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Let a1,…,an, not all equal to 0 , be such that ∑i=1nai=0. Show that there is a permutation i1,…,insuch that ∑j=1naijaij+1<0.
Hint: Use the probabilistic method. (It is interesting that there need not be a permutation whose sum of products of successive pairs is positive. For instance, if n=3, a1=a2=-1, and a3=2, there is no such permutation.)

Short Answer

Expert verified

∑j=1n-1aijaij+1<0

Step by step solution

01

:Concept Introduction

The question asks to show that there is a permutation i1,…,insuch that ∑j=1nai,ai,+1<0where a1,…,anare not all equal to 0and ∑i=1nai=0.

02

Explanation

The question asks to show that there is a permutation i1,…,insuch that ∑j=1nai,jai,+1<0where a1,…,anare not all equal to 0and ∑i=1nai=0

03

Explanation

Let I1,…,Inbe a random permutation that is equally likely to be any of the n ! Permutations Then:
Ealjalj+1=∑kEaljalj+1∣Ij=kPIj=k

04

Explanation

=1n∑kak·Ealj+1∣Ij=k
=1n∑kak·∑iai·PIj+1=i∣Ij=k

05

Explanation

=1n·(n-1)∑kak∑i=kai
=1n·(n-1)∑kak-ak

06

Explanation

Where the final equality followed from the assumption that ∑i=1nai=0. Since the preceding shows that
E∑j=1n-1aljalj+1<0

07

Final Answer

Therefore, one can conclude that there must be some permutation i1,…,infor which
$$∑j=1n-1aijaij+1<0

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