By BREZINSKI
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Extra info for Padé-Type Approximation and General Orthogonal Polynomials
Sample text
If qn (x) = x n , it is exactly Nuttall's compact formula. If qn(x)=Pn(x), then we obtain a result which is Chapter 1 36 closely related to the matrix interpretation of Pade approximants. This result will be given later on. 4. t 2k ~ . f(t)-[k -l/k]f(t) = Pk(t) i~ di+kt' where ~ = C(XiPk(X)) = bOci +b1Ci+l + ... +bkCi+k' Proof. Obvious since di = 0 for i = 0, ... 1. 5 is still valid and the next formula too. 3 also holds. • i=O i=O This property will be proved in Chapter 3. New properties are also satisfied by Pade approximants.
11. 1. Proof. Obvious since the Ok'S satisfy the same recurrence relation. 12. 0. Proof. Let us write the recurrence relation for P k + 1 and multiply it by Ok. Then write the recurrence relation for Ok+l and multiply it by Pk • If we subtract these two relations we get Pk(X)Ok+l(X) - OdX)Pk+1(X) = Ck+1[Pk-1(X)Qk (x)- Ok-l(X)Pk(x)]. Thus for k >0 Pk(X)Ok+l(X)- Ok(X)Pk+1(X) = Ck+1Ck··· C 2 [PO(X)01(X)- OO(X)Pl(X)] = Ak+1hk A1h o PO(X)Ol(X). Using the definitions of Al ho, Po and 0 1 we get the result.
12. 4. k Ok(X)P~+1(x)- Ok+1(x)PHx) = Ak+1hk L hi 1Pi (X)Oi(X), i=O O~(X)Pk+l(X)- 0~+1(X)Pk(X) = -A k+1hk k L hi1Pi (X)Oi(X), i=O Proof. 13 and dividing by x - t, we get Ok(t) Pk+1(X) - Pk+1(t) Ok+1(t) Pk(x)- Pk(t) x-t x-t k = Ak+1hk L hi 1Pi (X)Oi(t). i=O Let x - t and the first part of the corollary follows. 12 we get the second result. • Applying c to the first of these relations we get C(OkP~+1) = C(Ok+1 P O· Let k Hdx, t) = L hi1Pi(X)Oi(t), i=O then Let Mk(t) = (Ok«t» Pk t Nk(t) = ( Ok-l«t»), Akt+ Bk -Ck Pk - 1 t 1) 0' for for k = 0,1, ...