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Math Help - Solve A(B^-x) < (B-x)!

  1. #1
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    Solve A(B^-x) < (B-x)!

    I've been stuck for ages on this

    Solve A(B^-x) < (B-x)!

    B and A are constants, specifically,

    A = (365! / 0.25)
    B = 365
    x is an integer

    halp
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  2. #2
    Newbie karthiknadig's Avatar
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    Are you looking for the range of values x for which A(B^-x) < (B-x)! holds good? If so then the range is
    x <= -32
    x >= 32

    for values of x between -31 to 31 (included) it is invalid. Did it iteratively.
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  3. #3
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    Thanks for your effort.

    I'd actually already solved it iteratively using a calculator, but the people setting the question wanted a more analytic solution

    The trick is to use Stirling's (simple) formula, ln (x!) approximately equal to x.ln (x) - x

    I still didn't get a solution before I ran out of time.

    the range you came up with though was correct, which at least shows I was on the right track (I think!)
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  4. #4
    Newbie karthiknadig's Avatar
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    I'll try with the Stirling's formula.

    I used inequality in this form while iterating.

    \forall x \leq 0 Find x such that \displaystyle\prod_{i=0}^{|x|}\frac{365}{365 + i}
    and
    \forall x > 0 Find x such that \displaystyle\prod_{i=0}^{|x|}\frac{365 - i}{365}
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  5. #5
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    blimey! I havn't used iterative processes for years... I'm an undergrad physicist and

    If it helps, the problem is thus:

    What is the minimum number of people you need in a room for there to be a 75% chance that two people share the same birthday.

    my workings:

    P(n people do NOT share the same birthday) = 1 x (1- 1/365) x (1 - 2/365) x ... x ( 1 - n/365)

    here the first factor is the first person entering the room, etc

    we are looking for n such that

    PRODUCT'SUM' from i=0 to i = n-1
    (1 - i/365) = 0.25 *

    LHS = PRODUCT'SUM' 365^n (365 - i)
    = 365 ^ n [365 ! / n !]

    which, fitting this all into the * equation, gives us

    365 ! / 0.25 = 365^-n * n!

    hmm, perhaps I made an error when I last did it... anyway this was where I got stuck.


    though do remember that I have already handed my solutions in, I would still like to see the solution
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  6. #6
    Newbie karthiknadig's Avatar
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    If you are trying to solve it, as it is, I suggest you not to use Stirling's formula. 365! \approx e^{1796} but from Stirling's you get 365! \approx e^{1788}.

    Another way to find n is to use Taylor series expansion e^x \approx 1 + x for the probabilities as each person enters the room.

    Substitute,
    e^{\frac{-1}{365}} \approx 1 - \frac{1}{365}

    e^{\frac{-2}{365}} \approx 1 - \frac{2}{365}
    .
    e^{\frac{-(n-1)}{365}} \approx 1 - \frac{(n-1)}{365}

    Probability of n people not sharing birthday becomes,
    \bar{P} = e^{\frac{-(0+1+2+...+(n-1))}{365}}

    which can be further reduced since,
    1+2+...+(n-1) = \frac{n(n-1)}{2}

    Now,
    \bar{P} = e^{\frac{-n(n-1)}{2 \times 365}}

    Since,
    P = 1 - \bar{P} = 1 - e^{\frac{-n(n-1)}{2 \times 365}}

    Using \ln simplify,
    n^2 - n = 2 \times 365 \times \ln{(\frac{1}{1 - P})}

    For P = 0.75, you get n \approx 32

    Hope this helps.
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  7. #7
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    That's great, thankyou
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