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

  1. Oct 24, 2012 #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.
     
  2. jcsd
  3. Oct 24, 2012 #2

    mathman

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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.
     
  4. Oct 25, 2012 #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: Oct 25, 2012
  5. Oct 25, 2012 #4

    pwsnafu

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    It's just a telescoping sum. Did you try?
     
  6. Oct 25, 2012 #5
    Ya,
    I thought maybe there is a way to derived it directly from xn-yn...
     
  7. Oct 25, 2012 #6

    MarneMath

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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.
     
  8. Oct 25, 2012 #7
    OK, thanks pals :)
     
  9. Oct 25, 2012 #8

    mathman

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