The generating function of the Jacobi polynomials is given by $$ \sum_{n=0}^{\infty} P_{n}^{(\alpha, \beta)}(z) t^{n}=2^{\alpha+\beta} R^{-1}(1-t+R)^{-\alpha}(1+t+R)^{-\beta} $$ where $$ R=R(z, t)=\left(1-2 z t+t^{2}\right)^{\frac{1}{2}}. $$ However, we need a similar result or alternative representations for the square of the Jacobi polynomial given by $$ \sum_{n=0}^{\infty}( P_{n}^{(\alpha, \beta)}(z))^2 t^{n} $$ or more specifically, $$ \sum_{n=0}^{\infty}( P_{n}^{(1, 0)}(z))^2 t^{n}. $$ I have researched this in detail, but have not found an answer. If someone could give me some hints, that would be really helpful.
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$\begingroup$ You can ("automatically") derive a (3rd order) recurrence relation, for example using a guessing package, but it does not look very nice. Would this be sufficient for you? $\endgroup$– Martin RubeyFeb 11, 2022 at 8:17
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$\begingroup$ Sage says ${}_3F_2\left( \begin{matrix} \tfrac32, \tfrac32, 1 \\ 2, 2 \end{matrix} \Big\vert\, 4tx^2 - 8tx + 4t \right)$. Comment, not answer, because I haven't sanity-checked it. $\endgroup$– Peter TaylorFeb 11, 2022 at 8:50
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$\begingroup$ @PeterTaylor I don't see how your formula could make sense. If you pick out the coefficient of $t^n$ you get a multiple of $(x-1)^{2n}$, not the square of a Jacobi polynomial. $\endgroup$– Hjalmar RosengrenFeb 11, 2022 at 9:59
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$\begingroup$ @MartinRubey Thank you for the comment. I am afraid I don't know much about what you told me, but the ideal expression is a simple mathematical expression, something like the Elliptic integrals or Hypergeometric function that Peter Taylor has provided. $\endgroup$– KaneFeb 11, 2022 at 12:10
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$\begingroup$ @PeterTaylor Thank you for the comment. If possible, can you share the sage code with us? $\endgroup$– KaneFeb 11, 2022 at 12:16
3 Answers
I don't know if this is useful to you, but Bailey gave the identity \begin{multline*}\sum_{n=0}^\infty\frac{n!(\alpha+\beta+1)_n}{(\alpha+1)_n(\beta+1)_n}\,P_n^{(\alpha,\beta)}(x)P_n^{(\alpha,\beta)}(y)\,t^n\\ =\frac 1{(1+t)^{\alpha+\beta+1}}\sum_{m,n=0}^\infty\frac{(\alpha+\beta+1)_{2m+2n}}{m!n!(\beta+1)_m(\alpha+1)_n}\left(\frac t{4(1+t)^2}\right)^{m+n}((1+x)(1+y))^m((1-x)(1-y))^n.\end{multline*} The original reference is J. London Math. Soc. 13 (1938), 8-12, but I copied it from Stanton, Proc. Amer. Math. Soc. 80 (1980), 398–400. When $\beta=0$ and $x=y$, this gives an expression for $$\sum_{n=0}^\infty P_n^{(\alpha,0)}(x)^2\,t^n$$ as a double series.
Here is a way to guess an expression - it is a bit uneven, because fricas has no builtin jacobiP and sage has apparently trouble transferring a long list to fricas. But it works, at least if you fix $b=0$:
sage: S.<a,b,x> = QQ[]
sage: l = [S(jacobi_P(n, a, 0, x)) for n in range(30)]
sage: frec = fricas.guessPRec(l)[0]; frec
... # omitted to save space
sage: frec.name()
'sage74'
sage: fricas("guessPRec [(eval(sage74, ['n], [n])::FRAC POLY INT)^2 for n in 0..60]")
... # huge output of the form p3(n) f(n + 3) + p2(n) f(n + 2) + p1(n) f(n + 1) + p0(n) f(n)
If you fix $a = 1$, things become slightly nicer, but not much:
sage: l = [S(jacobi_P(n, 1, 0, x))^2 for n in range(50)]
sage: frec = fricas.guessPRec(l)[0].getEq().sage(); frec
