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riddles >> medium >> Another Sequence
(Message started by: ThudanBlunder on Apr 18th, 2007, 8:21am)

Title: Another Sequence
Post by ThudanBlunder on Apr 18th, 2007, 8:21am
35, 45, 60, X, 120, 180, 280, 450, 744, 1260

Title: Re: Another Series
Post by Grimbal on Apr 18th, 2007, 9:25am
Roughly 80 or 81?

Title: Re: Another Series
Post by ThudanBlunder on Apr 18th, 2007, 10:15am

on 04/18/07 at 09:25:50, Grimbal wrote:
Roughly 80 or 81?

Yes, it's a number in the 80's.

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.

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.

Title: Re: Another Sequence
Post by ThudanBlunder on Apr 19th, 2007, 3:38pm

on 04/19/07 at 15:20:00, 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.

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]!!


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]?  :-[

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)

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].

Title: Re: Another Sequence
Post by ThudanBlunder on Apr 19th, 2007, 9:28pm

on 04/19/07 at 21:15:36, Eigenray wrote:
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].

Well done!

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)?

Title: Re: Another Sequence
Post by Eigenray on Apr 22nd, 2007, 3:35pm

on 04/21/07 at 06:10:11, Ulkesh wrote:
At the risk of sounding stupid, you need L'Hopital's theorem (pardon my bad French spelling)?

Are you really using l'Hopital to compute

limx->0 (f(x)-f(0))/x

as f'(0)?

Title: Re: Another Sequence
Post by Ulkesh on Apr 22nd, 2007, 4:26pm

on 04/22/07 at 15:35:44, Eigenray wrote:
Are you really using l'Hopital to compute

limx->0 (f(x)-f(0))/x

as f'(0)?


Yes, that seems correct given your f(x). Is there a problem?

Title: Re: Another Sequence
Post by Eigenray on Apr 22nd, 2007, 6:44pm
How do you define f'(0)?

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)

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:
By l'Hopital's rule,

limx->0  [f(x) - f(0)]/x  =  f'(0).


?

Title: Re: Another Sequence
Post by towr on Apr 23rd, 2007, 1:19pm

on 04/23/07 at 11:52:13, Eigenray wrote:
Let me rephrase.  Do you see anything wrong with the following sentence:


?
Well, one might expect instead
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.

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.


Title: Re: Another Sequence
Post by towr on Apr 23rd, 2007, 1:49pm

on 04/23/07 at 13:41:49, 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.

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)

Title: Re: Another Sequence
Post by Aryabhatta on Apr 23rd, 2007, 1:59pm

on 04/23/07 at 13:49:15, towr 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)


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.

Title: Re: Another Sequence
Post by towr on Apr 23rd, 2007, 2:20pm

on 04/23/07 at 13:59:06, 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.
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..

Title: Re: Another Sequence
Post by Aryabhatta on Apr 23rd, 2007, 4:32pm

on 04/23/07 at 14:20:13, 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..


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

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.

Title: Re: Another Sequence
Post by towr on Apr 24th, 2007, 12:42am

on 04/23/07 at 17:05:21, Icarus wrote:
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
Isn't it 1 everytime sin(x)=0 ?

Title: Re: Another Sequence
Post by Grimbal on Apr 24th, 2007, 12:43am

on 04/23/07 at 13:41:49, 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).

Ah, OK.  Now I see what he was after.

Title: Re: Another Sequence
Post by rmsgrey on Apr 24th, 2007, 6:34am

on 04/24/07 at 00:42:26, towr wrote:
Isn't it 1 everytime sin(x)=0 ?

For that matter, isn't it bounded between e and 1/e?

Title: Re: Another Sequence
Post by towr on Apr 24th, 2007, 8:05am

on 04/24/07 at 06:34:21, rmsgrey wrote:
For that matter, isn't it bounded between e and 1/e?
No, because (x + cos x sin x) can grow unbounded. I think you missed the x term there.

Title: Re: Another Sequence
Post by Eigenray on Apr 24th, 2007, 1:20pm

on 04/23/07 at 13:41:49, 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).

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.

Title: Re: Another Sequence
Post by Eigenray on Apr 24th, 2007, 1:36pm

on 04/23/07 at 17:05:21, Icarus wrote:
f/g diverges


on 04/24/07 at 00:42:26, towr wrote:
Isn't it 1 everytime sin(x)=0 ?


on 04/24/07 at 06:34:21, rmsgrey wrote:
For that matter, isn't it bounded between e and 1/e?


on 04/24/07 at 08:05:33, towr wrote:
No, because (x + cos x sin x) can grow unbounded. I think you missed the x term there.

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.

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)

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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