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Algebra
6 Feb 2008 17:52:27 IST
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6 Feb 2008 21:44:58 IST
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@Neeraj: I was going to say this when my connection timed out-
The approach is good, only you have pretty confidently taken log on both sides without heed to whether the arguments are +ve or not. And not knowing complex analysis and logs of -ve numbers in complex form.
That is why I am waiting for a watertight approach.
So guys/gals dont lose heart! There still are points for the taking
The approach is good, only you have pretty confidently taken log on both sides without heed to whether the arguments are +ve or not. And not knowing complex analysis and logs of -ve numbers in complex form.
That is why I am waiting for a watertight approach.
So guys/gals dont lose heart! There still are points for the taking
6 Feb 2008 21:49:08 IST
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nahi neeraj bhai, there are finicky mathematicians who will take you to task for such oversights (our friend Konichiwa has caught me n times being careless). Math = Rigour in arguments.
That's why I think this business of having JEE in objective format is not a good idea.
That's why I think this business of having JEE in objective format is not a good idea.
7 Feb 2008 09:04:00 IST
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Well this is not a very different method, but anyway...
In general if 
, are the roots of a polynomial 
of degree n, then 
this is very easy to see: let
then 
now differentiate with respect to x to get the result. so, for example, if
,
then
, and 
thus 
, are the roots of a polynomial of degree n, then this is very easy to see: let
then now differentiate with respect to x to get the result. so, for example, if
, then
, and thus ps. you can do this by another method, but its pretty tedious. You can look at the case e.g n=3 or n=4 and then generalize.Clear the denominators, simplify, and group on the left. using that sums of roots, products of pairs of roots, etc are -1,+1, -1 , + 1 etc from coefficients of P(x) combined with P(1) = n+1, you can prove it.
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take log...
log(x-x1) + log (x-x2) +.....log(x-xn) = log(x^n+x^(n-1)+...+1)
now differentiate both sides...
1/x-x1 + 1/x-x2 +..+1/x-xn = nx^(n-1)+(n-1)x^(n-2)+...+1 /(x^n+x^(n-1) +...+1)
put x=1...
required value = (1+ 2 +...+n )/ (1+1+....+1) --->n+1 times
so required value = n/2 (using 1+2+....+n = n(n+1)/2)