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Topic: power of cosine function (Read 11883 times) |
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comehome1981
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power of cosine function
« on: Oct 25th, 2010, 8:40pm » |
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A question as follows:
It is clear that
[cos(k/n)]^(n^2) ----> exp{-k^2/2} as n goes to infinity
Does this hold when k=n/2 or n or some fraction of n?
Another question:
Does one know the estimate of cos(x) as x --> pi/2
Thanks for any tips!
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| « Last Edit: Oct 26th, 2010, 12:21pm by comehome1981 » |
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towr
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Re: power of cosine function
« Reply #1 on: Oct 26th, 2010, 1:17am » |
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on Oct 25th, 2010, 8:40pm, comehome1981 wrote:
It is clear that
[cos(k/n)]^(n^2) ----> exp{-k^2} as n goes to infinity |
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I've tried graphing it for a few k, and I must say, it's anything but clear. [cos(k/n)]^(n^2) / exp{-k^2} doesn't seem to convergence on 1.
[edit]Ah, wolframalpha says it should be exp(- 1/2 k^2)[/edit]
Quote:| Does one know the estimate of cos(x) as x --> pi/2 |
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Isn't it just zero? cos(pi/2)=0
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| « Last Edit: Oct 26th, 2010, 1:19am by towr » |
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pex
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Re: power of cosine function
« Reply #2 on: Oct 26th, 2010, 1:32am » |
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on Oct 26th, 2010, 1:17am, towr wrote:| Ah, wolframalpha says it should be exp(- 1/2 k^2) |
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Taking logs and using L'Hopital twice also gives this result 
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pex
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Re: power of cosine function
« Reply #3 on: Oct 26th, 2010, 1:36am » |
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on Oct 25th, 2010, 8:40pm, comehome1981 wrote:| Does this hold when k=n/2 or n or some fraction of n? |
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No, it doesn't. If k = cn, clearly k/n = c and you're taking the limit of [cos(c)]n^2. This limit is obviously either 1 (if cos(c)=1), nonexistent (if cos(c)=-1), or zero (otherwise).
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towr
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Re: power of cosine function
« Reply #4 on: Oct 26th, 2010, 5:20am » |
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on Oct 26th, 2010, 1:36am, pex wrote:
No, it doesn't. If k = cn, clearly k/n = c and you're taking the limit of [cos(c)]n^2. This limit is obviously either 1 (if cos(c)=1), nonexistent (if cos(c)=-1), or zero (otherwise). |
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But the limit of exp{- 1/2 k^2} = exp{- 1/2 (cn)^2} would also be 0. So the two expressions might still converge. (Were it not that as far as I can tell they generally don't.)
[edit]
If you want
[cos(c)]n^2 -> exp{- 1/2 (cn)^2}
then you must have
cos(c) -> exp{- 1/2 c^2}
Since c is constant, the expressions on both sides have to be equal, and thus c must be 0. (Which means it falls under the original case, since k=cn is constant for c=0.)
[/edit]
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| « Last Edit: Oct 26th, 2010, 5:29am by towr » |
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comehome1981
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Re: power of cosine function
« Reply #5 on: Oct 26th, 2010, 12:24pm » |
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For the second part, I meant
is there a formula for the error bound for
|cos(x)| when x closes to pi/2
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pex
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Re: power of cosine function
« Reply #7 on: Oct 27th, 2010, 12:31am » |
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on Oct 26th, 2010, 5:20am, towr wrote:If you want
[cos(c)]n^2 -> exp{- 1/2 (cn)^2}
then you must have
cos(c) -> exp{- 1/2 c^2}
Since c is constant, the expressions on both sides have to be equal, and thus c must be 0. (Which means it falls under the original case, since k=cn is constant for c=0.) |
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I don't think we can simply cancel the exponent n2 when throwing infinities around: if you just want both sides to tend to zero, it is sufficient that c is not a multiple of pi. (But that's not something I would use the "->" notation for...)
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towr
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Re: power of cosine function
« Reply #8 on: Oct 27th, 2010, 12:49am » |
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Well, they do both tend to zero, generally; but I figure we're interested in asymptotic behaviour, i.e. that one function approaches the other as n gets larger (rather than that they both approach a common limit).
So in that case, their quotient should tend to 1; and then it seems to me you can just cancel the n2 factors, because they don't qualitatively change the asymptotic behaviour.
If a(x)x/b(x)x goes to 1 as x increases, then a(x)/b(x) must go to 1 as x increases (a necessary, but not sufficient condition). But in our case a(x) and b(x) are constants, and if they're not equal, then the expression goes to 0 or +/-infinity.
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| « Last Edit: Oct 27th, 2010, 12:56am by towr » |
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comehome1981
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Re: power of cosine function
« Reply #9 on: Oct 27th, 2010, 9:18pm » |
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Thanks for all you who replied.
All answers are helpful. I think I got some idea now.
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