Thread: ord function and convergent in Qp

1. 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

2. Re: ord function and convergent in Qp

Hey Sonifa.

What is the ordp function for those unfamiliar with it?

3. Re: ord function and convergent in Qp

Originally Posted by chiro
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

4. 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

5. Re: ord function and convergent in Qp

Originally Posted by SlipEternal
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????

6. Re: ord function and convergent in Qp

I don't know if $ord_\infty (a)$ is defined. But, $| a |_\infty = \begin{cases}0 & \text{if }a=0 \\ \dfrac{1}{|a|} & \text{otherwise}\end{cases}$

7. Re: ord function and convergent in Qp

Ok, i get it, thanks for ur reply