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Math Help - ord function and convergent in Qp

  1. #1
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    ord function and convergent in Qp

    Solve the following :

    a) Show that
    ordp((p^n)!)=1+p+p^2+p^3+....+p^(n-1)

    b)For which values of p does the
    following series converge in
    Qp?
    1)1+(15/7)+(15/7)^2+(15/7)^3+......
    2)1!+2!+3!+4!+....


    For
    a) I wanna to count how many terms of (p^n)! containing the factor p but I
    failed using my way.

    For b) 1) I tried to use the definition of
    convergent in Qp but when i got the geometric series then it is complicated to
    analyse p and 2) I have no idea

    Can someone help me ?? many
    thanks







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  2. #2
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    Re: ord function and convergent in Qp

    Hey Sonifa.

    What is the ordp function for those unfamiliar with it?
    Thanks from LimpSpider
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  3. #3
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    Re: ord function and convergent in Qp

    Quote Originally Posted by chiro View Post
    Hey Sonifa.

    What is the ordp function for those unfamiliar with it?
    ordp(k)=e if p^e|k but p^(e-1)not|k
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  4. #4
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    Re: ord function and convergent in Qp

    Consider the set \{1,2,\ldots,p^n\}. Every p integers is divisible by p. There are \dfrac{p^n}{p} = p^{n-1} such numbers. Every integer divisible by p^2 is also divisible by p, so you need to cound those numbers an additional time (since the factorial takes a product of all of those numbers). There are \dfrac{p^n}{p^2} = p^{n-2} integers with p^2 as a factor. And in general, there are p^{n-m} integers with p^m as a factor. So, add that all up:

    \text{ord}_p\left( p^n\right)! = \sum_{m = 1}^n p^{n-m}

    For b) hint: (1) the answers should be 3 and 5. (2) all p

    Edit: for b) (1), I think p=\infty works, too
    Last edited by SlipEternal; November 17th 2013 at 05:40 PM.
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  5. #5
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    Re: ord function and convergent in Qp

    Quote Originally Posted by SlipEternal View Post
    Consider the set \{1,2,\ldots,p^n\}. Every p integers is divisible by p. There are \dfrac{p^n}{p} = p^{n-1} such numbers. Every integer divisible by p^2 is also divisible by p, so you need to cound those numbers an additional time (since the factorial takes a product of all of those numbers). There are \dfrac{p^n}{p^2} = p^{n-2} integers with p^2 as a factor. And in general, there are p^{n-m} integers with p^m as a factor. So, add that all up:

    \text{ord}_p\left( p^n\right)! = \sum_{m = 1}^n p^{n-m}

    For b) hint: (1) the answers should be 3 and 5. (2) all p

    Edit: for b) (1), I think p=\infty works, too
    many thanks, but i cannot catch the case for b)1 if p=infinite, what is definition for ordp(a) if p is infinite????
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  6. #6
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    Re: ord function and convergent in Qp

    I don't know if ord_\infty (a) is defined. But, | a |_\infty = \begin{cases}0 & \text{if }a=0 \\ \dfrac{1}{|a|} & \text{otherwise}\end{cases}
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  7. #7
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    Re: ord function and convergent in Qp

    Ok, i get it, thanks for ur reply
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