Re: Q Legendre orthogonal polynomials mistake
- To: mathgroup at smc.vnet.net
- Subject: [mg67907] Re: Q Legendre orthogonal polynomials mistake
- From: Roger Bagula <rlbagula at sbcglobal.net>
- Date: Wed, 12 Jul 2006 05:06:13 -0400 (EDT)
- References: <e8nt4n$k7c$1@smc.vnet.net> <e8qg8f$o63$1@smc.vnet.net> <e8tbj5$227$1@smc.vnet.net> <e8vtqe$ssc$1@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Paul Abbott,
Well it is nice to know it is built in.
My written documentation is older ( version3).
In that book only the P function is documented.
The Q was added in either version 4 or version 5.
Trying to run that code pulled down my system yesterday.( crash, burn...
yeah)
My point was that my toral inverse polynomials
f[x_.n_]=x^n*LengendreP[n,1/x]
are new and different than LengendreQ[n,x].
I'd appreaciate you guys stop saying my questions are "nonsense."
I view some of the respomses as irresposible too.
Now that I know it is built in I can try a limited 1dimensional system using
the Q polynomials and see how that works.
See the post below that seems to have gotten lost!
Roger Bagula
Paul Abbott wrote:
>In article <e8vtqe$ssc$1 at smc.vnet.net>,
> Roger Bagula <rlbagula at sbcglobal.net> wrote:
>
>>I got the ArcTan mixed with ArcTanh:
>>
>>(*Jahnke and Emde, Page 111*)
>>W[x_, n_] = Sum[LegendreP[m - 1, x]*LegendreP[n - m, x]/m, {m, 1, n}]
>>Q[n_, x_] = ArcTanh[x]*LegendreP[n, x] - W[x, n]
>>p0 = Table[Q[n, x], {n, 0, 5}]
>>norm = Table[(1/Integrate[p0[[n]]*p0[[n]], {x, -1, 1}])^(1/2), {n, 1, 6}]
>>p = Table[p0[[n]]*norm[[n]], {n, 1, 6}]
>>Inm = Table[N[Integrate[p[[n]]*p[[m]], {x, -1, 1}]], {n, 1, 6}, {m,
1, 6}]
>>MatrixForm[Inm]
>>
>>A way to check is the Integral (Abramowitz and Stegun page 337 gives):
>>Table[Integrate[LegendreP[n, x]* Q[m, x], {x, 1, Infinity}], {m, 0, 5},
>>{n, 0, 5}]
>>
>>Table[(m-n)/(m+n+1), {m, 0, 5}, {n, 0, 5}]
>
>
>I do not understand your question. First, note that LegendreQ is
>built-in. Next, note that there are 3 types of Legendre function (the
>3rd type is required here) and that the integral on Abramowitz and
>Stegun page 337, which is online at
>
> http://www.convertit.com/Go/ConvertIt/Reference/AMS55.ASP
>
>is only valid with conditions on the indices.
>
>As a check of Abramowitz and Stegun 8.14.1,
>
> Table[1/((m - n) (m + n + 1)) ==
> Integrate[LegendreP[n, x]*LegendreQ[m, 0, 3, x], {x, 1, Infinity}],
> {n, 0, 5}, {m, n + 1, 5}] // Flatten // Union
>
> {True}
>
>Cheers,
>Paul
>
>_______________________________________________________________________
>Paul Abbott Phone: 61 8 6488 2734
>School of Physics, M013 Fax: +61 8 6488 1014
>The University of Western Australia (CRICOS Provider No
00126G)
>AUSTRALIA http://physics.uwa.edu.au/~paul
>
>
Posted to that thread before but not up.
-------- Original Message --------
Subject: [mg67907] interesting non- orthogonal polynomials
From: Roger Bagula <rlbagula at sbcglobal.net>
To: mathgroup at smc.vnet.net
References: <e8nt4n$k7c$1 at smc.vnet.net> <e8qg8f$o63$1 at smc.vnet.net>
<e8tbj5$227$1 at smc.vnet.net>
These inversions are probably a new set of polynomials?
The idea is to do the 1/x type inversion that
is invariant in tori to the Legendre polynomials.
The spherical harmonics that result aren't orthogonal to each other.
They are also non-orthogonal on domain {1, Infinity}
They aren't the traditional Arctan[x] Q[n] functions associated with
Legendre P{n]'s.
Mathematica doesn't have a function for those.
For the regular orthogonals the normalizaion is
Integrate[LegendreP[n, x]*LegendreP[m,x],{x,-1,1}]=delta[n,m]*2/(2*n+1)
http://www.du.edu/~jcalvert/math/legendre.htm
Most energy type calculastion are made with the second type
LegendreP[n,m,x]
tpolynomials that are used in Schrödinger calculations.
(* toral inverse polynomials*)
q0 = Table[ExpandAll[x^n*LegendreP[n, 1/x]], {n, 0, 5}]
normq = Table[(1/Integrate[q0[[n]]*q0[[n]], {x, -1, 1}])^(1/2), {n, 1, 6}]
q = Table[q0[[n]]*normq[[n]], {n, 1, 6}]
Inm = Table[N[Integrate[q[[n]]*q[[m]], {x, -1, 1}]], {n, 1, 6}, {m, 1, 6}]
MatrixForm[Inm]