Wikipedia:Reference desk archive/Mathematics/2006 July 9

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[edit] Satisfying Poll Results Involving "The Average"

On a webpoll I recently saw, participants were asked to rank how well they drove, in comparison to everyone else who filled the poll. Options were "Average", "A bit better than average", "Considerably better than average", "Much better than average", "A bit worse than average", "Considerably worse than average", and "Much worse than average". Unsurprisingly, more people supposed that they were better than average than supposed they were worse, though every option had some takers. But some people seemed to interpret this as being "some people clearly think they are better drivers than they are," and it seems to me that is not necessarily true.

After all, consider a simpler quiz of three options "Average," "Better than average", and "Worse than average", and we rank people's driving out of 100 as a way of deciding an average. If 90 people drive at a 75 level, 5 people drive at a 68 level, and 5 people drive at a 10 level, then the average is 68, and 90% of people drive better than this, 5% drive at this, and 5 drive worse. Those could well have been our more-above-average-than-below results.

This got me thinking. Is it possible, given -any- distribution of people's guesses (satisfying certain requirements), to construct such a theoretical sample that would satisfy the result? Clearly, it could not be true if everyone said they were better than average, but it seems to me (I made a half-hearted attempt, but I can't prove it) that as long as you have an open-ended category on the underpopulated side of average that has at least one taker (e.g. a category like "Much worse than average" that covers every really bad driver, and there is no limit on how bad a driver can be, and there's at least one taker), then you can satisfy any possible set of results. This might not even be necessary. Anybody have any thoughts on the matter? What general rule can we construct about such a thing? Maelin 13:20, 9 July 2006 (UTC) (Minor edit for clarity 13:33, 9 July 2006 (UTC))

If you have a sample of n people, with mean \frac{\sum x_i}{n}, and you add one more person to the sample, you can get any new mean y that you like by taking x_{i+1}=(n+1) y - {\sum x_i}.
Still, while people might be jumping to a logically unsound conclusion from your webpoll, the Lake Wobegon effect does seem like a plausible interpretation. It's also unclear what respondents think the word "average" means. If they think it refers to the median, not the mean, then your logical loophole disappears. -- Avenue 14:50, 9 July 2006 (UTC)
Yes, I strongly suspect that by "better than average", they mean "better than half of the drivers", which is really the median, not the average. StuRat 17:24, 9 July 2006 (UTC)

[edit] volume of concrete

   Hello,
   My name is matt and I am helping my dad figure out a volume problem with our concrete.
   We are pouring gypcrete in our house and the contracter said that 1 1/2 inches of crete would do 
   18 square feet. My dad only wants to pour the thickness to 1 1/8 and hopefully have enough to do 
   our room that is 20 square feet.
                                       How do we figure this problem?
                                        Thank You so much
One thing that might help is to get them into the same units, instead of using both inches and feet. 1 square foot = 12x12 = 144 square inches. You could work out the volume of the concrete you have, then find out to what depth you can fill 20 square feet (2,880 square inches) of floor with it. If you can fill higher than you need to, you have more than enough. The volume of a rectagular prism (box shape) is depth x area of bottom. Black Carrot 17:55, 9 July 2006 (UTC)
If you really want to help your dad (and not your grade), you should advise him to rely on the expertise of the contractor. The contractor is being paid to do the job properly, and might be expected to know his job. Using a thinner layer may prove to be false economy. If that too-thin layer causes problems in the future, the cost will be much more than buying more "crete" now. Issues include strength, noise control, and fire protection. There may also be requirements in building codes for minimum thickness.
Q. "How do I stick beans up my nose? A. Don't! --KSmrqT 20:53, 9 July 2006 (UTC)
Another pretty simple answer is that if you're making a 1 1/8" (7/8") slab instead of 1 1/2" (12/8"), you'll cover 12/7 as much area; if the original covered 18 square feet, the thinner slab would cover nearly 31 square feet. Of course, none of this says anything about, as User:KSmrq said, the advisability of going against the contractor's advice. grendel|khan 00:13, 14 July 2006 (UTC)
Grendel, I think you need some coffee, 1 1/8 is 9/8, not 7/8. StuRat 22:53, 14 July 2006 (UTC)
And this, kids, is why you should never attempt basic arithmetic after spending twelve-hours on shift and then throwing your brain in the blender of learning to use threading with Java. That's my story, and I'm stickin' to it. grendel|khan 03:19, 15 July 2006 (UTC)

[edit] two different group laws

I have two different group laws on a manifold, but it turns out that they're not so different after all. The manifold is R3 and the two group laws are

(a,b,c)*_1(d,e,f) = \left(a+d,b+e,c+f+\frac{1}{2}(ae-bd)\right)

and

(u,v,w) * 2(x,y,z) = (u + x,v + y,w + z + uy).

It turns out that these two groups have the same Lie algebra, and in fact the two operations can be understood in terms of the exponential map from this Lie algebra. The Lie algebra is of course also homeomorphic to R3. Let Q, P, E be a basis for the Lie algebra, and in terms of the exponential map we have:

(a,b,c) = eaQ + bP + cE

and

(u,v,w) = euQevPewE

so we see that the differences in the group law come from the choice to first combine the vectors, then exponentiate, or vice versa. The two group laws are thus really consequences of two different choices of parametrizations for the group. The relationship between the two is given by the Baker-Campbell-Hausdorff formula, which comes to

e^{uQ}e^{vP}e^{wE} = e^{uQ+vP+(w+\frac{1}{2}uv)E}.

In case you haven't recognized it, the groups described are (both?) the Heisenberg group and their Lie algebra are the canonical commutation relations of quantum mechanics. This group is a particularly nice place to analyze my question due to the simplicity of BCH, but it could be asked of any (nonabelian) Lie group, I suppose.

