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 Forum index » Archive » Archive: General » ARG: Ocular Effect
[RABBITHOLE] Ocular Effect
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Rolerbe
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Joined: 01 Mar 2005
Posts: 330
Location: North America

Kanashimi No Kage wrote:
The value of e is calculated as follows: Limit as N goes to infinity of (1+(1/n))^n
This simplifies to (1+(0))^infinite.


Cute and clever. Flawed, of course, in that the baby has been thrown out with the bathwater, but cute nonetheless. Thanks for putting it in. I hadn't seen that before.

What about the sum of Zeros's argument?
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PostPosted: Fri Jul 14, 2006 10:58 am
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HappyFunNorm
Boot

Joined: 05 Jun 2005
Posts: 54
Location: Roswell, GA

Howdy. I asked a Math teacher friend of mine to try and explain this a little better... teachers are often good at that sort of thing Smile He also came up with a posible meaning for the number sequese in th URL, though he doubs his discovery is relavant.

----------------------------------------------------------------------------
Suggestions that an infinite number of zeros sum to 1 aside, the fibonacci
sequence should, indeed, have only a single 0. As you pointed out, each
term is the sum of the 2 previous, so two zeros as a seed would lead to
nothing but zeros. It absolutely doesn't technically start with an
infinite number of zeros.

The arguments that an infinite number of zeros add up to one largely fail
because they're using 'infinity' as a number-- it isn't, not exactly.

>The infinite sums of zeros works like this: You take N terms of the
>fraction (1/N) and add them together, this will always come to (N/N)
>which is 1. You can extend this infinitely till you get [1/(infinity)]
>which is 0,

He's confusing the _limit_ as n goes to infinity of 1/n with 1/infinity.
You can't divide by infinity, just like you can't divide by zero. It's
undefined.

>yet an infinite amount of these these still exhibit (N/N)
>behavior.

What he's really talking about is the lim[n->inf] of n(1/n). Limits are
evaluated in very specific ways. You can simplify inside the limit, and
you do get

lim[n->inf] n/n, which certainly is 1

You cannot, however, take the limit of this part over here and then take
the limit of that part over there. in other words, you can't say
lim[n->inf] n(1/n) = lim[n->inf] n(lim[n->inf]1/n) in some sort of fit of
half-assed distribution.

That's what he's doing when he says 1/inf is zero-- taking a limit. But
that n that just approached infinity is the same n outside the
parentheses. They either both approach infinity or they don't. So, you
can't claim that
lim[n->inf] n(1/n) == lim[n->inf] n(0) which is really what he's trying to
do.

lim[n->inf] n(0) = 0, of course.

As yanka pointed out:
> That still seems flawed... because technically, though the fraction
> 1/N -> 0 as N -> infinity, as long as you're not at infinity, you're
> adding up an almost infinite number of almost infinitely small terms,
> and that sum -> 1. So, you don't really have a bunch of 0s adding up to
> 1 - you have a huge number of tiny terms wanting to add up to 1.

and in fact you will never actually be _at_ infinity. Again, it's not
exactly a number.

The distinction is important: consider induction. To remind you, proof by
induction consists of showing that something is true for a foundation case
(n=0 or n=1 for instance) and then showing that, if it is true for n, it
is also true for n+1. This process shows taht it is true for all _finite_
integer n greater than the foundation case (foundation case isn't the
normal term used here, but the normal term escapes me).

For instance, consider the claim that any number of balls greater than or
equal to two can be divided into sets of 2 balls and/or sets of 3 balls
with none left over (a fairly trivial claim, and this isn't the cleanest
way to prove it, but whatever).

For the foundation case, n=2 is trivially divided into a single set of 2
balls.

For the induction step, assume a colletion of n balls can be thusly
divided. I need to show, based on that assumption, that n+1 can also be
properly divided. Well, for n+1 divide the n balls properly, then add the
extra to a group of 2 forming a group of 3. No groups of 2? Take one from
a group of three (leaving a group of 2 behind) and pair it with the extra.
Ta da.

Induction is the principle that tells us that such a proof covers
n = 2, 3, 4,... . But! While we can make n as big as we want
and be covered by that proof, it's still got to be a _finite_ number of
balls (because no matter how many times you add 1, you've got a finite
number. "Until you add 1 an infinite number of times", but you can't
really do that.)

This proof tells us about all integral numbers of balls greater than or
equal to 2, but doesn't tell us a thing about an infinite set of balls.
We've learned about an infinite number of possibilities, but none of them
involve an infinite number of balls.

Kanashimi No Kage:
> The value of e is calculated as follows: Limit as N goes to infinity of
> (1+(1/n))^n This simplifies to (1+(0))^infinite.