(2*n^4 + 25*n^3 + 113*n^2 + (8*n^6 + 132*n^5 + 882*n^4 + 3047*n^3 + 5727*n^2 + 5544*n + 2160)*x + 216*n + 144)*f(n + 3) + (2*n^4 - (32*n^6 + 496*n^5 + 3136*n^4 + 10328*n^3 + 18642*n^2 + 17451*n + 6615)*x^3 + 19*n^3 - (24*n^4 + 244*n^3 + 902*n^2 + 1431*n + 819)*x^2 + 64*n^2 + (8*n^6 + 124*n^5 + 782*n^4 + 2561*n^3 + 4576*n^2 + 4212*n + 1557)*x + 90*n + 45)*f(n + 2) - (2*n^4 - (32*n^6 + 464*n^5 + 2736*n^4 + 8392*n^3 + 14122*n^2 + 12369*n + 4410)*x^3 + 21*n^3 - (24*n^4 + 236*n^3 + 842*n^2 + 1289*n + 714)*x^2 + 79*n^2 + (8*n^6 + 116*n^5 + 682*n^4 + 2079*n^3 + 3461*n^2 + 2973*n + 1022)*x + 125*n + 70)*f(n + 1) - (2*n^4 + 15*n^3 + 38*n^2 + (8*n^6 + 108*n^5 + 582*n^4 + 1593*n^3 + 2322*n^2 + 1701*n + 490)*x + 39*n + 14)*f(n)
sage: f = function("f")
sage: var("n")
n
sage: frec.coefficient(f(n+3)).factor()
(4*n^2*x + 16*n*x + 15*x + 1)*(2*n + 3)*(n + 4)^2*(n + 3)
sage: frec.coefficient(f(n+2)).factor()
-(4*n^2*x^2 + 20*n*x^2 - n^2 + 21*x^2 - 5*n + 2*x - 5)*(4*n^2*x + 24*n*x + 35*x + 1)*(2*n + 3)*(n + 3)
sage: frec.coefficient(f(n+1)).factor()
(4*n^2*x^2 + 20*n*x^2 - n^2 + 21*x^2 - 5*n + 2*x - 5)*(4*n^2*x + 16*n*x + 15*x + 1)*(2*n + 7)*(n + 2)
sage: frec.coefficient(f(n)).factor()
-(4*n^2*x + 24*n*x + 35*x + 1)*(2*n + 7)*(n + 2)*(n + 1)^2
Here is an expression that involves elliptic functions and a portion represented as an integral that (I think) cannot be reduced to elliptic functions.
$$ \sum_{n=0}^\infty P^{(1,0)}_n(x)^2 t^n = \frac{1}{(1-x)^2}\big( F(x,t)+\frac{1}{t} (F(x,t)-1) \big) -\frac{2x}{t}\big(F(x,t)-1\big) + \frac{1+x}{1-x}\frac{G(x,t)}{t} $$ where $$F(x,t) = \sum_{n=0}^\infty P_n(x)^2t^n$$ and has a closed-form in elliptic functions from Bailey's formula $$ \tag{B} \sum_{n=0}^\infty P_n(x) \, P_n(y) t^n = r \cdot {}_2F_{1}\big(1/2,1/2;1;-4\sqrt{(1-x^2)(1-y^2)} \,t \, r^2\big)$$ $$ r= \big(1+t(t-2\sqrt{(1-x^2)(1-y^2)} - 2xy)\big)^{-1/2} $$ I've found two integral representations for $G(x,t):$ $$G(x,t)=\frac{2x}{\pi (1-x^2)}\int_0^t \frac{du}{u(1-u)}\Big( \frac{(u-1)^2}{(1+u)^2-4ux^2} \,E\big(\frac{4u(x^2-1)}{(u-1)^2}\big) - K\big(\frac{4u(x^2-1)}{(u-1)^2}\big) \Big) $$ and $$G(x,t)=\frac{-4 t}{\pi \, x}\int_0^1 du\sqrt{\frac{u}{1-u}}\frac{1}{\rho(u;x,t)} \Big((1+t+ \rho(u;x,t))^{-1} + (1-t+ \rho(u;x,t))^{-1} \Big)$$ $$ \text{with } \rho(u,x,t)=((1-t)^2+4tu(1-x^2))^{1/2} $$ In the first integral, $E$ and $K$ are elliptic functions in Mathematica notation.
The following is a proof sketch: $$ P^{(1,0)}_n(x) = \frac{P_n(x)-P_{n+1}(x)}{1-x} $$ where the $P_n(x)$ are the ordinary Legendre polynomials. Square this expression within the original sum, and there are three sums involving summands with $(P_n)^2,$ $(P_{n+1})^2$ and $P_n P_{n+1}.$ The 'perfect squares' can be dealt with eq. (B), although an index shift is needed for the $(P_{n+1})^2.$ This gives two of the $F(x,t)$ terms in the answer. The cross-term series can be dealt with a relationship among Legendre polynomials $$\sum_{n=1}^\infty P_n(x)P_{n-1}(x)t^n= \sum_{n=1}^\infty P_n(x)\Big(xP_n(x) + \frac{1-x^2}{n} \frac{d}{dx}P_n(x) \Big)t^n $$ Again, we have a 'perfect square' case that can be dealt with eq. (B). The last sum to deal with is $$ \sum_{n=1}^\infty P_n(x) \frac{d}{dx} P_n(x) \frac{t^n}{n} =\frac{1}{2} \frac{d}{dx} \sum_{n=1}^\infty P_n(x)^2 \frac{t^n}{n} $$ The first integral representation I gave for $G(x,t)$ involves integrating eq. B (specialized to y=x) with respect to $t.$ Then do the derivative with respect to $x$ afterwards. The second integral relationship started from an integral expression for the square of a Legendre polynomial, Gradshteyn and Rhyzik 7.137.5. Again, after getting the integral, differentiate with repsect to $x.$
I doubt the integrals simplify to elliptic functions, but would love to be proved wrong.