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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? Tags division, partial, quotient Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post guranbanan Calculus 2 June 14th, 2013 12:43 AM pomazebog Calculus 4 January 29th, 2012 03:06 AM tinynerdi Abstract Algebra 4 April 24th, 2010 01:24 PM bigli Abstract Algebra 3 February 13th, 2009 11:24 AM ricsi046 Real Analysis 7 December 31st, 1969 04:00 PM

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