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March 24

How to display surds on the TI-84 Plus

Does anyone know? Is there a program which I can download which does it?--124.171.116.21 (talk) 02:32, 24 March 2010 (UTC)[reply]

the volume of a solid bounded by the lines x=0, y=0, z=0 and the plane x+y+z=9

Just checking webwork ... isn't the volume 729/2? I'm attacking this from several angles -- e.g. double integration just plain ole geometric sense (it should be half the cube 9x9x9). But the system says I'm wrong ... John Riemann Soong (talk) 09:01, 24 March 2010 (UTC)[reply]

If you had

x+y\leq 9,\ 0\leq z\leq 9

you'd get a prism which is half the cube. Since you have

x+y+z\leq 9

you get a pyramid which is one third of the prism, or only 1/6 of the cube. Integration also gives 121.5. -- Meni Rosenfeld (talk) 09:21, 24 March 2010 (UTC)[reply]

You've got a pyramid one of whose faces is half of the square. The volume of a pyramid is 1/3 × base × height. So it's 1/3 × (half the square) × (height) = (1/3) × (1/2) × (volume of the cube). Michael Hardy (talk) 14:25, 24 March 2010 (UTC)[reply]

Poincare Conjecture and Ricci flow

Perelman used Ricci flow in proposing a proof of Poincare's conjecture.

Eddington in about 1934 said he proved, or seemed to prove, (as Kumar seemed to earlier) that if Pythagoras' theorem is extended to four or more dimensions then the sign of the fourth or further term is negative. For example, time in Einstein's metric contributes with the opposite sign to the space terms, ds squared = (say) dx1 squared + dx2 squared + dx3 squared - dt squared. Eddington said he did this to say something about Dirac's spin. Eddington's proof (?) involved group theory, but I'm not sure of its elimination of alternatives, for I only saw a semi-popular account of it.

Does this modify or limit the applicability of the Ricci flow method to Perelman's proof of Poincare's conjecture? Is Poincare's conjecture itself affected by the different sign on the fourth dimension, in a metric?

As a University of Sydney student, I had not much spare time to study or ask this.

122.152.132.156 (talk) 10:25, 24 March 2010 (UTC)[reply]

I should preface this by saying I don't know a whole lot about this topic. But the Poincare conjecture is purely a topological statement. It doesn't depend on any metric, even if a particular metric was used to prove it.

Also, the standard definition of a metric requires it be positive definite. Rckrone (talk) 17:47, 24 March 2010 (UTC)[reply]

Agree with above (I'm also not an expert here). Your question sounds similar to Marilyn vos Savant's infamous objection (later retracted) to Wiles' proof of Fermat's last theorem. She objected that the proof uses hyperbolic geometry which isn't allowed because the "real word" uses Euclidean geometry (which is only mostly true). The fact is that mathematicians use whatever abstract notions are appropriate for the setting, regardless of any considerations of the "real world", which is irrelevant in the context of an abstract proof. Staecker (talk) 22:41, 24 March 2010 (UTC)[reply]

The "metric" with -dt^2 is called the Minkowski metric which as Rckrone says is not a true metric in the usual mathematical sense. The geometric symmetry group of Minkowski space is the Poincaré group, if that's what you're thinking of. AFAIK it doesn't have anything to do with the Poincaré conjecture in topology. 66.127.52.47 (talk) 07:04, 25 March 2010 (UTC)[reply]

Beads on a necklace

If you had n beads, all different and unique, then would the number of unique orderings on the necklace, after allowing for rotations and reflection, be n!/2n ? 78.149.167.173 (talk) 22:09, 24 March 2010 (UTC)[reply]

No. If we plug in 2 beads, we get 2!/2(2) = 2/4 = 1/2. Obviously there isn't half a combination. Can you figure out what's wrong with your formula ? StuRat (talk) 00:49, 25 March 2010 (UTC)[reply]

The formula gives the correct answer for n=3 at least. 84.13.22.69 (talk) 14:09, 25 March 2010 (UTC)[reply]

You can cheat using the search function on wikipedia 9or google) and you'll find Necklace (combinatorics) Dmcq (talk) 01:24, 25 March 2010 (UTC)[reply]

