I have a burning question, man.
August 27, 2011 6:10 PM   Subscribe

Before it was a website, Ask A Mathematician / Ask A Physicist was two guys sitting in the desert at Burning Man, presuming to answer (almost) any question that happened to occur to whomever happened to appear at our stand.

E.g. What if the earth were a cube instead of a sphere?
Or, What is the value of 0^0?
Or even Does the double slit experiment disprove the existence of God?

Send your questions to AskAMathematician@gmail.com but we’ll be at burning man, and out of contact for the next week or so, so if it takes an especially long time for us to answer emails, it’s nothing personal.
posted by Obscure Reference (42 comments total) 21 users marked this as a favorite
You realize that the way you framed this makes it look like a self-link, right?
posted by asterix at 6:19 PM on August 27, 2011

Personally I'm partial to the guys at the Bad Advice booth.
posted by mannequito at 6:26 PM on August 27, 2011 [1 favorite]

Wow --- this argument (previously on MetaFilter) will never die.
posted by escabeche at 6:35 PM on August 27, 2011

[Added some italics to make it clearer this is quoting the linked site. That's really super important when pulling a first-person quote for post text so folks don't think you're suddenly spammin' Metafilter. Carry on.]
posted by cortex (staff) at 6:47 PM on August 27, 2011

Ooh, now it's my turn to say:
posted by Dr. Eigenvariable at 7:12 PM on August 27, 2011

George Bergman used to do/does this on Cal Day at Berkeley--sets up a table outside Evans Hall and takes questions.

I should know better than to read the comments on sites like these. Amazingly, no one tries to argue strenuously against the order of operations. Of course, someone does manage to spell Turing Türing, which amuses me for the presumably unintentional Cryptonomicon reference.
posted by hoyland at 7:28 PM on August 27, 2011

I don't know what I should think about this. Maybe I'll ask a ninja.
posted by smirkette at 8:56 PM on August 27, 2011

Q: How many mathematicians/physicists does it take to screw in a light bulb?

Physicist: The computers capable of accurately doing this simulation haven’t been invented (yet). So we’ve fallen back on some reasonable approximations, like massless light bulbs and spherical physicists. So far, it looks like physicists can’t pick up light bulbs, but two physicists can break a bulb between them.

The Mathematician, of course, provides a complete proof.
posted by Nomyte at 9:11 PM on August 27, 2011 [1 favorite]

How can anyone who has thought about it for more than ten seconds not understand that .9999 repeating is equal to 1? 1/3 + 1/3 + 1/3.
posted by gerryblog at 9:49 PM on August 27, 2011

People who don't believe .9999 repeating is equal to 1 also don't believe .3333 repeating is equal to 1/3.
posted by escabeche at 9:54 PM on August 27, 2011

But that's crazy too!
posted by gerryblog at 9:55 PM on August 27, 2011

Why? I would say that it's wrong, but not crazy, and that in fact the reasons it's wrong are rather subtle.
posted by escabeche at 9:57 PM on August 27, 2011

Just do the long division; 1/3 comes out to .3333 repeating.
posted by gerryblog at 10:03 PM on August 27, 2011

I think the problem is the repeating. The idea is that if it's repeating it's never actually complete and so it's not exactly equal. I'm not sayin' I believe that, I'm just sayin' ...
posted by DaddyNewt at 11:03 PM on August 27, 2011

Maybe I'll go to Dolores Park and set up a table for those who were Left Behind in the annual rapture. "Couldn't make it to Burning Man? Still want to ask a mathematician your burning questions?"

Or I could burn an AskMe on this: "what questions would you ask a mathematician if you got to ask one some questions?" -- but that would be chatfilter, and swiftly deleted, I assume.
posted by madcaptenor at 11:14 PM on August 27, 2011

People who don't believe .9999 repeating is equal to 1 also don't believe .3333 repeating is equal to 1/3.

Someone should ask Ron Paul about this.
posted by neuron at 11:39 PM on August 27, 2011 [1 favorite]

The idea is that if it's repeating it's never actually complete and so it's not exactly equal

If it isn't complete, then you'd agree there is some non-zero distance or difference between 0.999... and 1, right?

If this difference is greater than zero, then if you subtract this from one, do you get 0.999... back? Or do you get back a number that has to be smaller than 0.999...?
posted by Blazecock Pileon at 11:47 PM on August 27, 2011 [1 favorite]

I still don't get it. Please explain it with a shitty analogy.
posted by ryanrs at 2:10 AM on August 28, 2011 [2 favorites]

My mind is fuzzy but I think the shorthand for it is infinity agnosticism or a-infinityism or something like that. It logically follows from the premise that a real finite universe does not contain any real infinities.

Some folks who have a temperament which inclines them to debate find this concept worthy of argument.

.99999... is short for something that you could not make with all the paper, pencils, and time in the world, i.e. not really defined, ergo the left side of your equation does not really mean anything.

