RD-A149 384 ON THE SPANS OF POLYNOMIALS AND THE SPANS OF A 1/1LAGUERRE-POLYA-SCHUR SEGUE..(U) WISCONSIN UNIV-MRDISONMATHEMATICS RESEARCH CENTER I J SCHOENBERG OCT 84

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UNIVERSITY OF WISCONSIN - MADISONMATHEMATICS RESEARCH CENTER

ON THE SPANS OF POLYNOMIALS AND THE SPANS OF ALAGUERRE-POLYA-SCHUR SEQUENCE OF POLYNOMIALS

I. J. Schoenberg

Technical Summary Report #2757

October 1984 0

ABSTRACT

If P(x) = c(x-x1 ) *.° (x-xn ) is a real polynomial having only real

zeros such that xn _ xn_ 1 __<, X1 xn < x1 (n 1 3), then following

R. M. Robinson, A. Meir and A. Sharma, we call S(P) x- xn the span of

P(x). Meir and Sharma conjectured in 1969 that

S(p(k)) > S(P)((n-k)(n-k-1)jI 2 for k 1,...,n-2ffi n(n-1 ) "

with equality only for polynomials of the form

x +x n-2P(x) c(x-x )(x - (x-x

1 2 n

We introduce the quantity NS(P) = S(P)//n(n-1) and call it the normalized(k)span of P(x), using the notation N NS(p(k)), (k - 0,...,n-2). We thenn

prove in Theorem 1 the Meir-Sharma conjecture in the form

N(0) (1 (n-2

< N < < N(n-2) (n > 4)n = n - - n .

We also show that these inequalities may assume only one of two forms

described in Theorem 1.

In 1§3 to 6 we apply these results to the infinite sequences of

polynomials introduced by Polya and Schur in a famous paper from 1914.

AMS (MOS) Subject Classifications: 12D10, 30D15

Yey Words: Polynomials bavinq only real zeros, Polya-Schur sequences ofpolynomials

Work Unit Number 3 - Numerical Analysis and Scientific Computing

Sponsored i-, the United States Army under Contract No. DAAG29-80-C-0041.

. .. o .. "

SIGNIFICANCE AND EXPLANATION

If P(x) =c(x-x 1 ) *e x-xn) is a real polynomial such that

xn xn- :50*! x1, xn < x1, (n 1 2), then following R. M. Robinson, A. Meir

and A. Sharm'a, we call S(P) = x,- the span of P(x). -We prove:Fa

conjecture of Meir and Sharma from 1969 determining the least value of the

*span of-W4he derivative P'(x) for n 3, if xy and Y, are kept fixed.

-7 '-

Tools used are the Descartes rule of signs and the inequality A,4 H between

the arithmetic and harmonic mean. we also apply-.tt results to the infinite

sequences of polynomials introduced by G. Polya and I. Schur in a famous paper

from 1914. A Polya-Schur sequence of polynomials is of the form

POWx = 1, Pi(X)g***,Pn(x)i***

_x x+(n xn-I +( n-2with Pn~x 1 ~n+ )a 1 1 ()a 2x . an, such that all polynoutials

have only real zeros and satisfying

PA(x) = fPn...(x) for n =1,2,...

Distribution/ _S

Availability CodesAvail and/or

Dist special

The responsibility for the wording and views expressed in this descriptivesumm~ary lies with M RC, and not with the author of this report.

L

-. . . . . . . . .. . . . . . . . . . .. . . . . . . . . . . . .

I. J. Schoenberg

P(x) =c(x-xl) ... (x-xn.)

is a polynomial having only real zeros such that

(11)xn xn-l X1 , Xn < x1 , (n 2)

then, following R. 14. Robinson, A. Meir and A. Sharma, we call

(1.2) S(P) xl- Xyn

the span of P(x). Clearly, the sequence of the spans of P(x) and its derivatives

(1.3) 5 (P(k)) ( k 0 ,1,...,n-2)

form a non-increasing sequence.

