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February 12

Matrix property

If:

A^{T}=-A

What is this property called? It's something -symmetric but I can't quite remember what the prefix is.

Thanks in advance. 128.86.152.139 (talk) 01:52, 12 February 2009 (UTC)[reply]

Skew-symmetric matrix. Algebraist 02:02, 12 February 2009 (UTC)[reply]

Ah yes, genius! Thanks. 128.86.152.139 (talk) 02:11, 12 February 2009 (UTC)[reply]

Points in [0,1] whose decimal expansions contain only the digits 4 and 7

I've posted this question on the math portal talk section and was told the answer, but I tried and don't know how to prove it.

Let E be the set of all x in [0,1] whose decimal expansion contains only the digits 4 and 7. How is it closed?

If x is in [0,1] and not in E, it'll have a digit different from 4 and 7. Then I tried to find a neighborhood of x that's disjoint from E, but it's difficult as there are many cases each requiring separate treatment. Can anyone offer a proof? —Preceding unsigned comment added by IVI JAsPeR IVI (talk • contribs) 13:27, 12 February 2009 (UTC)[reply]

If it has a digit different that 4 and 7, then it will have a first such digit. You can do what you like to digits after that and always stay outside E. Does that help? --Tango (talk) 13:55, 12 February 2009 (UTC)[reply]

As always with decimal expansions, there's the annoying matter of non-uniqueness to be dealt with. Algebraist 13:59, 12 February 2009 (UTC)[reply]

You have the right idea; just show that the complement is open. There will be several cases, because you have to worry about numbers like 0.3999999... and 0.474740000... . Proofs of statements that refer to decimal digits are always difficult because of the non-uniqueness. It is much easier to prove this statement for Baire space (set theory), and that space is sufficiently similar to the real line to guide intuition. — Carl (CBM · talk) 14:03, 12 February 2009 (UTC)[reply]

Why would you have to worry about that? He says "only 4 and 7." Anyway, I would use convergent sequences. If a convergent sequence in [0,1] consists of numbers containing only 4 and 7, it converges to a number made of only 4 and 7. The set is closed. Black Carrot (talk) 14:11, 12 February 2009 (UTC)[reply]

Well, you would need to either know that fact about pointwise convergence of the decimal expansion, or prove it. And in general one cannot say that if a sequence x converges to y then any sequence of decimal expansions of x converges pointwise to any given decimal expansion of y. This is the usual headache with decimal expansions. — Carl (CBM · talk) 14:16, 12 February 2009 (UTC)[reply]

I guess it depends on whether we count trailing 0's as digits. If we do, then both recurring 9's expansions and terminating expansions contain digits other than 4 and 7, so there's no problem. If we don't, then 0.6999... and 0.7 are in different categories despite being the same number. --Tango (talk) 14:19, 12 February 2009 (UTC)[reply]

Even in the former case, you have to worry (briefly) about this in your approach: which is the first digit not 4 or 7 in 7/10? Algebraist 14:24, 12 February 2009 (UTC)[reply]

If doesn't matter. You don't need to know which is the first such digit, just that it exists. Just call it the nth digit and get on with it. --Tango (talk) 14:39, 12 February 2009 (UTC)[reply]

Hence the briefness of the worry. Algebraist 14:46, 12 February 2009 (UTC)[reply]

If a decimal expansion has n digits after the point there are n2^(n-1) possible sequences of digits comprising exclusively 2 digits. But there is no limit to increasing n. Therefore the set E is infinite. Cuddlyable3 (talk) 14:22, 12 February 2009 (UTC)[reply]

Yes. So what? Algebraist 14:24, 12 February 2009 (UTC)[reply]

(I added a descriptive title.) I think this is pretty easy with just two cases. For a nonterminating decimal (which has no alternate terminating expansion), find the first illegal digit and choose a neighborhood small enough that that digit doesn't vary. For a terminating decimal... you fill in the blank. -- BenRG (talk) 14:26, 12 February 2009 (UTC)[reply]

The non-uniqueness of decimal expansion is definitely a plague. I've thought about for a bit more and came up with this: Let x be outside of E, and the first digit different from 4 and 7 be α at the nth place. The next digit must be a digit from 0 to 9 (since we are using base 10). If it's 1-8, the neighborhood 1/10ⁿ⁺¹ with center at x, has no element in common with E (because if we add any amount less than 1/10ⁿ⁺¹ to x it will not change the digit α (even if its 0.(... ...)α8999999... ..., we must add or subtract something with absolute value strictly less than 0.00000... ...01 - the digit 1 is at n+1th place - to it and so it can't get to 0.(... ...)α999999... ... = 0.(... ...)α+100000... ... (or if α is already 9 α+1 will contribute to the digit on its left)

