By Emil Grosswald (auth.)

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**Example text**

If a = l-n, equation (26) becomes y" _ n-I y' z b Z 2 n z and has the general solution y = --~ + ClZ+C2+bz log z. = 0, So, 17 This is a polynomial n only if b = 0, when it reduces to y = ClZ +c2, also of no further interest. Whenever such cases arise, not zero, or a negative integer. we shall tacitly or explicitly assume that a is CHAPTER 3 RECURRENCE RELATIONS 1. 5) = -K~+I + (~/z)K v. it follows that Kn+i/2(z) = (~/2)i/2e-Zz-n-i/2@ (z) and c o = (2n)'/2nn! • n " If we substitute this in (i) for Kv(z), we obtain after routine simplifications (see [17]) (5) @n+l(Z) = (2n+l)@n(Z) + Z2gn_l(Z).

Is based on two lemmas, the first of which is indeed well-known. LEMMA i. 4) in [ZO]). t k-n (k). ~. (n)(t_v) = v=0 It 28 k n k-n Z (k) xt = ( x + l ) k = ( x + l ) n ( x . l ) k - n = Z (n) xv Z ck-n)xWw t=O v=O w=O Proof. k Z xt t=0 k n n k-n Z (~) (k-n)w : Z xt Z (v) (t_v) v+w=t t=0 v=0 O~v~n O~wgk-n and the lemma follows from the comparison of the coefficients of x t in the first and last member. The function LEMMA 2. f(x) (23 n = (_l)v (v) r(k+~+x+l) Z r (n+x) v=O (k+v+x) ( k - l + v + x ) .

I)n (v)y v=0 Hence, by Leibniz' rule, the (n-k-l)-th derivative is of the form asy n+x-l-s.. n-t If-y) and vanishes for y = i, because n-t > n-(n-k-l) = s+t=n-k-i k+l > 1. By taking for a a natural integer and, in particular a = 2, we obtain from Theorem I the following corollaries: COROLLARY 2. (see [ 6 8 ] ) . For i n t e g r a l a > O, 30 Mk(Yn(Z;a,b);p) = (_b)k+l COROLLARY 3. (see [68]). k! (k+n+a-l)' (k-n) ! " k! a = 2, Mk[Y n (z;2,b);p) = (-b)k+l (k+n+l) ~ [k-n) : For (_l)n(b/z) n e-b/z n!