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21-356 Lecture 8

# 21-356 Lecture 8 - N A multi-index is a vector =!1!N(N0 The...

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Theorem 29 (Multinomial Theorem) Let x =( x 1 ,...,x N ) " R N and let n " N . Then ( x 1 + ··· + x N ) n = ! " multi-index, | " | = n n ! ! ! x " . Proof. The proof by induction on N o r N =2 ,bytheb inom ia ltheorem ( x 1 + x 2 ) n = n ! k =0 " n k # x k 1 x n ! k 2 = n ! k =0 n ! k !( n # k )! x k 1 x n ! k 2 = ! " 1 + " 2 = n n ! ! 1 ! ! 2 ! x " 1 1 x " 2 2 = ! " | " | = n n ! ! ! x " , where in the third equality we have deFned ! 1 := k , ! 2 := n # k and in the fourth ! := ( ! 1 , ! 2 ) . Assume the result is true for 2 ,...,N # 1 and let’s prove it for N .G i v e n x x 1 N ) " R N ,u s ingth eindu c t ionhypo th e s i sfo r M and for 18

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M = N # 1 ,wehave ( x 1 + ··· + x N ) n =(( x 1 + + x N ! 1 )+ x N ) n = n ! k =0 n ! k !( n # k )! ( x 1 + + x N ! 1 ) k x n ! k N = n ! k =0 n ! k n # k )! ! # " ( N 0 ) N ! 1 , | # | = k k ! # ! y # x n ! k N = n ! k =0 ! # " ( N 0 ) N ! 1 , | # | = k n ! ( n # k )! 1 # ! y # x n ! k N = ! " multi-index, | " | = n n ! ! ! x " where y := ( x 1 ,...,x N ! 1 ) " R N ! 1 and in the last equality we set ! := ( # 1 ,..., # N ! 1 ,n # k ) ,sothat ! != # n # k )! and y # x n ! k N = x " . Given an open set U \$ R N ,f o rev e ryn onn e g a t iv ein t e g e r m " N 0 ,w e denote by C m ( U ) the space of all functions that are continuous together with their partial derivatives up to order m .W ese t C # ( U ):= # \$ m =0 C m ( U ) .
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21-356 Lecture 8 - N A multi-index is a vector =!1!N(N0 The...

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