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Topic: Another Sequence (Read 2834 times) |
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ThudnBlunder
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Another Sequence
« on: Apr 18th, 2007, 8:21am » |
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35, 45, 60, X, 120, 180, 280, 450, 744, 1260
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| « Last Edit: Apr 19th, 2007, 3:37pm by ThudnBlunder » |
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Grimbal
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Re: Another Series
« Reply #1 on: Apr 18th, 2007, 9:25am » |
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Roughly 80 or 81?
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ThudnBlunder
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Re: Another Series
« Reply #2 on: Apr 18th, 2007, 10:15am » |
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on Apr 18th, 2007, 9:25am, Grimbal wrote:
Yes, it's a number in the 80's.
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ThudnBlunder
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Re: Another Series
« Reply #3 on: Apr 19th, 2007, 9:16am » |
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Actually, I should ask for a simple continuous function which generates the series.
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| « Last Edit: Apr 20th, 2007, 3:32am by ThudnBlunder » |
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Grimbal
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Re: Another Series
« Reply #4 on: Apr 19th, 2007, 3:20pm » |
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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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ThudnBlunder
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Re: Another Sequence
« Reply #5 on: Apr 19th, 2007, 3:38pm » |
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on Apr 19th, 2007, 3:20pm, Grimbal wrote: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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I sinc you are being funny. No, it is not a trignometric function.
(BTW, I have corrected the thread title.)
MEGAhint: X = 83.1776....!!
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| « Last Edit: Apr 20th, 2007, 12:50pm by ThudnBlunder » |
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Eigenray
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Re: Another Sequence
« Reply #6 on: Apr 19th, 2007, 7:34pm » |
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Can you be more specific? 
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ThudnBlunder
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Re: Another Sequence
« Reply #7 on: Apr 19th, 2007, 8:00pm » |
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on Apr 19th, 2007, 7:34pm, Eigenray wrote:
Yes
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Eigenray
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Re: Another Sequence
« Reply #8 on: Apr 19th, 2007, 9:15pm » |
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Aha... it's much clearer once you divide by 120. Then you just have (2n-1)/n, with the missing value at n=0.
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ThudnBlunder
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Re: Another Sequence
« Reply #9 on: Apr 19th, 2007, 9:28pm » |
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on Apr 19th, 2007, 9:15pm, Eigenray wrote:| Aha... it's much clearer once you divide by 120. Then you just have (2n-1)/n, with the missing value at n=0. |
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Well done!
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Ulkesh
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Re: Another Sequence
« Reply #10 on: Apr 21st, 2007, 6:10am » |
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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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Eigenray
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Re: Another Sequence
« Reply #11 on: Apr 22nd, 2007, 3:35pm » |
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on Apr 21st, 2007, 6:10am, Ulkesh wrote:| At the risk of sounding stupid, you need L'Hopital's theorem (pardon my bad French spelling)? |
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Are you really using l'Hopital to compute
limx->0 (f(x)-f(0))/x
as f'(0)?
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Ulkesh
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Re: Another Sequence
« Reply #12 on: Apr 22nd, 2007, 4:26pm » |
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on Apr 22nd, 2007, 3:35pm, Eigenray wrote:
Are you really using l'Hopital to compute
limx->0 (f(x)-f(0))/x
as f'(0)? |
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Yes, that seems correct given your f(x). Is there a problem?
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Eigenray
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Re: Another Sequence
« Reply #13 on: Apr 22nd, 2007, 6:44pm » |
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How do you define f'(0)?
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Grimbal
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Re: Another Sequence
« Reply #14 on: Apr 23rd, 2007, 2:09am » |
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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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Eigenray
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Re: Another Sequence
« Reply #15 on: Apr 23rd, 2007, 11:52am » |
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Let me rephrase. Do you see anything wrong with the following sentence:
Quote:By l'Hopital's rule,
limx->0 [f(x) - f(0)]/x = f'(0). |
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?
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towr
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Re: Another Sequence
« Reply #16 on: Apr 23rd, 2007, 1:19pm » |
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on Apr 23rd, 2007, 11:52am, Eigenray wrote:Let me rephrase. Do you see anything wrong with the following sentence:
? |
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Well, one might expect instead
lim x 0 f(x)/g(x) = lim x0 f'(x)/g'(x)
but g(x)=x, and g'(x)=1, and lim x0 f'(x)=f'(0)
So we have: lim x0 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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| « Last Edit: Apr 23rd, 2007, 1:23pm by towr » |
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Aryabhatta
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Re: Another Sequence
« Reply #17 on: Apr 23rd, 2007, 1:41pm » |
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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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| « Last Edit: Apr 23rd, 2007, 1:47pm by Aryabhatta » |
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towr
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Re: Another Sequence
« Reply #18 on: Apr 23rd, 2007, 1:49pm » |
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on Apr 23rd, 2007, 1:41pm, Aryabhatta wrote:| Eigenray, I am not sure what the premises of Lhospital rule are, but does it assume the existence of the derivative at the limit? In which case, there will be some circular logic. |
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L'Hospital's rule says so much as, if f(x) and g(x) are both 0 or both +/-
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 +/- you can recurse. However that might never yield anythign usefull.
(More at http://mathworld.wolfram.com/LHospitalsRule.html)
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| « Last Edit: Apr 23rd, 2007, 1:52pm by towr » |
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Aryabhatta
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Re: Another Sequence
« Reply #19 on: Apr 23rd, 2007, 1:59pm » |
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on Apr 23rd, 2007, 1:49pm, towr wrote:
L'Hospital's rule says so much as, if f(x) and g(x) are both 0 or both +/-
then if lim f'(x)/g'(x) exist, we have
lim f(x)/g(x) = lim f'(x)/g'(x)
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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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towr
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Re: Another Sequence
« Reply #20 on: Apr 23rd, 2007, 2:20pm » |
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on Apr 23rd, 2007, 1:59pm, Aryabhatta 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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If g'(x) is 0 as well, I don't see a problem (because then f'(x)/g'(x) wouldn't exist in that case and we already couldn't apply the rule for that reason).
In the case g'(x) isn't 0, then g(x) would be rather odd, I think..
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Aryabhatta
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Re: Another Sequence
« Reply #21 on: Apr 23rd, 2007, 4:32pm » |
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on Apr 23rd, 2007, 2:20pm, towr wrote:
If g'(x) is 0 as well, I don't see a problem (because then f'(x)/g'(x) wouldn't exist in that case and we already couldn't apply the rule for that reason).
In the case g'(x) isn't 0, then g(x) would be rather odd, I think.. |
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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 
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| « Last Edit: Apr 23rd, 2007, 4:34pm by Aryabhatta » |
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Icarus
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Re: Another Sequence
« Reply #22 on: Apr 23rd, 2007, 5:05pm » |
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Consider the example
f(x) = x + cos x sin x and g(x) = esin x(x + cos x sin x) as x . f/g diverges, while f'/g' 0.
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towr
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Re: Another Sequence
« Reply #23 on: Apr 24th, 2007, 12:42am » |
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on Apr 23rd, 2007, 5:05pm, Icarus wrote:Consider the example
f(x) = x + cos x sin x and g(x) = esin x(x + cos x sin x) as x . f/g diverges |
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Isn't it 1 everytime sin(x)=0 ?
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Grimbal
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Re: Another Sequence
« Reply #24 on: Apr 24th, 2007, 12:43am » |
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on Apr 23rd, 2007, 1:41pm, Aryabhatta wrote:| 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). |
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Ah, OK. Now I see what he was after.
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