December 27

I asked this before and maybe the question was ignored due to the holidays...

Is there an algorithm (like for the simplex method in linear programming) to reduce polynary equations to minimum form? 71.100.6.153 (talk) 02:06, 27 December 2009 (UTC) [reply]

I believe that the problem is your use of "polynary". That is not a well-defined word. The meaning depends on which group of people is using it. If you just made it up, please define what you mean by it. Otherwise, define what the group of people you got it from intend for it to mean. It literally means "pertaining to many" in the way that binary means "pertaining to two". So, it means "can have many values". Most equations can have many values. Most variables can have many values. Therefore, the usage is very important to make sense of the question. -- kainaw™ 03:04, 27 December 2009 (UTC)[reply]

My intended meaning of polynary is identical to the phrases "multiple state variable" or "many stated variable" or "poly-stated variable" in the same sense that binary pertains to two. My usage covers binary variables and any variable with discrete and finite number of states. My usage does not include fractions directly since probability values can be normalized to percentages and percentages are meaningless beyond a few decimal places which can be rounded or truncated to result in only integer values. 71.100.6.153 (talk) 15:44, 27 December 2009 (UTC) [reply]

A polynary equation is a logical equation with variables which may have two states although binary is the specific term used to refer to such an equation as trinary is the specific term used to refer to an equation of variables made up of three states. The algorithm I am looking for, however, should be applicable to all discrete and finite stated equations to include binary, trinary and beyond. 71.100.6.153 (talk) 00:11, 28 December 2009 (UTC) [reply]

In short, a polynary variable is any variable having a finite number of discrete states. 71.100.6.153 (talk) 00:20, 28 December 2009 (UTC) [reply]

In established terms polynary refers not to natural numbers but to a "...more recent notion..." of nominal numbers. (see Use of nominal numbers) 71.100.1.76 (talk) 19:25, 31 December 2009 (UTC) [reply]

First of all, when I say that a set A is "bigger than" another set B, it means that B is entirely contained in A and that A and B are not equal. So B is a proper subset of A. Working an the interval [0,1] for example, I know that starting from a single point, we can work our way up to the Cantor set which is uncountable and still has measure zero. My question is what is the "biggest" subset of [0,1] that still has measure zero. All of [0,1] has measure 1 obviously.

I have the same question regarding countable sets (countable means finite or countably infinite) in [0,1] for example. Cantor showed (using his usual ingenious arguments) that starting from a single point, we can work our way up to the algebraic numbers which are countable. Is there any set "bigger" than the set of all algebraic numbers in [0,1] which is still countable? Thanks! -Looking for Wisdom and Insight! (talk) 02:56, 27 December 2009 (UTC)[reply]

Adding a single point (or any countable set) to a set which is countable/measure zero yields a new set which is again countable/measure zero. You'll need some better questions to get more interesting answers. Algebraist 03:07, 27 December 2009 (UTC)[reply]

I think that the OP is looking for the existence of a subset of [0,1] containing the algebraic numbers within [0,1], maximal with respect to the property that it is countable and has measure zero (you might as well omit the "measure zero" from the OP's question since any such countable set has measure zero). In this case, however, there is no such maximal set, for the reason given by algebraist (if M is any such countable set, ${\displaystyle M\cup \{x\}}$ for ${\displaystyle x\in [0,1]\setminus M}$ contains M and is countable, so M cannot be maximal with respect to the property of having measure zero). --PST 04:08, 27 December 2009 (UTC)[reply]

You know what I meant and I was afraid of this answer. I thought about that too but that is not what I was looking for. So what is the largest subset of [0,1] which is countable/has measure zero?-Looking for Wisdom and Insight! (talk) 03:37, 27 December 2009 (UTC)[reply]

There is no such countable set for the reason given by algebraist above. There is no such set having measure zero for exactly the same reason (if M is any such set having measure zero, ${\displaystyle M\cup \{x\}}$ for ${\displaystyle x\in [0,1]\setminus M}$ contains M and has measure zero, so M cannot be maximal with respect to the property of having measure zero). --PST 04:11, 27 December 2009 (UTC)[reply]

Basically, although many aspects of mathematics are firmed on intuition, which formalizm consolidates (and is thus often tacit), in this case formalizm is important to obtain some interesting intuition. --PST 04:14, 27 December 2009 (UTC)[reply]

The question is analogous to asking the existence of a "largest number"; unless you add extra assumptions to your definition of "number", you cannot obtain a meaningful answer. In this case, you will need to add additional assumptions to your definition of "set". --PST 04:17, 27 December 2009 (UTC)[reply]