Don't even get them started on 0^0.
posted by bukvich at 5:59 AM on August 28, 2011

In case ryanrs is being serious, rather than try to decide what 1/∞ means, lets take the first method they give and add some whitespace to clarify a few things:
  x = 0.9… (10  × x = 10    × 0.9…)
10x = 9.9… (10x - x =  9.9… - 0.9…)
 9x = 9    ( 9x ÷ 9 =  9    ÷ 9)      
  x = 1
I hope that's better for you.
posted by wobh at 6:20 AM on August 28, 2011 [1 favorite]

Also, you can do this trick with any repeating decimals to get the fraction version.
posted by wobh at 6:25 AM on August 28, 2011

(and any non-repeating decimal too)
posted by wobh at 6:34 AM on August 28, 2011

Note to math tea partiers: Mathematics is about logic, definitions and proofs. An infinite decimal is defined in a particular way. It's the limit of the sequence of finite decimals. Since you can get arbitrarily close to 1 by making your decimal of 9s sufficiently long, .9999... has the value 1 as a limit by definition. If this does not jibe with your intuition, it's your intuition that is wrong. What's more, you'll find there is a lot more of mathematics that is counter-intuitive. In related news, the earth goes around the sun, though it appears otherwise.

I found the cubicle earth question interesting. (Blatant attempt to underail.)

Also, how the speed of light can be constant, and yet slow down passing through different materials.

You might want to check them out.
posted by Obscure Reference at 8:26 AM on August 28, 2011 [5 favorites]

Regarding the 0.999... is equal to 1 issue, it is interesting how people's intuitions are so split. I wonder how Cantor's diagonal argument comes to bear on the issue. That is to say, the 'distance' between 0.999... and 1 is of a more powerful cardinality than the natural numbers we use for counting this distance.
posted by ageispolis at 10:52 AM on August 28, 2011 [1 favorite]

It's clear that whether 0.9999... = 1 depends on your philosophy of mathematics- and most mathematicians will simply take a pragmatic perspective and go with the assumptions that allow them to make other proofs. Calculus is also based on the fiction of infinitessimals but I've heard that people find it occasionally useful.

One does wonder whether sticking to a properly logically justified mathematics where every mathematical form has a genuine meaning would be the best way to solve some problems, like for instance the pattern of prime numbers.
posted by leibniz at 1:06 PM on August 28, 2011

It's clear that whether 0.9999... = 1 depends on your philosophy of mathematics

Not really your philosophy; rather, it depends on your axioms.
posted by Philosopher Dirtbike at 1:11 PM on August 28, 2011

leibniz: ". . . Calculus is also based on the fiction of infinitessimals but I've heard that people find it occasionally useful.

posted by Obscure Reference at 1:26 PM on August 28, 2011 [1 favorite]

In case ryanrs is being serious

Ah, no. It's all perfectly clear to me. Of course 0.9999... = 0^0 because a^x > 0^x.
posted by ryanrs at 5:14 PM on August 28, 2011 [1 favorite]

You know what else equals 0.999999?

- 2.723.14i
posted by ryanrs at 5:38 PM on August 28, 2011

I have a math question that occurred to me the other day when I was shuffling a deck of cards.

You shuffle a deck of cards and stand in a field. Each suit represents a compass direction, and the value of the card represents a number of meters. You pull cards from the deck one by one, each time walking in the direction of the suit for a number of meters dictated by the value of the card.

Find f(x)=y, where x is the number of cards pulled and y is the average distance from the origin for each possible combination of cards after the given number.

f(52) = 0, because after you've drawn every card they'll all cancel out, and I'm relatively sure the graph will be symmetrical around x=13.5.

Anyway, it really did just pop into my head the other day and I want to know how you'd solve it.
posted by Rinku at 9:41 PM on August 28, 2011 [3 favorites]

I don't have an answer for you, but this is a specific case of what is generally known as a random walk; it's possible you could track down something relevant using that as a keyword.
posted by cortex at 9:58 PM on August 28, 2011

Anyway, it really did just pop into my head the other day and I want to know how you'd solve it.

I don't know how you would go about actually solving it, but I wrote a quick python program to simulate it (A=1 unit, K=13 units) and these are the results I got with 10,000 trials:

