2 Orthogonal polynomials over $\left[ -1,\, +1\right] \protect$

2.1 Starting from a self-adjoint operator

Orthogonal polynomials can be introduced in many ways. Our choice is to start from a linear operator $\mathrm{Aut}\left( . \right)$ that acts over some superset of the set $\mathbb{R}\left[ x\right]$ (the real polynomials). The $F_{n}$ are the eigenvectors of $\mathrm{Aut}\left( . \right)$, and a scalar product is chosen for $\mathrm{Aut}\left( . \right)$ being self-adjoint and, therefore, the $F_{n}$ being an orthogonal family.

In the following, $\alpha \in \mathbb{N}$ is a constant (indexing the families), $n\in \mathbb{N}$ is the degree of $F_{n}$ (indexing the polynomials of a given family), $x\in \mathbb{R}$ is the main variable, $z\in \mathbb{C}$ is the $z$-transform variable, and $A,\, X,\, Z$ are the functions $\left( \right) \mapsto \alpha$, $\left( \right) \mapsto x$, $\left( \right) \mapsto z$.

An interesting choice is $\mathrm{Aut}\left( f \right) =\left( X^{2}-1\right) \mathrm{D}^{2}\left( f\right) +\left( A+1\right) \, X\, \mathrm{D}\left( f\right)$, i.e. $\mathrm{Aut}\left( f \right) \left( x\right) =\left( x^{2}-1\right) f''\left( x\right) +\left( \alpha +1\right) \, x\, f'\left( x\right)$, since $\alpha =0$ leads to the Chebyshev's, $\alpha =1$ to the Legendre's and $\alpha =2$ to the Gegenbauer's polynomials. The corresponding eigenvalues are obtained by equating the leading coefficients of both sides, leading to :

$$\lambda \left( n\right) =n\left( n-1\right) +n\, \left( \alpha +1\right) =n\left( n+\alpha \right)$$ (2)

Operator $\mathrm{Aut}\left( . \right)$ can be written as $\mathrm{Aut}\left( f \right) \left( x\right) =\frac{1}{\mu \left( x\right) }\f... ...}}{\mathrm{d}\, x ^{ }}\left( -Fac\left( x\right) \, f'\left( x\right) \right)$ and an identification gives $\mu \left( x\right) =\left( 1-x^{2}\right) ^{\left( \alpha -1\right) /2}$ and $Fac\left( x\right) =\left( 1-x^{2}\right) ^{\left( \alpha +1\right) /2}=\left( 1-x^{2}\right) \mu \left( x\right)$. Defining $\left\langle . \vert. \right\rangle$ as

$$\left\langle f \vert g \right\rangle =Cte\, \int _{-1}^{+1}f\left... ...\qquad \mu \left( x\right) =\left( 1-x^{2}\right) ^{\left( \alpha -1\right) /2}$$ (3)

we have $\left\langle f \vert\mathrm{Aut}\left( g \right) \right\rangle =Cte\, \int _{-... ...{+1}Fac\left( t\right) \, f'\left( t\right) \, g'\left( t\right) \: \mathrm{d}t$ (since the product term vanishes) and $\mathrm{Aut}\left( . \right)$ is self-adjoint as it has to be.

The choice of $Cte=1$ in EQ. 3 leads to $\left\langle 1 \vert 1 \right\rangle =2^{\alpha }\frac{\left( \frac{\alpha }{2... ...( \frac{\alpha }{2}-\frac{1}{2}\right) !\div \left( \frac{\alpha }{2}\right) !$, i.e. $\pi$ when $\alpha =0$ (Chebyshev), $2$ when $\alpha =1$ (Legendre), $\frac{1}{2}\pi$ when $\alpha =2$ (Gegenbauer). Therefore, the choice

$$Cte=2^{-\alpha }\frac{\alpha \, !}{\left( \frac{\alpha -1}{2}\right) !^{\, 2}}$$ (4)

is an interesting one. It should however be reminded that most of formulas are stating about proportionality, canceling the constant, and that normalizing all the $F_{n}$ (of a given family) by $\left\langle F_{n} \vert F_{n} \right\rangle =1$ is not a good idea since it introduces square roots depending on $n$ (further discussion about normalization in ???).

