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A comparison on binomial expansion

  1. Dec 6, 2009 #1
    Could anyone help me on this question? Is it true that
    [tex]\sum_{k=n+1}^{2n}\left(\begin{array}{c}
    2n\\k\end{array}\right)x^{k}\left(1-x\right)^{2n-k}\leq2x[/tex]
    for any [tex]x\in(0,1)[/tex] and any positive integer [tex]n[/tex]?

    Any help on that will be greatly appreciated!
     
    Last edited: Dec 6, 2009
  2. jcsd
  3. Dec 6, 2009 #2

    Gib Z

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    Well, the LHS seems to be the "second half" of the terms of the expansion of ( x+ (1-x) )^(2n) = 1.

    This gives that the LHS is [tex] 1 - \sum_{k=0}^n \left(\begin{array}{c}
    2n\\k\end{array}\right)x^{k}\left(1-x\right)^{2n-k}[/tex].

    I can't see how thats incorrect, but then I note the inequality doesn't seem to hold anymore for when x is small, for the LHS would seem to tend to 1 whilst the RHS goes to zero. Maybe I'm going crazy..
     
  4. Dec 6, 2009 #3
    When k=0, [tex]x^k(1-x)^{2n-k}[/tex] goes to 1 as x goes to 0, so [tex]1-\sum_{k=0}^{n}\left(\begin{array}{c}
    2n\\
    k\end{array}\right)x^{k}\left(1-x\right)^{2n-k}[/tex] tends to 0 as x goes to 0 because the other terms with k>0 has a factor x. But thanks anyway.
     
    Last edited: Dec 6, 2009
  5. Dec 6, 2009 #4

    Gib Z

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    Well if that is true, haven't you shown that the LHS is negative? I must have done something completely wrong as it seems my idea implies the original LHS must be negative..

    Today obviously isn't my day. Sorry for wasting your time !
     
  6. Dec 6, 2009 #5
    The LHS is always positive for [tex]x\in(0,1)[/tex]. That is fine.
     
  7. Dec 7, 2009 #6
    In fact, I have checked the cases of n=1 through 1000, and it holds for all the cases. But I still don't know how to show it in general. Any other help?
     
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