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October 20

Ergodic splitting in L^p

Moved to Talk:Ergodic theory --pma (talk) 17:23, 21 October 2009 (UTC)[reply]

Common Complementary subspaces in finite dimensional vector spaces

Hi all! I was wondering if anyone could perhaps point me to a (simple-ish if possible please!) proof, or explain one to me, that 2 same-dimensional subspaces of a finite-dimensional vector space have a common complementary subspace - I've been staring at my page for about an hour or so now and as the time of day would suggest I'm getting nowhere - I previously managed to show that for any subspace $U\subset \mathbb {F} ^{n}$ , there is a subset $I\subset {1,\,2\,...n}$ for which the subspace $W=<\{e_{i}:\,i\,\in \,I\}>$ is complementary to $U$ in $\mathbb {F} ^{n}$ (this is part ii, that was part i!), but I'm not sure to what extent that's related. I've been told 'I may wish to consider the case where the subspaces have dimension 1 less than the space first', yet still all I have is a page of garbled symbols and 2AM nonsense-mathematics.

Please help! Many thanks - Typeships17 (talk) 00:49, 20 October 2009 (UTC)[reply]

I don't understand the question. Take the simple case of Euclidean 3-space. Treat the x-axis as a 1-dimensional subspace. Is the complementary subspace the yz-plane? In that case, the complementary subspace of the y axis is the xz-plane, which is not the same. 66.127.54.181 (talk) 05:39, 20 October 2009 (UTC)[reply]

66 - there are a lot of complementary subspaces to a given subspace. Typeships - try following the suggestion first: let A and B be subspaces of V of codimension one. Then the union of A and B is not equal to V (why?) so there's a vector v not in AuB. Then the span of v complements A and complements B. For the general case, use the same idea to find a vector v not in A union B. Then repeat, starting with new subspaces span(A,v) and span(B,v)... Tinfoilcat (talk) 08:59, 20 October 2009 (UTC)[reply]

Word Probability Question

I would rather not say where this question comes from. I realize it's not a reference-desk type question exactly, but perhaps it's not totally inappropriate. It should not have an answer that is very sensitive to the dictionary used (though it is limited to English or at least the Roman alphabet). Here is the situation: A certain five-letter word stands in relationship to certain specific three- and six-letter words that the five-letter word and each of the other words seperately are both prime in the first bases in which there are enough digits for the pair of words in a non-case-sensitive way. What fraction of word selections--draw the three words uniformly from a dictionary limited to the requisite lengths--meet this pair of coincidence pairs? Let's say the words are hypothetically HORSE, DOG and RABBIT. Then what would be necessary for the statement to be true would be for (17)(24)(27)(28)(14) to be prime in both bases 29 and 30, while (13)(24)(16) is prime in base 29 and (27)(10)(11)(11)(18)(29) is prime in base 30. I just pose this here because I figure there is a fair chance someone has at hand a good easy solution--a database dictionary and the ability/tools for doing the simulations and primality tests.Julzes (talk) 12:21, 20 October 2009 (UTC)[reply]

Sorry, this is unclear. Is (17)(24)(27)(28)(14) an encoding of HORSE? How? It does match the old A=1, B=2... scheme. And what do you mean by "prime in base X"? Primeness is a property of the number, not of any particular base. Do you want to interpret the numerical encodings as numbers written in different bases, i.e. the string (17)(24)(27)(28)(14) would represent two different numbers? --Stephan Schulz (talk) 12:52, 20 October 2009 (UTC)[reply]

I think we're using an extension of hexadecimal notation here, so a positional notation with A=10, B=11 etc, to various possible bases. Algebraist 12:55, 20 October 2009 (UTC)[reply]