2.2 The general three terms formula

In this paragraph, the general three terms formula for orthogonal polynomial is recalled.

1. A first formula follows from $\left\langle x\, F_{n} \vert F_{n-1} \right\rangle =\left\langle F_{n} \vert x\, F_{n-1} \right\rangle$. Since $x\, F_{n-1}-\frac{\mathrm{cd}\left( F_{n-1} \right) }{\mathrm{cd}\left( F_{n} \right) }F_{n}$ belongs to $\mathrm{Span}\left( F_{n-1},\, F_{n-2},\, \cdots ,\, 1 \right)$ we have

   $$\left\langle x\, F_{n} \vert F_{n-1} \right\rangle =\frac{\mathrm... ...}{\mathrm{cd}\left( F_{n} \right) }\left\langle F_{n} \vert F_{n} \right\rangle$$ (5)

   
   

2. From the obvious fact that $F_{n+1}$ belongs to $\mathrm{Span}\left( x\, F_{n,}\, F_{n},\, F_{n-1},\, \cdots ,\, 1 \right)$, we deduce the existence of coefficients such that $F_{n+1}=\hat{a}\, x\, F_{n}+\hat{b}\, F_{n}+\hat{c}\, F_{n-1}+\sum _{k<n-1}d_{k}\, F_{k}$. All the $d_{k}$ must vanish since $0=\left\langle F_{k} \vert F_{n+1} \right\rangle =\hat{a}\left\langle x\, F_{k} \vert F_{n} \right\rangle +d_{k}\left\langle F_{k} \vert F_{k} \right\rangle$ and $\mathrm{dg}\left( x\, F_{k} \right) <n$. Coefficient $\hat{b}$ vanishes too when the $F_{n}$ have a definite parity (that of $n$). Then $\hat{a}$ is obtained from inspection of the leading coefficients, and $\hat{c}$ from $0=\left\langle F_{n-1} \vert F_{n+1} \right\rangle =\hat{a}\left\langle x\, F_... ...t F_{n} \right\rangle +\hat{c}\left\langle F_{n-1} \vert F_{n-1} \right\rangle$ together with EQ. 5. To summarize :

   $$F_{n+1}=\hat{a}\, x\, F_{n}+\hat{b}\, F_{n}+\hat{c}\, F_{n-1}\qqu... ...\rangle }{\left\langle F_{n-1} \vert F_{n-1} \right\rangle } \end{array}\right.$$ (6)

   
   

3. EQ. 6 starts from the $\mathrm{cd}\left( . \right)$ and the $\left\langle . \vert. \right\rangle$ to obtain the $\hat{a},\, \hat{b},\, \hat{c}$. Reverting the process leads to the recurrence :

   $$\left\langle F_{n} \vert F_{n} \right\rangle =-\frac{\hat{c}\, \m... ...hrm{cd}\left( F_{n-1} \right) }\left\langle F_{n-1} \vert F_{n-1} \right\rangle$$ (7)

   
   
   where the $\mathrm{cd}\left( F \right)$ can be chosen at will and the $\hat{a},\, \hat{c}$ must be fixed.

2.3 Closed formula for the $F_{n}$

From the very definition of $\mathrm{Aut}\left( . \right)$ and EQ. 2, we have

$$\left( x^{2}-1\right) F''_{n}\left( x\right) +\left( \alpha +1\ri... ...n}\left( x\right) -n\, \left( n+\alpha \right) F_{n}\left( \mathrm{x}\right) =0$$ (8)

Knowing that $F_{n}$ has a definite parity, we substitute $F_{n}\left( x\right) =\sum c\left( j,\, n\right) x^{n-2j}$ and obtain $c\left( j+1,\, n\right) =-\frac{\left( n-2j-1\right) \, \left( n-2j\right) }{2\left( 2n-2j-2+\alpha \right) \left( j+1\right) }c\left( j,\, n\right)$ . As it should be, the numerator vanishes when $j=floor\left( n/2\right)$ avoiding terms with negative exponents.

