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 October 25th, 2013, 03:56 AM #1 Member   Joined: Sep 2013 Posts: 93 Thanks: 2 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.
 October 25th, 2013, 05:14 AM #2 Math Team     Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory 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.
 October 26th, 2013, 03:56 AM #3 Member   Joined: Sep 2013 Posts: 93 Thanks: 2 Re: Partial quotient division but in that equation you have 4 unknown
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.

 October 29th, 2013, 08:41 AM #5 Member   Joined: Sep 2013 Posts: 93 Thanks: 2 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)
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?

 October 29th, 2013, 10:16 AM #7 Member   Joined: Sep 2013 Posts: 93 Thanks: 2 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.
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
.

 October 29th, 2013, 11:51 AM #9 Member   Joined: Sep 2013 Posts: 93 Thanks: 2 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 ?
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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