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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/((3n2)*(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/((3n2)*(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  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Partial quotient division Quote:
 
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  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Partial quotient division Quote:
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  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Partial quotient division Quote:
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  
Math Team Joined: Mar 2012 From: India, West Bengal Posts: 3,871 Thanks: 86 Math Focus: Number Theory  Re: Partial quotient division Quote:
 

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