By recurrence, we obtain :

$$c\left( k,\, n\right) =\left( -\frac{1}{4}\right) ^{k}\left( \fra... ...( n-2k\right) !\, \left( n+\frac{1}{2}\alpha \right) !}\: c\left( 0,\, n\right)$$ (9)

We have pointed out the factor $\left( \frac{\alpha +2n}{\alpha +2n-2k}\right)$ since it becomes ambiguous when $\alpha =0$ and $n=0$. To obtain unambiguous values, the following convention must be used : all variables but $\alpha$ are substituted first, and the expression is simplified while $\alpha$ remains formal. After what $\alpha =0$ is substituted. For example, we substitute $k=n/2$, obtaining $\left( \frac{\alpha +2n}{\alpha +n}\right) \, \frac{\left( \frac{1}{2}n+\frac{... ...}{\left( \frac{1}{2}n\right) !\, \left( n+\frac{1}{2}\alpha \right) !\, 2^{n}}$, then $n=0$, obtaining the obvious $c\left( 0,0\right) =1\times c\left( 0,0\right)$.

Remembering that $F_{n}\left( x\right) =\sum c\left( j,\, n\right) x^{n-2j}$, a summation gives :

$$F_{n}\left( x\right) =x^{n}\, \mathrm{hypergeom}\left( \left[ \fr... ...}{2}-n\right] ,\, \frac{1}{x^{2}}\right) \times \mathrm{cd}\left( F_{n} \right)$$ (10)

2.4 Normalization

It remains now to choose the $\mathrm{cd}\left( F_{n} \right)$. The key point is not a global scalar factor, affecting all of polynomials, but the relative weights of these polynomials, since these weights are affecting the simplicity of recurrence relations, and the form of the generating function. For that reason, the choice $\left\langle F_{n} \vert F_{n} \right\rangle =1$ has bad as possible, since it introduces square roots depending on $n$.

The most obvious choice is $\mathrm{cd}\left( F_{n} \right) =1$ leading to unitary polynomials, denoted $\widehat{F_{n}}$ in what follows. Putting $Term\left( n,\, k\right) =x^{n-2k}\, \frac{c\left( k,\, n\right) }{c\left( 0,\, n\right) }$, we have $\widehat{F_{n}}\left( x\right) =\sum _{k}Term\left( n,\, k\right)$. Following the $A=B$ trick, let us now define $\mathrm{Oper}$ and $Key$ by :

$$\begin{array}{ccc} \mathrm{Oper}\left( \Phi \left( n,\, k\ri... ...\left( n-2k+1\right) \left( n-2k+2\right) }\, x^{2} \end{array}$$

It can be checked that

$$\mathrm{Oper}\left( Term\left( n,\, k\right) \right) =\left( Term... ...ght) \left( n,\, k+1\right) -\left( Term\times Key\right) \left( n,\, k\right)$$

By telescopic summation, we obtain $\mathrm{Oper}\left( \sum _{k}Term\left( n,\, k\right) \right) =0$ and therefore

$$\begin{array}{rcl} \left( n+1\right) \, \left( n+\alpha \righ... ...ha \right) \widehat{F_{n+2}}\left( x\right) & = & 0 \end{array}$$ (11)

Substituting $x=1$ in this relation, we obtain a recurrence over the $\widehat{F_{n}}\left( 1\right)$, with initial values $\widehat{F_{0}}\left( 1\right) =1$ and $\widehat{F_{1}}\left( 1\right) =1$ since $\widehat{F_{0}}=1$ (obvious) and $\widehat{F_{1}}\left( x\right) =x$ (parity). That recurrence can be solved into $\widehat{F_{n}}\left( 1\right) =\frac{2\, \Gamma \left( \frac{1}{2}\right) \, ... ...1}{2}\alpha +\frac{1}{2}\right) \, \Gamma \left( \frac{1}{2}\alpha +n\right) }$.

