of nonnegative integers, and letr = r1 + · · ·+ rn and s = s1 + · · ·+ sm. We define FA(r,s) andGA(r,s) by

FA(r,s) = [ys11 ys2

2 · · · ysmm ](

1 +m∑

i=1

ai1yi

)r1

· · ·(

1 +m∑

i=1

ainyi

)rn

and for r ≥ s,

GA(r,s) = [x r11 x r2

2 · · · xrnn ]( n∑

j=1

xj

)r−s( n∑j=1

a1jxj

)s1

· · ·( n∑

j=1

amjxj

)sm

.

If r < s then GA(r,s) = FA(r,s) = 0.

It’s easy to show that(r

r1, r2, . . . , rn

)FA(r,s) =

(r

r − s, s1, s2, . . . , sm

)GA(r,s).

Let A = (aij) and B = (bij) be two m × n matrices. Let M be the(n + m)× (n + m) matrix (

J Bt

A 0

),

where J is an n × n matrix of ones and 0 is an m ×m matrix ofzeros, and let Z be the (n + m)× (n + m) diagonal matrix withdiagonal entries x1, . . . , xn, y1, . . . , ym. Then∑

r,s

FA(r,s)GB(r,s)xr11 · · · x

rnn ys1

1 · · · ysmm = 1/det(I − ZM).

Proof: Use MacMahon’s Master Theorem.

So if A = B then each coefficient of 1/det(I − ZM) is a positiveinteger times the square of a polynomial in the aij , and if the aijare real numbers then 1/det(I − ZM) has nonnegativecoefficients.

Let A = (aij) and B = (bij) be two m × n matrices. Let M be the(n + m)× (n + m) matrix (

J Bt

A 0

),

where J is an n × n matrix of ones and 0 is an m ×m matrix ofzeros, and let Z be the (n + m)× (n + m) diagonal matrix withdiagonal entries x1, . . . , xn, y1, . . . , ym. Then∑

r,s

FA(r,s)GB(r,s)xr11 · · · x

rnn ys1

1 · · · ysmm = 1/det(I − ZM).

Proof: Use MacMahon’s Master Theorem.

So if A = B then each coefficient of 1/det(I − ZM) is a positiveinteger times the square of a polynomial in the aij , and if the aijare real numbers then 1/det(I − ZM) has nonnegativecoefficients.

Let A = (aij) and B = (bij) be two m × n matrices. Let M be the(n + m)× (n + m) matrix (

J Bt

A 0

),

where J is an n × n matrix of ones and 0 is an m ×m matrix ofzeros, and let Z be the (n + m)× (n + m) diagonal matrix withdiagonal entries x1, . . . , xn, y1, . . . , ym. Then∑

r,s

FA(r,s)GB(r,s)xr11 · · · x

rnn ys1

1 · · · ysmm = 1/det(I − ZM).

Proof: Use MacMahon’s Master Theorem.

So if A = B then each coefficient of 1/det(I − ZM) is a positiveinteger times the square of a polynomial in the aij , and if the aijare real numbers then 1/det(I − ZM) has nonnegativecoefficients.

Example

Take A = B =[a b

], and write x for x1, y for x2, and z for y1.

Then the matrix is

I −

x 0 00 y 00 0 z

1 1 a1 1 ba b 0

=

1− x −x −ax−y 1− y −by−az −bz 1

with determinant

1− x − y − a2xz − b2yz + (a− b)2xyz.

Thus the coefficients of

11− x − y − a2xz − b2yz + (a− b)2xyz

are positive integers times squares of polynomials in a and b.

This is really the same as Ismail and Tamhankar’s result:replace x with (1− λ)x and y with λy and set a =

√λ/(1− λ)

and b = −√

(1− λ)/λ.

Setting a = 1,b = −1 gives

C(x , y , z) =1

1− x − y − xz − yz + 4xyz.

Unfortunately, in other cases there does not seem to be ananalogous nice specialization so the general theorem doesn’tseem to give generalizations of the rational functions A(x , y , z)and B(x , y , z).

This is really the same as Ismail and Tamhankar’s result:replace x with (1− λ)x and y with λy and set a =

√λ/(1− λ)

and b = −√

(1− λ)/λ.

