Legendre Polynomials and an Inequality for a Combinatorial Sum

Abstract

Let $P_n$ be the Legendre polynomial of degree n. We use an estimate for the ultraspherical polynomials and a gamma function inequality to prove that $\left|{\int }_{-1}^x{P}_n\left(t\right)dt\right|<\sqrt{{\displaystyle \frac{2}{\pi }}}{\displaystyle \frac{1}{n^{3/2}}}(n\ge 1;-1\le x\le 1)$ and we apply this result to obtain the combinatorial inequality $\left|{\displaystyle \sum _{k=0}^n}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{\left(n+k-1\right)/2}{n}}\right){\displaystyle \frac{x^{k+1}}{k+1}}-{\displaystyle \frac{1}{n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{n/2}{n}}\right)\right|<\sqrt{{\displaystyle \frac{2}{\pi }}}{\displaystyle \frac{1}{2^n{n}^{3/2}}}(n\ge 1;-1\le x\le 1).$ The factor $\sqrt{2/\pi }$ given in both inequalities is the best possible.

1. Introduction and Statement of the Main Results

The classical Legendre polynomials $P_n\left(x\right)$$(n=0,1,2,\dots )$, well-known for their applications in various fields of mathematics and physics, can be defined by the formula

$$P_n\left(x\right)=2^n{\displaystyle \sum _{k=0}^n}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{(n+k-1)/2}{n}}\right)x^k,$$

(1)

where, as usual,

$$\left({\displaystyle \genfrac{}{}{0pt}{}{a}{0}}\right)=1,\left({\displaystyle \genfrac{}{}{0pt}{}{a}{n}}\right)={\displaystyle \frac{a(a-1)\dots (a-n+1)}{n!}}(n\ge 1).$$

These polynomials have been studied intensively by many authors. Detailed information on this subject is given, for example, in the monographs Andrews et al. [1], Ismail [2], Milovanović et al. [3], and Szegö [4]. In the literature, we can find several inequalities involving the Legendre polynomials. The following is Turán’s inequality. It was first published by Szegö [5] in 1948:

$$P_{n-1}\left(x\right)P_{n+1}\left(x\right)<P_n{\left(x\right)}^2(n\ge 1;-1<x<1).$$

In 2023, Alzer and Volkmer [6] compared $P_n$ with the Chebyshev polynomial of the first kind, $T_n$. They proved

$${\displaystyle \frac{1-T_n\left(x\right)}{1-P_n\left(x\right)}}<{\displaystyle \frac{2n}{n+1}}(n\ge 3;-1<x<1).$$

The upper bound is sharp.

Our work has been inspired by an interesting paper published by Fejér [7] in 1926. He presented remarkable inequalities involving $P_n\left(x\right)$ and related polynomials. One of the results states that there exists a real number c (which does not depend on n and x) such that

$$\left|{\int }_{-1}^x{P}_n\left(t\right)dt\right|\le {\displaystyle \frac{c}{n^{3/2}}}(n=1,2,3,\dots ;-1\le x\le 1).$$

(2)

Fejér remarked that (2) was also proved by Stieltjes. It is natural to ask: what is the best possible (i.e., the smallest) constant c in (2)? Our first theorem gives an answer to this question.

Theorem 1.

For all integers $n\ge 1$ and real numbers $x\in [-1,1]$, we have

$$\left|{\int }_{-1}^x{P}_n\left(t\right)dt\right|<{\displaystyle \frac{c}{n^{3/2}}}$$

(3)

with the best possible constant

$$c=\sqrt{{\displaystyle \frac{2}{\pi }}}=0.79788\dots$$

Remark 1.

Our proof of Theorem 1 shows that the following slightly sharper inequality holds:

$$\left|{\int }_{-1}^x{P}_n\left(t\right)dt\right|\le {\displaystyle \frac{2}{\sqrt{\pi }}}{\displaystyle \frac{1}{n(n+1)}}\sqrt{{\displaystyle \frac{n+1}{2}}}<\sqrt{{\displaystyle \frac{2}{\pi }}}{\displaystyle \frac{1}{n^{3/2}}}.$$

An application of Theorem 1 and the representations

$$(2n+1){\int }_{-1}^x{P}_n\left(t\right)dt=P_{n+1}\left(x\right)-P_{n-1}\left(x\right)={\displaystyle \frac{2n+1}{n(n+1)}}(x^2-1)P_n^{\prime }\left(x\right),$$

(4)

(see [7]) leads to the following inequalities.

Corollary 1.

