Comment author: aspera 13 October 2013 03:29:33AM 1 point [-]

Thanks to whomever moved this to Discussion. From the FAQ, I wasn't sure where to put it. This is better, in retrospect.

Comment author: VincentYu 13 October 2013 12:59:47AM *  1 point [-]

Another small typo: Under "The Finite Problem", n_{A,s} and n_{B,s} should be n_{A,t} and n_{B,t} instead.

Comment author: aspera 13 October 2013 03:27:39AM 0 points [-]

Thanks! Do you guys want to copy edit my journal papers? ;)

Comment author: DanielLC 13 October 2013 02:09:49AM 3 points [-]

There is no additional structure. It's not as if we can come upon two pairs of spheres, and notice while the end result is the same, they're the limits of two different processes, and therefore different choices are better. There is only the infinite case. If you want to consider sequence that converge to it, there's no clear way to decide which sequence to look at.

Limits help you if you're looking at an extreme value. If the limit as the population goes to infinity is that a random sample of X of them will give you Y confidence on a poll, then you can just use that if there's a large population. If you're dealing with the limit itself, it doesn't always help. You can start with a square, and then cut little squares off of the corners, and then more squares off of those corners etc. until you approach a circle. The perimeter will always be four times the length, but this won't be true of the circle.

In this problem, you can get literally any answer if you take the limit appropriately, so once you've decided on the right answer, there is some way to get to it with the limit, but deciding that the right answer is one that a limit converges to helps you not at all.

The problem isn't a sequence of finite cases. It's just the infinite case all by itself.

Comment author: aspera 13 October 2013 03:27:14AM 0 points [-]

You're completely right! As stated, the problem is ill posed, i.e. it has no unique solution, so we didn't solve it.

Instead, we solved a similar problem by introducing a new parameter, \alpha. It was useful because we gained a mathematical description that works for very large n and s, and which matches our intuition about the problem.

It is important to recognize, as you point out, that that taking limits does not solve the problem. It just elucidates why we can't solve it as stated.

Comment author: buybuydandavis 12 October 2013 11:44:21PM 1 point [-]

Very nice work.

I feel it accomplishes too much for most people, though, so that they might not get everything out of it.

Great as a demonstration of Jaynes' point about paradoxes generate by failure to identify the limiting process for proposed infinities. But also great as a demonstration of how "doing the math" can resolve a lot of these philosophical debates which just don't specify problem conditions sufficiently because they don't explicitly write them down in mathematical form.

I think a lot of the self referential problems many like around here (Omega stuff, as an example) would be similarly dissolved by the latter.

Comment author: aspera 13 October 2013 03:23:32AM 0 points [-]

I agree that it's a lot to cover, but I wanted to work a full example. We talk a lot on LW about decision analysis and paradoxes in the abstract, but I'm coming from a math/physics background, and it's much more helpful for me to see concrete examples. I assume some other people feel the same way.

Self-referential problems would be an interesting area to study, but I'm not familiar with the techniques. I suspect you're right, though.

Comment author: Gabriel 12 October 2013 09:44:34PM 2 points [-]

Nice.

Bug reports:

Note that even though t is 1, we'll end up with a unit error if we don't carry it around.

I think you meant r here.

And you have flipped the sign of \Delta U_{linear}.

Comment author: aspera 12 October 2013 10:36:48PM 1 point [-]

Fixed. Thanks for reading so closely. It's amazing how many little mistakes can survive after 10 read-throughs.

Comment author: Decius 09 October 2013 10:39:44PM 2 points [-]

Conclusions that involve infinity cannot generally be generalized to any finite solution; this seems like a 'each monkey now has two bananas' moment.

Comment author: aspera 10 October 2013 10:58:30PM 0 points [-]

By the way, are you talking about this meme, or is there another problem with monkeys and bananas?

Comment author: CronoDAS 10 October 2013 10:00:33PM *  4 points [-]

The Ross-Littlewood Paradox is amusing.

You have an infinite collection of balls in a storeroom, labeled with the natural numbers (1, 2, 3, etc.) and a vase that can hold any number, or all, of them.

At each integer time T, starting with T=1, you take the 10 lowest numbered balls out of the storeroom and put them in the vase, and then take the lowest numbered ball out of the vase and destroy it. So at any finite time T, there are 9T balls in the vase and all the balls labeled with a number less than or equal to T have been destroyed.

Now, because the number of balls at any given time T is given by 9T, in the limit as T approaches infinity, there are infinitely many balls in the vase. On the other hand, because every ball has a time T at which it will be destroyed, the limit of the set of balls in the vase as T approaches infinity is the empty set. So at T = infinity, you have an empty vase that contains infinitely many balls.

The moral of the story is to be careful what limits you take, because taking two different limits can give two different answers even if they seem like they're measuring the same thing.

(Can this be used as an argument for the existence of nonstandard numbers?)

Comment author: aspera 10 October 2013 10:55:39PM 1 point [-]

Great problem, thanks for mentioning it!

I think the answer to "how many balls did you put in the vase as T->\infty" and "How many balls have been destroyed as T->\infty" both have well defined answers. It's just a fallacy to assume that the "total number of balls in the vase as T->\infty" is equal to the difference between these quantities in their limits.

Comment author: aspera 10 October 2013 09:40:21PM *  4 points [-]

My parents stopped me from skipping a grade, and apart from a few math tricks, we didn't work on additional material at home. I fell into a trap of "minimum effort for maximum grade," and got really good at guessing the teacher's password. The story didn't change until graduate school, when I was unable to meet the minimum requirements without working, and that eventually led me to seek out fun challenges on my own.

I now have a young son of my own, and will not make the same mistake. I'm going to make sure he expects to fail sometimes, and that I praise his efforts to go beyond what's required. No idea if it will work.

Comment author: Decius 10 October 2013 08:07:40PM *  0 points [-]

Um... -inf and +inf are not real numbers. (Noting that your function as described is undefined at -inf.)

In addition, the definition of continuous restricts it to points which exist on an open interval; if the limit from below and limit from above are equal to the value at X, then the function is continuous on an open interval containing X. How do you determine the limit as X approaches +inf from above?

Comment author: aspera 10 October 2013 08:18:05PM 0 points [-]

MrMind explains in better language below.

Comment author: MrMind 10 October 2013 12:48:45PM 5 points [-]

How did you make those wonderful graphs?

Comment author: aspera 10 October 2013 04:56:24PM 6 points [-]

The plots were done in Mathematica 9, and then I added the annotations in PowerPoint, including the dashed lines. I had to combine two color functions for the density plot, since I wanted to highlight the fact that the line s=n represented indifference. Here's the code:

r = 1; ua = 1;ub = -1; f1[n, s] := (ns - s^2r ) (ua - ub); Show[DensityPlot[-f1[n, s], {n, 0, 20}, {s, 0, 20}, ColorFunction -> "CherryTones", Frame -> False, PlotRange -> {-1000, 0}], DensityPlot[f1[n, s], {n, 0, 20}, {s, 0, 20}, ColorFunction -> "BeachColors", Frame -> False, PlotRange -> {-1000, 0}]]

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