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May 28

Season

If a sporting league consisted of n teams, which each played g games, how many games would there be in total? —Preceding unsigned comment added by 76.69.241.74 (talk) 04:12, 28 May 2008 (UTC)[reply]

Rather than giving you the answer, let's look at it this way... Major League Baseball has 30 teams, each of which are scheduled to play 162 games. According to Major League Baseball season, that comes out to 2430 scheduled games per season. How does one get that figure? Alternatively, try to break it down into a scenario such as this: you have 6 teams, each of which has to play the other teams exactly once each. How many situations can you form based on this? I hope this leads you in the right direction. --Kinu ^t/_c 04:32, 28 May 2008 (UTC)[reply]
Hint: You'll need to consider how many teams take part in each game. --Tango (talk) 10:13, 28 May 2008 (UTC)[reply]

I initially though n*g/2, but someone told me, and he seems quite knowledgable in sports, that there are 1335 games in an NHL season (30 teams and 82 games apiece). So is he wrong (predicted value=1230)? —Preceding unsigned comment added by 76.69.241.74 (talk) 03:04, 29 May 2008 (UTC)[reply]

The regular season does hold 1230 contests, as the formula (which you are correct on) dictates. I'm assuming the difference (105 games) your friend is including is the maximum number of playoff games (7 game series) that could be played in the postseason: 8 first round series, 4 conference semifinal series, 2 conference final series, and 1 Stanley Cup final series; 7*(8+4+2+1)=105. --Kinu ^t/_c 04:22, 29 May 2008 (UTC)[reply]

Mappings

$S={1,2,3,...,n}\,$ .

What are the number of mappings from $S\to S$ ?

My book gives the straight answer as $n^{n}\,$ but I was wondering how to arrive at that conclusion, I assume by induction. If there are n mappings from 1 to S, and n mappings from 2 to S, and n mappings from 3 to S, and so on then there are n * n * n e.t.c mappings from S to S, but this doesn't convince me. Any thoughts? Damien Karras (talk) 14:08, 28 May 2008 (UTC)[reply]

There's no need to use induction: your reasoning seems perfectly sound to me. To reword it only slightly, there are n choices for what 1 maps to, n choices for 2, etc, and these choices are independent, so the total number is ~~n*n~~

n^{n}\;\!

. Try doing it explicitly for n = 2 or 3 and it might become clearer. AndrewWTaylor (talk) 14:31, 28 May 2008 (UTC)[reply]

Actually, any function from S to itself can be considered as an ordered n-tuple (the i-th entry in the tuple tells you what element i gets mapped to). Count the number of such n-tuples. –King Bee ^{(τ • γ)} 16:23, 28 May 2008 (UTC)[reply]

I'm pretty sure Andrew meant

n^{n}\;\!

, with n*n being either a typo or part of a larger description which was cut short due to lack of space. -- Meni Rosenfeld (talk) 16:46, 28 May 2008 (UTC)[reply]

Oops, you're quite right Meni, I did indeed mean

n^{n}\;\!

. Thanks for the correction. (I've amended my previous comment.) AndrewWTaylor (talk) 19:14, 28 May 2008 (UTC)[reply]

Diagonalizing a matrix.

Regarding diagonalizing matrices, I know that a matrix P whose columns are linearly independent eigenvectors will always diagonalize a matrix by taking P^-1AP, and that a matrix is diagonalizable iff it has a full set of linearly independent eigenvectors. What I can’t figure out is if those are the only such matrices. Specifically, if at least one of the columns of P is not in an eigenspace of an eigenvalue of A, is it possible that P^-1AP is a diagonal matrix?

More generally, if one has a linear transformation of vector spaces $T:V\to W$ , does the corresponding result hold for infinite dimensional vector spaces? GromXXVII (talk) 16:55, 28 May 2008 (UTC)[reply]

The columns of P must be eigenvectors of A. Let

P^{-1}AP=D=\mathrm {diag} (\lambda _{1},\ldots ,\lambda _{n})

. For every i, denote

P_{i}

the ith column of P, and

\epsilon _{i}=AP_{i}-\lambda _{i}P_{i}

. Then

\lambda _{i}e_{i}=De_{i}=P^{-1}APe_{i}=P^{-1}AP_{i}=P^{-1}(\lambda _{i}P_{i}+\epsilon _{i})=\lambda _{i}(P^{-1}P)_{i}+P^{-1}\epsilon _{i}=\lambda _{i}e_{i}+P^{-1}\epsilon _{i}\;\!

. Therefore

P^{-1}\epsilon _{i}=0

,

\epsilon _{i}=0

and

AP_{i}=\lambda _{i}P_{i}

.

I don't know about the infinite dimensional case. -- Meni Rosenfeld (talk) 17:31, 28 May 2008 (UTC)[reply]

Set Notation

I have a set of classes, $C$ , and a set of documents, $D$ . I want to define a relation, $B$ , that says which documents belong to which classes.

I then want to sum over all documents of a particular class, $c$ .

I think the relation can be defined either with

    $(d,c)\in B$

or

    $B\subset D\times C$

My question is, how can I express the subset of documents belonging to class $c$ ?

My guess was something like

    $\sum _{d\in \{d|\exists c,(d,c)\in B\}}{\ldots }$

Thanks, 80.175.114.193 (talk) 21:59, 28 May 2008 (UTC)[reply]

I don't think you want the "there exists c" bit - that means any c, I think you mean a specific c, so you just want {d|(d,c) in B}, or, a shorter notation: {d|dBc} (using B in the same way we use =, <, > or whatever). --Tango (talk) 22:08, 28 May 2008 (UTC)[reply]

Ah yes, that makes lots of sense. It's surprising how simple it is when you see it. Proberts2003 (talk) 22:26, 28 May 2008 (UTC) [Same user as 80.175.114.193 above][reply]

Also, the statement "

d\in \{d|dBc\}

" is identical to just "

dBc\;\!

", so you could also simply have

\sum _{dBc}^{}(\cdots )

. It makes it less clear that you are summing over different ds, though. -- Meni Rosenfeld (talk) 22:51, 28 May 2008 (UTC)[reply]

I believe the usual notation for that would be

\sum _{d\in D,\;dBc}\ldots

—Ilmari Karonen (talk) 22:57, 28 May 2008 (UTC)[reply]

(ec) Or you could even define D(c) = {d|(d,c) ∈ B} and then just write your sum over d ∈ D(c). You could even use that notation exclusively, starting with a definition of D(c) as a function of c and writing d ∈ D(c) instead of (d,c) ∈ B or dBc. Of course, if you also need the sets of classes for a particular document, C(d) = {c|(d,c) ∈ B}, then going with the relational notation as suggested by Tango may be simpler. —Ilmari Karonen (talk) 22:57, 28 May 2008 (UTC)[reply]

Or, better yet, since c is supposed to be a class, why not simply define it as the set of documents it contains? Then you could simply write d ∈ c. —Ilmari Karonen (talk) 23:00, 28 May 2008 (UTC)[reply]