For all integers $n\ge 1$ and real numbers $x\in [-1,1]$, we have

$$|P_{n+1}\left(x\right)-P_{n-1}\left(x\right)|<2\sqrt{{\displaystyle \frac{2}{\pi }}}{\displaystyle \frac{n+1/2}{n^{3/2}}}\mathit{and}(1-x^2)\left|P_n^{\prime }\left(x\right)\right|<\sqrt{{\displaystyle \frac{2}{\pi }}}{\displaystyle \frac{n+1}{\sqrt{n}}}.$$

Many remarkable formulas for combinatorial sums have been published. An extensive list of 500 identities is given in Gould [8]. We also refer to Egorychev’s book [9]. The aesthetic appeal of these formulas is certainly a reason for the tremendous interest in this field. A second reason might be that these sums have applications in number theory, the analysis of algorithms, and other areas. In contrast to the numerous identities, there are only few inequalities involving combinatorial sums. Some of them are given in Mitrinović ([10] Section 3.1). Here, we show that (3) can be used to obtain a new inequality for a combinatorial sum.

Theorem 2.

For all integers $n\ge 1$ and real numbers $x\in [-1,1]$, we have

$$|{\displaystyle \sum _{k=0}^n}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{(n+k-1)/2}{n}}\right){\displaystyle \frac{x^{k+1}}{k+1}}-{\displaystyle \frac{1}{n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{n/2}{n}}\right)|<c^{\ast }{\displaystyle \frac{1}{2^n{n}^{3/2}}}$$

(5)

with the best possible constant $c^{\ast }=\sqrt{2/\pi }$.

From (5) with $2n$ instead of n we get the following result.

Corollary 2.

For all integers $n\ge 1$ and real numbers $x\in [-1,1]$, we have

$$\left|{\displaystyle \sum _{k=0}^{2n}}\left({\displaystyle \genfrac{}{}{0pt}{}{2n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n+(k-1)/2}{2n}}\right){\displaystyle \frac{x^{k+1}}{k+1}}\right|<{\displaystyle \frac{1}{2\sqrt{\pi }}}{\displaystyle \frac{1}{4^n{n}^{3/2}}}.$$

In the next section, we collect some helpful formulas. Proofs of Theorem 1 and Theorem 2 are given in Section 3.

2. Preliminaries

The Jacobi polynomials are defined by

$$P_n^{(\alpha ,\beta )}\left(x\right)={\displaystyle \frac{1}{2^n}}{\displaystyle \sum _{k=0}^n}\left({\displaystyle \genfrac{}{}{0pt}{}{n+\alpha }{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{n+\beta }{n-k}}\right){(x-1)}^{n-k}{(x+1)}^k$$

and the ultraspherical (or Gegenbauer) polynomials are given by

$$C_n^{\left(\lambda \right)}\left(x\right)={\displaystyle \frac{{\left(2\lambda \right)}_n}{{(\lambda +1/2)}_n}}{P}_n^{(\lambda -1/2,\lambda -1/2)}\left(x\right),$$

(6)

where

$${\left(a\right)}_0=1,{\left(a\right)}_n=a(a+1)\dots (a+n-1)(n\ge 1)$$

is the Pochhammer symbol. It follows that

$$P_n\left(x\right)=C_n^{(1/2)}\left(x\right)=P_n^{(0,0)}\left(x\right).$$

(7)

The Jacobi polynomials satisfy the differential formula

$${\displaystyle \frac{d}{dx}}{P}_n^{(\alpha ,\beta )}\left(x\right)={\displaystyle \frac{1}{2}}(n+\alpha +\beta +1)P_{n-1}^{(\alpha +1,\beta +1)}\left(x\right);$$

(8)

see Olver et al. ([11], 18.9.15). Durand [12] proved that

$${\left(sin\theta \right)}^{2\lambda -1}\left|C_n^{\left(\lambda \right)}(cos\theta )\right|\le {\displaystyle \frac{\Gamma (n/2+\lambda )}{\Gamma \left(\lambda \right)\Gamma (n/2+1)}}(\lambda \ge 1;0\le \theta \le \pi ).$$

(9)

Moreover, we need two results on the gamma function. The inequality

$${\left({\displaystyle \frac{x}{x+s}}\right)}^{1-s}\le {\displaystyle \frac{\Gamma (x+s)}{x^s\Gamma \left(x\right)}}(x>0;0<s<1)$$

(10)

is due to Wendel [13] and the limit relation

$$\underset{x\to \infty }{lim}{x}^{b-a}{\displaystyle \frac{\Gamma (x+a)}{\Gamma (x+b)}}=1$$

(11)

can be found in Abramowitz and Stegun ([14], p. 257).

3. Proofs

Proof of Theorem 1.

