Here we exhibit alternative proofs of the identities given by Persson-Strang and (Huat-Chan)- -Wan-Zudilin for the Legendre polynomials. Besides, we show the connection between the Lanczos derivative and these polynomials via the Rangarajan-Purushothaman’s formula.

Keywords: (Huat-Chan)-Wan-Zudilin’s property, Legendre polynomials, Persson-Strang’s identity, Rangarajan-Purushothaman’s expression, Lanczos derivative

The Legendre’s polynomials¹ $P_n\left(x\right),-1\le x\le 1,$  can be defined via the following recurrence relation:^2–4

$\left(n+1\right)P_{n+1}=\left(2n+1\right)x{P}_n-n{P}_{n-1},P_0=1,P_1=x,n=1,\mathrm{2...},$   (1)

hence:

$P_2=\frac{1}{2}\left(3x^2-1\right),P_3=\frac{1}{2}\left(5x^3-3x\right),P_4=\frac{1}{8}\left(35x^4-30x^2+3\right),\mathrm{...}$   (2)

These polynomials also are determined univocally through the conditions;^5,6

${\int }_{-1}^1{P}_m\left(x\right)P_n\left(x\right)dx=0,m\ne n,P_n\left(1\right)=1,\forall n,$   (3)

therefore:

${\int }_{-1}^1{x}^m{P}_n\left(x\right)dx=0,m<n,$   (4)

and the Laplace’s integral formula^3–5,7 gives an alternative way to generate the expressions (2):

$P_n\left(x\right)=\frac{1}{2^n}{\int }_{-\pi }^{\pi }{\left(x+\sqrt{x^2-1}\cos \beta \right)}^{n}d\beta ,n=0,1,\mathrm{2...}$   (5)

or equivalently:

$P_n\left(x\right)=\frac{1}{2^n}{\displaystyle {\sum }_{k=0}^{\frac{n}{2}}{\left(-1\right)}^k}\left({}_k^n\right)\left({}_n^{2n-2k}\right)x^{n-2k}.$   (6)

Here we have interest in the value of the following integral:

$Q\left(m\right)\equiv {\int }_{-1}^1\frac{1}{x}{P}_{2m+1}\left(x\right)dx,m=01,2,\mathrm{..}$   (7)

then from (6) with $n=2m+1:$

$Q\left(m\right)=\frac{1}{2^n}{\int }_{-1}^1{\displaystyle {\sum }_{k=0}^m{\left(-1\right)}^k}\left({}_k^n\right)\left({}_n^{2n-2k}\right)x^{2m-2k}dk=\frac{1}{4^m}A\left(m\right),$   (8)

where $A\left(m\right)$ can be calculated via the method of Petkovsek-Wilf-Zeilberger,^8–18 in fact:

$A\left(m\right)\equiv {\displaystyle {\sum }_{k=0}^m\frac{{\left(-1\right)}^k\left(2n-2k\right)!}{k!\left(n-k\right)!\left(n-2k\right)!\left(n-2k\right)}}=\frac{\left(2n\right)!}{n{\left(n!\right)}^2}{\displaystyle {\sum }_{k=0}^{\infty }{t}_k,t_k}=\frac{{\left(-1\right)}^{k}n{\left(n!\right)}^2\left(2n-2k\right)!}{\left(2n\right)!k!\left(n-k\right)!\left(n-2k\right)!\left(n-2k\right)},$   (9)

Therefore $\frac{t_{k+1}}{t_k}=\frac{{\left(k-m-\frac{1}{2}\right)}^2\left(k-m\right)}{\left(k-m+\frac{1}{2}\right)\left(k-2m-\frac{1}{2}\right)\left(k+1\right)},$  hence:

$A\left(m\right)\equiv \frac{\left(2n\right)!}{n{\left(n!\right)}^2}3F_2\left(-m,-m-\frac{1}{2},-m-\frac{1}{2};-m+\frac{1}{2},-2m-\frac{1}{2};1\right)=\frac{{\left(-1\right)}^m{4}^n{\left(m!\right)}^2}{2\left(n!\right)},n=2m+1,$   (10)

where it was applied the following value of the hypergeometric function in (10):

$3F_2\left(\right)=\frac{{\left(-16\right)}^{m}n!{\left(m!\right)}^2}{\left(4m+1\right)!}.$   (11)

then (8) and (10) imply the result:

$Q\left(m\right)=\frac{2{\left(-4\right)}^m{\left(m!\right)}^2}{\left(2m+1\right)!}.$   (12)

