Looking for an idea for proving inequality, probably using binomial theorem.

  • Thread starter Polychoron
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  • #1
Guess what? I just got my new calculus book last week! ^^

The book opens with the definition of the real numbers by Dedekind and goes to prove properties of this numbering system such as The supremum axiom and others.
At the end of the chapter are about 30 exercises without their solutions; some practice of the chapter and some algebraic preparation for the next chapter deals with limits (mainly inequalities).

I want to consult with you about one inequality:
Prove that if x>y>0 and n>1 natural number than
n(x-y)yn-1< xn - yn < n(x-y)xn-1

I managed to elegantly prove the left part and it goes like that:
[tex]x^{n}=[y+(x-y)]^{n}=\sum_{k=0}^{n}\binom{n}{k}y^{k}(x-y)^{n-k}=y^{n}+\sum_{k=0}^{n-1}\binom{n}{k}y^{k}(x-y)^{n-k}[/tex]
from here:
[tex]x^{n}-y^{n}=\sum_{k=0}^{n-1}\binom{n}{k}y^{k}(x-y)^{n-k}=(x-y)[\sum_{k=0}^{n-2}\binom{n}{k}y^{k}(x-y)^{n-k-1}+\binom{n}{n-1}y^{n-1}]>(x-y)ny^{n-1}[/tex]
As we were asked to show.

About the right inequality, well, it's not working so nicely...
When I try the same concept but slightly different, I have a hard time showing the last transition (where the "<" sine comes to play)

I got:
[tex]x^{n}-y^{n}=(x-y)[\sum_{k=0}^{n-2}\binom{n}{k}x^{k}(y-x)^{n-k-1}+\binom{n}{n-1}x^{n-1}][/tex]

Ans it stuck.
 

Answers and Replies

  • #2
mathman
Science Advisor
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xn - yn = (x-y)(xn-1 + xn-2y +.... + xyn-2 + yn-1)

Since there are n terms inside the parenthesis, the inequalities follow immediately.
 
  • #3
Nice.

I agree that if the identity is right then the inequalities immediately derive from it. But can you prove it?
Of course I can see it by open the parenthesis, I simply don't fan of identities that came out of nowhere...
 
Last edited:
  • #4
pwsnafu
Science Advisor
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But can you prove it?

It's just a telescoping sum. Did you try?
 
  • #5
Ya,
Of course I can see it by open the parenthesis, I simply don't fan of identities that came out of nowhere...
I thought maybe there is a way to derived it directly from xn-yn...
 
  • #6
MarneMath
Education Advisor
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I guess if you wanted to prove what mathman used you could use the polynomial reminder theorem, but it's fairly obvious and I think common knowledge. Perhaps induction would be easier, either way, it's straight forward enough, if you desire to do so.
 
  • #7
OK, thanks pals :)
 
  • #8
mathman
Science Advisor
7,942
496
Nice.

I agree that if the identity is right then the inequalities immediately derive from it. But can you prove it?
Of course I can see it by open the parenthesis, I simply don't fan of identities that came out of nowhere...

Use the geometric series summation formula with a = xn-1 and r = y/x
 

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