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Title: Another Sequence Post by ThudanBlunder on Apr 18th, 2007, 8:21am 35, 45, 60, X, 120, 180, 280, 450, 744, 1260 |
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Title: Re: Another Series Post by Grimbal on Apr 18th, 2007, 9:25am Roughly 80 or 81? |
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Title: Re: Another Series Post by ThudanBlunder on Apr 18th, 2007, 10:15am on 04/18/07 at 09:25:50, Grimbal wrote:
Yes, it's a number in the 80's. |
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Title: Re: Another Series Post by ThudanBlunder on Apr 19th, 2007, 9:16am Actually, I should ask for a simple continuous function which generates the series. |
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Title: Re: Another Series Post by Grimbal on Apr 19th, 2007, 3:20pm f(x)*35 + f(x-1)*45 + f(x-2)*60 + f(x-3)*X + f(x-4)*120 + f(x-5)*180 + f(x-6)*280 + f(x-7)*450 + f(x-8)*744 + f(x-9)*1260 where f(x) = sin(x·pi)/(x·pi) or something like that. |
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Title: Re: Another Sequence Post by ThudanBlunder on Apr 19th, 2007, 3:38pm on 04/19/07 at 15:20:00, Grimbal wrote:
I sinc you are being funny. No, it is not a trignometric function. (BTW, I have corrected the thread title.) MEGAhint: [hide]X = 83.1776....[/hide]!! |
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Title: Re: Another Sequence Post by Eigenray on Apr 19th, 2007, 7:34pm Can you be more [link=http://bootes.math.uqam.ca/cgi-bin/ipcgi/lookup.pl?Submit=GO+&number=83.1776&lookup_type=browse]specific[/link]? :-[ |
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Title: Re: Another Sequence Post by ThudanBlunder on Apr 19th, 2007, 8:00pm [quote author=Eigenray link=board=riddles_medium;num=1176909685;start=0#6 date=04/19/07 at 19:34:34]Can you be more [link=http://bootes.math.uqam.ca/cgi-bin/ipcgi/lookup.pl?Submit=GO+&number=83.1776&lookup_type=browse]specific[/link]? :-[/quote] Yes (http://bootes.math.uqam.ca/cgi-bin/ipcgi/lookup.pl?Submit=GO+&number=83.17766&lookup_type=browse) |
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Title: Re: Another Sequence Post by Eigenray on Apr 19th, 2007, 9:15pm Aha... it's much clearer once you [hide]divide by[/hide] 120. Then you just have [hide](2n-1)/n[/hide], with the missing value at [hide]n=0[/hide]. |
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Title: Re: Another Sequence Post by ThudanBlunder on Apr 19th, 2007, 9:28pm on 04/19/07 at 21:15:36, Eigenray wrote:
Well done! |
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Title: Re: Another Sequence Post by Ulkesh on Apr 21st, 2007, 6:10am Hmm... I'm sure I've seen something similar a number of years ago. At the risk of sounding stupid, you need L'Hopital's theorem (pardon my bad French spelling)? |
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Title: Re: Another Sequence Post by Eigenray on Apr 22nd, 2007, 3:35pm on 04/21/07 at 06:10:11, Ulkesh wrote:
Are you really using l'Hopital to compute limx->0 (f(x)-f(0))/x as f'(0)? |
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Title: Re: Another Sequence Post by Ulkesh on Apr 22nd, 2007, 4:26pm on 04/22/07 at 15:35:44, Eigenray wrote:
Yes, that seems correct given your f(x). Is there a problem? |
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Title: Re: Another Sequence Post by Eigenray on Apr 22nd, 2007, 6:44pm How do you define f'(0)? |
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Title: Re: Another Sequence Post by Grimbal on Apr 23rd, 2007, 2:09am Isn't it f'(x) computed for x=0? f(x) = 2x f'(x) = ex·ln(2)·ln(2) = 2x·ln(2) f'(0) = 20·ln(2) = ln(2) |
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Title: Re: Another Sequence Post by Eigenray on Apr 23rd, 2007, 11:52am Let me rephrase. Do you see anything wrong with the following sentence: Quote:
? |
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Title: Re: Another Sequence Post by towr on Apr 23rd, 2007, 1:19pm on 04/23/07 at 11:52:13, Eigenray wrote:
lim xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif0 f(x)/g(x) = lim xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif0 f'(x)/g'(x) but g(x)=x, and g'(x)=1, and lim xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif0 f'(x)=f'(0) So we have: lim xhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif0 f(x)/x = f'(0) That still leaves the -f(0) which is in fact 0, but it's a bit superfluous. It would rather be the definition of the derivative in 0 than an application of l'Hospital's rule, although it gives the same result. |
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Title: Re: Another Sequence Post by Aryabhatta on Apr 23rd, 2007, 1:41pm I think Eigenray is pointing to the fact that it seems odd to use LHospital's rule to just see that the limit is f'(0). The given limit is the definition of f'(0). Eigenray, I am not sure what the premises of Lhospital rule are, but does it assume the existence of the derivative at the value which x tends to? In which case, there will be some circular logic in appplying LHospital's rule. |