 On the other hand, from (6) for $n=2m+1:$

${[\frac{P_{2m+1}\left(x\right)}{x}]}^2=\frac{1}{2^n}{\displaystyle {\sum }_{k=0}^m{\left(-1\right)}^k\left({}_k^n\right)}\left({}_n^{2n-2k}\right)x^{2m-2k}\frac{P_{2m+1}\left(x\right)}{x},$

where we can integrate in the interval $[-1,1]$  and apply the properties (4) and (12) to obtain the relation:

${\int }_{-1}^1{[\frac{P_{2m+1}\left(x\right)}{x}]}^{2}dx=\frac{{\left(-1\right)}^m}{2^n}\left({}_k^n\right)\left({}_n^{2n-2k}\right)Q\left(m\right)=2,m=0,1,2,\mathrm{...}$   (13)

deduced by Persson-Strang;¹⁹ Amdeberhan et al.²⁰ generalized the identity (13) in the form:

${\int }_{-1}^1{[\frac{P_1\left(x\right)-P_1(0)}{x}]}^{2}dx=2[1-{\beta }^2\left(l\right)],l=0,1,2,\mathrm{...}$   (14)

such that:

$\beta \left(l\right)=\{2-1\left({}_{1/2}^l\right) { }_{, if l is even}^{0, if is odd}$   . (15)

 Remark. - In (6) we may use $x=\frac{b}{\sqrt{b^2-4c}}$  to obtain:

$P_n\left(\frac{b}{\sqrt{b^2-4c}}\right)=\frac{1}{{\left(b^2-4c\right)}^{n/2}}{\displaystyle {\sum }_{j=0}^{\text{n/2}}\frac{b^{n-2j}{\left(-4c\right)}^j}{2^{n}j!}}R\left(n\right)$   (16)

where:   

$R\left(n\right)\equiv {\displaystyle {\sum }_{k=j}^{\text{n/2}}\frac{{\left(-1\right)}^k\left(2n-2k\right)!}{\left(2n-2k\right)!\left(k-i\right)!\left(n-k\right)!}}=\frac{{\left(-1\right)}^j{2}^{n-2j}n!}{j!\left(n-2j\right)!},0\le j\le n/2$   (17)

thus (16) and (17) imply the interesting identity of (Huat-Chan)-Wan-Zudilin:^21,22

${\left(b^2-4c\right)}^{n/2}{P}_n\left(\frac{b}{\sqrt{b^2-4c}}\right)={\displaystyle {\sum }_{j=0}^{n/2}\left({}_{2j}^n\right)\left({}_j^{2j}\right)}{b}^{n-2}{}^j c^j$   (18)

We may indicate two useful relations:^23,24

${[P_n\left(x\right)]}^2={\displaystyle {\sum }_{k=0}^n\left({}_k^n\right) \left({}_n^{n+k}\right) \left({}_k^{2k}\right)}{(-\frac{1-x^2}{4})}^k , n=0,1,2,\mathrm{...}$   (19)

$f_{-1}^1{x}^m{P}_n\left(x\right)dx=\frac{2^{n+1}}{m+1}.\frac{\left(\frac{m+n}{{}_n^2}\right)}{\left({}_n^{m+n+1}\right)},m-n=0,2,4,\mathrm{...}$   (20)

We emphasize the importance of the method of Petkovsek-Wilf-Zeilberger to obtain (10) and (17).

Rangarajan-Purushothaman^25,26 obtained the following generalization of the Lanczos derivative:^27,28

$f^{\left(m\right)}\left(x\right)=li{m}_{\epsilon \to 0}\frac{\left(2m+1\right)!!}{2{\epsilon }^{m+1}}{\displaystyle {\int }_{-\epsilon }^{\epsilon }{P}_m}\left(\frac{t}{\epsilon }\right)f\left(x+t\right)dt,m=1,2,\mathrm{...}$   (21)

involving the Legendre polynomials.