In Ell Meir and Sharma consider the special case of the polynomials of the form

x +x n-2(1.4) P~x) - c(x-x c)rx -- -- 1 (x-xn) , (n 1 2, xn < X,

We call it the class of centered polynomials and denote it by the symbol .we also sayn

that P(x) in centered at the Point (xl+xn)/2.

The sequence of spans (1.3) for centered polynomials is easily determined. Without

loas of cenerality we may assume that xl 1, x = -1, and therefore

(15)P(x) - l)xn-2 (x+1) = n -xn-2

Their derivatives are

=~k fl(fl1) ... (flk+1,Xn k -(n-2)(n-3) ... (nk)xnk-2

hmence

(1.6) p(k)(x) n1 -X (n-k)(n-k-l))xn-k-2, k 0..n2(n-k)! n(n-1) x k=O..-2

This is a centered polynomial having the span

(1.71 S(p(k)) - 2 (n-k)(n-k-1)_ =Sp) /(n-k)(n,-k-1)n(n-1) VF n(-n-1)

This equality for centered polynomials suggested to Meir and Sharma the conjecture that for

Sponsored by the United States Army under Contract No. DAAG29-80-C.-0041.

%7.

.. ,,- ; .- .- - ,-. .. . , . .. . . -. -. -..--. .. ... . .. ... .- , .- .-. ' . -,, . . -_ ; :_. -- - "-.-'' - - - -

any P(x) of degree n 2 3, having only real zeros, we have the inequality

(.)S(P(k)) S (p) /(n-k)(n-k-1) for k 1,...,n-2 Sn(n-1)

with the equality sign if and only if P(x) is centered. This they prove in (1, page 530]

for the largest value of kc = n-2 only. The main point, however, is to establish (1.8)

for the least value of k = 1, for then the result so obtained can he applied successively

to P, P",...,p~n-2) I proving the complete conjecture. This is what we are going to do.

The inequality (1.8) suggests normalization of the span S(P): w'e call the quantity

(1.9) NS(P) S(P)

/n(n-1)

the normalized span of P(x), and to simplify notations we write

(1.10) NnO = NS(P), N (k) = NS(p(k))n n

We m~ay now state the following theorem whose second part goes beyond the Meir-Sharma

conjecture.

Th~rr~ 1. 1. If P(x) is a polynomial of degree n whose zeros satisfy (1.1),

*then for the normalized spans of its derivatives we have that

) < NM < ... < N ( )< N(n 2 ) .>n = n = =n =n(n )

2. These inequalities can assume only one of the two forms: Either

(1.12) NI (0) = (1 7n-1 1

f r (n -

*(1.13) F (0 N1Nn n

The equations (1.13) hold only if P(x) is centered, i.e. of the form (1.4). if P(x) is

*not centered, then (1.12) holds. In (1.12) we have the equality sign in the last

inequality, if and only if P if )x) is centered, hence is a cubic having its three zeros

* in an arithmetic progression.

The inequalities (1.11) are equivalent with the Meir-Sharma conjecture, while (1.12),

o t(1.13), go beyond it; they will he derived below by oescartes rule of signs.

Proof of the inequalities (1.11). It suffices to prove the first one

*(1.14) N S(P) ()- -(0) (n-)

(1.14n°)

n Nn (n x 3)

as we simply iterate it to "et (1.11).

-2-

-. 9 r. rerw r

We are also to show that if n 3 then

n n

if and only if P(x) is centered.

Let

(1.16) Y n-2 yj be the zeros of P'(x)

The inequality (1.14) then amounts explicitly to

.'nIn-1n v'(-1)n-2)

or

(1.17) 1i Yn-1 n 31 -Xn

To prove this vs also need the new polynomial

(1.18) P.(x) -(x-x 1 n(~~n(~

wherex+X +...+x

(1.19)2 3 n-I(1.19)n-2

We will also show that in (1.17) we have the equality sign if and only if P*(x) is

centered, hence whenx +x

(1.20) 1 2

We also need the derivative of p*(x), and let

n-3

(1.21) o- < F . ... E < n1 be the zeros of P,(x)

The inequaliy (1.17) will be established as soon as we prove the two inequalities

(1.22) yI -I n i rn I

and

(1.23) '1 - n-I /'n (xl-xn)

we observe first that

(1.24) Yn-I < yI

for if yn1=y, their comm'on value being - 0, say, then P'(x) =c*x , an

therefore P(x) - c( Xn - xn)/nV which can not have only real zeros, as we assume that

n 3.