Thanks for the explanation, everyone!-Looking for Wisdom and Insight! (talk) 04:29, 27 December 2009 (UTC)[reply]

Maybe more interesting answers can be found if we also consider description complexity. "Rational numbers" fails because a much bigger set (algebraics) can be obtained with an equivalently short description. "Algebraic numbers and ${\displaystyle 1/\pi }$" fails because the added complexity of specifying ${\displaystyle 1/\pi }$ is not justified by the increase of one element. In other words, that extra element does not "belong" in this set. So, is there a countable set which is significantly larger then algebraics and yet similarly simple? Is there some quantification of these notions, under which an optimal set can be found? -- Meni Rosenfeld (talk) 06:00, 27 December 2009 (UTC)[reply]

Perfect, like if all elements in a set share a common property. That is definitely a better wording of my question.-Looking for Wisdom and Insight! (talk) 07:02, 27 December 2009 (UTC)[reply]

The Computable numbers are countable and the only way you'll write a number that's not amongst them is by doing something like throwing a dice for each digit. Dmcq (talk) 22:55, 27 December 2009 (UTC)[reply]

Well, it depends on what you mean by the way, will I be man enough to ignore the barbarism a dice??? Probably not. At least I can rant about it in small text. by write such a number. Of course you can't really "write" it (because it would take too long) but you can certainly specify one, with no ambiguity whatsoever. For example, consider the number in binary representation that has a zero at position n if the Turing machine with Goedel number n halts, and a 1 otherwise. --Trovatore (talk) 23:01, 27 December 2009 (UTC)[reply]

The wikipedia article dice allows it for the singular form but yes die is better. A set larger than the algebraic numbers that includes practically everything before the 20th century can be got by using the hypergeometric series to generate extra numbers. Dmcq (talk) 23:12, 27 December 2009 (UTC)[reply]

We could extend that idea to the set of all numbers for which there exists a formula (in the first-order language of set theory) which your favourite set theory proves to uniquely define the number in question. That set's countable, and it contains every number you're ever likely to talk about. Algebraist 02:13, 28 December 2009 (UTC)[reply]

I'm not quite sure why you want to drag in provability. What exactly do you mean by saying that the theory proves that the formula describes "the number in question"? You can't talk about the number in question until you have a definition for it.

So presumably what you really mean is that the theory proves that the formula defines a unique real, and then consider the set of all reals that (Platonistically) satisfy some such formula. But then what does the proof add? Why not just consider the set of all reals that Platonistically are the unique real satisfying some formula, whether or not you can prove that there's a unique real satisfying the formula? --Trovatore (talk) 02:59, 28 December 2009 (UTC)[reply]

By the way have you come across the Vitali set? Dmcq (talk) 13:29, 29 December 2009 (UTC)[reply]

Largest countable this that and the other

This isn't exactly responsive to what the original poster asked, but it seems similar enough to mention. There are results from effective descriptive set theory that certain lightface pointclasses have largest countable members. For example there is a largest countable ${\displaystyle \Pi _{1}^{0}}$ set of reals. A set of reals is ${\displaystyle \Pi _{1}^{0}}$ if it's the complement of a ${\displaystyle \Sigma _{1}^{0}}$ set; a set is ${\displaystyle \Sigma _{1}^{0}}$ if it's the union of a collection of (open) intervals with rational endpoints, where some computer program, given infinite time, can list all the pairs of rational endpoints involved.

(So a ${\displaystyle \Pi _{1}^{0}}$ set is always closed, but being ${\displaystyle \Pi _{1}^{0}}$ is a stricter notion than being closed. The sense in which it is closed has to be somehow "effective" or "computable". For example there are uncountably many closed sets, but there are only countably many ${\displaystyle \Pi _{1}^{0}}$ sets, because there are only countably many computer programs.)

There's a classic paper by Donald A. Martin on this, in Proceedings of the Cabal Seminar, called Largest countable this that and the other.

I don't know how to characterize the largest countable ${\displaystyle \Pi _{1}^{0}}$ set, but the largest countable ${\displaystyle \Sigma _{2}^{0}}$ set is simply the set of all reals that are in Goedel's constructible universe. Obviously this requires something beyond ZFC; you have to be able to show that that set is countable. But I think the very modest assumption of zero-sharp suffices. I don't know whether you can prove in ZFC alone that there exists a largest countable set in any of these pointclasses. --Trovatore (talk) 23:14, 27 December 2009 (UTC)[reply]

