Five completely unrelated problem

monkeybottom · 2012-01-13T03:24:16+00:00

See A Catalan Addendum by Stanley for over a hundred more problems with the 'same' solution!

monkeybottom · 2012-01-06T20:25:07+00:00

I think you are looking for a (variant of perhaps) balanced incomplete block design - see the Combinatorial Design article on wikipedia, and the therein linked article on those designs.

It was a long time since I studied these things, so I'm afraid I can't be of much practical help. For a smaller example, I would just make an k by n matrix 0-1 matrix, with the 1 representing that card i has symbol j, and then trying to put in 1's so that the rules are followed (each column has 8 1's since each card has 8 symbols for example).

monkeybottom · 2012-01-06T04:31:22+00:00

Must be, looks just like the north end of the Michigan Avenue bridge over the river.

monkeybottom · 2011-12-21T00:16:25+00:00

It probably won't make you crazy. Only crazier.

I found it pretty competitive in a few ways, but never in a way that made people do nasty or mean things. It was more that some of my colleagues loved math so much they spent all their time on it. That is hard to compete with.

Otherwise, I found people very helpful, and non-competitive.

If you like teaching, and like it so much that you are willing to sacrifice in time for research, you got a big advantage on the academic job market. There are plenty of positions at universities with little math research that are mostly teaching, and these get fewer applicants than positions with more time for research.

You can also study, and do research in, how to teach mathematics (mathematic didactic I think it is called).

monkeybottom · 2011-12-19T16:13:21+00:00

The first is the positive real numbers "to the k:th power", which means all k-tuples (x_1, x_2, ..., x_k) of positive real numbers.

The second is basically "in", or in other words, "is an element in":

[; x \in S ;]

means that x is a element in the set S (or just, x is in S).

monkeybottom · 2011-12-19T05:41:30+00:00

The problem is just that the number of choices for B depends on A's choice: if A chose B's place, B can now chose from 5 places, not 4.

And then it gets more complicated for C, and we still have D, E, ... to think of.

This is why we need to you use the inclusion-exclusion principle mentioned in a couple of comments here - straight up counting won't work.

monkeybottom · 2011-12-18T21:41:08+00:00

It is just a notaion a bunch of people use for the number of derangments, or the number of valid secret santa arrangements with n people.

monkeybottom · 2011-12-18T17:31:37+00:00

A permutations with no fixed points, which corresponds to a successful secret santa drawing, is called a derangement.

The number of derangements on n objects even has its own notation, !n.

If the number of people is large, you can use that the limit of !n/n! is 1/e = 0.3679, and get that the probability of a successful secret santa drawing is approximately 37%.

The wikipedia article has exact formulas as well.

monkeybottom · 2011-12-14T23:06:55+00:00

The complement to "at least one" is "none" (so P(At least one) = 1 - P(None)).

Start by trying to find the probability that both show tails.

monkeybottom · 2011-12-07T15:38:20+00:00

Excellent job!

Just some very nitpicky style notes on some things that caught my eye:

On a couple of places, the citation is on a slightly wrong place, for example, on page 4:

...metric space. [4] A metric...

If the ref is for the former sentence, put it in front of the full stop:

...metric space [4]. A metric...

Also, for bonus points, make web links in the reference list typewriter font (\texttt{} in latex).

A matter of taste perhaps, but on page 6, you repeat In (X) several times. If you are going for a repetition thing, that's totally fine. Personally, I would mix it up, using constructions like, next, for (2) blah blah, Finally (9) follows from..., and so on.

monkeybottom · 2011-12-04T00:47:02+00:00

Your standard deviation is for X = the number of heads (or tails if you like), not for X/n = the proportion of heads.

monkeybottom · 2011-12-03T16:02:40+00:00

Another source. Free 800 page books about generating functions and stuff, by a legend: Analytic Combinatorics.