An other choice is to define $F_{n}$ by $F_{n}\left( 1\right) =1$ and put therefore $\mathrm{cd}\left( F_{n} \right) =1\div \widehat{F_{n}}\left( 1\right)$, i.e.

$$\mathrm{cd}\left( F_{n} \right) =\left( \frac{\alpha +n}{\alpha +... ...lpha !\, 2^{n}}{\left( \alpha +n\right) !\, \left( \frac{1}{2}\alpha \right) !}$$ (12)

$$c\left( k,\, n\right) =\, \left( -1\right) ^{k}\, 2^{n-2k}\, \lef... ...{1}{2}\alpha \right) !\, \left( \alpha +n\right) !\, \left( n-2k\right) !\, k!}$$ (13)

With that choice, $\alpha =0$ gives the Chebyshev polynomials $T\left( n,\, x\right)$, $\alpha =1$ the Legendre polynomials $P\left( n,\, x\right)$, while $\alpha =2$ gives $F_{n}\left( x\right) =\frac{1}{n+1}G\left( n,\, x\right)$, where $G\left( n,\, x\right)$ is the nth Chebyshev of second king. Here again, we put $Term\left( n,\, k\right) =x^{n-2k}\, c\left( k,\, n\right)$ and obtain

$$\begin{array}{ccc} \mathrm{Oper}\left( \Phi \left( n\right) ... ... x^{2}}{\left( n-2k+1\right) \left( n-2k+2\right) } \end{array}$$

By telescopic summation, we obtain $\mathrm{Oper}\left( \sum _{k}Term\left( n,\, k\right) \right) =0$ i.e. :

$$\left( n+\alpha \right) \, F_{n+1}\left( x\right) -\left( 2n+\alpha \right) \, x\, F_{n}\left( x\right) +n\, F_{n-1}\left( x\right) =0$$ (14)

A comparison of EQ. 6 and EQ. 14 leads to $\hat{a}=\frac{2n+\alpha }{n+\alpha }$ and $\hat{c}=-\frac{n}{n+\alpha }$. Substituting into EQ. 7 we obtain $\left\langle F_{n+1} \vert F_{n+1} \right\rangle =\frac{\left( \alpha +2n\righ... ...t) \, \left( \alpha +2n+2\right) }\left\langle F_{n} \vert F_{n} \right\rangle$ and, by recurrence from $\left\langle 1 \vert 1 \right\rangle =1$ :

$$\left\langle F_{n} \vert F_{n} \right\rangle =\left( \frac{\alpha +n}{\alpha +2\, n}\right) \, \frac{\alpha !\, n!}{\left( \alpha +n\right) !}$$ (15)

As expected, EQ. 15 cancels to $1$ when substituting $n=0$ and to $1/2$ when substituting $\alpha =0$ (and, therefore assuming $n\neq 0$).

2.5 Generating functions

The generating function $Sg_{\alpha }\left( x,\, z\right)$ of the $F_{n}$ is defined by $Sg_{\alpha }\left( x,\, z\right) =\sum _{\mathbb{N}}z^{n}\, F_{n}\left( x\right)$. As it is well known, a recurence equation over the $F_{n}$, like EQ. 14, induces a differential équation over $Sg_{\alpha }\left( x,\, z\right)$. Here, we have :

$$\left( z^{3}-2x\, z^{2}+z\right) \frac{\partial }{\partial z}Sg_{... ... +\left( z^{2}+x\, z-a\, x\, z+a-2\right) Sg_{\alpha }\left( x,\, z\right) -a+2$$

together with the initial condition $Sg_{\alpha }\left( x,\, z\right) =1+x\, z+\mathrm{O}\left( \mathrm{z}^{2}\right)$. We obtain :

$Sg_{0 }\left( x,\, z\right)$	$$\frac{1-x\, z}{1-2x\, z+z^{2}}$$
$Sg_{1 }\left( x,\, z\right)$	$$\frac{1}{\sqrt{1-2x\, z+z^{2}}}$$
$Sg_{2 }\left( x,\, z\right)$	$$\frac{\arctan \left( \frac{z\, \sqrt{1-x^{2}}}{1-x\, z}\right) }{z\, \sqrt{1-x^{2}}}$$
$Sg_{3 }\left( x,\, z\right)$	$$\frac{2\sqrt{1-2x\, z+z^{2}}}{z^{2}\, \left( 1-x^{2}\right) }-2\, \frac{1-x\, z}{z^{2}\, \left( 1-x^{2}\right) }$$
$Sg_{4 }\left( x,\, z\right)$	$$\frac{3}{2}\, \frac{\left( 1-2\, x\, z+z^{2}\right) \, \arctan \... ...2}\right) ^{3/2}}-\frac{3}{2}\, \frac{1-x\, z}{z^{2}\, \left( 1-x^{2}\right) }$$