Now we come to my question. Are the two groups the same? The most obvious map between them doesn't seem to be a homomorphism, unless I'm mistaken. On the other hand, aren't they just two different parametrizations for the same group, and therefore shouldn't they be isomorphic? I have one reference which calls the first group the Heisenberg group and the second group the polarized Heisenberg group and never mentions an isomorphism between them (which seems telling in its ommission). I have several other references which call the second group the Heisenberg group and make no mention of the first, and one which calls the first the Heisenberg group and makes no mention of the second. Right now the Wikipedia article has both groups, but in an inconsistent way, and it is in attempting to fix that article that I have gotten myself stuck on this issue and have come to you folk in suppliance.

I could ask the same question about any Lie group, I think. For example, for the rotation group, I think it would sound like this: write the group law for rotatations in terms of their Euler angles, and then write the group law in terms of the single axis of rotation. Is there an isomorphism between the two groups? -lethe talk + 21:47, 9 July 2006 (UTC)

Perhaps it is enough in this case to observe that both groups are finite-dimensional, simply connected, and have the same Lie algebra. Thus each is the same universal covering group (up to isomorphism), yes? Also, the groups are path-connected and not compact, so the exponential map should be your friend in constructing an isomorphism. --KSmrqT 23:13, 9 July 2006 (UTC)
Yes, this is of course correct. There is only one connected simply-connected Lie group to any Lie algebra, so we know at the outset that the two groups must be isomorphic. Furthermore, your suggestion that the exponential map should point to this isomorphism also sounds eminently reasonable. Either the "obvious" map between the two groups is not the correct map, or else my calculation about the map contains mistakes. I guess I will repeat the calculation. I'll post back my findings. Thank you for your input. -lethe talk + 00:28, 10 July 2006 (UTC)
I've repeated the calculation, and actually the obvious map is a homomorphism. The groups are isomorphic, as they should be. Nothing more to say about this, I guess, except that I should be more careful, triple-check my calculations before I come to RD with silly questions. Thanks again. -lethe talk + 13:48, 10 July 2006 (UTC)
Classifying smooth actions by real Lie groups is not the same as classifying real vector fields on Rn or classifying (real or complex) Lie groups themselves (which is not quite the same as classifying Lie algebras). There are many interesting questions still be answered in these areas. One interesting point is that Lie got into this from the left hand endpoint and wound up at the right hand endpoint, which is easier. I'm sure Lethe already knows all about this stuff, but the books
  • Olver, Peter J. (1991). Applications of Lie groups to Differential Equations, second, New York, NY: Springer-Verlag. ISBN 0-387-95000-1. 
  • Olver, Peter J. (1995). Equivalence, Invariants, and Symmetry. Cambridge: Cambridge University Press. ISBN 0-521-47811-1. 
offer a nice discussion. Any fan of symmetry, Lie groups, representation theory, reflection groups, the ADE classification, invariant theory, differential equations of mathematical physics and Cartan's method of classification of (semi)-Riemannan manifolds up to local isometry will benefit from these books, I should think! ---CH 02:21, 14 July 2006 (UTC)
Oh, BTW, Lethe, do you know how the Heisenberg group arises in the Bianchi classification? OK, this isn't very profound :-/ Sometimes people ask about the difference between the Bianchi classification and Thurston's classification. ---CH 02:24, 14 July 2006 (UTC)

[edit] Oi, barkeep, wheres my quid

3 guys in a bar, buy three drinks, cost comes to £30.

Bartender refunds £5 for miscalculation, the three men take £1 each, and give the bartender back £2 for honesty.

Each man spent £10 - £1 = £9

the three men spent £9 * 3 = £27

plus the £2 the tender had, comes to £29.

Where is the other pound?

There are several issues with this problem, the whole thing originates from there drinks costing £8⅓ each. Which is the first problem, I figured thats where the extra pound is, by I still can't explain the incosistency with the amount going in and the amount going out. Philc TECI 23:28, 9 July 2006 (UTC)

I believe the issue is you added when subtraction is needed. The three men did indeed spend £27, which includes a £2 gratuity (£25 for drinks, £2 for the barkeep). Thus adding £27 and £2 is meaningless--a more meaningful quantity is £27 - £2, the total spent minus the tip, which yields the cost of the drinks. --TeaDrinker 00:03, 10 July 2006 (UTC)
There seems to be a large hole in this problem. Is the £30 the price with or without the miscalculation? -- He Who Is[ Talk ] 01:22, 10 July 2006 (UTC)

[Edit Conflict] :This seems to be one of the most commonly repeated questions on this desk (because of the various ways to word it I'm not sure you could search for it, but I wouldn't be surprised if it was asked at least once a month) - see Missing dollar paradox. Confusing Manifestation 01:19, 10 July 2006 (UTC)

There is NO missing £1. What you should do is ask "who has each of the original £30?" The answer then is: the cahier has £25, the three men have £1 each (£3) and the bar-tender has £2, so all 30 are accounted for - no 'missing' £1.
Hmm fair enough, makes sense, it all in the wording. Philc TECI 17:07, 10 July 2006 (UTC)


Start from realizing that the three drinks cost £25. Then see the cash flow:

  The three guys have The cashier has The bartender has
Initially 10+10+10 0 0
The three guys pay 0 30 0
and the change goes back 5 25 0
the bartender gets his tip 1+1+1 25 2
The money actually spent 1+1+1 25+2 = 27

Note two things:

  1. No penny is missing – the sum in every row is the same £30.
  2. The £27 sum is what the three guys have paid – and it already contains the £2 tip.

So it makes no sense to add 27+2. Such sum does not appear anywhere in the balance.
 :) --CiaPan 19:24, 10 July 2006 (UTC)