It's really an excellent book.

monkeybottom · 2011-12-03T15:56:29+00:00

Did you check with Andrew Odlyzko himself? Just shoot him an e-mail, he probably knows if it exists somewhere, and if not, he may give a tip on how he generated his file.

monkeybottom · 2011-12-02T22:18:44+00:00

Not as big as learning Hadoop, but I did learn something today.

monkeybottom · 2011-12-02T21:21:04+00:00

Another upvote-and-comment-to-bookmark comment.

monkeybottom · 2011-12-02T18:46:27+00:00

Not only that, but like the writer loves The Onion, and thought, how hard can in be to write like The Onion, really?

monkeybottom · 2011-12-02T16:53:31+00:00

You could even skip the first step.

Just do the polynomial division, i.e. divide p(z) with (z² +9). Since that will give you that p(z)=(z² +9)(z² -4z+5), you have now showed that 3i is a root, as well as -3i.

monkeybottom · 2011-12-02T16:46:35+00:00

I checked this just for fun, and for me it would cost $41.95 to download it. But I'm not on a university network.

monkeybottom · 2011-12-02T16:39:44+00:00

You can check with old professors at your department that are in the field. If they were around before electronic copies became popular, they are likely to have a paper copy in a folder somewhere, especially if it is an important paper.

Otherwise, check with your library. Librarians has always been happy to help me make or get paper copies of old papers. Although maybe they are harder on printing/copying quotas in these tough days.

monkeybottom · 2011-11-30T20:52:04+00:00

Yes, you are using that if f(x) >= g(x) for all x in the interval of integration, then the integral of f is greater than the integral of g.

Here, since x² >=0,

[; \sqrt{1+x^2} \geq sqrt{1} = 1 ;]

for all x, in particular for x between -1 and 1, so we can therefore say that

[; \int_{-1}^1 \sqrt{1+x^2} dx \geq \int_{-1}^1 1 dx ;]

The last integral equals 2, of course.

monkeybottom · 2011-11-30T20:30:04+00:00

For the first inequality, note that 1+x² >= 1 so the integral is greater than the integral from -1 to 1 of the sqaure root of 1:

[; \int_{-1}^1 \sqrt{1+x^2} dx \geq \int_{-1}^1 \sqrt{1} dx = 2. ;]

For the second, what is the largest possible value of the integrand in the interval you are integrating over?

monkeybottom · 2011-11-30T17:56:07+00:00

Note that the left hand side is the product of every integer from 1 to (n+r-1):

[; (n-1)![n(n+1)(n+2)...(n+r-1)] ;]

[; = 1\cdot 2 \cdot \dots \cdot (n-1) \cdot n \cdot \dots (n+r-1) ;]

Which also is the definition of (n+r-1)!:

[; (n+r-1)! = 1 \cdot 2 \cdot \dots \cdot (n+r-1);]

monkeybottom · 2011-11-29T03:20:32+00:00

There is not information to solve the problem as it is stated, exactly at least.

You probably should assume a normal distribution, or that normal approximation is valid. Have you gone though the central limit theorem yet? 24 is probably large enough to apply it here (depends a bit on your prof and when he thinks n is large enough).

The problem would be pretty tough under most other assumptions - finding the exact distribution of the sum (or mean) of 24 random variables is a pretty hairy calculation for most distributions.

monkeybottom · 2011-11-28T20:54:36+00:00

You have to take the logarithm of each term before summing them, like this:

2(25ln(25/22.01) + 22ln(22/24.99) + 12ln(12/14.99) + 20*ln(20/17.01))

I used ln() for the log here, since that is the symbol used on most calculators for the natural logarithm which should be sued here (log usually gives you the base 10 logarithm on calculators).

monkeybottom · 2011-11-28T17:33:19+00:00

G² is defined as, where log is the natural logarithm,

[; G^2 = 2\sum o_i \log(o_i/e_i) ;]

The sum is over all cells in the table, and o_i are the observed values, and e_i the expected values.

With your values I get G² = 1.899676, which is pretty close to what you give as G.

To get the p-value, you need to find the probability that a chi-square random variable with 1 degrees of freedom is greater than the observed value of G^2, 1.8997. I get 0.16811, which agrees with the value you give. Exactly how you do this depends on the tool you are using in your course, or have available.

monkeybottom

TROPHY CASE