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Title: Re: Another Sequence Post by towr on Apr 23rd, 2007, 1:49pm on 04/23/07 at 13:41:49, Aryabhatta wrote:
L'Hospital's rule says so much as, if f(x) and g(x) are both 0 or both +/-http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif then if lim f'(x)/g'(x) exist, we have lim f(x)/g(x) = lim f'(x)/g'(x) If f'(x) and g'(x) are both 0 or both +/- http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif you can recurse. However that might never yield anythign usefull. (More at http://mathworld.wolfram.com/LHospitalsRule.html) |
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Title: Re: Another Sequence Post by Aryabhatta on Apr 23rd, 2007, 1:59pm on 04/23/07 at 13:49:15, towr wrote:
Hmm... Not sure if that is entirely incorrect. I think we also require g'(x) to be non-zero in an interval around the point in question. |
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Title: Re: Another Sequence Post by towr on Apr 23rd, 2007, 2:20pm on 04/23/07 at 13:59:06, Aryabhatta wrote:
In the case g'(x) isn't 0, then g(x) would be rather odd, I think.. |
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Title: Re: Another Sequence Post by Aryabhatta on Apr 23rd, 2007, 4:32pm on 04/23/07 at 14:20:13, towr wrote:
Yes, but it might so happen that when g'(c) is 0, f'(c) is 0 and the Lim f'/g' actually exists (maybe by recursing like you said) and in this case it might not necessarily be the same as Lim f/g Anyway, we should not be polluting this thread with irrelevant information ;D |
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Title: Re: Another Sequence Post by Icarus on Apr 23rd, 2007, 5:05pm Consider the example f(x) = x + cos x sin x and g(x) = esin x(x + cos x sin x) as x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif. f/g diverges, while f'/g' http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif 0. |
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Title: Re: Another Sequence Post by towr on Apr 24th, 2007, 12:42am on 04/23/07 at 17:05:21, Icarus wrote:
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Title: Re: Another Sequence Post by Grimbal on Apr 24th, 2007, 12:43am on 04/23/07 at 13:41:49, Aryabhatta wrote:
Ah, OK. Now I see what he was after. |
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Title: Re: Another Sequence Post by rmsgrey on Apr 24th, 2007, 6:34am on 04/24/07 at 00:42:26, towr wrote:
For that matter, isn't it bounded between e and 1/e? |
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Title: Re: Another Sequence Post by towr on Apr 24th, 2007, 8:05am on 04/24/07 at 06:34:21, rmsgrey wrote:
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Title: Re: Another Sequence Post by Eigenray on Apr 24th, 2007, 1:20pm on 04/23/07 at 13:41:49, Aryabhatta wrote:
Right. It's not wrong, exactly, just backwards. (I don't hate l'Hopital as much as [link=http://groups.google.com/groups/search?q=l%27hopital+robin+chapman]some people[/link].) A classic example is: by l'Hopital, limx->0 sin(x)/x = sin'(0)/1 = cos(0) = 1, even though the standard proof that sin'(x) = cos(x) uses the fact that sin(x)/x -> 1. |
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Title: Re: Another Sequence Post by Eigenray on Apr 24th, 2007, 1:36pm on 04/23/07 at 17:05:21, Icarus wrote:
on 04/24/07 at 00:42:26, towr wrote:
on 04/24/07 at 06:34:21, rmsgrey wrote:
on 04/24/07 at 08:05:33, towr wrote:
I assume you're all talking about the ratio f(x)/g(x) = e-sin(x) (for x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/ne.gif 0). It oscillates between 1/e and e, hence it diverges (because it doesn't converge). I'm confused by the confusion. |
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Title: Re: Another Sequence Post by towr on Apr 24th, 2007, 2:59pm Oops, I seem to have switched from considering the ratio to considering just g(x) |
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Title: Re: Another Sequence Post by Icarus on Apr 24th, 2007, 5:51pm I got that example from the Wikipedia article on L'Hopital's rule. Though it seems towr is now past his confusion, let me spell it out for others: f(x) = x + cos x sin x = x + (1/2)sin 2x g(x) = esin x(x + cos x sin x) = esin xf(x) f(x)/g(x) = e-sin x, which oscillates back and forth between 1/e and e as x http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gif http://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/infty.gif, so f/g diverges. f'(x) = 1- sin2 x + cos2 x = 2cos2 x g'(x) = cos x esin x(x + cos x sin x) + esin x(2cos2 x) = cos x esin x(x + cos x sin x + 2cos x). f'/g' = (2cos x)/esin x(x + cos x sin x + 2cos x). Since the top is bounded while the bottom grows without bound (because of the x term), limxhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/to.gifhttp://www.ocf.berkeley.edu/~wwu/YaBBImages/symbols/subinfty.gif f'/g' = 0. This shows that L'Hopital's Rule can fail when the denominator is allowed to be 0 in every neighborhood of the point. (To apply the same example at finite points, you can compose it with a transformation such as x = 1/|t|.) |
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