Monday, August 13, 2018

abstract algebra - Polynomial with n real roots



Let P(x)=xn+an1xn1++a1x+1 where ai are nonnegative and real. Assume P has n real roots.


Prove P(2)3n.


I thought I had a good idea about rewriting P as (xα1)(xαn). The fact that the roots are real means that you can order them. Choose the largest root, αk, then 2αk would have the smallest such value among the αi. And so we would have P(2)(2αk)n.


But I haven't been able to think of a way to compare it with 3n.


Answer



The proof follows from the following Lemma:


Lemma If $0

Proof: (2+a)(2+b)3(2+ab)4+2a+2b+ab6+3ab022a2b+2ab02(a1)(b1)


QED Lemma


Now, lets solve the problem. Let bi=α1, and we can assume without loss of generality that b1b2...bn.



As b1..bn=1 it follows that bn1b1...bk1


Then, by repeadely applying the previous lemma, we get


(2+b1)(2+b2)(2+b3)...(2+bn)3(2+b1b2)(2+b3)...(2+bn)32(2+b1b2b3)...(2+bn)....3k1(2+b1b2b3...bk)(2+bk+1...(2+bn)...3n1(2+b1b2b3...bn)=3n


Much simpler second solution:


If b1,..,bn are non-negative numbers then by the AM-GM inequality:


2+bi=1+1+bi33bi


Therefore (2+b1)(2+b2)(2+b3)...(2+bn)3n3b1b2...bn=3n


No comments:

Post a Comment

analysis - Injection, making bijection

I have injection f:AB and I want to get bijection. Can I just resting codomain to f(A)? I know that every function i...