Sorry, I had a total brain freeze. Actually most of what I write above is correct, except that I should be talking about the largest countable ${\displaystyle \Pi _{1}^{1}}$ and ${\displaystyle \Sigma _{2}^{1}}$ sets (note the superscript 1 instead of 0). The definition I gave for ${\displaystyle \Pi _{1}^{0}}$ is correct, but not relevant, because I should have been talking about ${\displaystyle \Pi _{1}^{1}}$ sets, which are much more complicated (as sets) and somewhat harder to explain in elementary terms. The largest countable ${\displaystyle \Sigma _{2}^{1}}$ set (not ${\displaystyle \Sigma _{2}^{0}}$) is the set of all reals in Goedel's constructible universe. --Trovatore (talk) 01:34, 28 December 2009 (UTC)[reply]

PST's response made me think of a (I hope) interesting question. Let X be the collection of all countable subsets of [0,1], partially ordered by set inclusion. Since there's no maximal element, by Zorn's lemma there exists a chain with no up bound. Can someone exhibit such a chain? I'm thinking about embedding all countable ordinals into [0,1] in some way... Money is tight (talk) 13:05, 29 December 2009 (UTC)[reply]

"In some way" — well put. Of course there is an injection from the set of all countable ordinals into [0,1]. If f is such an injection, then letting A_α be {f(β) | β < α}, then the A_α are your desired chain. Good example.

I don't know exactly what you're requiring when you say "exhibit" such a chain, though. It's consistent with ZF, without the axiom of choice, that no such f exists. In fact it follows from the axiom of determinacy, AD, that there is no such f. AD of course is inconsistent with the axiom of choice, or to put it another way, false.

In this context people tend to claim that there is no "explicit" such f. You see this claim all over the WP math articles when discussing objects whose existence can't be proved without AC. It's a problematic claim, as it tends to depend on what you mean by "explicit", which is a term that has no precise accepted definition. It may also depend on such things as whether V=HOD, which (unlike V=L) is not currently known to be ruled out by any accepted large cardinal hypotheses. --Trovatore (talk) 21:19, 29 December 2009 (UTC)[reply]

I want to do math but I'm terrible with calculations. Is it possible to do math without calculating/solving complicated equations with formulas? If so, how? My friend's a math person and he says that when he does math he thinks and rarely does calculate. He says he researches algebra, but I thought algebra was about calculating??? He also says that math people don't do arithmetic with numbers. My world has collapsed! Help! Or maybe my friend's wrong. Can't calculators solve math? Why does he research it? What sorts of math are there (my friend says there's lots)? Please tell me how I can do math without my calculator! —Preceding unsigned comment added by 122.109.239.199 (talk) 09:54, 27 December 2009 (UTC)[reply]

See Mathematics. Some areas of mathematics (such as abstract algebra, as opposed to high-school algebra) don't involve arithmetic calculations at all. Some do, and application of these fields can benefit tremendously from the use of a calculator. Being comfortable around numbers and equations is an important skill for the mathematically literate. -- Meni Rosenfeld (talk) 11:49, 27 December 2009 (UTC)[reply]

So you should first learn how to do computations, and then you should learn how not to do computations. --pma (talk) 12:28, 27 December 2009 (UTC)[reply]

Good advice. --PST 12:44, 27 December 2009 (UTC)[reply]

"Your world has collapsed"? That should not be the case! Almost everyone is told that they are wrong about something, at some point in their lives; such is an important experience. With regards to computations, I have a couple of remarks. Firstly, I think that the mathematical brain, by default, has ability to do computations; that is, if you are able to do the "mathematics without computations", you should have the ability to do the "mathematics with computations". Basic computations really do not require extreme intelligence to carry out, save silly human errors. On the other hand, much of mathematics is not really concerned with computations; rather, it is concerned with deeper problems which may require computations (at some stage) to solve. In abstract algebra, for instance, one uses substructures to encapsulate computations abstractly (ideals in ring theory, subgroups in group theory, etc...). In another branch of mathematics, topology, much of the goal is to attain a strong intuition of closeness; although seemingly simple, this is very deep. Differential topology sometimes employs differential geometry for this purpose. Hope this answers your question(s). However, if this is an attempt to mock mathematics, please do not; mathematics has developed tremendously over the past few hundred years, and not surprisingly, it is difficult for many to comprehend the extent of this development (I should add that it was even difficult for me to comprehend during my first exposure to formal mathematics). --PST 12:43, 27 December 2009 (UTC)[reply]

Why do you want to do maths if you don't have some feel for what it is about? Being very good at calculating isn't that important though it is quite useful. I've known what maths is since I was a child stringing up a frame when I realized I didn't get a circle and I wondered how to arrange the pegs to get a circle. I can't see any basis for your desire to do maths. Dmcq (talk) 13:12, 27 December 2009 (UTC)[reply]