This suffers from the same sorts of problems as the earlier stuff. He is,
essentially, trying to take the limit in stages-- taking the limit of 1/n
and then tossing 0 in its place and taking the limit of that stuff raised
to the nth power. Qualitatively, it is
(1 + SomethingGoingTowardsZero)^SomethingGoingTowardsInfinity
In fact, I wouldn't object to that characterization. But
SomethingGoingTowardsZero is not necessarily zero. In this case it
definitely isn't-- 1/n assymptotically approaches zero, which means it
gets as close to zero as you'd like, but is never actually equal to zero.
As before, (1 + SomethingGoingTowardsZero) is going to be always something
just a little bit more than one.

The central claim:

DrumMajor:
> In an obscure mathematical effect an infinite amount of zeros will add
> to one

Uhm. No. An infinite number of zeros added together will generally be
undefined (the idea is that you can't add an infinite number of anything)
or at best, zero.

An aside: we do sometimes consider the extended real numbers-- that's the
real numbers (infinity is not one) along with 2 extra elements-- pos_inf
and neg_inf. These are special elements, and we define their behavior.
It's almost like they become catchalls for those things that aren't really
defined, but they're larger than all rest.
http://planetmath.org/encyclopedia/ExtendedRealNumbers.html
is a n accessible discussion of some of the details.

As for 00112358, sloanes (an online encyclopedia of integer sequences) has
it as part of a number of sequences:
http://tinyurl.com/jckhl
However, it only starts one:
http://www.research.att.com/~njas/sequences/A092362
"Number of partitions of n^2 into squares greater than 1" (note that the
list starts with n=0).

That is, for a number n, the nth term in the list tells you the number of
ways to write n^2 as a sum of only squares greater than 1.

For example, n = 5 gives 3-- there are 3 ways to write 5^2 = 25 as a sum
of squares greater than 1:

25 = 25
25 = 9 + 16
25 = 4 + 4 + 4 + 4 + 9

However, I rather doubt this is what the webpage is about.
-------------------------------------------------------------------------------

PostPosted: Fri Jul 14, 2006 1:29 pm
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Rolerbe
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Joined: 01 Mar 2005
Posts: 330
Location: North America

Nice summary -- Thanks!

*Rolerbe sleeps easy again knowing the fate of the (largely mathematical) universe is secure.
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PostPosted: Fri Jul 14, 2006 1:44 pm
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Kanashimi No Kage
Greenhorn

Joined: 11 Jul 2006
Posts: 3

Rolerbe wrote:
Kanashimi No Kage wrote:
The value of e is calculated as follows: Limit as N goes to infinity of (1+(1/n))^n
This simplifies to (1+(0))^infinite.


Cute and clever. Flawed, of course, in that the baby has been thrown out with the bathwater, but cute nonetheless. Thanks for putting it in. I hadn't seen that before.



Flawed?
I fail to see what is flawed about it.
reference: page 245 of Calculus Concepts & Contexts by James Stewart, Published by Thomson, Brooks and Cole
http://search.barnesandnoble.com/bookSearch/isbnInquiry.asp?r=1&isbn=0534409865

"e=lim (x->0) (1+x)^(1/X)"
"e=lim (n->infinite) (1+(1/n))^n"

Though it may be hypocritical to say it, I'd appreciate it if we could get back to working out what we can about the ARG, instead of arguing over DrumMajor's choice of words. THe point he was trying to make is that math is subject to occasional holes in logic that just work out the way that they do because of convention.
Beyond that, if we are going to dwell on the mathimatical content, can we have constructive input? I realize my word carries no weight seeing as that I am new, but I was led to believe that the ARG community is as a whole supportive instead of the punitive nature I am feeling currently.

PostPosted: Fri Jul 14, 2006 7:48 pm
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Phaedra
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Joined: 21 Sep 2004
Posts: 4033
Location: Here, obviously

Kanashimi No Kage wrote:
I realize my word carries no weight seeing as that I am new, but I was led to believe that the ARG community is as a whole supportive instead of the punitive nature I am feeling currently.


Your word carries weight equivalent to the strength of your argument. How long you've been here has nothing to do with it.

And in general, the ARG community is very supportive. I don't see anything "punitive" here -- just people who disagree. But disagreement != flaming. I think maybe you're reading animus into people's comments that isn't actually there.

(That said, for some reason whenever a math discussion crops up here, it seems to get slightly tetchy. Moreso than politics, even. Shocked )
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PostPosted: Fri Jul 14, 2006 8:02 pm
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colin
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Joined: 13 Oct 2003
Posts: 810
Location: Australia

Can we argue off a common base, that everyone has access to? Like WikiPedia which shows that Kanashimi No Kage is correct (at least for lim->inf)

From now on I want to see mathematical proofs, not just "it doesn't do that" Twisted Evil

They say if you include one formula in a book, it halves the sales. This ARG must has lost most it's audience by now.

