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Recurrence equations and their classical orthogonal polynomial ...

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318 W. Koepf, D. Schmersau / Appl. Math. Comput. 128 (2002) 303–327<br />

the falling factorials xn :¼ xðx 1Þ ðx n þ 1Þ satisfy<br />

Dx n ¼ nx n 1 :<br />

It turns out that they are connected with the Charlier <strong>polynomial</strong>s by the<br />

limiting process<br />

limð<br />

1Þ<br />

l!0 n l n c ðlÞ<br />

n<br />

ðxÞ ¼lim<br />

l!0 ðx n þ 1Þ n 1F1<br />

¼ð 1Þ n ðx n þ 1Þ n ¼ x n ;<br />

n<br />

x n þ 1 l<br />

where we used the hypergeometric representation given in [18, (2.7.9)].<br />

Note, however, that other than in the differential equation case the above<br />

type of difference equation is not invariant under general linear transformations,<br />

but only under integer shifts. We will have to take this under consideration.<br />

The <strong>classical</strong> discrete <strong>orthogonal</strong> <strong>polynomial</strong>s satisfy a recurrence equation<br />

(1)<br />

pnþ1ðxÞ ¼ðAnx þ BnÞpnðxÞ Cnpn 1ðxÞ<br />

with An; Bn <strong>and</strong> Cn given by Theorem 1.<br />

Similarly as in the continuous case, this information can be used to generate<br />

an algorithm to test whether or not a given holonomic recurrence equation has<br />

<strong>classical</strong> discrete <strong>orthogonal</strong> <strong>polynomial</strong> solutions. Obviously the first three<br />

steps of this algorithm agree with those given in Algorithm 1.<br />

Algorithm 2. This algorithm decides whether a given holonomic three-term<br />

recurrence equation has <strong>classical</strong> discrete <strong>orthogonal</strong> <strong>polynomial</strong> solutions, <strong>and</strong><br />

returns <strong>their</strong> data if applicable.<br />

1. Input: A holonomic three-term recurrence equation<br />

qnðxÞpnþ2ðxÞþrnðxÞpnþ1ðxÞþsnðxÞpnðxÞ ¼0 ðqnðxÞ; rnðxÞ; snðxÞ 2Q½n; xŠÞ:<br />

2. Shift: Shift by maxfn 2 N0 j n is zero of either qn<br />

essary.<br />

1ðxÞ or snðxÞg þ 1ifnec-<br />

3. Rewriting: Rewrite the recurrence equation in the form<br />

pnþ1ðxÞ ¼tnðxÞpnðxÞþunðxÞpn 1ðxÞ ðtnðxÞ; unðxÞ 2Qðn; xÞÞ:<br />

If either tnðxÞ is not a <strong>polynomial</strong> of degree one in x or unðxÞ is not constant<br />

with respect to x, return ‘‘no <strong>orthogonal</strong> <strong>polynomial</strong> solution<br />

exists’’; exit.<br />

4. Linear transformation: Rewrite the recurrence equation by the linear transformation<br />

x 7! ðx gÞ=f with (as yet) unknowns f <strong>and</strong> g.<br />

5. St<strong>and</strong>ardization: Given now An; Bn <strong>and</strong> Cn by<br />

pnþ1ðxÞ ¼ðAnxþ BnÞpnðxÞ<br />

define<br />

Cnpn 1ðxÞ ðAn; Bn; Cn 2 QðnÞ; An 6¼ 0Þ;

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