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New bounds on the Hermite polynomials

a r X i v :m a t h /0401310v 1 [m a t h .C A ] 23 J a n 2004

NEW BOUNDS ON THE HERMITE POLYNOMIALS

ILIA KRASIKOV

Abstract.We shall establish two-side explicit inequalities,which are asymp-totically sharp up to a constant factor,on the maximum value of |H k (x )|e ?x 2

/2,on the real axis,where H k are the Hermite polynomials.

1.Introduction

We refer to [7]for the required de?nitions and basic properties of the Hermite polynomials H k (x )mentioned in the sequel.There are a few known upper bounds

on the function (H k (x ))2e ?x 2

,see e.g.[1],mainly obtained as a specialization of a more general case of the Laguerre polynomials.Some recent results can be found in [6].However,it seems that all presently known inequalities are larger by the factor k 1/6than the true asymptotic value.The aim of this note is to establish the following two-side bounds,which are sharp up to constant factors.Theorem 1.Let

M k =(2k )

1/6

max x

(H k (x ))2e ?x 2

then for k ≥6,

C k exp

4(2k )1/3?9

where

C k =2k

√

16k 2?16k +6k !(k ?1)!2k ?1((k ?1)/2)!2

for k odd.

The constants are not,of course,best possible and can be improved at the cost of more extensive calculations.

The asymptotic of the Hermite polynomials in the transition region,where the max-imum of the function (H k (x ))2e ?x 2

is attained,is given by the following classical formula [7].For x =

√2k !k ?1/6 A (z )+O (k ?2/3) 2

.Here A (z )is the Airy function which can be de?ned by means of the Bessel functions

J νof the ?rst kind,

A (z )=πξ

(2ξ3)+J 11991Mathematics Subject Classi?cation.33C45.

2I.KRASIKOV

whereξ= 3.

The absolute maximum of A(z)is attained for z=1.46935...,and is equal to 1.1668....Therefore,asymptotically,the optimal x is

√

C k

=0.715452...

Thus,for su?ciently large k,our upper bound(2

dx H k(x)e?x2/2=0,just means t(x)=x.A quick inspection of the graph of

the function t(x),consisting of k+1decreasing branches,and the corresponding straight line reveals that their last intersection occurs for some x=ω>x kk.Thus, the absolute maximum of the function(H k(x))2e?x2is attained for x=ω.

We will deduce Theorem1from the following bounds established in[3]. Theorem2.For x2<2k?3

2y(2y?3)2

Moreover the inequality is sharp in a sense that

(H k(x)2e?x2≥C k F k(y)/G k(y),

for all the roots of the equation

(2)xy(2y?3)H k(x)=(2y2?4y+3)H k?1(x),

that is at a point between any two consecutive zeros of H k.

We need the following technical result.

POLYNOMIALS3 Lemma3.For k≥2,the function v1(y)=F k(y)G k(y)decreases in y for y≥2. The function v2(y)=F k(y)/G k(y)decreases in y for y≥3(2k)1/3.

Proof.It is easy to check by the substitution y:=y+2,that for y≥2,the denominator of F k(y)is positive.Now,v1is a decreasing function since for k≥2,y≥2,

G?2

k (y)

dv1

y3(2y?3)3(4y4?12y3+9y2+10ky?12k)2<0,

where A(k,y)is a polynomial with deg k A=2,deg y A=10.Indeed one can check that A(k,y)>0,as it is transformed into a polynomial with only positive terms by the substitutions k:=k+2,y:=y+2.

Similarly,for v2we have

G2k(y)dv2

y3(2y?3)3(4y4?12y3+9y2+10ky?12k)2<0,

where B(k,y)is a polynomial with deg k B=2,deg y B=10.It is left to check that the substitution

y:=y+3(2k)1/3,k:=k+2,

transforms B(k,y)it into a polynomial with only positive terms.We omit the details.

Let g=g(x)be a real polynomial with only real zeros x1,...,x k.The following inequality is called the Laguerre inequality,

(3)g′2?gg′′

(x?x i)2

>0,

and will be our main technical tool in this note.It is worth noticing that Theorem 2was established by applying a higher order generalization of(3).