Using (6), (7) and (8) gives

$$P_n^{\prime }\left(x\right)={\displaystyle \frac{d}{dx}}{P}_n^{(0,0)}\left(x\right)={\displaystyle \frac{n+1}{2}}{P}_{n-1}^{(1,1)}\left(x\right)=C_{n-1}^{(3/2)}\left(x\right).$$

(12)

Let $\left|x\right|\le 1$. Applying (4) and (12) leads to

$$\left|{\int }_{-1}^x{P}_n\left(t\right)dt\right|={\displaystyle \frac{1}{n(n+1)}}(1-x^2)|P_n^{\prime }\left(x\right)|={\displaystyle \frac{1}{n(n+1)}}(1-x^2)\left|C_{n-1}^{(3/2)}\left(x\right)\right|,$$

(13)

and from (9) and (10) (with $x=n/2$ and $s=1/2$) we obtain

$$(1-x^2)\left|C_{n-1}^{(3/2)}\left(x\right)\right|\le {\displaystyle \frac{\Gamma (n/2+1)}{\Gamma (3/2)\Gamma \left(\right(n+1)/2)}}={\displaystyle \frac{2}{\sqrt{\pi }}}{\displaystyle \frac{\Gamma (n/2+1)}{\Gamma \left(\right(n+1)/2)}}\le {\displaystyle \frac{2}{\sqrt{\pi }}}\sqrt{{\displaystyle \frac{n+1}{2}}}.$$

(14)

Combining (13) and (14) gives

$$\left|{\int }_{-1}^x{P}_n\left(t\right)dt\right|\le {\displaystyle \frac{2}{\sqrt{\pi }}}{\displaystyle \frac{1}{n(n+1)}}\sqrt{{\displaystyle \frac{n+1}{2}}}<\sqrt{{\displaystyle \frac{2}{\pi }}}{\displaystyle \frac{1}{n^{3/2}}}.$$

This settles (3) with $c=\sqrt{2/\pi }$. It remains to prove that this constant is sharp. We set $n=2r+1$ and apply (4). This gives

$$n^{3/2}\left|{\int }_{-1}^0{P}_n\left(t\right)dt\right|={\displaystyle \frac{{(2r+1)}^{3/2}}{4r+3}}|P_{2r+2}\left(0\right)-P_{2r}\left(0\right)|.$$

(15)

Since

$$P_{2r}\left(0\right)={\displaystyle \frac{{(-1)}^r}{\sqrt{\pi }}}{\displaystyle \frac{\Gamma (r+1/2)}{\Gamma (r+1)}},$$

it follows that

$$|P_{2r+2}\left(0\right)-P_{2r}\left(0\right)|={\displaystyle \frac{1}{2\sqrt{\pi }}}(4r+3){\displaystyle \frac{\Gamma (r+1/2)}{\Gamma (r+2)}}.$$

(16)

From (15) and (16) we get

$$n^{3/2}\left|{\int }_{-1}^0{P}_n\left(t\right)dt\right|={\displaystyle \frac{1}{2\sqrt{\pi }}}{\left(2+{\displaystyle \frac{1}{r}}\right)}^{3/2}{r}^{3/2}{\displaystyle \frac{\Gamma (r+1/2)}{\Gamma (r+2)}}.$$

Applying (11) gives

$$\underset{r\to \infty }{lim}{(2r+1)}^{3/2}\left|{\int }_{-1}^0{P}_{2r+1}\left(t\right)dt\right|={\displaystyle \frac{1}{2\sqrt{\pi }}}·2^{3/2}·1=\sqrt{{\displaystyle \frac{2}{\pi }}}.$$

(17)

The proof of Theorem 1 is complete. □

Next, we offer two proofs for the following combinatorial identity, which plays an important role in the proof of Theorem 2.

Lemma 1.

For all integers $n\ge 1$, we have

$${\displaystyle \sum _{k=0}^n}{\displaystyle \frac{{(-1)}^{k+1}}{k+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{(n+k-1)/2}{n}}\right)={\displaystyle \frac{1}{n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{n/2}{n}}\right).$$

(18)

Proof. 

We define

$$A_n\left(x\right)={\displaystyle \sum _{k=0}^n}{\displaystyle \frac{1}{k+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{(n+k-1)/2}{n}}\right)x^{k+1}.$$

From (1) we obtain

$$A_n^{\prime }\left(x\right)=2^{-n}{P}_n\left(x\right).$$

(19)

Using (19) and (4) gives

$$-A_n(-1)=A_n\left(0\right)-A_n(-1)={\int }_{-1}^0{A}_n^{\prime }\left(t\right)dt=2^{-n}{\int }_{-1}^0{P}_n\left(t\right)dt=-2^{-n}{\displaystyle \frac{P_n^{\prime }\left(0\right)}{n(n+1)}}=-{\displaystyle \frac{1}{n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{n/2}{n}}\right).$$

This leads to (18). □

Proof. 