Skill with numbers is likely to develop with practice. The more you work with arithmetic and elementary algebra, the more proficient you will become. A solid basis in those areas is needed to progress to more advanced mathematics. Many of the branches of mathematics which do not work with computations are are too complex to be meaningful without the foundations. —Anonymous Dissident^Talk 13:29, 27 December 2009 (UTC)[reply]

I want to become a mathematician but I hate equations and formulas, and my friend says that's not necessary for mathematics. You tell me something, and probably that was what my friend said but I wasn't paying attention. Thanks! But why do people research mathematics? People research physics because it's important to know about the universe. People research medicine because it helps our survival. What's the use of mathematics without calculating??? I want to do math because my math teacher said to me when I was in high school that it's purpose is to shape the world, and that mathematicians are society's core, because of their mind boggling human-calculator ability. And if math's not about numbers, what is it about? My high school teacher said that math is about understanding different properties about numbers, such as prime and composite and he has a math degree. Many thanks for the responses. Please tell me why people think math though! It hurts my brain and isn't fun. But I need it to be part of society's finest. —Preceding unsigned comment added by 122.109.239.199 (talk) 13:48, 27 December 2009 (UTC)[reply]

You want to become a mathematician but you hate some aspects of it; it hurts your brain and isn't fun. Have you considered a career in masochism?→→86.160.104.185 (talk) 16:15, 27 December 2009 (UTC)[reply]

Please don't be disrespectful to the OP; he or she doesn't understand what mathematics is about and would like to better understand it, I'm sure you can provide some insight that would be useful to him or her. Eric. 131.215.159.171 (talk) 21:59, 27 December 2009 (UTC)[reply]

By now it should be considered very unlikely that the OP is asking these questions in good faith. 86's response is appropriate. -- Meni Rosenfeld (talk) 10:32, 28 December 2009 (UTC) [reply]

Considering the OP's more recent questions, you are probably right. However there is still no use to mocking him or her. Eric. 131.215.159.171 (talk) 01:40, 29 December 2009 (UTC)[reply]

It's possible you will find Lockhart's Lament interesting; it's an article about the teaching of high school mathematics in the US. Although it does not spend much time explaining what mathematics is, it does address some misconceptions of mathematics and explains what mathematics is not. Eric. 131.215.159.171 (talk) 22:05, 27 December 2009 (UTC)[reply]

The best way to judge whether you can become a mathematician is probably to simply find out if more time spent with it still leaves you feeling dislike toward it. In the meantime, you will become at least good enough at math to make a living doing something with math that doesn't involve actually being a mathematician. This just seems like a reasonable piece of advice/opinion for the OP. One thing about mathematics that is appealing and repellent at the same time is that it involves abstract objects directly and entirely, and its only connection to the 'real world' is through the filter of modelling. Doing mathematics per se involves ignoring the real world for extended stretches of time, but there is a dividend personally and in the large for doing so that makes it worthwhile (virtually always, despite some mathematics seeming ten steps removed from reality).Julzes (talk) 05:12, 28 December 2009 (UTC)[reply]

Let {x_n} be a sequence with ${\displaystyle \lim _{n\rightarrow \infty }{\frac {x_{n}}{n}}=g}$ with g>0. How to prove the fact that ${\displaystyle \lim _{n\rightarrow \infty }{\sqrt[{n}]{x_{n}}}=1}$? --84.62.197.235 (talk) 11:24, 27 December 2009 (UTC)[reply]

First do it in the case ${\displaystyle x_{n}:=n}$ (to this end you may write the inequality of arithmetic and geometric means with the n numbers ${\displaystyle a_{1}=\dots =a_{n-2}:=1,a_{n-1}=a_{n}:={\sqrt {n}})}$. For the more general case observe that ${\displaystyle {\frac {g}{2}}n\leq x_{n}\leq 2g\,n}$ for large n; then use the former case and a sandwich argument. --pma (talk) 11:42, 27 December 2009 (UTC)[reply]

(Edit Conflict) Let ${\displaystyle g>0}$ and let ${\displaystyle x_{n}=gn}$ for all ${\displaystyle n\in \mathbb {N} }$. Do you accept that ${\displaystyle \lim _{n\rightarrow \infty }{\sqrt[{n}]{x_{n}}}=1}$ (Proof: ${\displaystyle \lim _{n\rightarrow \infty }{\sqrt[{n}]{x_{n}}}=\lim _{n\rightarrow \infty }{\sqrt[{n}]{g}}{\sqrt[{n}]{n}}=\lim _{n\rightarrow \infty }{\sqrt[{n}]{n}}\cdot \lim _{n\rightarrow \infty }{\sqrt[{n}]{g}}=1\cdot 1=1}$)? Now apply this intuition to prove your claim. --PST 11:44, 27 December 2009 (UTC)[reply]