We also assume to start with that

-3-

% %.

which by Rolle's theorem implies that

(1.26) Yi- > 0

our next objective is to find an explicit expression for the quadratic polynomial

(xfl)(x-ln-1), which will turn out to be in terms of x1 , xn, and r* Indeed, by

differentiation of (1.18) we find

P4(x) = (xE )Xn) + (n-2)(x-xl)(x-xn) + (x-xl)(x-r)l

showing that r1and T)n1are the zeros of the second factor. Tnis is the quadratic

Q(x) =n(x-n 1 )(X-nn-I)

=(x=E)(x-x n + (n-2)(x-x I)(x-xn + (x-x )(x-F)(1.27)n

2 X1 +Xn=nx -2((n-1) 2- + r)x + (n-2)x x + (x I+X )

*The discrimninant h2 -ac of 0(x) = ax2 + 2bx + c is easily evaluated and we find for

difference n~ n- = i!b-ac the expression

(1.28) ~ 1 n n-1 = /3 (xl-xn) + 222n

If we remove the second non-negative term under the square root sign we already prove

the inequality (1.23), hence

(1.29) 1 -n _ x-xI n-1 n n

Proof of (1.22). We will first prove that

(1.30)n, Y

From the identity

P'(x) n(1.31)Px) 1xx

and P*(y,) =0, we obtain

n-i11 1 =0.(1.32)

Here we sh-ould first observe that our assumption (1.25) implies (1.26) and therefore that

6.all cenominators in (1.32) are positive. For the special case of the polynomial P.(x) -

-4-

. . . .. . . . .

. . . . . . . . . . .

r.

defined by (1.18) and (1.19), the equation (1.32) becomes

1 n-2 i 0SXl-n I r 1-C n-_X n

We shall prove now if we replace in (1.33) the quantity n, by Y1 , the left side of

(1.33) becomes non-negative, hence

(1.34) 1 n-2 I > 0xl-Y 1 yl-( Yl-Xn =

Indeed, observe that the arithmetic mean A of the n - 2 positive quantities

(1.35) yl x2 , Yl - ... - xn-1

is

(1.36) A y - '

while their harmonic mean H is given by

n-1 I n-2(1.37) -1 .

2 Y 1 Xi H

However, by the classical inequality

H, A(1.3A) H A ,-

and (1.36) we derive from (1.37) that"i n-1-.

n I1 n-2(139 Y-X >l

2 yI-xi

Therefore (1.32) shows that(1.40) 1 1 n-2

X1 -Y1 Y1=xn = yl-.

I claim that (1.40) shows that (1.30) holds: From (1.27) we have the identity in x

(1.41) 1 n-2 ,(x)X-X x- x-x n (xl-x)(x-)(x-xn )

If we set here x yl we obtain from (1.40) that

I n-2 1 __ __)0

xI-yl yl- [ yl"x (x 1 -Yl)(yl-&)(ylx) 0

and therefore

(1.42) (y 1 a 0

Since Q(x) c(x-rI)(x-nn- 1 1 and by Rolle [ x2 < yl < x1 , we conclude that

(1.43) Yl A nl

On replacing x by -x we similarly find that

; .' ..; .•. ... ... ,.. . -.. . -. . . . .. . . . ... . . . . .-. . . .-..-.. .- ., . . . .. ... .. . . . . . ... .. -. • , . . . -. . -. % % . % o •% % " ' ." J... ........ .-.. ... ... .-.. ... .' .. .... ..- '% ."" " " "-' "-. -' " °.J

-''" . ,' "- i . "":/"" '" "-".''. . ''"'"",".'"""."-" "'"."''.'."'- .""""."".' """ """ ". ." -"."' ." . "'"". .