• [ $T\left( n,x\right)$] $\left( \alpha =0\right)$ the nth Chebyshev polynomial of the first kind.
• [ $P\left( n,x\right)$] $\left( \alpha =1\right)$ the nth Legendre polynomial.
• [ $U\left( n,x\right)$] $\left( \alpha =2\right)$ the nth Gegenbauer polynomial (Chebyshev of the second kind).
• [ $Do\left( n,x\right)$] $\left( \alpha =3\right)$ the nth Douillet polynomial.

Putting $n=4$, we get

$\alpha$	$T\; :\; \alpha =0$	$P\; :\; \alpha =1$	$U\; :\; \alpha =2$	$Do\; :\; \alpha =3$
$F\left( 4,\, x\right)$	$8x^{4}-8x^{2}+1$	$\frac{35}{8}x^{4}-\frac{15}{4}x^{2}+\frac{3}{8}$	$\frac{1}{5}\left( 16x^{4}-12x^{2}+1\right)$	$\frac{21}{8}x^{4}-\frac{7}{4}x^{2}+\frac{1}{8}$
$F\left( 4,\, 1\right)$	$1$	$1$	$1$	$1$

It should be noticed that the usual normalization for the Gegenbauer polynomials is $U_{n}\left( 1\right) =n$, since the generating function will be simplified to $\frac{1}{1-2\, x\, z+z^{2}}$.

2.6 About derivatives

Many properties can be derived about the derivatives $D\left( F_{n}\right)$. A straightforward summation issued from Eq (9) leads to

$$F'_{n}\left( x\right) =n\, x^{n-1}\, \mathrm{hypergeom}\left( \le... ...}{2}-n\right] ,\, \frac{1}{x^{2}}\right) \times \mathrm{cd}\left( F_{n} \right)$$ (16)

Since $\frac{\partial }{\partial x}\, x^{n}\, \mathrm{hypergeom}\left( [v,\, -\frac{1... ...n-1}\, \mathrm{hypergeom}\left( [v,\, -\frac{1}{2}n+1],\, [w],\, x^{-2}\right)$, a comparison with Eq (10) gives:

$$F'_{\alpha ,\, n}=\frac{a\left( n+a-1\right) }{a}F_{\alpha +2,\, n-1}$$ (17)

Therefore, the $F'_{n}$ have (obviously) the required degree, and are orthogonal for another scalar product, namely the scalar product corresponding to $\alpha +2$.

From EQ. 14 and 17, we obtain a relation of the form $C\times F'_{n+1}\left( x\right) -C\times \, x\, F'_{n}\left( x\right) +C\times F'_{n-1}\left( x\right) =0$, while differentiating EQ. 14 gives another equation, involving $F_{n}$ and the three derivatives. An straigthforward elimination gives the well-known matricial recurrence:

$$\left( \begin{array}{c} F_{n+1}\left( x\right) \\ F'_{n+1}\left... ...{array}{c} F_{n}\left( x\right) \\ F'_{n}\left( x\right) \end{array}\right)$$

together with the linear formula:

$$\displaystyle F'_{n+1}\left( x\right) =\frac{\left( n+1\right) \,... ...left( \alpha +n\right) \, \left( \alpha +n-1\right) }\, F'_{n-1}\left( x\right)$$ (18)

We are now able to expand a given $F'_{m}\left( x\right)$ over the $F_{n}\left( x\right)$, i.e. obtain the coefficients in $F'_{m}\left( x\right) =\sum _{k}\, \delta \left( m ,\, k \right) {}F_{k}\left( x\right)$. We have $\delta \left( m ,\, m-1 \right) =\frac{m\, \left( \alpha +2\, m-2\right) }{\alpha +m-1}$ and a recurrence gives:

$$\delta \left( m ,\, k \right) =\left( \frac{\left( \alpha +m\righ... ...ght) !\, m!}{\left( \alpha +m\right) !\, k!}\qquad ;\qquad m-k\, odd\, positive$$ (19)

