By Professor Dr. Arnold F. Nikiforov, Professor Dr. Vasilii B. Uvarov, Sergei K. Suslov (auth.)

Whereas classical orthogonal polynomials look as ideas to hypergeometric differential equations, these of a discrete variable become options of distinction equations of hypergeometric variety on lattices. The authors current a concise creation to this idea, providing even as equipment of fixing a wide type of distinction equations. They follow the speculation to varied difficulties in clinical computing, chance, queuing conception, coding and data compression. The ebook is an improved and revised model of the 1st version, released in Russian (Nauka 1985). scholars and scientists will discover a helpful textbook in numerical research.

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4). 15) u(x) . tm we have only to observe that LlTm(x) and Ll2u(x) are independent of x. tm-l + T' + (m - l)u". tk-l) = ,\ + mT' + tm(m - l)u" . 5. 4) in the self-adjoint fonn. 4) by a function g(x). 8) we obtain u(x)g(x)Ll\7y(x) + T(x)g(x)Lly(x) = u(x)g(x)Ll\7y(x) + \7y(x + l)Ll[u(x)g(x)] = Ll[u(x)g(x)\7y(x)] . As a result, Eq. 4) can be reduced to the fonn Ll[u(x)g(X)\7y(x)] + ,\g(x)y(x) =0 . 18) Similarly Eq. 19) where the function gm(x) satisfies the equation Ll[u(x)gm(X)] = Tm(X)gm(x) . 20) Like Eq.

An important special case of the Hahn polynomials are the Chebyshev polynomials of a discrete variable tn(x) = h~'O)(x, N), introduced in Ref. [TI], for which e(x) = 1. b) Let a(x) = x (x + ,1), a(x) + T(X) = (,2 - x )(/'3 - x) . 5) will be satisfied if ,1 > -1, ,2> N In this case putting I' = - 2, ,3 ,1, v = ,2 - ~x)= (I' > = N - 1. N + 1 we obtain C r(x + l)r(x + I' + l)r(N + v - x)r(N - x) -1, v> -1) . 9) with C are denoted by h~'v)(x, N). When m, v = Mp - m, m 1 C = M! (M - Mp)! 9) coincides with the hypergeometric distribution known in probability theory [K26].

2. 1) corresponding to the different degrees of the polynomial a(x). 1. Let a(x) be a polynomial of the second degree. We consider the following cases. a) Let a( x) :;:: x ('Yl - x), a( x) + r( x) :;:: (x + 'Y2)( /3 - x) . Here 'Yl, 'Y2, 'Y3 are constants. 5) will be satisfied if we take 'Yl :;:: N + 0'. , 'Y2 :;:: ,B + 1 (0'. > -1, ,B > -1), 'Y3:;:: N - 1. 1) assumes the form e(x + 1) e(x) :;:: (x + ,B + l)(N - 1 - x) (x + I)(N + 0'. - 1 - x) . 6) A solution of this equation is + 0'. \ (0'. > -1, ,B > -1) .