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October 25th, 2013, 03:56 AM   #1
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Partial quotient division

Hello,
can someone tell me how can i learn this?I just don't see what i should do with the quotients to divide them into partial quotients.
For example:
1/(n*(n+3))
or 1/((3n-2)*(3n+1))
etc
Please don't just tell me the solution i want to know how to treat these quotients.I need this to calculate infinite sums like sum( 1/((3n-2)*(3n+1)) ) n=1 to infinity.
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October 25th, 2013, 05:14 AM   #2
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Re: Partial quotient division

Well, the basic idea is to solve the diophantine equation mA + nB = 1 over Z[m, n] where mn is the denominator of the fraction you want to split.
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October 26th, 2013, 03:56 AM   #3
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Re: Partial quotient division

but in that equation you have 4 unknown
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October 26th, 2013, 04:36 AM   #4
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Re: Partial quotient division

Quote:
Originally Posted by ricsi046
but in that equation you have 4 unknown
2 unknowns, actually. (m, n) is the denominator pair.
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October 29th, 2013, 08:41 AM   #5
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Re: Partial quotient division

sorry for the late answer,but if i have a fraction like this:
1/(n*(n+1)*(n+2)) i can make it like this 1/((n^2*n)*(n+2))
then i have this equation
(n^2*n)*x + (n+2)*y = 1
and how do i solve it?we have learnt only a bit about diophantine equations(solved equations like 21x+31y=121)
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October 29th, 2013, 09:54 AM   #6
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Re: Partial quotient division

Quote:
Originally Posted by ricsi046
1/(n*(n+1)*(n+2)) i can make it like this 1/((n^2*n)*(n+2))
It's better to partially factorize it when in the 3rd form. I will show you how to do it by a simple example :

1/((3n + 1)*(3n - 2)). If one considers it factors as A/(3n + 1) - B/(3n - 2), then A = B = 1 gives 3n - 2 - 3n + 1 = -1 in the numerator. So, you have the decomposition now.

Do you see how it is done? Can you do a few exercises by yourself?
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October 29th, 2013, 10:16 AM   #7
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Re: Partial quotient division

in my example i got A=1,B=1/n is that correct?
so 1/((n^2*n)*(n+2)) = (1/(n^2*n))-((1/n)/(n+2)) = (n + 2 - n - 1)/(n^2*n)*(n+2) = 1/((n^2*n)*(n+2)).
Isn't it a problem that B is not a fixed number?I need to work with series and create telescopic sums,thats why i need partial quotient division.
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October 29th, 2013, 11:13 AM   #8
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Re: Partial quotient division

Quote:
Originally Posted by ricsi046
Isn't it a problem that B is not a fixed number?
Yes, it would. That's why said that you need to use 1/(n(n+1)(n+2)) rather than 1/((n^2+n)(n+2)).

By using the former, one arrives at the diophantine form (n + 1)(n + 2)A + n(n + 2)B + n(n + 1)C = 1. Put n = -1 to have B = -1, n = 0 to have A = 1/2 and n = -2 to get C = 1/2.

Balarka
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October 29th, 2013, 11:51 AM   #9
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Re: Partial quotient division

ah i see,so i can have many letter(A,B,C,D,...) not only 2.
Now i don't understand why is it (n + 1)(n + 2)A + n(n + 2)B + n(n + 1)C = 1 ?
why not n*A + (n+2)*B + (n+2)*C =1 ?
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October 29th, 2013, 12:19 PM   #10
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Re: Partial quotient division

Quote:
Originally Posted by ricsi046
why not n*A + (n+2)*B + (n+2)*C =1 ?
If your partial decomposition is of the form A/n + B/(n+1) + C/(n+2), Don't you have (n + 1)(n + 2)A + n(n + 2)B + n(n + 1)C = 1?
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