(1.44) Yn-1 n-1-

Now (1.43) and (1.44) show that the inequality (1.22) holds. 5

Having established (1.22) and (1.23), it is clear the our main inequality (1.17) is

also proved.

There remains the question: When do we have the equality sign in (1.17)?

Equality in (1.17) implies equality in both (1.22) and (1.23). However, we know that .

equality in (1.22) implies that the terms of the sequence (1.35) are all equal to eac'

other, and so by (1.19) we have

(1.45) x2 = =...= x. 1 =

Moreover, equality in (1.23) implies that the second term under the square root sign in

(1.28) must vanish so that

xl+xn(1.46) 1 =.2

Now (1.45) shows that P(x) = P.(x), while (1.46) shows that P(x) is a centered P

polynomial

x1 +x n-2p(x) = (X-X1 )(x ----- ) (x-x)2 n

Can we eliminate the additional assumption (1.25)? The answer is affirmative: The Ix1, x2,...,x n are independent, and all the other quantities, such as yi., nil n 2

depend continuously on them. Therefore the string of inequalities (1.11), conjectured by

Meir and Sharma, is estahlished.

2- .:

We assume first that

(2.1) P(x) e TnWe then know from (1.5) implying (1.7), the equations (1.13) must hold.

We now assume that

(2.2) P(x) In"

I claim that the Inequalities (1.12) must hold.

PWe distinguish two cases:

-6-

. . . . . . . . . . . -'- - . . . .,-

.:-°*x7" Y- %

1. If n 4 we have

(2.3) N() 1 2

with the first being "less than" sign, because by (2.2) P(x) is not centered.

2. If n 5 then none of the n - 4 22lnomals4

(2.4) P'(x), Pn(X),...,P n4 (x) L

can be centered, so that we have

(2.5) N (0 ) N (1 ) . < Nn nn n n

Proof: Let us assume that one of the polynomials (2.4) is centered, e.g.

P(k)(x) e 1 ni where

(2.6) 1 k n-4

and let us reach a contradiction to our assumption (2.2). We may assume by a change of

scale that

p(k)(x) -(X-I)xn-k- 2 (x+1) xn-k - n-k-2

and therefore, by integration, that

(2.7) p(k-i)(x) = .... ,n-k+1 xn-k-1 + Cn-k+i n-k-i

i and let us show that necessarily

(2.8) C 0

Indeed, let us assume that C > 0, when (2.7) shows, by Descartes rule of signs, that for

* the number of zeros in (0,-) of P(k-1) we have

(2.9) z(0,-) £2

Moreover, fromn-k+1 n-k-i

(2.10) ~ ( k-i) (-x) - -) n-k+i (-1) rn-k-i 1n-k+i n-k-i

we conclude that

(2 if n -k is odd

1. if n -k is even

In any case we have that

(2.12) z-) 4

while (2.7), having only real zeros, should have at least, by

n-'k+i n-(n-4)+i 5

-7-

five zeros, in contradiction with (2.12). d

Likewise, if C < 0 we have

z(0,-) 1

and from (2.10)

if n - k is odd

2if n -k is even

hence again Z(-,-) 4, leading to the same contradiction. The conclusion, that

n-k)(X implies that Pe n-k+1'e may be iterated until we reach the

conclusion that P(x) e n n in contradiction to our assumption (2.2).

Let

(3.1) F(z) -- (. s0 nI ,( 0 =1

be a formal real power series. Multiplying by ex we obtain

P (X)n n

(3.2) F~z)ext = I -

which generates a so-called Appell sequence of polynomials

(33 ~(x) _ xn + (n)5 ,n- +.. a (P0 (x) =1, n =1,2....

* having the property that

(3.4) PA(x) -n Pn-l(x) jo

A fundamental theorem of Polya and Schur from 1914 (21 states that

(3.5) all PnCx) have only real zeros

if and only if the function F'(z) is entire and belongs to the class &2of entire

functions

2 -

(3.6) F(z) e _Y +A'Z ( + z zek.1

such that

(3.7) Y > 0, 0 <Y + A2 <-k

- A."