Setting a = 1,b = −1 gives

C(x , y , z) =1

1− x − y − xz − yz + 4xyz.

Unfortunately, in other cases there does not seem to be ananalogous nice specialization so the general theorem doesn’tseem to give generalizations of the rational functions A(x , y , z)and B(x , y , z).

This is really the same as Ismail and Tamhankar’s result:replace x with (1− λ)x and y with λy and set a =

√λ/(1− λ)

and b = −√

(1− λ)/λ.

Setting a = 1,b = −1 gives

C(x , y , z) =1

1− x − y − xz − yz + 4xyz.

Unfortunately, in other cases there does not seem to be ananalogous nice specialization so the general theorem doesn’tseem to give generalizations of the rational functions A(x , y , z)and B(x , y , z).

A product generating functionWe find also that (writing i for r1, j for r2, and k for s1), we havemore generally

11− x − y − a1b1xz − a2b2yz + (a1 − a2)(b1 − b2)xyz

=∑i,j,k

Fi,j,k (a1,a2)Gi,j,k (b1,b2)x iy jzk ,

whereFi,j,k (a1,a2) =

∑m

(im

)(j

k −m

)am

1 ak−m2

and

Gi,j,k (b1,b2) =∑

m

(km

)(i + j − k

i −m

)bm

1 bk−m2 .

AlsoFi,j,k (a1,a2)/Gi,j,k (a1,a2) =

i! j!k ! (i + j − k)!

.

The coefficients on the left side can be written as a double sum.

The identity we obtain by equating this double sum with theright side is equivalent to a formula of Watson:

(β − α)n

(β)n

∑i,j

(−n)i+j(α)i+j

i! j! (β)i(1− n − β + α)j(xy)i((1− x)(1− y)

)j

= 2F1

(−n, αβ

∣∣∣ x)

2F1

(−n, αβ

∣∣∣ y)

(The left side is Appell’s F4.)

The coefficients on the left side can be written as a double sum.

The identity we obtain by equating this double sum with theright side is equivalent to a formula of Watson:

(β − α)n

(β)n

∑i,j

(−n)i+j(α)i+j

i! j! (β)i(1− n − β + α)j(xy)i((1− x)(1− y)

)j

= 2F1

(−n, αβ

∣∣∣ x)

2F1

(−n, αβ

∣∣∣ y)

(The left side is Appell’s F4.)

A conjecture

We may write

B(x1, x2, x3) =1

1− x1 − x2 − x3 + 4x1x2x3

=1

1− e1 + 4e3,

where en is the nth elementary symmetric function in x1, x2, x3.

In four variables, (1− e1 + 4e3)−1 does not have nonnegative

coefficients, but Koornwinder (1978), and later Gillis, Reznick,and Zeilberger (1983), proved that in four variables(1− e1 + 4e3 − 16e4)

−1 has nonnegative coefficients.

Can we extend this to more variables?

A conjecture

We may write

B(x1, x2, x3) =1

1− x1 − x2 − x3 + 4x1x2x3

=1

1− e1 + 4e3,

where en is the nth elementary symmetric function in x1, x2, x3.

In four variables, (1− e1 + 4e3)−1 does not have nonnegative

coefficients, but Koornwinder (1978), and later Gillis, Reznick,and Zeilberger (1983), proved that in four variables(1− e1 + 4e3 − 16e4)

−1 has nonnegative coefficients.

Can we extend this to more variables?

A conjecture

We may write

B(x1, x2, x3) =1

1− x1 − x2 − x3 + 4x1x2x3

=1

1− e1 + 4e3,

where en is the nth elementary symmetric function in x1, x2, x3.

In four variables, (1− e1 + 4e3)−1 does not have nonnegative

coefficients, but Koornwinder (1978), and later Gillis, Reznick,and Zeilberger (1983), proved that in four variables(1− e1 + 4e3 − 16e4)

−1 has nonnegative coefficients.

Can we extend this to more variables?

Conjecture.1

1− e1 + 4e3 − 16e4

has nonnegative coefficients in infinitely many variables.

Checked for terms up to degree 21 (using John Stembridge’sSF package for Maple).

Conjecture.1

1− e1 + 4e3 − 16e4

has nonnegative coefficients in infinitely many variables.

Checked for terms up to degree 21 (using John Stembridge’sSF package for Maple).