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#1 Posted 10:11pm 13-03-10 LIMITHere goes my first question in TargetIIT :):):):):):) Find the following limits - 1 > lim [ [frac]( n ) ![/]n[p] n[/p][/frac] ] 2 > lim [ [frac]( n - 1 ) ![/]n[p] n - 1.5[/p][/frac] ] e[p] n[/p] In both cases n tends to infinity .
" BLACK HOLES RESULT FROM GOD DIVIDING THE UNIVERSE BY ZERO !!!!!! " ------ GALLARDO Edited on 10:12pm 13-03-10 |
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#2 Posted 10:14pm 13-03-10 Re: LIMITi hope [.] doesn't mean greatest integer function??????
"Every physicist thinks that he knows what a photon is...I spent my life to find out what a photon is and I still don't know it!!!"- _Einstein_ |
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#3 Posted 10:18pm 13-03-10 Re: LIMIT:)
"Every physicist thinks that he knows what a photon is...I spent my life to find out what a photon is and I still don't know it!!!"- _Einstein_ |
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#4 Posted 10:23pm 13-03-10 Re: LIMIT1) ans is 0
"Every physicist thinks that he knows what a photon is...I spent my life to find out what a photon is and I still don't know it!!!"- _Einstein_ |
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#5 Posted 10:26pm 13-03-10 Re: LIMIT[im]http://codecogs.izyba.com/gif.latex?\frac{n!}{n^{n}}=\frac{n(n-1)(n-2)(n-3)(n-4)....3.2.1}{n.n.n.n.n.....n.n.n}=\frac{n}{n}.\frac{n-1}{n}.\frac{n-2}{n}......\frac{3}{n}.\frac{2}{n}.\frac{1}{n}[/im] lim n->∞ => 1/n->0...hence the result!![1][1]
"Every physicist thinks that he knows what a photon is...I spent my life to find out what a photon is and I still don't know it!!!"- _Einstein_ |
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#6 Posted 10:30pm 13-03-10 Re: LIMITAns 1 [im]http://codecogs.izyba.com/gif.latex?\frac{n!}{n^{n}}%20=%20\frac{n(n-1)(n-2)...1}{n^{n}}%20=%20\frac{n^{n}{(1-\frac{1}{n})(1-\frac{2}{n})(1-\frac{3}{n})...}}{n^{n}}%20=%200..at..%20x%20-%20infinty[/im]
Edited on 01:02am 14-03-10 |
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#7 Posted 10:34pm 13-03-10 Re: LIMITthat is clear!!the LOT tougher thing!![1][1]
"Every physicist thinks that he knows what a photon is...I spent my life to find out what a photon is and I still don't know it!!!"- _Einstein_ |
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#8 Posted 10:36pm 13-03-10 Re: LIMIT2nd one apparently is either 0 or infinite!! not got any strong proof yet tho!![2][2]
"Every physicist thinks that he knows what a photon is...I spent my life to find out what a photon is and I still don't know it!!!"- _Einstein_ |
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#9 Posted 10:39pm 13-03-10 Re: LIMITchecking my basics!! lim (e[p]x[/p]-1)/x = 1 right?? or is this wrng???i mean is it e[p]x-1[/p] or e[p]x[/p]-1??? x->0
"Every physicist thinks that he knows what a photon is...I spent my life to find out what a photon is and I still don't know it!!!"- _Einstein_ |
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#10 Posted 00:13am 14-03-10 Re: LIMITno answer is surely 0 govind
"Every physicist thinks that he knows what a photon is...I spent my life to find out what a photon is and I still don't know it!!!"- _Einstein_ |
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#11 Posted 00:43am 14-03-10 Re: LIMITinfinite!! [hide]still unsure tho!![/hide]
"Every physicist thinks that he knows what a photon is...I spent my life to find out what a photon is and I still don't know it!!!"- _Einstein_ |
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#12 Posted 01:04am 14-03-10 Re: LIMITActually the answer's zero.
First is first , and the rest are no-where . |
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#13 Posted 01:07am 14-03-10 Re: LIMITAtleast that is what I think -- havn't been able to prove yet .
