Infinite sumsDate: 6/26/96 at 2:23:26 From: Anonymous Subject: Toughie Given the function f(x)= x^2/(1+x^2) find the sum f(1/1)+f(2/1)+...+f(n-1/1)+f(n/1) f(1/2)+f(2/2)+...+f(n-2/2)+f(n/2) f(1/n)+f(2/n)+...+f(n-1/n)+f(n/n) for any positive integers "N". Date: 6/26/96 at 11:47:1 From: Doctor Anthony Subject: Re: Toughie I assume from this that you require the sum of ALL the terms made up of n rows and n columns. Note first that f(x) + f(1/x) = x^2/(1+x^2) + (1/x^2)/(1+(1/x^2)) = x^2/(1+x^2) + 1/(1+x^2) = (x^2+1)/(1+x^2) = 1. If we consider the terms laid out in n rows and n columns, then the leading diagonal has terms like f(1/1), f(2/2)....f(n/n), each equal to f(1) = 1/2. So the terms in the leading diagonal add up to n/2. Now having removed the leading diagonal, you are left with n^2-n terms. These can be paired off, with each pair contributimg 1 to the total. So other terms contribute (n^2-n)/2. The overall total is therefore (n^2-n)/2 + n/2 = n^2/2. For n even we have the very simple result for n^2/2. Thus if n = 2 total is 4/2 = 2 if n = 4 total is 16/2 = 8 if n = 6 total is 36/2 = 18 and so on. n ODD. if n = 1 total is 1^2/2 = 1/2 n = 3 total is 9/2 = 4 +(1/2) n = 5 total is 25/2 = 12 +(1/2). With n ODD we shall always have a half as part of the sum. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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