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February 2

Twelvefold way

We recently went over the twelvefold way. I calculated two of the boxes in a way that my professor doesn't necessarily agree with but I got the correct answer both times and I think it makes perfect sense. It's not for a grade, so not a big deal but I just want to make sure my thinking makes sense.

In the case of N indistinguishable, X distinguishable, and arbitrary function, I got the answer ${\binom {n+x-1}{n}}$ . Then, to find the answer for a surjective function (same conditions on N and X), I thought of putting 1 ball in each of the x urns to start out. The balls are indistinguishable and there must be at least 1 in each urn so I don't see any problem. Then you have n - x balls and no restrictions so it's like the arbitrary case with n - x balls and x urns, so you get the answer ${\binom {n-1}{n-x}}$ . My professor says he's not sure about double counting certain ways to do it. I don't think that comes into affect here. All that really matters is how many balls end up in the various urns. I did a similar solution for the case where N and X are both indistinguishable. That is, I put 1 ball in each urn and was left with n - x balls and x urns so I plugged these into the arbitrary formula, and again I got the correct answer. StatisticsMan (talk) 02:31, 2 February 2011 (UTC)[reply]

please explain the question ,what you have to find out,permutation,combination or others . — Preceding unsigned comment added by True path finder (talk • contribs) 02:42, 2 February 2011 (UTC)[reply]

If you click the link, it explains it all. The point is you have a set N and a set X and you want to count the number of functions from N to X under various conditions. The twelve ways come because the elements of N can be distinguishable or undistinguishable, similarly with X, and then the functions are either arbitrary, or require that they are injective, or require that they are surjective. So, 2 * 2 * 3 = 12 different types here. And, we want to count the number of such functions in each of the 12 cases. StatisticsMan (talk) 02:46, 2 February 2011 (UTC)[reply]

I think your explanation for

\textstyle {n-1 \choose n-x}

is valid. —Bkell (talk) 03:22, 2 February 2011 (UTC)[reply]

Quick request for proof

Could someone please show me a quick proof of why a normal (normal: $TT^{*}=T^{*}T$ ) bounded linear operator T: $X\mapsto X$ on Hilbert space X, every element of the spectrum is an approximate eigenvalue - i.e. $\exists (x_{n})_{n=1}^{\infty }$ such that ( $T-\lambda I$ ) $x_{n}\,\to \,0$ ? I feel like the proof should be fairly quick and obvious - I know it is true for self-adjoint operators (such as $TT^{*}=T^{*}T$ ), but that doesn't help me in a way I can see, though I suspect it is the reason behind the proof. Would someone mind showing me quickly why this is the case? Thankyou! :) Otherlobby17 (talk) 04:36, 2 February 2011 (UTC)[reply]

Do you have the spectral theorem for bounded normal operators available? If so, that proves your property in much the same way as for self-adjoint operators. –Henning Makholm (talk) 10:54, 2 February 2011 (UTC)[reply]

Yes, though I found another way to solve it anyway - thankyou! Otherlobby17 (talk) 16:07, 2 February 2011 (UTC)[reply]

Line segment

In Line_segment#Definition it says that a line segment from a=(x1,y1) to b=(x2,y2) can be parametrized as {a+tb: t in [0,1]}. What is the proof for this? Thanks-210.212.167.113 (talk) 04:37, 2 February 2011 (UTC)[reply]

That's given as the definition of "line segment" so can't be proven (and doesn't require proof).—msh210℠ 07:59, 2 February 2011 (UTC)[reply]

That's not quite correct; what you wrote is the line segment from a to a+b. The line segment from a to b is parametrized by {a + t(b-a): t in [0,1]}. When you see b-a, you should think of it as the vector from a to b. (The vector from one point to another is given by (endpoint)-(startpoint).) So, the line segment parametrization tells us we start at a (when t=0), then add some fraction of the vector from a to b. 146.186.131.121 (talk) 12:06, 2 February 2011 (UTC)[reply]

Er, yeah, quite right.—msh210℠ 19:14, 2 February 2011 (UTC)[reply]

Stats and probability

What is a good, rigorous text that covers stats and probability (together, of course) from the very basics up? Thanks. 24.92.70.160 (talk) 05:23, 2 February 2011 (UTC)[reply]

I've been using David Stirzaker's Probability and random variables: A beginner's guide. It doesn't cover statistics, but it's perfect as an introduction to probability. —Anonymous Dissident^Talk 12:54, 2 February 2011 (UTC)[reply]

I'm a fan of the book I used in my first stats class, it covers probability and probability calculations as well as most applications of statistical inference. It does not get into specifics of probability theory. The book is "Statistics: Concepts and Methods" by authors D. Monrad, W. Stout, and E. Harner. Cliff (talk) 17:35, 4 February 2011 (UTC)[reply]

Sectitive and Quantative Numbers

Before you say anything this is just the theory of an 11 year old boy. First I must introduce you to sectitive and quantative numbers. Picture a cup, with the brim resembling infinity and an ordinary liquid, say water, resembling positive and negative numbers. There is water in the cup, but it can only be filled to the brim at the most. So, that means you can have any number, ranging from infinity to negative infinity, in the cup. Let's say you have the same cup, but liquid helium being sectitive and quantative numbers. Liquid helium is in the cup, but slowly creeps out, defying viscosity. The liquid helium on the outside of the cup is the sectitive and quuantative numbers. From now on I will refer to them as the "extreme numbers." The extreme numbers went above the brim and beyond, making them larger than infinity. Sectitive numbers and quantitive numbers are opposites, sectitive numbers are smaller than negative infinity and quantative numbers are larger than positive infinity, therefore, the term "extreme numbers." The sign of a sectitive number is a division sign and a quantative sign is a multiplication sign. (Please excuse me, since I do not have multiplication and division signs on my keyboard, I will use % for division and x for multiplication.) So, quantative 25, or 25 over infinity, is x25. Sectitive 25, or 25 below negative infinity, is %25. But even the extreme numbers must have an ending point, and I call that omnity. the symbol for omnity is an upside down micro sign. (See micrometer for symbol.) So that concludes my definition of extreme numbers, I hope you liked my theory. —Preceding unsigned comment added by 205.237.144.164 (talk) 07:45, 2 February 2011 (UTC)[reply]

There's a good deal of real mathematical study of numbers "greater than infinity" and "less than negative infinity". See the article on ordinal numbers for a start.—msh210℠ 07:53, 2 February 2011 (UTC)[reply]

Also hyperreal numbers. -- Meni Rosenfeld (talk) 08:28, 2 February 2011 (UTC)[reply]

Conformal Cyclic Cosmology might be fun for you to wrap your mind round too. Dmcq (talk) 09:47, 2 February 2011 (UTC)[reply]

Okay, so you've told us what the words and symbols for your new numbers are, but how do they behave? What can we do with them? Can we add and subtract? Multiply? What is ×38 plus -500? What is 2 times ÷13? Is that the same as ÷13 plus ÷13? –Henning Makholm (talk) 09:57, 2 February 2011 (UTC)[reply]

Yes, we'd like to see how they work. I hope the OP comes back to tell us. SemanticMantis (talk) 17:18, 3 February 2011 (UTC)[reply]