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July 11

Simple Probability

Suppose that there have been 20 no-hitters in baseball history. What is the probability that there's been at least one on each day of the week?

The case for 7 no-hitters is simple enough, but I'm having trouble extending this to 8+ days. Help would be greatly appreciated. 74.15.137.192 (talk) 00:33, 11 July 2010 (UTC)[reply]

I only have time for this quick answer: I wonder if Schroedinger's method would help with this one? Michael Hardy (talk) 02:16, 11 July 2010 (UTC)[reply]

Isn't this what Stirling numbers of the second kind are for?

(S(20,11)*7!)/(7^{20})

should do it. —Preceding unsigned comment added by 203.97.79.114 (talk) 04:52, 11 July 2010 (UTC)[reply]

Shouldn't it be S(20,7)*7!/7^20? (There are seven days of the week, right?) 74.15.137.192 (talk) 09:22, 11 July 2010 (UTC)[reply]

This is the Coupon collector's problem. Robinh (talk) 19:27, 14 July 2010 (UTC)[reply]

ordinals

I'm interested mostly in countable ordinals in this question, in fact computable ones, I think. Is there a concept of a limit ordinal that represents a fixed point of the (informally) biggest operation seen so far? By that I mean ω is such an ordinal, and maybe 2ω and ω^ω are as well (not sure about further iterations of exponentiating), and the next thing after that is ε₀, but something like ω³+ω²·2+ω·3 would not be, even though it's a limit ordinal. The next thing after ε₀ would be ε₁ or maybe Γ₀. I don't mean a singular ordinal since there's only one of those for each cardinal. Am I making any sense? Maybe what I'm looking for is the places where various reasonable ordinal notations run out of gas. Thanks. 71.141.88.179 (talk) 01:16, 11 July 2010 (UTC)[reply]

The question does not make much sense without specifying rigorously which operations you take into consideration. It's fairly trivial to cook up an artificial ordinal notation system which "runs out of gas" at ω³ + ω²·2 + ω·3.—Emil J. 13:13, 12 July 2010 (UTC)[reply]

Thanks, I'm just asking if there is a way to formalize the usual informal way of explaining ordinals. You know, you've got 1,2,3...ω,ω+1,...,ω·2,...ω·3...ω²...ω³...ω^ω...ω^{ω^ω}...ε₀...

For each place that "..." appears in the above, what comes next? Obviously I've missed some intermediate ones, but the idea is that at each stage we've iterated some newly concocted ordinal operation and found a fixed point for it, as opposed to just finding the nearest limit ordinal by iterating the successor operation. Maybe each one corresponds to the order type of some natural combinatorial structure? 71.141.88.179 (talk) 19:28, 12 July 2010 (UTC)[reply]

Simplifying Radical Expressions

Hello. How would you simplify ${\sqrt {11+6{\sqrt {2}}}}$ to $3+{\sqrt {2}}$ ? If you let ${\sqrt {11+6{\sqrt {2}}}}=a+b$ , how can you identify 11 as a sum of squares, $a^{2}+b^{2}$ and not anything else? Thanks in advance. --Mayfare (talk) 03:18, 11 July 2010 (UTC)[reply]

You can check the identity by letting

a={\sqrt {11+6{\sqrt {2}}}},b=3+{\sqrt {2}}

, computing a² and b², and checking whether they are equal.

11=3^{2}+({\sqrt {2}})^{2}

, a perfectly good sum of two squares in the extension field

\mathbf {Q} [{\sqrt {2}}]

.

71.141.88.179 (talk) 03:29, 11 July 2010 (UTC)[reply]

See Nested radical for the conditions where this can be done for nested square roots. Dmcq (talk) 11:25, 11 July 2010 (UTC)[reply]

In general, to find

{\sqrt {p+q{\sqrt {2}}}}

where p and q are rational you proceed as follows:

{\sqrt {p+q{\sqrt {2}}}}=r+s{\sqrt {2}}

\Rightarrow p+q{\sqrt {2}}=r^{2}+2rs{\sqrt {2}}+2s^{2}

\Rightarrow p=r^{2}+2s^{2}{\text{ ; }}q=2rs

\Rightarrow 2r^{4}-2pr^{2}+q^{2}=0

\Rightarrow r^{2}={\frac {p\pm {\sqrt {p^{2}-2q^{2}}}}{2}}{\text{ ; }}s^{2}={\frac {p\mp {\sqrt {p^{2}-2q^{2}}}}{4}}

The identification of coefficients in line 2 not only requies p and q to be rational, but also r² and s² must also be rational, and so

{\sqrt {p^{2}-2q^{2}}}

must be rational too. Gandalf61 (talk) 17:26, 11 July 2010 (UTC)[reply]