Here we have excluded the trivial case when y 0 and all k 0, when

Pn(x) - (x+A)n for all n.

What can we say about the normalized spans of the polynomials Pn(x)? We recall from

(1.4) that we denote by Tn the class of centered polynomialsnnX1 +X n-2 "'-

(3.8) P(x) = c(X-X1 )(x --- -- ) (x-x) (n 3, Xn < xj) "

From our Theorem 1 we can easily derive our

jb&2j 2. Let F(z) e &2 . For the sequence of normalized spans

(3.9) NS(Pn) (n = 2,3,...

we have one of the following two alternatives: Fither

(3.10) NS(P2 ) I NS(P 3 ) > NS(P4 ) >...> NS(P n ) >...

or else

(3.11) NS(P 2 ) - NS(P 3 ) -...= NS(Pn) -...I

The second alternative (3.11) holds only if P4 (x) e 14' which implies that

(3.12) Pnlx) e I for n = 3,4,...n

In (3.10) we have the e!Mality sign among the first two terms if and only if P3 (x) e 3'-

which means that P3 (x) is a cubic having Its zeros in arithmetic progression.

Proof: 1. Let us consider Pn(x) for some n 4. If Pn(x) e "n' then we know,

by (3.4), and Theorem 1 that

NS(P2)o= NS(P 3 ) ..... NS(Pn)

On the other hand we know from (2.6), (2.7) and (2.8), that all integrals of P4(x),

having only real zeros, are centered. This qiv-s the remaining eauations NS(P n ) =

NS(Pn+i) *...

2. If

(3.13) P4 (x) TT4

then, by Theorem 1, we have NS(P 3 ) > NS(P4 ). We also know that no integral of P4 (x)

having only real zeros, can be centered, Hence (3.13) implies that Pn (x) 7n n if n > 4.

This shows that NS(P4 ) > NS(P 5 ) >.... Finally, we have NS(P2 ) = NS(P 3 ) or

NS(P 2 ) > NS(P 3 ), depending on whether P3 (x) is centered or not.

"-i:;-~~~~ ~~~~~.......--.. ".. ......'- ........ :.. .. . .. .. .... -... .. . .-. - . . ..... .. .... . •

Let Hn(x) be the Hermite polynomials as defined by Szeg6 [3, page 105]:

* 3.14) H n(x) = 2xHnl(x) - 2(n-l)Hn_ 2 (x), for n 2,3,... "

.itn H0)x1 = 1, HI(x) = 2x. Also H2 (x) = 4x2 - 2 and H3 (x) = x - 12x. We see that

* 'x is centered, while Hn(k) can not be centered if n 4, because (H (x)I

form ann .'.

-qr-n1 ,naI system of polynomials and can therefore not have multiple zeros. It follows

wdt -e K'Ave

NS(H 2 ) = NS(H 3 ) > HS(H 4 ) >...> NS(Hn) >..

"'Wat is t'e limit of the decreasing sequence (3.15) of positive numbers? We can easily

SiO that

Inli NS(Hn) = 0

n +0

using an est.mate of the largest zero of Hn(x) found in SzegB [3, page 119, (6.2.18)].

Szeq6 shows that the largest zero of Hn(x) does not exceed d = (/ (n-1)/ln+2, which

n

implies that

2NS(Hn) 2 _ d + 0

/n(n-1)n

4. On the vanishinq of consecutive coefficients of F(z).

We wish to prove the

"heorem 3. If in the expansion (3.1) of the function (3.6) we have

f4.1) a, $ 0 (m • 1)

(4.2) a+1= 0, am+2= 0

an 0 for all n > m

Pro,,f: We shall use the property of the Descartes rule of signs to the effect that

ra c]ynomial havinq only real zeros the number w of variations of sign of its

,cefticients is equal to the number of its positive zeros. We consider the three poly-

I

.~~~~~ .. . . . .