The problem is fairly easy, and this article doesn't make it any easier at all. -- Meni Rosenfeld (talk) 09:58, 25 March 2010 (UTC)[reply]

You seem to be quite right, you'd need to know how to solve the problem above long before you ever turned to that article! Dmcq (talk) 14:30, 25 March 2010 (UTC)[reply]

The way I got the n!/2n was, if you had a line of n unique and different objects, then the number of different ways of ordering them would be n!. Join the ends together, then the number of different ways would be reduced by n as you could rotate the string of n beads into n positions, and by a further 2 because you could flip the necklace over. What is wrong with that reasoning? 84.13.22.69 (talk) 14:06, 25 March 2010 (UTC)[reply]

In fact that's quite reasonable but it still goes wrong for 0, 1 or 2 beads! :) For more than 2 you're quite right, you don't get exactly the same pattern flipping the necklace over. Dmcq (talk) 14:30, 25 March 2010 (UTC)[reply]

And the reason is that flipping the necklace over doesn't change the order, with 2 beads or less. StuRat (talk) 17:58, 25 March 2010 (UTC)[reply]

So is the formula correct provided that n>2 ? 84.13.22.69 (talk) 14:45, 25 March 2010 (UTC)[reply]

Indeed. -- Meni Rosenfeld (talk) 15:00, 25 March 2010 (UTC)[reply]

Would it be worth a mathematician noting that on the Bracelet (combinatorics) page, since itys not obvious to the casual reader that the complicatated formula given does reduce down to the above with the limitations given? 84.13.34.56 (talk) 13:14, 26 March 2010 (UTC)[reply]

I gave it a shot, using the material from this Q. Note that I'm not a mathematician. However, I think the original version having been written by a mathematicians was the problem, as this results in articles only readable by other mathematicians. StuRat (talk) 13:43, 26 March 2010 (UTC)[reply]

Unfortunately, you added it to the wrong article. Your example is about (what our articles refer to as) bracelets, not necklaces. Algebraist 13:46, 26 March 2010 (UTC)[reply]

Thanks, I moved it here: Bracelet_(combinatorics)#Example. Although, I don't care for the terms "necklace" and "bracelet", I much prefer "reversible necklace" and "nonreversible necklace", since those terms actually describe the difference. But, if the mathematicians have chosen the vague terms, we may be stuck with them. StuRat (talk) 13:56, 26 March 2010 (UTC)[reply]

So, if we allow rotation, but not flipping it over, what they call a "necklace", is the formula then n!/n (which is the same as (n-1)!) ? What are the restrictions on the range of n (obviously it can't be zero) ? StuRat (talk) 14:00, 26 March 2010 (UTC)[reply]

Except that the beads on a necklace are not assumed to have pairwise different colours, so this formula has nothing to do with the

N_{k}

and

M_{k}

counting functions in the article. The same problem is with bracelet counting by the way, so the example you added there is quite misleading, n!/2n has nothing to do with

B_{n}(n)

.—Emil J. 14:13, 26 March 2010 (UTC)[reply]

Please explain what "pairwise different colours" means, so we can bridge the gap. StuRat (talk) 14:20, 26 March 2010 (UTC)[reply]

Let n = 3, for example. The 3!/3 = 2 things you counted are 012 and 210 (as cyclic orders). The

N_{3}(3)=11

necklaces are 000, 111, 222, 001, 002, 110, 112, 220, 221, 012, and 210. Is it clear?—Emil J. 14:26, 26 March 2010 (UTC)[reply]

It looks like my example assumes each bead is unique (which is stated in the first sentence), while, in general, this assumption is not made. That seems OK to me, as long as the assumption is stated clearly. I can clarify it a bit more, if you like. StuRat (talk) 14:35, 26 March 2010 (UTC)[reply]

I added an example to the necklace article, too: Necklace_(combinatorics)#Example. Is it OK ? Specifically, is calling it a "2D" necklace right, since it closes back on itself, or should it still be considered 1D ? StuRat (talk) 15:34, 27 March 2010 (UTC)[reply]