PostPosted: Fri Jul 14, 2006 9:06 pm
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Rolerbe
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Joined: 01 Mar 2005
Posts: 330
Location: North America

I don't think anyone (myself included) has been 'punative' or, in fact, disparaging in any way. And, although this is clearly not the forum to debate mathematics itself, I thought it was a natural question arising from the original postings about sums of zeros and products of ones.

Part of the enjoyment of the challenges that arise in ARG's or puzzle trails is not just the single minded pursuit of the answer of the moment that allows the next game veil to be lifted, but also the collateral learning to be had in the byways along the trail.

No one dispute's that the limit as n->inf of (1+1/n)^n = e. But that's NOT the same thing as saying the product of an infinite number of 1's = e, even when promoted by the argument given, since the effect is entirely due to the 1/n contribution (even in the limit). That's what I meant by throwing the baby out with the bathwater, and which as much better commented on by the post that followed. Nevertheless, I enjoy the humor, for lack of a more precise word for it, of such mathematical curiosities. Whether you view it as a 'flaw' or not, it leads to greater insight.

It was a good thing until some of you got pissed off. And it was on its way out (winding down). It's not really such an imposition to scroll past side discussions that you see as detours, and a little patience shows that they (usually) close themselves off pretty quickly.

So, personally, I'd still like to see the sum of zero's argument -- just for fun. It can even go in its own thread so as not to 'bother' those not curious about it.
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PostPosted: Fri Jul 14, 2006 10:05 pm
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dontpanic42
Veteran


Joined: 22 Jun 2006
Posts: 147
Location: naples, florida

that said can we slow down on the math now, my head is really starting to spin. i'm really bummed though, monday when the clock dies down i'll be at college orientation, hopefully i can get my psp to connect so i can read what happens.

PostPosted: Fri Jul 14, 2006 10:29 pm
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limeWraith
Guest


I did some searching and figured out that July 17 was the opening day of Disneyland, and the second address is probably just one they registered when they realized oculareffect.com was registered to Disney. So, almost definitely disney-related.

PostPosted: Fri Jul 14, 2006 11:55 pm
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limeWraith
Guest


Also, the other websites on the server are all religion-oriented, so it is probably the Fallen ARG.

PostPosted: Sat Jul 15, 2006 12:31 am
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DrumMajor
Boot


Joined: 18 Oct 2004
Posts: 54
Location: Colorado Springs

I made the mistake of dumbing things down. I removed the discussion of limits and how they actually work to put things in ley terms, which causes some inheriant logical flaws. That coupled with some of my own origional logical flaws (yes even I am wrongs sometimes) caused confusion. But it shouldn't cause bickering. Lets all just keep it friendly and enjoy the game!
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PostPosted: Sat Jul 15, 2006 2:33 am
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Giskard
Sassypants


Joined: 07 Oct 2003
Posts: 2066
Location: Chicago

With that said, I've moved this thread to the "ARGs With Potential" section in preparation for monday, when there will probably be a lot more reasons for discussion Smile

There are some posting guidelines for this forum, so please read through them for a bit so everyone knows what can and cannot be done here. Basically, this forum allows us to have multiple threads on the same game, which I suggest we start doing after the countdown has finished. Using the tag [OCULAR] before new posts is probably the best way to identify the threads belonging to this game's discussion.

If it turns out there's even more space needed when the game goes live, we'll quickly decide if the game warrants an own section.

In the mean time, enjoy : Very Happy
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PostPosted: Sat Jul 15, 2006 5:53 am
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colin
Entrenched

Joined: 13 Oct 2003
Posts: 810
Location: Australia

looks like they tried kicking things off on youtube
OMG, like this chick found teh ocular site on July 03, 2006 look here. I'm going to put this vid on my mySpace, it's like totally freaky.

We could play guess the actor, but i'm trying to resist.

woohoo! 600 posts

PostPosted: Sat Jul 15, 2006 6:53 am
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limeWraith
Guest


On the website for Fallen, there is a link to oculareffect.http://abcfamily.go.com/fallen/index.html There is an option in the lower right hand corner to "Look through the oculus", which directs you there.

PostPosted: Sat Jul 15, 2006 1:03 pm
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limeWraith
Guest


[EDIT] Warning, only look below if you want to find out stuff that is not really meant to be found out this way! It may spoil the countdown for you! --Giskard

Original post:

Spoiler (Rollover to View):
When Monday comes, look at this, this, and this. They are all on the same server as 00112358.net, and are all currently 403 forbidden, which will probably change when the countdown finishes.





[EDIT] Fixed URL's and spoilered post --Giskard

PostPosted: Sat Jul 15, 2006 1:14 pm
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