-10--

. . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . ...-

P (x) = + (In)ax m .I a 0

m+2 M+2 + Im+2 2P (x) x +. )alx +...+( )ax + am42 1 1 in mn 3

having only real zero. The next polynomial is

(4.5) Pm+3(x) - xm+3 + "m+3 )ax2 . m+3 )x m M m+3

We consider the numbers of variations of sign

(4.6) w . v(1,aj .... ,am), w' = ( -) , -1 -l ,...,pam)

By our introductory remark we have that for P3 (x)

(4.7) Zm(0,-) = w, Zm(- ,0) = w', and therefore

(4.8) W + w' mS

We now consider P3+3(x). If we assume for the moment that

(4.9) am+3 > 0

then

Z .m+3(0,)

= w, Zm+3 (--,0) " w' + 1, if am > 0

and

Zm+3(0,-) -w + 1, Zm+3(-,0) w', if a. < 0

From (4.9) and (4.8), it follows that

(4.10) Z.4 3 (-@,) - w + w' + 1 = i + 1 ,

in contradiction with the reality of all zeros of PIm+3•

We get the same contradiction (4.10) if & < 0. Our conclusion is that :: 0.

This proves Theorem 3 if we apply the above reasoning to Pm+k, if am+k should be

the first non-vanishing coefficient. S

This shows that the assumptions (4.1) and (4.2) imply that F(z) reduces to a poly-

nomial. This is what we assume in our next section.

S -.

-11-

0, -'o

0o ~

-I- """q "

(5.1 F~)=1 +a 1 aF~~~z) ~ + ifz - M (am 0 )

and therefore

(5.2) P (X) = + ..a.' 1 + (l if n

We raise the question: Can we decide which case of the dichotomy (3.10) or (3.11) of

Theorem 2 applies to the normalized spans of the polynomials (5.2)?

We first observe that without loss of generality we may assume that a, 0. From

(3.6) and (3.1) we find that

2 k

However, the effect of the value of 4 is trivial: If we multiply (3.2) by e-A we

obtain

(X-A )z pn(A Zn

0 n

*showing that the factor e in (3.6) merely translates the x-axis by the amount 5

However, the property of a polynomial being centered is invariant if we replace x by

x - A. for this reason we shall assume that

*(5.4) A 0 and by (5.3), that a1, 0,

wh.:) a2 . -(2y + A 2 ) <0

We may now easily answer the question concerning the dichotomy between (1.10) and

(1.11) by provingi

4. For the case when (5.1) holds we have for the normalized spans of the

polynomials (5.2), that (1.11) holds. Fence

*(5.6) NS(p2 ) NS(P3 ) =.* NS(Pn)

if and only if in (5.1) we have m -2, hence

*(5.7) F(z) 1 + 1'r2 (a2 < 0)

-12-

l~o "

-°el .

If m 3 we have the inequalities (3.10), hence

(5.8) NS(P2 ) a NS(P3 ) > NS(P4 ) >..'> NS(Pn) )...

Proof: By Theorem 2 we know that (3.10) holds, hence (5.8), if and only if

(5.9) P4 (x) - x + 6a2x + 4a3x + a4

is not centered. From a1 - 0 it is clear that if P4 is centered, it can only he

centered at point x - 0. S

If a4 0 0, the P4 (x) has no vanishing zero, and can therefore not be centered at

0. If a4 = 0, hence a3 0 0, then

P4 (x) - (x3

+ 6a2x + a3 )x

The cubic factor has three real zeros all 0 0, and so P4 (x) can not be centered at 0,

where it should have a double zero. Therefore P4 (x) if not centered, and Theorem 4 is

established.

The last question to be discussed is as follows.

We may and do assume that

(6.1) a1 = 0

This implies that the zeros of(62 m + n) xm-2 . (n) a ) (m a n)

(6.2) Pn(x) - x x 2 a-2 m * ."

have the sum 0, it follows that the noint 0 is between the extreme zeros of Pn(x).

Therefore the span of Pn(x) does not change if we omit the factor xn-m. We therefore

have

n m-2 nS(Pn(x)) - S(Xm + () s2x +...+ ) am

If we now replace x by nx in the polynomial on the right side, then its zeros shrink by

the factor 1/n, and its span is similarly contracted. Therefore we restore the equation

by multiplying by n, obtaining

S(Pn(x)) - nS(n"xm

+ nm- 2

(") a2 xm- 2

...+ n%-m

(n) am) t

-13-

%.%.