Let us now compute the scalar products $\left\langle F'_{m} \vert F'_{n} \right\rangle$, assuming that $m$ and $n$ have the same parity and $m\leq n$. Let us define $p$ as $floor\left( m/2\right)$, so that $m=2p$ (even case) or $m=2p+1$ (odd case). We have :

$$\left\langle F'_{m} \vert F'_{n} \right\rangle =\left\{ \begin{ar... ... F_{2k} \vert F_{2k} \right\rangle & \left( m\, odd\right) \end{array}\right.$$

Both sums are giving the same closed form, namely

$$\left\langle F'_{m} \vert F'_{n} \right\rangle =\frac{m\, \left( ... ...frac{\alpha !\, n!}{\left( \alpha +n\right) !}\qquad ;\qquad n-m\in 2\mathbb{N}$$ (20)

2.7 About the Chebyschev's generating function

In the Chebyschev case ($\alpha =0$), we have $$Sg_{ }\left( x,\, z\right) =\frac{1-x\, z}{1-2x\, z+z^{2}}$$. All the $\left\langle F_{m} \vert F_{n} \right\rangle$ can be obtained together by computing the two-variables series $\mathbf{S}_{1}\doteq \left\langle Sg\left( x,\, y\right) \vert Sg\left( x,\, z\right) \right\rangle$, i.e. $$\mathbf{S}_{1}=\frac{1}{\pi }\int _{-1}^{1}\frac{\left( 1-t\, y\... ...{2}\right) \left( 1-2t\, z+z^{2}\right) }\frac{\: \mathrm{d}t}{\sqrt{1-t^{2}}}$$. Changing $x$ into $\cos u$, and using $$I_{z}\doteq \int _{0}^{\pi }\frac{1}{1-2\cos u\, z+z^{2}}\, du=\frac{\pi }{1-z^{2}}$$, we obtain $$\mathbf{S}_{1}=\frac{\left( 1-z^{2}\right) \, \left( y\, z^{2}-2... ...right) }{4\pi \, \left( 1-z\, y\right) \, \left( y-z\right) }I_{y}+\frac{1}{4}$$, i.e.

$$\displaystyle \mathbf{S}_{1}=\frac{1}{2}\frac{2-z\, y}{1-z\, y}=1+\frac{1}{2}z\, y+\frac{1}{2}z^{2}\, y^{2}+\frac{1}{2}z^{3}\, y^{3}+\cdots$$

This result embodies the well-known results on orthogonality and normalization of the Chebyschev polynomials.

The decompositions of the $F'_{m}$ over the $F_{n}$ are given by $\mathbf{S}_{2}\doteq \left\langle \frac{\partial }{\partial x}Sg\left( x,\, y\right) \vert Sg\left( x,\, z\right) \right\rangle$, i.e. $$\mathbf{S}_{2}=\frac{y\, \left( 1-y^{2}\right) }{\pi }\int _{0}^... ...ft( 1-2\cos u\, y+y^{2}\right) ^{2}\, \left( 1-2\cos u\, z+z^{2}\right) }\, du$$. We obtain

$$\mathbf{S}_{2}=\frac{\left( 1+y^{2}-2y^{3}\, z\right) \, y}{\left... ...) y^{3}+4\left( z+z^{3}\right) y^{4}+5\left( 1+z^{2}+z^{4}\right) y^{5}+\cdots$$

showing that $\left\langle F'_{m} \vert F_{n} \right\rangle =m$ when $m-n\in 2\mathbb{N}+1$. Computing $$\int \mathbf{S}_{2}\frac{1}{y}dy$$, gives the more concise form $$\frac{y}{\left( 1-y\, z\right) \left( 1-y^{2}\right) }=$$ $y+z\, y^{2}+\left( 1+z^{2}\right) \, y^{3}+\left( z+z^{3}\right) \, y^{4}+\left( 1+z^{2}+z^{4}\right) \, y^{5}+\cdots$