Why does ${\sqrt {a^{2}-b^{2}c}}$ have to be an integer if some nested radical, ${\sqrt {a+b{\sqrt {c}}\ }}$ , can be denested into a sum of surds? Thanks again in advance. --Mayfare (talk) 20:35, 12 July 2010 (UTC)[reply]

Angle

Hi, How is the angle=arc/radius?Max Viwe | Wanna chat with me? 04:49, 11 July 2010 (UTC)[reply]

Radian should get you started on this. -- Meni Rosenfeld (talk) 05:22, 11 July 2010 (UTC)[reply]

Automorphism group of a Hilbert space

What is the largest subgroup of the diffeomorphism group of a 2n-1 sphere that is holomorphic when the sphere is taken to be the unit sphere of an n-dimensional complex space? 74.14.108.160 (talk) 06:58, 11 July 2010 (UTC)[reply]

What exactly are you looking for here? It doesn't make sense to say that a subgroup of maps that are also holomorphic? Do you mean a subgroup of holomorphic maps? Are you considering the diffeomorphism group of the sphere (as an abstract manifold) or the subgroup of the diffeomorphism group of the ambient space which preserves the sphere (as a submanifold of the ambient space)? •• Fly by Night (talk) 19:47, 14 July 2010 (UTC)[reply]

Yes, I meant to say the subgroup whose members are holomorphic maps rather than calling the subgroup itself holomorphic. The sphere was meant to be an abstract manifold and was only embedded in the complex space in order to give diffeomorphisms of the sphere a holomorphicity criterion. 76.67.74.129 (talk) 07:54, 16 July 2010 (UTC)[reply]

Mathematical induction

The article we have on mathematical induction seems to imply that assumption is an integral part of the process. This doesn't appear right to me. Statements made about an arbitrary k during the inductive step are only allowable because the existence of at least one suitable k is proven in the basis step. No speculation or assumption is done. Now, to the right person, there is no difference between "Let k be an integer for which the statement holds..." and "Assume the statement holds for some k..."; but to another reader, it's the difference between speculation and logical argument. The prevalence of the word assume in textbooks, at least over here, is part of the difficulty people seem to have with the legitimacy of induction. Am I being too picky or misunderstanding things, or is this something that should be looked at? Thanks for the help. —Anonymous Dissident^Talk 13:33, 11 July 2010 (UTC)[reply]

I always used to say "suppose it is true for some k" ... I also explained (once, not every time) that use of a logical contrapositive makes the argument clear and finite (rather than infinite induction). Dbfirs 22:00, 11 July 2010 (UTC)[reply]

The article explains the concept in several different ways so I think most readers will understand at least one version. My experience teaching this is that it's something of a pons asinorum in that there are always going to be a few people who have a very difficult time getting it. Keep in mind though that Wikipedia isn't intended to teach the subject from scratch, so a completely foolproof explanation isn't the goal.--RDBury (talk) 23:13, 11 July 2010 (UTC)[reply]

It's always a bit hard to find the right wording to explain this point when teaching induction, but once you see it it's not a difficult concept at all. The issue is that, yes, you make an assumption, but only temporarily. In logical jargon the assumption is discharged before the end of the proof, and doesn't count as an assumption for the proof as a whole.

To prove

\forall nP(n)

, you have to prove two things: First, you have to prove

P(0)\,\!

; how you do that doesn't concern us here. More to the point, you need to prove

\forall k\left(P(k)\implies P(k+1)\right)

.

How do you do that? Well, how do you prove

A\implies B

in general? You assume

A\,\!

and conclude

B\,\!

. The assumption of

A\,\!

is truly an assumption, but only for the length of time that you're in the little subproof of

B\,\!

. Once you're done with that,

A\implies B

has been proved and the assumption is discharged (that is, it's no longer an assumption).

Exactly the same deal for the assumption of

P(k)\,\!

when proving

P(k+1)\,\!

.

(Aside to Dbfirs: I don't think saying "suppose" rather than "assume" helps at all; they're synonyms.) --Trovatore (talk) 22:46, 11 July 2010 (UTC)[reply]

Not if you say "suppose we can find a value of k for which the statement is true". It conveys a different impression of the argument, avoiding the (unfounded) suspicion that we are assuming what we set out to prove. Dbfirs 07:26, 12 July 2010 (UTC)[reply]

Hmm, possibly. But they still have to deal with the proof-within-a-proof thing, and specifically that the proof-within-a-proof may have different assumptions than the whole proof, one way or another.

