Jacobi polynomials

[Herb]

I'm looking for the asymptotic behavior, just one term, of the nth
Jacobi polynomial with (1,-1) superscript and argument x which is less
than -1. Like, x= -3/2 would be typical. I need this behavior for
fixed x and n large. I have looked at several papers on the subject
but they seem to give varying results that are not purely real
numbers, or that oscillate. Any help would be appreciated.

[Roland Franzius]

Herb schrieb:

A Mathematica Plot is shows n -> Log[JacobiP[n,1,-1,-3/2]]/(-3/2)^n is
linear in (n,500,100) with coefficients

tb = Table[JacobiP[n, 1.0, -1.0, -3/2]/(-3/2)^n, {n, 500, 1000, 1}] //
Log;

Fit[tb, {1, x , x^2 , Log[x]}, x]

273.522+ 0.556084 n + 4.15696*10^-7 n^2 - 0.00262757 Log[n]

Its of course essential to use integer n only because the complex
approximations used for large n probably converge inside the unit disk
only and give pure nonsense for arguments outside.

--

Roland Franzius

[jum...@aussiemail.com.au]

In the following P(n,a,b,x) is the Jacobi polynomial P^(a,b)_n(x) and
P(n,x) the Legendre polynomial P_n(x) (=P(n,0,0,x)).

Theorem: The asymptotic behavior of P(n,1,-1,x) for n>>1 and fixed
x<-1 is

(1) P(n,1,-1,x) ~ ( f(n-1,x) - x f(n,x) ) / (1-x)

with f(n,x) = (-1)^n (-x+y)^(n+1/2) / sqrt(2*Pi*n*y) and y = sqrt
(x^2-1).

Remark: For x=-3/2 this is better than 1% accurate already for n=8.

Proof: (1) follows at once from

(2) (1-x) P(n,1,-1,x) = P(n-1,x) - x P(n,x) (for n>=1),

(3) P(n,x) ~ f(n,x) (for n>>1 and fixed x<-1).

Proof of (2): We start from the generating function
<http://functions.wolfram.com/05.06.11.0001.01>

sum_{n>=0} P(n,a,b,x) z^n = 2^(a+b) (1-z+R)^(-a) (1+z+R)^(-b) / R

with R = sqrt(1-2*x*z+z^2). For a=1 and b=-1 this gives

(1-x) sum_{n>=0} P(n,1,-1,x) z^n = (1-x) (1-z+R)^(-1) (1+z+R) / R

= R^(-1) (1-x) [1-(z+R)^2]/[(1-z)^2-R^2] = (z-x+R)/R = z/R - x/R - 1

Taking coefficients of z^n for n>=1 yields (2).

Proof of (3): We assume x<-1 and use the integral representation
<http://functions.wolfram.com/05.03.07.0001.01>

P(n,x) = int_0^Pi ( x - y cos(t) )^n dt / Pi

= (-1)^n/Pi int_0^Pi exp(n g(t)) dt

where g(t) = ln(-x + y cos(t)), which has a single maximum in [0,Pi]
at t=0 with g(0) = ln(-x+y) and g''(0) = -y/(-x+y) < 0. For n>>1
Laplace's method thus yields

P(n,x) ~ (-1)^n/Pi exp(n g(0)) int_0^infty exp(n g''(0) t^2 / 2) dt

= (-1)^n/Pi (-x+y)^n sqrt(Pi)/2 sqrt(2)/sqrt(-n g''(0)) = f(n,x)

which establishes (3).

Cheers,
Icke Selba