First is first , and the rest are no-where . |
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#14 Posted 01:34am 14-03-10 Re: LIMITi think 2nd answer is 1 {but i am only 95% sure } |
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#15 Posted 10:36am 14-03-10 Re: LIMITOKKK , here is the solution --- There is something called STIRLING'S APPROXIMATION , which states that for large enough n , we are allowed to write n ! = n[p] n[/p] e[p] - n[/p] [sqrt]2 Π n[/sqrt] or , [frac]( n - 1 ) ! x n[/] n[p] n[/p][/frac] e[p] n [/p] = [sqrt]2 Π [/sqrt] x n[p] . 5[/p] or , [frac]( n - 1 ) ![/]n[p] n - 1 . 5[/p][/frac] e[p] n[/p] = [sqrt]2 Π[/sqrt] or , k = [sqrt]2 Π[/sqrt] ( suppose ) Hence , lim[ss]n--> ∞ [/ss] k = [sqrt]2 Π[/sqrt] [hide]Actually , the expression that I have written for stirlings approximation is approximately correct , and the real expression is pretty big and involves big - O notation of Gauss and many other complicated things which is again very tough to grasp at once - but the result won't differ anyway . If you want to know more , just use WIKIPEDIA [/hide]
" BLACK HOLES RESULT FROM GOD DIVIDING THE UNIVERSE BY ZERO !!!!!! " ------ GALLARDO |
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#16 Posted 11:24am 14-03-10 Re: LIMITthat approx has been used a lot of times here..[1] so dont worry [6] |
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#17 Posted 6:47pm 14-03-10 Re: LIMIT@gallardo first of all r u a student....? ur ways doesnt seem ur r student becuz if i know correctly stirling approximation is neither in any olympiad topic and obviously not in jee.....so how u know abt it...? btw how will u evaluate this using stirling approx. ;) [im]http://codecogs.izyba.com/gif.latex?\lim_{n\to\infty}\;\;\;\left(\frac{n!\;\cdot\;n^{2007}}{2008\cdot2009\cdot2010\cdot2011\cdot2012\cdot\;\;\;\cdots\;\;\;\cdot%20(2007+n)}\right)[/im] goodluck :) 
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#18 Posted 7:06pm 14-03-10 Re: LIMITRearranging , the givrn term =[frac]n ! 2007 ! n[p]2007[/p][/]( n + 2007 ) ![/frac] =[frac]{ n[p]n[/p] n[p]2007[/p] } { 2007 ! } { [sqrt]2 pi n[/sqrt] } { e[p] - n [/p] }[/]( n + 2007 )[p]{ n + 2007 }[/p] { e[p] - n - 2007[/p] } { [sqrt] 2 pi ( n + 2007 ) [/sqrt] }[/frac] =[frac]{ 2007 ! } { n[p] n + 2007[/p] } { .......}[/]{ ...... }[/frac] = [frac] { 2007 ! }[/] { [ n + 2007 ] / n }[p] n + 2007[/p] { e[p] - n - 2007 + n [/p] } [sqrt][ n + 2007 ] / n [/sqrt][/frac] =[frac]{ 2007 ! }[/]{ 1 + [ 2007 / n ] } [p]n + 2007[/p] { e [p]- 2007[/p] } { [sqrt] 1 + [ 2007 / n ][/sqrt] }[/frac] = [frac]2007 ![/]e [p]- 2007[/p] e [p]2007[/p][/frac] = 2007 ! , after putting the limit .
" BLACK HOLES RESULT FROM GOD DIVIDING THE UNIVERSE BY ZERO !!!!!! " ------ GALLARDO Edited on 7:11pm 14-03-10 |
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#19 Posted 7:17pm 14-03-10 Re: LIMITyup...tahts correct... so u know much of higher real analysis.... gr8 :)
Edited on 7:17pm 14-03-10 |
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#20 Posted 7:28pm 14-03-10 Re: LIMITI think 1-[frac](n-1)[/]n[/frac] is not equal to ∞ as n →∞ since, 1-[frac]∞[/]∞[/frac] = 1-1 =0 is this correct
We dont shine in our life inspite of our hardwork, mainly bcoz we NEGLECT EVERY LITTLE THING. |
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#21 Posted 01:08am 15-03-10 Re: LIMITWOWWWW !!!!!! How did Gallardo do the generalisation , i mean the question ?????? When I fist saw the qs, , I was left completely dumbstruck !!! Gallardo bhai , KUDOSSSSS
First is first , and the rest are no-where . |
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#22 Posted 01:10am 15-03-10 Re: LIMITSTIRLINGS APPROXIMATION -- pretty interesting thing though :o :o !!
First is first , and the rest are no-where . |
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