To simplify the formulas in the sequel we will use the substitution k=27m12?1

k2+1

)3/2m?3.

Proof.We put f(x)=H k(x),g(x)=f(x)?f′(x)/q,where q is a parameter independent on x.Observe that g(x)has only real zeros as well.Therefore,by(3),

U(t,x,q)=

q2(g′2?gg′′)

2k?1.Choosing q=0,we have U(x,x,0)=2(2k2?(k+1)x2)≥0,hence

x≤ k+1<√

4I.KRASIKOV

The optimal value of q=q0corresponds to the case when the discriminant?of (4),considered as a quadratic in q,is zero.That is,practically,one has to solve the system U(x,x,q0)=0,?=0,yielding

x=(m4?19m4?3

3)3/2m?3<

√

2k?1)=?

2(9m8?15m4+1)

)3/2m?3.

Now we can prove the upper bound of Theorem1.

Lemma5.Let k≥6,then

M k<2

8 1+45

3m2

=y0>2.

Thus,the conditions of Lemma3are ful?lled and it is left to estimate F k(y0)G k(y0). The required upper bound on G k(y0)follows from

2k?y0

(6m4?9m2?2)2≤1

4m2?9

<14(2k)1/3

Straightforward calculations also yield

F k(y0)<2

(2k)?1/6,

and the result follows.

To demonstrate the lower bound of Theorem1we observe that by H′k(x)= 2kH k?1(x),equation(2)can be rewritten as

2kxy(2y?3)

POLYNOMIALS5 F k(x k?1,k?1)/G k(x k?1,k?1)and show that2k?x2k?1,k?1>3(2k)1/3,the condi-tion imposed by Lemma3.The last claim is justi?ed by the following lemma which maybe of independent interest.

Theorem6.For k>2the largest zero x kk of H k(x)satis?es

x kk>√

(2k)?1/6

In particular,2k?x2

k?1,k?1

>3(2k)1/3,for k≥3.

Proof.We use the method of[5]based on the so-called Bethe ansatz equations. First we shall prove that for any x>x kk,

(5)2k?x2<1

Using the di?erential equation(1)to exclude higher derivatives we get f′2?ff′′

f2=

k i=11

(x?x ik)2<

(x kk?x ik)2.

The last sum is equal to

2k?x2kk?2

f2?

4s8?6s4+4s2+3?3

4s ,

and the result follows.

The second claim is a matter of simple calculations.

It is worth noticing that the obtained result is quite precise,as

x kk<

√

6I.KRASIKOV

Lemma7.

M k>

G k(y)

for

x=√

(2k?2)?1/6.

One can check that it has the only minimum0.44265...>27/61,for k=46,(notice that the asymptotic value,for k→∞,is0.4586...,and only slightly better).This completes the proof.

References

[1]M.Abramowitz,I.A.Stegun,Handbook of Mathematical Functions,Dover,New York,1964.

[2]T.Erdelyi,A.P.Magnus,P.Nevai,Generalized Jacobi weights,Christo?el functions,and Ja-

cobi polynomials,SIAM J.Math.Anal.25(1994),602-614.

[3]W.H.Foster and I.Krasikov,Explicit bounds for Hermite polynomials in the oscillatory region,

LMS https://www.wendangku.net/doc/7a17047709.html,put.Math.,Vol.3(2000)307-314.

[4]I.Krasikov,On zeros of polynomials and allied functions satisfying second order di?erential

equation,East J.Approx.,9(2003)51-65.

[5]I.Krasikov,On extreme zeros of classical orthogonal polynomials,submitted.

[6]M.Michalska,J.Szynal,A new bound for the Laguerre polynomials,https://www.wendangku.net/doc/7a17047709.html,put.Appl.Math.

133(2001)489-493.

[7]G.Szeg¨o,Orthogonal Polynomials,Amer.Math.Soc.Colloq.Publ.,v.23,Providence,RI,

1975.

Department of Mathematical Sciences,Brunel University,Uxbridge UB83PH United Kingdom

E-mail address:mastiik@https://www.wendangku.net/doc/7a17047709.html,