Step 1 average distance: 6.9975
Step 2 average distance: 9.76083808561
Step 3 average distance: 11.8740711581
Step 4 average distance: 13.6162611895
Step 5 average distance: 15.0246986109
Step 6 average distance: 16.4108286162
Step 7 average distance: 17.5323296396
Step 8 average distance: 18.5337551344
Step 9 average distance: 19.3944321421
Step 10 average distance: 20.2918795282
Step 11 average distance: 21.002914909
Step 12 average distance: 21.6251535604
Step 13 average distance: 22.2202698262
Step 14 average distance: 22.7451498521
Step 15 average distance: 23.2175681276
Step 16 average distance: 23.6485611721
Step 17 average distance: 24.011697802
Step 18 average distance: 24.4815097847
Step 19 average distance: 24.8037178263
Step 20 average distance: 25.1007989239
Step 21 average distance: 25.3057892742
Step 22 average distance: 25.5702856747
Step 23 average distance: 25.7226920013
Step 24 average distance: 25.7984966336
Step 25 average distance: 25.9507731391
Step 26 average distance: 25.9750234559
Step 27 average distance: 25.8956472974
Step 28 average distance: 25.8267870461
Step 29 average distance: 25.7384883598
Step 30 average distance: 25.4716587665
Step 31 average distance: 25.283767572
Step 32 average distance: 25.0233176984
Step 33 average distance: 24.8184074868
Step 34 average distance: 24.5395667756
Step 35 average distance: 24.2457471158
Step 36 average distance: 23.9033569416
Step 37 average distance: 23.5177744252
Step 38 average distance: 22.9145408642
Step 39 average distance: 22.3438012051
Step 40 average distance: 21.8065908756
Step 41 average distance: 21.1550101318
Step 42 average distance: 20.4439191752
Step 43 average distance: 19.6512283606
Step 44 average distance: 18.748104668
Step 45 average distance: 17.6455031287
Step 46 average distance: 16.5227362821
Step 47 average distance: 15.274395987
Step 48 average distance: 13.7873627845
Step 49 average distance: 12.0826706761
Step 50 average distance: 9.86076179004
Step 51 average distance: 7.0187
Step 52 average distance: 0.0
posted by burnmp3s at 9:57 AM on August 29, 2011 [2 favorites]

Very interesting! So you'd naturally guess that there are N cards of each suit, the expected distance maxes out at time about 2N and has value about 2N.
posted by escabeche at 11:07 AM on August 29, 2011

The "maxes out at time 2N" is obvious by symmetry, since the total amount of time is 4N. The "value about 2N", I don't see a quick hand-wavy argument for. If it were later in the semester I'd assign it as a homework problem to my students, and report back after they did it. As it is I might just have to do it myself.

it might be easier to prove things about the square of the distance from the origin than the distance itself. square roots are evil.
posted by madcaptenor at 1:23 PM on August 29, 2011

Very cool. When I said I thought the averages would by symmetrical around 13.5, I meant 26.5, halfway through the deck. Having two cards left to get to the origin should be the same as walking two times with two cards, and so on - the averages for cards 26 and 27 should be the same too, I think.

The plateau in the simulation is interesting - for the middle half of the game, the average is really consistent.
posted by Rinku at 1:39 PM on August 29, 2011

So you'd naturally guess that there are N cards of each suit, the expected distance maxes out at time about 2N and has value about 2N.

The max at 2N is consistent, but the exact 2N value with 13 ranks is just a coincidence:

1 ranks, step 2 average: 0.944411816953
2 ranks, step 4 average: 2.14105429749
3 ranks, step 6 average: 3.49578227248
4 ranks, step 8 average: 5.1008280504
5 ranks, step 10 average: 6.8196908308
6 ranks, step 12 average: 8.76241421219
7 ranks, step 14 average: 10.7296170187
8 ranks, step 16 average: 12.9640279126
9 ranks, step 18 average: 15.417654851
10 ranks, step 20 average: 17.7826620831
11 ranks, step 22 average: 20.2665065186
12 ranks, step 24 average: 23.0446526851
13 ranks, step 26 average: 25.9749676718
14 ranks, step 28 average: 28.6379580508
15 ranks, step 30 average: 31.6451471868
16 ranks, step 32 average: 34.4382666256
17 ranks, step 34 average: 37.7294319527
18 ranks, step 36 average: 41.004842197
19 ranks, step 38 average: 44.6593220325
20 ranks, step 40 average: 48.3297875569
posted by burnmp3s at 1:53 PM on August 29, 2011

Ha ha, that's what I get for arguing by "induction from one case."
posted by escabeche at 2:11 PM on August 29, 2011 [3 favorites]

I have a truly marvelous argument that the expectation of the distance at time 2N, with N ranks, is asymptotic to some constant times N3/2, which this box is too small to contain.

(Actually, it would probably fit here, but I'm not totally convinced, so I don't want to write it out and look like an idiot.)
posted by madcaptenor at 4:43 PM on August 29, 2011

Yeah, it looks like cN^{3/2} to me too, and the data is consistent with that, but the constant my half-assed computation suggested was about .362, while the data above suggests that (if N^{3/2} is indeed the right order of growth) the constant is a little over 1/2. My half-assed computation consisted of pretending that you were just using each card independently with probability 1/2 and not forcing the total number of cards drawn to be 2N; seemed non-terrible enough to get the right power of N but I can see why it wouldn't be reliable as to the constant.
posted by escabeche at 6:32 PM on August 29, 2011

That's roughly what I did too, although I didn't bother to work out the constant.
posted by madcaptenor at 7:32 PM on August 29, 2011

Whereas I went the extra mile to work out the incorrect constant.
posted by escabeche at 8:43 PM on August 29, 2011

Whoa, the cubical Earth thing is a LOT more interesting than I thought (and also a lot more interesting than the tired, flat-earther .999... vs 1 thing):
The vast majority of the Earth would take the form of vast, barren expanses of rock, directly exposed to space. If you were standing on the edge of a face, and looked back toward the center, you’d be able to clearly see the round bubble of air and water extending above the flat surface. I strongly suspect that it would be pretty.
posted by DU at 4:57 AM on August 30, 2011 [1 favorite]

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