If $f\left(x\right)=1,$  then (21) implies the property:

$f_{-1}^1{P}_n\left(u\right) du=0, n=2,4,6,\mathrm{...}$   (22)

From (21) for $f\left(x\right)=x^N:$

$f_{-1}^1{P}_n\left(u\right)u^k du=0, k<n,$   (23)

$f_0^1{P}_n\left(u\right)u^{n}du=\frac{n!}{\left(2n+1\right)!!}=\frac{2^n{\left(n!\right)}^2}{\left(2n+1\right)!},n=0,2,\mathrm{...}$   (24)

On the other hand, we know the relations:

$f_0^1{P}_{2l}\left(u\right)u^{m}du=\frac{{\left(-1\right)}^l\Gamma \left(l-\frac{m}{2}\right)\Gamma (\frac{m+1}{2})}{2\Gamma \left(-\frac{m}{2}\right)\Gamma \left(l+\frac{m+3}{2}\right)},m>-1,$   (25)

$f_0^1{P}_{2l+1}\left(u\right)u^{m}du=\frac{{\left(-1\right)}^l\Gamma \left(l+\frac{1-m}{2}\right)\Gamma (1+\frac{m}{2})}{2\Gamma \left(1+2+\frac{m}{2}\right)\Gamma \left(\frac{1-m}{2}\right)},m>-2,$   (26)

thus (24) can be deduced from (25) and (26) for $m=n=2l$  and $m=n=2l+1,$  respectively.

 We have the following Schmied’s formula (2005):²⁹

$u^m={\displaystyle {\sum }_{l=m,m-2,\mathrm{...}}^{}\frac{m!\left(2l+1\right)}{2\frac{m-1}{2}\left(\frac{m-1}{2}\right)!\left(m+l+1\right)!!}}{P}_l\left(u\right),$   (27)

which gives (20), and for $m=n$  implies (24).

 The Legendre polynomials can be written in terms of the Gauss hypergeometric function:

$P_n\left(0\right)=\frac{\left(2n-1\right)!!}{n!}{\displaystyle {\sum }_{k=0}^n\left({}_k^n\right) { }_2^{}{F}_1}\left(k-n,-n; -2n; 2\right)x^k,$   (28)

and we know the result:

${}_2^{}{F}_1\left(-n,-n;-2n;2\right)= \{{}_{ \frac{{\left(1-\right)}^{{}^{\frac{n}{2}}}n!\left(n-1\right)!!}{n!!\left(2n-1\right)!!} , n=2, 4, 6,\mathrm{...}}^{ 0,n=1, 3, 5,\mathrm{...}}$ ,  (29)

then from (28) and (29):

$P_n\left(0\right){=}_2^{}{F}_1\left(-n,n+1;1;\frac{1}{2}\right)= \{{}_{ \frac{{\left(1-\right)}^{{}^{\frac{n}{2}}}n!\left(n-1\right)!!}{n!!} , n=2, 4, 6,\mathrm{...}}^{ 0,n=1, 3, 5,\mathrm{...}}$ .  (30)

Finally, the expression:

$P_n\left(x\right)\equiv \frac{1}{2^n}{\left(-1\right)}^k{\left(1-x\right)}^k{\left(1+x\right)}^{n-k}{\left({}_k^n\right)}^2,$   (31)

and (30) imply the relation:

${\displaystyle {\sum }_{k=0}^n{\left(-1\right)}^k}{\left({}_k^n\right)}^2= \{{}_{ \frac{{\left(1-\right)}^{{}^{\frac{n}{2}}}{2}^n\left(n-1\right)!!}{n!!} , n=2, 4, 6, \mathrm{...}}^{ 0,n=1, 3, 5,\mathrm{...}}$   . (32)

Thus, we see that the Rangarajan-Purushothaman’s formula for the Lanczos derivative allows deduce some properties of Legendre polynomials, and it represents differentiation by integration. The $P_n\left(x\right)$ are orthogonal polynomials, hence Diekema-Koornwinder³⁰ consider that the name “orthogonal derivative” is adequate for (21).

Remark. - From (3) we have the property $P_n\left(1\right)=1\forall n,$ then (6) for $x=1$  gives the identity:

$2n={\displaystyle {\sum }_{k=0}^{\frac{n}{2}}{\left(-1\right)}^k\left({}_k^n\right)}\left({}_n^{2n-2k}\right);$   (33)

on the other hand, we know the relation:³¹

${\displaystyle {\sum }_{k=0}^n{\left(-1\right)}^k\left({}_k^n\right)}\left({}_n^{z+ky}\right)={\left(-y\right)}^n,y\ne 0,$   (34)

which for  and is equivalent to (33) because $\left({}_n^{2n-2k}\right)=0$ for $k>\frac{n}{2} .$

Finally, we consider that the publications^32–37 have useful relationship with the study realized in the present paper.

None.

The author declares there is no conflict of interest.

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