The scalar products $\left\langle F'_{m} \vert F'_{n} \right\rangle$ are given by $\mathbf{S}_{3}\doteq \left\langle \frac{\partial }{\partial x}Sg\left( x,\, y\right) \vert\frac{\partial }{\partial x}Sg\left( x,\, z\right) \right\rangle$, i.e $$\mathbf{S}_{3}=\frac{1}{\pi }\left( 1-y^{2}\right) \left( 1-z^{2... ...t) ^{2}\left( 1-2t\, z+z^{2}\right) ^{2}}\frac{\: \mathrm{d}t}{\sqrt{1-t^{2}}}$$. Some computations, involving $$\frac{1}{\pi }\int _{0}^{\pi }\frac{1}{\left( 1-2\cos u\, z+z^{2}\right) ^{2}}\, du=\frac{\left( 1+z^{2}\right) }{\left( 1-z^{2}\right) ^{3}}$$ lead to

$$\mathbf{S}_{3}=y\left( z+3z^{3}+5z^{5}\right) +4y^{2}\left( 2z^{2}+4z^{4}\right) +3z\, y^{3}+9y^{3}\left( 3z^{3}\right) +\cdots$$

i.e. to $\left\langle F'_{m} \vert F'_{n} \right\rangle =m\, n\, \min \left( m,\, n\right)$ when $m-n\in 2\mathbb{Z}$. This result becomes more evident when computing $$\int \int \mathbf{S}_{3}\frac{1}{y\, z}\, dy\, dz=\frac{y\, z\, ... ...z\right) }{\left( 1-z\, y\right) \left( 1-y^{2}\right) \left( 1-z^{2}\right) }$$ : symmetry is obvious, $m$ comes from $\int \mathbf{S}_{3}\frac{1}{y}\, dy$, $n$ from $\int \mathbf{S}_{3}\frac{1}{z}\, dz$, parity from the $1-y^{2}$ and $1-z^{2}$ and the $\min \left( m,\, n\right)$ from the $1-y\, z$ at denominator.

2.8 About the Legendre's generating function

In the Legendre case ($\alpha =1$), we have $\mathbf{S}_{1}=\frac{1}{2\sqrt{y\, z}}\ln \left( \frac{1+\sqrt{y\, z}}{1-\sqrt{y\, z}}\right)$, i.e. $$\mathbf{S}_{1}=1+\frac{1}{3}\, y\, z+\frac{1}{5}\, y^{2}\, z^{2}+\frac{1}{7}\, y^{3}\, z^{3}+\frac{1}{9}\, y^{4}\, z^{4}+\cdots$$ since $\frac{\mathrm{d}^{ }}{\mathrm{d}\, A ^{ }}\frac{1}{2A}\ln \left( \frac{1+A}{1-A}\right) =\frac{1}{1-A^{2}}$. Thus $\left\langle F_{n} \vert F_{n} \right\rangle =\frac{1}{2n+1}$. We have also $\mathbf{S}_{2}=\frac{y}{\left( 1-y^{2}\right) \left( 1-y\, z\right) }=y+y^{2}\... ...}\right) +y^{4}\left( z+z^{3}\right) +y^{5}\left( 1+z^{2}+z^{4}\right) +\cdots$ therefore $\left\langle F'_{m} \vert F_{n} \right\rangle =1$ when $m-n\in 2\mathbb{N}+1$ and $0$ otherwise. To conclude, we have $\mathbf{S}_{3}=\frac{y\, z\, \left( 1+y\, z\right) }{\left( 1-y^{2}\right) \left( 1-z^{2}\right) \left( 1-y\, z\right) ^{2}}$. From $\frac{A\left( 1+A\right) }{\left( 1-A^{2}\right) }=A+3\, A^{2}+5\, A^{3}+7\, A^{4}+9\, A^{5}+\cdots$ we can verify that $\left\langle F'_{m} \vert F'_{n} \right\rangle =\frac{m\left( m+1\right) }{2}$ when $n-m\in 2\mathbb{N}$.

It can be noticed that $\left\langle F'_{m} \vert F_{n} \right\rangle =\frac{\left( m+\alpha \right) \, m!\, \alpha !}{\left( m+\alpha \right) !}$ : the independance from $n$ (provided the parity and $n<m$) holds for all $\alpha$.