Algebraist is right here. For clarification, replace the words 'are both prime' by 'both represent primes' in the sentence beginning 'Here is the situation:', though I don't quite understand why the original wording was so hard to comprehend. The answer to Stephan's question is Yes.Julzes (talk) 20:28, 20 October 2009 (UTC) It's using an extension of the natural concept employed in hexadecimal, though theoretically one could have pairs of words with no letters beyond E, in which case the base would be a shortening of hexadecimal.Julzes (talk) 20:35, 20 October 2009 (UTC)[reply]

Oh, heck, I might as well come out with what I found. Call this numerology or mathematical theology or whatever. The words are ALLAH, GOD and YAHWEH, and I claim that the first thing I checked was whether ALLAH and GOD were both prime in base 25, and the second thing I checked was whether ALLAH and YAHWEH were both prime in base 35. After that, I confirmed my suspicion that GOD is not prime in base 35, and then I went on to do some interesting/strange stuff with the word BUDDHA which entailed reversing digits and exchanging digits with letters. The question I've ask here is to gauge the result that I got.Julzes (talk) 20:46, 20 October 2009 (UTC)[reply]

RATS! I just discovered--nobody has corrected me yet--that I did YAHXEH in base 35 rather than YAHWEH. I guess it has something to do with my getting 21 on all six lines of Chance in the Triple Yahtzee game I played on my 21st birthday. Still, I'm wondering about the result I got, but now it's contingent on there being available a one-digit offset to the six-letter word being a prime.Julzes (talk) 21:17, 20 October 2009 (UTC) Also, the one digit that is offset should not be the digit that determines the base, and one should allow that any subset of any repetitions of the letter/digit that is mistakenly offset should get the offset. It's a litle bit trickier to implement, I suppose.Julzes (talk) 21:49, 20 October 2009 (UTC)[reply]

To be honest, you ask the wrong question. As you have demonstrated in your last paragraph, you seem to be very willing (possibly even unconsciously) to tweak your numerological hypotheses if one of them fails. Thus, the question is not the probability of a certain match of data (post-hoc it's one), but the probability that you will find one set of stunning mathematical properties among all the possible codings and properties. Given that there is an infinite number of such codings and at least a very large number of interesting properties, the probability for that is bound to be 1, too. It's like picking a piece of straw from a huge bale and then asking what the chance for this particular kinky straw was. --Stephan Schulz (talk) 17:32, 21 October 2009 (UTC)[reply]

We've bumped heads before. You have to understand that I am asking a mathematical question here based upon my real experience of mediating mathematical results. I did indeed immediately calculate the value of the (non-)word YAHXEH. The fact that you don't like this claim and want to insist that I tweaked something after the fact does not amount to a proof that I am being dishonest. You obviously never did get very far with probability theory compared to me, so please don't pretend to try to help. I'm at the help desk right now for a real reason, and not on some whim. I may re-post this to the computer science desk, since it may be more up the alley of the people who assist there. You are betraying a simple bias. The probability for the event I claim is going to be quite low, and not close to 1. It's not going to compare with some of the other strange coincidences I've had to deal with in terms of probability, but this one actually is not so complicated that an answer to my question is made all that difficult by an inability to get it into mathematical terms of a clear question. I've presented a clear question with a (somewhat limited by choice of dictionary) clear answer.Julzes (talk) 21:17, 21 October 2009 (UTC)[reply]

I've not claimed that you are dishonest in any way. I do think you are misguided, though. I also suggest you refrain from guessing my level of competence from five lines of text and do take a look at Google Scholar. --Stephan Schulz (talk) 21:28, 21 October 2009 (UTC)[reply]

Whatever your real qualifications are, we have met before and you did not understand me any better then. This is not an opinion forum, so what you think about whatever is going on with my head and what I'm asking is really irrelevant and bothersome. I've asked a mathematical question. It might be easier on a dictionary researcher and on the terms of the problem if nouns in singular case is a limitation, but aside from tweaking the problem in that way and perhaps asking the question both where there is and is not a specification for the number of letters, I have a clear question with a clear answer.Julzes (talk) 21:35, 21 October 2009 (UTC)[reply]