If the digit after α is 0 or 9, we still use the neighborhood 1/10ⁿ⁺¹ with center at x but it gets a bit more difficult to demonstrate (and I'm not entirely sure it's correct due the decimal non-uniqueness), it's easier to do it with a diagram but frankly I really don't know how to use wikipedia. If anyone can point out any mistakes i made (which I'm pretty sure I did) please correct it. And also with regards to Baire Space isn't it uncountable? Because countably many cartesian product of natural numbers simply means the set of all functions from N to N, so why do they use ω^ω? Isn't that ordinal countable (from my intuition it's somewhat like the union of all finite cartesian products of of natural numbers - it's defined to be the supreme of ωⁿ, n run over the naturals, and ωⁿ is somewhat like Nⁿ. Homeomorphism is probably the word but that's just from casual reading). Sorry I'm not quite at your level yet.--IVI JAsPeR IVI (talk) 12:44, 14 February 2009 (UTC)[reply]

I think there is some confusing notation going on. ω is used both to represent the set of natural numbers and the first infinite ordinal (they are, after all, the same thing), but how you manipulate the symbol depends on which meaning you are giving it. If it's the set of natural numbers then ω^ω means the set of all functions from the natural numbers to themselves, which is uncountable. If it's an ordinal, then ω^ω means the limit of ωⁿ as n goes to infinity, which is a countable ordinal. I think the answer is a avoid using ω to refer to the set of natural numbers and just use it for the ordinal (I think that's the most common notation - I've only seen ω used for the natural numbers in rather old books). --Tango (talk) 13:06, 14 February 2009 (UTC)[reply]

The use of ω for the naturals is still common in logic and, I believe, in set theory. One slight advantage of this notation is that ω quite definitely includes 0, while with ℕ it's anyone's guess. Algebraist 14:27, 14 February 2009 (UTC)[reply]

Another benefit of using ω to refer to the set of finite ordinals is that it's very clear exactly which set is intended. This usage is extremely common in practice. The corresponding solution in practice to the issue Jasper and Tango mentioned is that you need to say so explicitly if you are using ordinal exponentiation (which is used somewhat rarely in practice). This leads to the following conventions:

ω^ω is the set of infinite sequences of natural numbers.
ω^<ω is the set of finite sequences of natural numbers.
[ω]^ω is the set of infinite sets of natural numbers.
[ω]^<ω is the set of finite sets of natural numbers.

— Carl (CBM · talk) 14:37, 14 February 2009 (UTC)[reply]

I tend to use

\mathbb {Z} _{>0}

and

\mathbb {Z} _{\geq 0}

to avoid the ambiguity. --Tango (talk) 15:42, 14 February 2009 (UTC)[reply]

Thanks. I thought the symbol ω (and anything containing ω that "looks" like elementry operations) was used exclusively for ordinals and arithmatic/exponentiation on ordinals.--IVI JAsPeR IVI (talk) 08:56, 15 February 2009 (UTC)[reply]

Book recommendations

Could someone recommend me easy to understand and interesting to read book(s) covering the following topics: Rodrigues' rotation formula, Clifford algebra, Rotation groups, Lie groups, Exponential map, etc. Thanks a ton! deeptrivia (talk) 18:39, 12 February 2009 (UTC)[reply]

/Wrt Clifford algebras, search the net for stuff from John Baez, his weekly column has two specials on it, and book recommendations. I would expect to find similar recommendations for the other things in similar places. --Ayacop (talk) 19:17, 12 February 2009 (UTC)[reply]

Simple Math

This is a really easy question compared to most of the ones here so I'm sure someone can help me.

How do I solve for 'm' in the following equation:

900 = 1500(0.95)^m

I will admit right now that this is from my homework but I have tried really hard but just can't get it. Thanks in advance.

This page isn't for homework problems. I suggest you reread the chapter in your textbook from which the problem originated, paying particular attention to the worked examples. Ray (talk) 23:35, 12 February 2009 (UTC)[reply]

Are you familiar with logarithms? If yes, this is easy. If no, you should read your textbook's section on them. Algebraist 23:36, 12 February 2009 (UTC)[reply]

And, if you haven't had logarithms in school yet, and don't care to learn them, either, this could also be solved by a trial-and-error approach. I'll get you started:

900/1500 = (1500/1500)(0.95)^m

0.6 = (1)(0.95)^m

0.6 = (0.95)^m

Now try a range of values for m:

m  (0.95)^m
-- ---------
 1  0.95
10  0.598737

Since 0.6 is between 0.95 and 0.598737, but much closer to 0.598737, we should try a value for m between 1 and 10, but much closer to 10. I'll try 9.9:

m      (0.95)^m
--     ---------
 1     0.95
 9.9   0.601816
10     0.598737

Since 0.6 is approximately halfway between 0.601816 and 0.598737, next try an m approximately halfway between 9.9 and 10. Continue with this process until you have the desired number of significant digits for m. StuRat (talk) 01:23, 13 February 2009 (UTC)[reply]

The trial-and-error approach that StuRat showed is also called "successive approximation" and at this link you can see it used in an electronic circuit. You have yet another method of solving your equation if you can still find a sliderule that has LL scales (and the ancient knowledge of how to use them). Cuddlyable3 (talk) 19:59, 13 February 2009 (UTC)[reply]