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August 14

Schläfli symbol

If the Schläfli symbol of a regular polyhedron is {p,q}, meaning its faces are p-gons and each vertex is surrounded by q faces, how many faces does it have? How is this answer obtained? Duoduoduo (talk) 12:26, 14 August 2013 (UTC)[reply]

Thats doing things the hard way round, easier to start from a polyhedron and find the Schläfli symbols. As there are only 5 convex Platonic solids, and 4 nonconvex Kepler-Poinsot polyhedra a simple lookup table is easiest.

If you wanted to do things the hard way, you could take k n-gons around a vertex. You should be able to work out all the angles and, assuming a unit edge length, the radius of the enclosing sphere. Then project one polygon onto the sphere, and find its area. Dividing the surface area of the sphere the area of the spherical polygon gives the number of faces.

There is a close relationship between Schläfli symbol and Schwarz triangle which may give the best explanation.--Salix (talk): 14:04, 14 August 2013 (UTC)[reply]

The answer can be found in our platonic solids article at the bottom of this section. A quick check shows that this doesn't work for the nonconvex solids though, but a generalised version could probably be derived using the equation here in place of the standard Euler equation (though p and q take into account the density of the face and vertex face, so only a correction factor for the overall density will be needed). MChesterMC (talk) 15:26, 14 August 2013 (UTC)[reply]

Thanks to both of you! Duoduoduo (talk) 17:07, 14 August 2013 (UTC)[reply]

Are there infinite axioms?

If you are counting only fundamental axioms, and not thing like 1 + 10 = 11, 1 + 1,000,000 = 1,000,001, which could lead to an infinite number of axioms. OsmanRF34 (talk) 14:25, 14 August 2013 (UTC)[reply]

Axioms for what? It's certainly possible to construct axiomatic systems with infinitely many axioms. Sławomir Biały (talk) 14:33, 14 August 2013 (UTC)[reply]

But that doesn't mean you should do that. The question is if you can derive all theorems with a limited set of axioms, in the same way that in physics you have the 5 types of energy that are at the basis of all the rest. OsmanRF34 (talk) 14:50, 14 August 2013 (UTC)[reply]

No, see here why not. Count Iblis (talk) 16:40, 14 August 2013 (UTC)[reply]

You should look at our article on Gödel's incompleteness theorems. The "Background" section explicitly discusses your question, and the section on the first theorem basically answers it. Looie496 (talk) 16:49, 14 August 2013 (UTC)[reply]

See also Peano axioms#First-order theory of arithmetic, which describes the induction schema as a countably infinite set of axioms. EdJohnston (talk) 16:53, 14 August 2013 (UTC)[reply]

This discussion risks conflating a couple of very different things. EdJohnston is talking about axiom schemata, which formally speaking comprise infinitely many axioms, but are quite reasonably thought of as a single axiom at an informal level. In general you can turn such an axiom schema into a single axiom by enriching the language, in a conservative extension of the theory (that is, the theory doesn't prove any new statements in the original language).

Looie496 is talking about failure of finite axiomatization in a much more fundamental way. The Gödel theorems imply that true arithmetic is not only not finitely axiomatizable in the above-mentioned formal sense — you can't even capture it by a set of axioms that can be output by an (idealized) computer program running forever. Such a computer program could be considered, not a finite set of axioms per se, but a finite amount of information that would effectively generate all (and only) arithmetic truths; the theorems imply that no such thing is possible. --Trovatore (talk) 21:02, 14 August 2013 (UTC)[reply]

I should add, however, that it is possible to characterize arithmetic truth by a finite set of axioms — if you consider their consequences in full second-order logic. The problem is that second-order logic has no effective proof system. That is, there is no way to write a fixed computer program that tells you whether a given statement is a second-order logical consequence of another given statement (even if the program is allowed to run forever if the statement is not a logical consequence). --Trovatore (talk) 21:06, 14 August 2013 (UTC)[reply]

A deceptively simple statistics question

Dear Wikipedians:

I am confronted with a problem whereby an analyst wants to show the mean change from last year in customer satisfaction is different from zero. With a random sample of 10 observations the test statistic is 2.07. What is the P-value?

I have looked up P(Z<-2.07) and found that it is 0.0192 so I have chosen "between 0.01 and 0.025". However the answer is "between 0.05 and 0.10".

I am curious how that answer came about. I also tried dividing the test statistic by square root of 10, to no avail.

Any help is much appreciated.

216.58.82.243 (talk) 17:42, 14 August 2013 (UTC)[reply]

Not sure exactly what you mean when you refer to "the test statistic". In any event, maybe this passage from Mean#Using the sample mean will help:

Often, since the population variance is an unknown parameter, it is estimated by the mean sum of squares; when this estimated value is used, the distribution of the sample mean is no longer a normal distribution but rather a Student's t distribution with n − 1 degrees of freedom.

Duoduoduo (talk) 18:07, 14 August 2013 (UTC)[reply]

As Duoduoduo has implied, when you are trying to calculate p-values and you have only ten observations you are in the domain of small-sample statistics. If you have observations made in two successive years, it's conceivable that a two-sample t-test is what's needed. You did not tell us what the ten observations were, or if you were doing a before-after experiment. If you obtained your 'test statistic' by running a standard program, you might tell us what program you were using, if you have five before and five after, or whatever. EdJohnston (talk) 20:34, 14 August 2013 (UTC)[reply]

It's kind of an inherent problem with the standard 1-5 satisfaction rating that the distribution doesn't have a useable model. It's common to have bimodal and even trimodal distributions to occur, so forget about any kind of discretized normal or binomial distribution. I suggest binning the results: 4-5=satisfied, 1-3=not satisfied. That way you can use a simple Bernoulli distribution and significance tests can be found in any stats book. The drawback here is that you're not really being honest with the people filling in the ratings who, for reason, might think that their 1-extremely dissatisfied rating means something different than a 3-neither satisfied or dissatisfied. Ideally the questions should be changed to simple yes/no responses but my experience has been that the people who make this type of decision, on whom such subtleties as statistical significance are usually lost, find it hard to see the merits. As a mathie I always find the 1-5 rating odd anyway; if it's a true/false question then the possible answers are true or false. If I say "The sum of the squares of the sides of the right triangle is equal to the square of the the hypotenuse; do you 5-strongly agree, 4-agree, 3-neither agree or disagree, 2-disagree, 1-strongly disagree?" what's the difference between a 4 and a 5. Maybe 4=The Pythagorean theorem is true and 5=The Pythagorean theorem is true and I'll punch anyone who thinks otherwise. --RDBury (talk) 12:29, 15 August 2013 (UTC)[reply]

I agree with the answers above, but just to restate the conclusion, you are using the wrong test. You are using a Z test when you should be using a paired t test (two-tailed). Looie496 (talk) 14:53, 15 August 2013 (UTC)[reply]