In a sense, Stephan Schulz is right. You have to go along with my claim that these were the first things checked to get to a claim of a low probability. If one were to present a straw to the public that had some odd bulges or something at very close to regular intervals and say "Hey, look! This is the very first straw I picked and it's like only one in 10,000 straws in this property that I happened to check first," then the public would have a right to question the statements that they were the first straws and properties looked at. The calculation contingent on accepting these claims would be taken in that light, but it would still be a 1/100 of a percent chance in that light. I'm going to go ahead and just get a rough measure on my own more or less by hand.Julzes (talk) 04:51, 22 October 2009 (UTC)[reply]

homework help plz?

how do i work 3√2x =4 solving for X —Preceding unsigned comment added by DanielTrox (talk • contribs) 19:38, 20 October 2009 (UTC)[reply]

Just to clarify, do you mean

3{\sqrt {2}}x=4

or

3{\sqrt {2x}}=4

? ~~ Dr Dec (Talk) ~~ 19:50, 20 October 2009 (UTC)[reply]

When asking for homework help, you should at least tell us what you have tried and/or what is confusing to you. Don't expect us just to give you an answer. --LarryMac | Talk 19:53, 20 October 2009 (UTC)[reply]

(ec) If ax = b and a ≠ 0 then it follows that x = b/a. If a√x = b and a ≠ 0 then it follows that x = (b/a)². ~~ Dr Dec (Talk) ~~ 19:57, 20 October 2009 (UTC)[reply]

Usually if you have a variable under a square root symbol, isolating the square root on one side of the equation and then squaring both sides is the recommended course of action. -- 128.104.112.179 (talk) 20:02, 20 October 2009 (UTC)[reply]

i was looking at it wrong, it was $3{\sqrt {2x}}=4$ but the 3 was a cubed symbel. so i cubed both sides then sqaured both side and divided by 2. and to clear up what larry mac said, i was asking how to work it not for the answer. Talk Shugoːː 20:18, 20 October 2009 (UTC)[reply]

A 3 and the root symbol together mean the cube root, so cubing is fine but the extra squaring isn't.→86.132.161.236 (talk) 14:10, 21 October 2009 (UTC)[reply]

Not really. If we write

{\sqrt[{3}]{x}}

then we mean the cube root of x. If we write

3{\sqrt {x}}

then we mean three multiplied by the square root of x. Both have "a 3 and the root symbol together". ~~ Dr Dec (Talk) ~~ 16:21, 21 October 2009 (UTC)[reply]

2D splines

How is spline interpolation (specifically cubic spline interpolation) for "true" multidimensional curves done? (e.g, for curves in 2D space where both the X and Y coordinate are parametric, like the example shown, rather than a "2D" spline formed by forcing the X coordinate to be equal to the spline parameter, like in File:Interpolation example spline.svg.) Is it simply that each coordinate is treated as an "independent" spline, all of which can then simply be graphed simultaneously against the common parameter t, or is it necessary to consider all coordinate values simultaneously when fitting the spline? Would there be any discontinuities introduced (for cubic splines, this would include discontinuities in slope and curvature) as a result of combining separate splines for each coordinate? (In addition to being mathematically dense, the Spline (mathematics) article only treats the one-dimensional, univariate case.) -- 128.104.112.179 (talk) 19:59, 20 October 2009 (UTC)[reply]

Yes you're right - it is simply that each coordinate is treated as an "independent" spline, all of which can then simply be graphed simultaneously against the common parameter t. RupertMillard (Talk) 21:09, 20 October 2009 (UTC)[reply]

Parametrization of a curve is not unique: e. g., you get the same curve if you replace t by t³. However, don't we get different splines then? It looks like there are better methods, giving the same splinization for any parametrization.—Pt_(T) 11:16, 26 October 2009 (UTC)[reply]