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C(n,n/2)/(2^(N+1)) > 1/(2*sqrt(n))

  1. Aug 5, 2011 #1
    1. The problem statement, all variables and given/known data

    For n even, C(n,n/2)/(2^(N+1)) > 1/(2*sqrt(n)).

    2. Relevant equations

    sqrt(2*pi*k) * k^k * e^-k < k! < sqrt(2*pi*k) * k^k * e^-k * (1 + 1/4*k)

    3. The attempt at a solution

    Have tried using the inequality to get lower bound for C(n,n/2), but get something smaller than 1/(2*sqrt(n)).
     
  2. jcsd
  3. Aug 6, 2011 #2

    Ray Vickson

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    If your inequality is *exactly* as written, it is false. Look at R(n) = 2*sqrt(n)*C(n,n/2)/2^(n+1). We can easily plot it, and see that it is strictly increasing, starting at R(1) = 1/sqrt(pi) = 0.6366 and rising asymptotically to R(infinity) = sqrt(2/pi) = 0.7979. In other words, we never have R(n) > 1, which your inequality gives. We can get simple bounds on R(n) by using bounds S(k) < k! < S(k)*exp(1/ 12k) [see Feller, Intro. to Probability for a simple proof], where S(k) = sqrt(2*pi*k)*k^k*exp(-k) is Stirling's formula. Then we get a lower bound on R(n) by using the lower bound in the numerator and upper bounds in the denominator, and vice-versa for an upper bound. In this way we get:
    sqrt(2/pi)*exp(-1/ 3n) < R(n) < sqrt(2/pi)*exp(1/ 12n). The upper bound is < 1 for all n >= 1.

    RGV
     
  4. Aug 11, 2011 #3
    Sorry for the mistake. This is from the top of page 17 in Approximation Of Functions By Rivlin,
    unless I'm reading it wrong. If you're familiar with the book, I'd be grateful for some help with this.
     
  5. Aug 11, 2011 #4

    Ray Vickson

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    I don't have the book, and since retiring and moving away from my old university I no longer have access to its books---only to some e-journals. What, exactly, is the statement you want to prove?

    RGV
     
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