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Differential Calculus
23 Feb 2008 09:22:07 IST
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So you thought you were good at numbers!
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Let S = r - 2r2 + 3r3 - 4r4+...
Sr = r2 - 2r3 + 3r4 -
Hence S(1+r) = r - r2 + r3 -....
or S(1+r) = r/(r+1)
or S = r/(r+1)2
Now taking limit as r
1, we get
1, we get r
1 S = 1/4
1 S = 1/4Also r1 S = r1 r - 2r2 + 3r3 - 4r4+... = 1 - 2 + 3 - 4 + ....
1 S = r1 r - 2r2 + 3r3 - 4r4+... = 1 - 2 + 3 - 4 + ....Now 1 - 2 + 3 - 4 + .... = (1+2+3+4+...) - 2(2+4+6+8+...)
= (1+2+3+4+...) - 4(1+2+3+4+...)
= -3(1+2+3+4+...)
Hence 1+2+3+4+... = -1/12.
What's going on here?!!!!!
Comments (23)
23 Feb 2008 11:14:19 IST
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Just check your working, I don't think you will be able to evaluate the limit when you mutilpy by r2. I mean if you try S(1-r2), you will get a geometric sum, with no limit when r tends to 1.
If you multiply by r3 and in general odd powers of r, you will get the limit as 1/4. So no probs about the value of the limit
23 Feb 2008 11:38:31 IST
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yes bhattji ..but pls correct me if wrong
actually u have used that
r-r^2+r^3 ...= r/r+1
but sum of infinite series only can b written like that
and for that r<1 and secondly there is some ambiguity
with 1-2+3 ... as it cud b positive for odd n
anyway that i think is not a problem except that r /r+1
i think is not right
actually u have used that
r-r^2+r^3 ...= r/r+1
but sum of infinite series only can b written like that
and for that r<1 and secondly there is some ambiguity
with 1-2+3 ... as it cud b positive for odd n
anyway that i think is not a problem except that r /r+1
i think is not right
23 Feb 2008 11:42:32 IST
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Actually, this is a famous result by a great mathematician. And an accepted result. But it made no sense to me. Apparently it does to a lot of guys. I wanted to see your reactions to it.
Just check: http://www.mathlinks.ro/viewtopic.php?t=78489
sboosy i compared the limit of two equivalent expressions when r tends to 1. I didnt equate them for r=1.
23 Feb 2008 11:52:02 IST
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sboosy: I thought that's a regular sum of geometric progression with c.r. = -r. and |r|<1
@apurva: I checked even with r2 you get the same limit using S(1+r2). So there's no problem with the limit at all. Sorry it takes me some time to do these evaluations. 1 bcos its so long since i touched such stuff and 2. i have a job you know!
23 Feb 2008 12:04:50 IST
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Lets call S as f(r)
clearly f(r)<0 for r>1 and f(r)>0 for r<1 (even no. of terms)
also f(r) depends on the number of terms,
for n tendin to infinity, the ans 1/4 is rt if its r<1
the summation formula holds good only if |r|<1
so the limit must be taken as r tends to 1 minus.
u r not justified in using it for r=1.
and importantly at r=1
, the function has value -3*n*(n+1)/2 as u suggested were n is d no of terms.
so it is finite n depds. on d no. of terms.
so i think its basically to do wid d discontinuity of the function
as discontinuous,
u are not justified in equating summation of some terms less dan 1 upto infin(wich turns out to be finite ) to anoder finite quantity!!
i.e
so fn is -ve for r>1 and even no.of terms.
fn is pos. for r>1 and odd no. of terms.
if u take r tends to 1 plus ull end up wid same limit as f(r) at r=1.
so LHL not = RHL for f(r).
wat u havdone will hold good in a case wen f(r) is continuous, doesn depend on no. of terms frm either side. (which im not sure is poss.)
i think das d prob here.
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it can also be multiplied with r^2 and in that case a different result is obtained
just an attempt
plz correct me if i m wrong