21-356 Lecture 8

# 1 2 assume the result is true for 2 n

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Unformatted text preview: 1 2 k=0 ! x&quot;1 x&quot;2 = 1 2 xk xn!k = 12 &quot; multi-index, |&quot;|=n n! &quot; x, !! where in the third equality we have deﬁned !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 . Given x = (x1 , . . . , xN ) &quot; RN , using the induction hypothesis for M = 2 and for 18 M = N # 1, we have n n (x1 + · · · + xN ) = ((x1 + · · · + xN !1 ) + xN ) n ! n! k = (x1 + · · · + xN !1 ) xn!k N k ! (n # k )! = k=0 n ! k=0 = n! k ! (n # k )! n ! ! ! # &quot;(N0 )N !1 , |# |=k k=0 # &quot;(N0 )N !1 , |# |=k = ! &quot; multi-index, |&quot;|=n k ! # n!k y xN #! n! 1 # n!k y xN (n # k )! # ! n! &quot; x !! where y := (x1 , . . . , xN !1 ) &quot; RN !1 and in the last equality we set ! := (#1 , . . . , #N !1 , n # k ), so that !! = # ! (n # k )! and y# xn!k = x&quot; . N Given an open set U \$ RN , for every nonnegative integer m &quot; N0 , we denote by C m (U ) the space of all functions that are continu...
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## This document was uploaded on 03/31/2014.

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