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watchmath

  • 5 years ago

Interesting problem! Let f(n) be the number of zeros in n! Prove that lim f(n)/n = 1/4 (as n approaches infinity)

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  1. anonymous
    • 5 years ago
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    We have the number of zeroes in n! :\[\sum_{i}^{}\lfloor n/5^{i} \rfloor\] this finds the "5's" and we have 5*2=10 which will give a zero. This is a start I geuss.

  2. watchmath
    • 5 years ago
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    Great!

  3. anonymous
    • 5 years ago
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    still trying to think about why the quotient goes to 1/4

  4. watchmath
    • 5 years ago
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    Now if we look at that series. Is there any relation between that "kind" of sereis(after we divide by n) and 1/4 ?

  5. anonymous
    • 5 years ago
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    I think we can approximate it as the geometric series: \[\sum_{i=1}^{n}(1/5)^{i}\] which has sum 5/4-1=1/4 when n goes to infinity.

  6. watchmath
    • 5 years ago
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    :) Now we just need to make it rigorous by figuring out how to open up that greatest integer function.

  7. anonymous
    • 5 years ago
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    show that the error goes to zero when we approximate this way?

  8. watchmath
    • 5 years ago
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    yes Note that the upper limit of the summation is \(\log_5n\), and this upper limit is somehow important.

  9. anonymous
    • 5 years ago
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    Well in the approximation the upper limit is infinity, so the excess that is being added is: \[\sum_{i=(\log_{5} n)+1 }^{\infty}(1/5)^{i}\] this goes to 0 as n goes to infinity. Is this right? Also I'm trying to think of how to deal with the error from roundoff using the floor function.

  10. watchmath
    • 5 years ago
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    Now quite because \(\lfloor n/5^i\rfloor/n\neq 1/5^i \) in general.

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