Complex Analysis and Algebraic Geometry by Grauert G. (ed.)

By Grauert G. (ed.)

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Yn) E F; we start with a calculation of syndromes: F; n Sj =Sj(y) = LYi

Prove that Krawtchouk polynomials have the following properties: (a) Pi(U)=L]=O(-qy(q-l)i-J(~=j)0); (b) Pi(U) =L]=o(-I Yi/-J (n-;+j) (~=;); (c) Pi(U) is polynomial of degree i in u, with leading coefficient (-q)iji! and constant tenn (~) (q - l)i; Bounds on Codes 39 (d) Orthogonality relations: Ita G) (q - I)' Pi (l)Pj(l) = qn(q - l)i (~) aij; (e) (q-l)'G)Pi(l) = (q-l)iG)P,(i); (t) Il=oPi(l)P,(j) = qnaij; (g) Recurrence: (i + 1)Pi+ 1(u) = (( n - i) (q - 1) + i - qu )Pi (u) -(q-l)(n-i+l)Pi-l, Po = 1,Pl (u) ={q - l)n - qu; (h) Iff(u) is a polynomial of degree t and t feu) = I aiPi(U), i=O then n ai = q-n If(j)Pj(i).

Consider a linear space Lm (r) of all polynomials of degree at most r in m variables over Fq • Fix a subset P = {YI, ... ,Yn} ~ F:; and consider the evaluation map: Ev : Lm(r) -+F;, fH- (f(yJ), ... ,J(yn)). Set C = Ev(Lm(r)). Finding out the parameters ofC is a rather difficult problem. For simplicity let us suppose that P = F:;, n = qm. The map Ev in general is not injective (in fact Ev(f) = Ev(fq) for everyf). Let L~ (r) be the space spanned by monomials of the form Ufl ... u~,m, 0 ::; Otj ::; q - I, L Otj = r.

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