| Author |
Message |
![[Post New]](/templates/default/images/icon_minipost_new.gif) 3 Feb 2008 18:15:46 IST
|
|
|
a1=(a+b)/2 , b1=(a1b)1/2, a2=(a1+b1)/2, b2=(a2b1)1/2 find a infinity and b infinty
|
|
|
|
3 Feb 2008 18:51:29 IST
|
|
|
They both are equal because at infinity there is no diff between inf & inf + 1 So in the first two relns put a( inf + 1 ) = a( inf ) & b ( inf + 1 ) = b ( inf ) U ll get the result .
|
this reply: 0 points
(with 0 
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
3 Feb 2008 18:58:03 IST
|
|
|
We can assume that b>a Looks like the only conclusion is that the two sequences converge towards each other as we always have an<bn<bn-1. Thus bn is a decreasing sequence bounded below by an. Also an is an increasing sequence bounded above by bn. A similar argument can be made for the case b<a. I would also like to point out that inf+1  inf does not lead us to conclude that a na n-1 etc.as n goes to inf
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
8 Feb 2008 13:00:05 IST
|
|
|
both are same.
the sequence will be a a1 a2 b2 b1 b
since it is sum to infinity, they are almost equal
|
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
8 Feb 2008 21:04:24 IST
|
|
|
Well done guys.
|
The most incomprehensible thing about the world is that it is
at all comprehensible. |
this reply: 0 points
(with 0
in 0 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
11 Feb 2008 13:50:27 IST
|
|
|
Well, I happened across this soln for the case a<b in a book. The author of the book says "By the way, this is a hard problem by any competition standard". Anyway here goes: We have already seen in my previous post that a<b means an<bn for all n. Part 1: Now an+1/bn+1 = an+1/ an+1bn = (an+1/bn ) = (an+bn)/2bn = (1+an/bn)/2 Since an/bn<1 Let an/bn = cos Then an+1/bn+1 = (1+cos)/2 = cos/2 Hence if a/b = cos then an/bn = cos(/2n) Part 2: Now consider (bn+1)2-(an+1)2 = (an+bn)/2 (bn-an+bn/2) = (bn2 - an2)/4 Hence we get (bn2 - an2) = (b2-a2)/2n. which gives bn(1 - an2/ bn2) = bn sin/2n = (b2-a2)/2n In the limit n  , if limit of the sequence {bn} is is l Then l/2n = (b2-a2)/2n or l = (b2-a2)/ where = cos-1(a/b) Hence limit {an} = limit {bn} = (b2-a2)/cos-1(a/b) Phew! One example, consider a sequence x0 = a<1, xn = (1+xn-1)/2 find the limit of 2n(1-xn2).
|
this reply: 12 points
(with 2
in 3 votes ) [?]
|
|
You have to be logged on to rate
|
|
|
|
|
Moderation Team
![[Up]](/templates/default/images/icon_up.gif "[Up]"){kind=icon}
