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Topic: power of cosine function (Read 11538 times) 

comehome1981
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power of cosine function
« on: Oct 25^{th}, 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!

« Last Edit: Oct 26^{th}, 2010, 12:21pm by comehome1981 » 
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towr
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Re: power of cosine function
« Reply #1 on: Oct 26^{th}, 2010, 1:17am » 
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on Oct 25^{th}, 2010, 8:40pm, comehome1981 wrote: It is clear that [cos(k/n)]^(n^2) > exp{k^2} as n goes to infinity 
 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 
 Isn't it just zero? cos(pi/2)=0

« Last Edit: Oct 26^{th}, 2010, 1:19am by towr » 
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pex
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Re: power of cosine function
« Reply #2 on: Oct 26^{th}, 2010, 1:32am » 
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on Oct 26^{th}, 2010, 1:17am, towr wrote:Ah, wolframalpha says it should be exp( 1/2 k^2) 
 Taking logs and using L'Hopital twice also gives this result


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Re: power of cosine function
« Reply #3 on: Oct 26^{th}, 2010, 1:36am » 
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on Oct 25^{th}, 2010, 8:40pm, comehome1981 wrote:Does this hold when k=n/2 or n or some fraction of n? 
 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 26^{th}, 2010, 5:20am » 
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on Oct 26^{th}, 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). 
 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]

« Last Edit: Oct 26^{th}, 2010, 5:29am by towr » 
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comehome1981
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Re: power of cosine function
« Reply #5 on: Oct 26^{th}, 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 27^{th}, 2010, 12:31am » 
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on Oct 26^{th}, 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.) 
 I don't think we can simply cancel the exponent n^{2} 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 27^{th}, 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 n^{2} 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.

« Last Edit: Oct 27^{th}, 2010, 12:56am by towr » 
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comehome1981
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Re: power of cosine function
« Reply #9 on: Oct 27^{th}, 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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