Let

$$S_n={\displaystyle \sum _{k=0}^n}{(-1)}^{k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{n+1}{k+1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{(n+k-1)/2}{n}}\right).$$

We show that for $n\ge 1$,

$$S_n=\left({\displaystyle \genfrac{}{}{0pt}{}{n/2}{n}}\right)$$

(20)

which is equivalent to (18). To prove (20) we consider two cases.

Case 1. n is even.

We set $n=2N$$(N\ge 1)$. Using

$${\displaystyle \sum _{k=0}^{[m/2]}}\left({\displaystyle \genfrac{}{}{0pt}{}{m+1}{2k+1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{x+k}{m}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{2x}{m}}\right),$$

see Gould ([8], 3.169), with $m=2N$ and $x=N-1/2$, we obtain

$$\begin{array}{cc}\hfill S_{2N}=& ({\displaystyle \sum _{{\displaystyle \genfrac{}{}{0pt}{}{k=0}{k\mathrm{even}}}}^{2N}}+{\displaystyle \sum _{{\displaystyle \genfrac{}{}{0pt}{}{k=0}{k\mathrm{odd}}}}^{2N}}){(-1)}^{k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2N+1}{k+1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{N+(k-1)/2}{2N}}\right)\hfill \\ \hfill =& -{\displaystyle \sum _{k=0}^N}\left({\displaystyle \genfrac{}{}{0pt}{}{2N+1}{2k+1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{N+k-1/2}{2N}}\right)+{\displaystyle \sum _{k=1}^N}\left({\displaystyle \genfrac{}{}{0pt}{}{2N+1}{2k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{N+k-1}{2N}}\right)\hfill \\ \hfill =& -\left({\displaystyle \genfrac{}{}{0pt}{}{2N-1}{2N}}\right)+0=0=\left({\displaystyle \genfrac{}{}{0pt}{}{N}{2N}}\right).\hfill \end{array}$$

Case 2. n is odd.

Let $n=2N-1$$(N\ge 1)$. We apply

$${\displaystyle \sum _{k=0}^{m+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2m+2}{2k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{x+k}{2m+1}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{2x+1}{2m+1}}\right),$$

see Gould ([8], 3.174), with $m=N-1$ and $x=N-3/2$. Then

$$\begin{array}{cc}\hfill S_{2N-1}=& ({\displaystyle \sum _{{\displaystyle \genfrac{}{}{0pt}{}{k=0}{k\mathrm{even}}}}^{2N-1}}+{\displaystyle \sum _{{\displaystyle \genfrac{}{}{0pt}{}{k=0}{k\mathrm{odd}}}}^{2N-1}}){(-1)}^{k+1}\left({\displaystyle \genfrac{}{}{0pt}{}{2N}{k+1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{N+k/2-1}{2N-1}}\right)\hfill \\ \hfill =& -{\displaystyle \sum _{k=0}^{N-1}}\left({\displaystyle \genfrac{}{}{0pt}{}{2N}{2k+1}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{N+k-1}{2N-1}}\right)+{\displaystyle \sum _{k=1}^N}\left({\displaystyle \genfrac{}{}{0pt}{}{2N}{2k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{N+k-3/2}{2N-1}}\right)\hfill \\ \hfill =& 0+\left({\displaystyle \genfrac{}{}{0pt}{}{2N-2}{2N-1}}\right)-\left({\displaystyle \genfrac{}{}{0pt}{}{N-3/2}{2N-1}}\right)=\left({\displaystyle \genfrac{}{}{0pt}{}{N-1/2}{2N-1}}\right).\hfill \end{array}$$

This completes the proof of (20). □

Proof of Theorem 2.

Let $n\ge 1$ and $x\in [-1,1]$. Applying (1), (3) and (18) gives

$$|{\displaystyle \sum _{k=0}^n}\left({\displaystyle \genfrac{}{}{0pt}{}{n}{k}}\right)\left({\displaystyle \genfrac{}{}{0pt}{}{(n+k-1)/2}{n}}\right){\displaystyle \frac{x^{k+1}}{k+1}}-{\displaystyle \frac{1}{n+1}}\left({\displaystyle \genfrac{}{}{0pt}{}{n/2}{n}}\right)|={\displaystyle \frac{1}{2^n}}\left|{\int }_{-1}^x{P}_n\left(t\right)dt\right|<\sqrt{{\displaystyle \frac{2}{\pi }}}{\displaystyle \frac{1}{2^n{n}^{3/2}}}.$$

Moreover, from the limit relation (17) we conclude that the constant factor $\sqrt{2/\pi }$ is the best possible. □