I think this is the same issue people have with proof by contradiction — there's a mental leap involved in "assuming" something that, come to find out, is not actually true. You see it over and over again in objections to, say, Cantor's diagonal argument; people will say "Cantor assumes you can list the real numbers! But you can't!", and not notice that that's actually the thing being proved. --Trovatore (talk) 07:46, 12 July 2010 (UTC)[reply]

"Suppose" and "Assume" are basically synonyms, so I don't see any real improvement. --Tango (talk) 09:24, 12 July 2010 (UTC)[reply]

This is exactly what the OP is talking about- to the uninitiated, these two words are not synonyms. In common language, "suppose" is somewhat of a hypothetical assertion of something which may or may not be true (this is the meaning mathematicians want), while "assume" means something is asserted without proof as being true in some wider or objective context (this is basically never what mathematicians mean). People beginning to read serious mathematics need to be taught that "assume" means "suppose", and it's natural for them to get confused otherwise. Staecker (talk) 11:58, 12 July 2010 (UTC)[reply]

Mathematical inductions consists of two steps: step 1: P(0), step 2: P(n)==>P(n+1). one of these is called the induction. A common misunderstanding is to call part 2 the induction. Logically it is part 1 which is the induction step. Induction is concluding from special cases. Bo Jacoby (talk) 07:44, 12 July 2010 (UTC).[reply]

I would say that the "induction" is putting the two steps together to reach a conclusion. Dbfirs 19:19, 12 July 2010 (UTC)[reply]

The word 'induction' is much older than the idea of 'mathematical induction'. So 'mathematical induction' got its name for a reason. That reason being that is contains an element of good oldfashioned philosophical induction. That element being P(0). Bo Jacoby (talk) 21:26, 12 July 2010 (UTC).[reply]

I'm skeptical of the claim that the name "induction" was applied because the method was thought of as philosophical induction from a single example. --Trovatore (talk) 23:40, 12 July 2010 (UTC)[reply]

Why else call it induction, if not because it contains an element of induction? (Although a single example is theoretically sufficient, in practice you also try P(1) and P(2) and P(3) to give a hint about the truth of the assertion before you work on the P(n)==>P(n+1) argument). Bo Jacoby (talk) 04:47, 13 July 2010 (UTC).[reply]

I too am skeptical of this. Do you have any evidence beyond your own inability to think of another explanation? Algebraist 09:33, 13 July 2010 (UTC)[reply]

Induction is concluding a general statement from observed special cases (and more generally, any form of reasoning that usually works but is not strictly guaranteed to be correct), but not from a single special case. One would check P(0), P(1), P(2), P(3), etc., and after reaching a sufficiently (depending on the application and desired level of confidence) high number and getting sufficiently bored, one would conclude that P(n) for every n by induction. Mathematical induction, which is indeed named after this "philosophical" induction (it does not actually have much to do with philosophy, it's the usual way of inferring general rules in natural sciences like physics), does the same in a sense, except that it involves a property (a proof of P(n) → P(n + 1)) which enables to generate an argument for all these special cases P(0), P(1), P(2), etc. uniformly (by chaining instances of the implication), and leads to the desired conclusion in a mathematically rigorous way. Claiming that the P(0) step is the induction is nonsense. As for calling P(n) → P(n + 1) the "induction step", this does not quite correspond to the origin of the term "induction" as described above, but that's irrelevant. In mathematics it is established terminology, with obvious etymology (it is the main step in what is called "(mathematical) induction", so we can as well call it the "induction step"), and there is nothing wrong with it.—Emil J. 13:21, 13 July 2010 (UTC)[reply]

Maybe it's language-dependent, but in Hebrew the two parts are often called "induction base" and "induction step" (בסיס האינדוקציה, צעד האינדוקציה), and I took the etymology to be: Both parts together constitute (mathematical) induction; the first is where we start, and the second is what allows us to step from each number to the next. -- Meni Rosenfeld (talk) 15:02, 13 July 2010 (UTC)[reply]

Gabriel's Horn

In Gabriel's Horn are the diagrams wrong? The text says x>=1 yet the diagrams seem to do x>0. -- SGBailey (talk) 14:16, 11 July 2010 (UTC)[reply]

Well the diagram does do some other shape than 1/x between 0 and 1, but it would be better to have a diagram that starts at x=1. I'd like one that showed the opening a bit so it looked more horn like rather than 2d. I'm surprised there's no reference to the problem of painting it. You only need a finite amount to fill the inside but would need an infinite amount to paint the outside. Dmcq (talk) 16:10, 11 July 2010 (UTC)[reply]

The outside may have infinite area, but a finite amount of paint of zero thickness would be required to cover it - just get two of them vertically. fill one with paint, fit the other inside then lift it out.→86.132.163.37 (talk) 15:46, 13 July 2010 (UTC)[reply]