*. . . . . . . . . .. . . . . . .

:-' . ".,-.". .... "..",'.. . . .. ... . ...- •...-• . :. -..--. "-

r-. 1

Dividing the riqht side polynomial by nm, we obtain

1•( 1- 11 ( 1 ). .( - n-ra+ )

n m-2 n nS(Pn(x)) - nS(x' + 21 a x +...+ am)21t 2 MaI'.,.

Dividing this equation by /n(n-1) and letting n * we obtain

5. For the generating function (5.1), satisfying (6.1) we have that

a a(6.3) lir NS(Pn(x)) - S(x m + 2 +...+ -M) .

21 mIn+-m

By (5.1) this limit can also be expressed as

(6.4) lim NS(Pn(x)) = S(x F(-))

A n For (5.1) we choose m = and2 2) 5z 4"

(6.5) z = (1 - 1--) 5 2 -2 22 1 4

On combining with (5.1) we find a1 = 0, a2 = -5/2, a3 = 0, a4 - 6. For the right side of

(6.4) we therefore have

(6.6) S(x4F(1/x)) = Sf(x -)(x - l) 22 212

as the extreme zeros are -1 and +1.

Let us find the largest zero x(n) of (6.2), hence of(6.7) Pn(x) - (x

4- (n) I x2 + (n) 6)Xn-4

2 2 4

From (6.4) and (6.6) we obtain after canceling the factor 2 on both sides

(n)lrn x I/Sn(n-1)= 1 ,n=-1

hence

(6.8) x1 n) n

Finding the largest zero of (6.7) for n - 7,8,9,10 we obtain the table

nI 8 9 10

x 6.93812 7.96069 8.98372 9.99771

which illustrates well the relation (6.8), or x n)/n + 1.

-14- " -I

....

I T -7

I. A. Meir and A. Sharma, Span of derivatives of polynomials, Amer. Math. Monthly, 74 *.-

(1967), 527-531.

2. G. Polya and 1. Schur, t;ber zwei Arten von Faklorenfolgen in der Theorie der

algebraischen Gleichungen. J. fUr die reine und angev. Math., 144 (1914), 89-113.

3. G. Szeg6, Orthogonal polynomials, Amer. Math. Soc., revised edition, 1959.

IJS/Jvs

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IS. SUPPLEMENTARY NOTES

Is. KEY WORDS (Continue an reverse side It n0466441 ad identify' by block nusmher)

Polynomials having only real zeros, Polya-Schur sequences of polynomials

20. AVISTRACT (Continue an reverse side 1 necessuary and Identaify by block nustbr)

If P(X) = c(x-x) ... (x-x) is a real polynomial having only real zeros

*such that x < x (n > 3), then following R. M. Robinson,n = n- -***-' xl n x

A. Meir and A. Sharma, we call S(P) =x - x the span of P(x). Meir and

* Sharma conjectured in 1969 that

(k) (n-k) (n-k-l) fo1/2..,-S(P )>S(P)( n(n-1)fo k l..,-

00 1 'SNT 1473 EDITIONp OF I Nov 419 is oesoLETe UNCLASSIFIED (continued)

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ABSTRACT (continued)

with equality only for polynomials of the form

SXl+xnn2 nPx) - c(x-x1 ) x - (x-x)

We introduce the quantity NS(P) S(P)//n(n-l) and call it the normalized spanak) (k)'-of P(x), using the notation N NS(Plk) (k - 0,...,n-2). We then prove in

nTheorem 1 the Meir-Sharma conjecture in the form

N(0) (1) (n-2)N < N < .. < N (n > 4)n n = - n '

We also show that these inequalities may assume only one of two forms describedin Theorem 1.

* In §§3 to 6 we apply these results to the infinite sequences of polynomialsintroduced by Polya and Schur in a famous paper from 1914.

I.

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