One of the problems with your approach is that ${\displaystyle x_{n}\not =gn}$ (Igny (talk) 15:33, 27 December 2009 (UTC))[reply]

It was not intended to be a formal proof; rather it was intended to guide the intuition (and my x_n was not intended to be the same sequence as the OP's x_n; I defined a specific sequence). There is really no purpose in proving the value of a limit unless you have some idea as to how to obtain the value. The sequence x_n that I defined, and its limit, motivates the OP's question, and provides some intuition. But of course, this is guided with the assumption that the OP did not have any idea as to how to solve the problem initially (so pma's response, in that case, might have been what the OP was looking for). --PST 01:16, 28 December 2009 (UTC)[reply]

My solution turned out to be in line with PST's. (I didn't see his before I worked out the problem.) Write ${\displaystyle x_{n}^{1/n}=n^{1/n}({x_{n} \over n})^{1/n}}$. The right goes to 1 since ${\displaystyle x_{n} \over n}$ doesn't blow up. (So, the existence of a limit is unnecessarily strong.) More conceptually, the hypothesis means ${\displaystyle x_{n}}$ has the same asymptotic behavior as ${\displaystyle n}$. So, it is quite natural that we have ${\displaystyle x_{n}^{1/n}\to 1}$. -- Taku (talk) 23:34, 30 December 2009 (UTC)[reply]

Let {x_n} be a sequence with ${\displaystyle \lim _{n\rightarrow \infty }x_{n}=g}$ with g>0. How to prove the fact that ${\displaystyle \lim _{n\rightarrow \infty }{\sqrt[{n}]{x_{n}}}=1}$? --84.62.197.235 (talk) 14:22, 27 December 2009 (UTC)[reply]

Since x_n converges to g>0, you have ε<x_n<1/ε for some number ε>0 and for all n large enough, hence also ε^1/n<x_n^1/n<1/(ε^1/n). As you can see, the sandwich principle allows you to reduce your problem to the case of x_n=c>0 a positive constant. Is that clear? Then, can you do it for the case x_n=c>0? Take a logarithm for instance. --pma (talk) 13:52, 28 December 2009 (UTC)[reply]

Could you give examples of things that are expensive to solve but easy to check. I'd like to know the best solve/check ratio even if it's not well defined, but just any examples would be useful. --93.106.33.14 (talk) 14:32, 27 December 2009 (UTC)[reply]

Anyone with a piece of paper and pencil can calculate a product of two integers to verify the solution to a factorization problem (given time and persistence that can be done even for numbers with thousands of digits). The factorization problem for big integers however requires a lot of computational power, and modern supercomputers start chocking at factorization of 100 digit numbers. (Igny (talk) 15:57, 27 December 2009 (UTC))[reply]

All NP-complete problems (probably) have this property. Algebraist 16:55, 27 December 2009 (UTC)[reply]

Here's a stab at a well-defined ratio: RSA numbers says that the RSA-200 factoring challenge was solved using "the equivalent of 75 years work for a single 2.2 GHz Opteron". The solution was found 14 years after the problem was presented. Verifying its correctness takes less than 3 milliseconds on my computer. So that's a ratio of at least 75*365*24*60*60*1000/3 = ~788 billion in the amount of processor time used. 98.226.122.10 (talk) 03:00, 28 December 2009 (UTC)[reply]

Can someone explain to me why RSA number RSA-576 can be factored but RSA-180 and RSA-190 has not been factored yet? Isn't 576 much bigger number than 180 or 190? 122.107.207.98 (talk) 22:39, 31 December 2009 (UTC)[reply]

As RSA numbers says, RSA-576 is a 576-bit number (see binary numeral system). RSA-180 and RSA-190 have 180 and 190 decimal digits. RSA-576 has 174 decimal digits. All the decimal and binary sizes are at RSA Factoring Challenge. The RSA challenge numbers become harder to factor the larger they are, but they are not always factored in increasing order. Some people choose to not go after the smallest and easiest unfactored number. Maybe they prefer the newer challenges with bit-sized numbers which may look more relevant to cryptography. Or maybe they like an impressive round record like RSA-200 with 200 digits. PrimeHunter (talk) 23:01, 1 January 2010 (UTC)[reply]

What's the most common way of writing the all-ones vector, that is, the vector, when projected onto each standard basis vector of a given vector space, has length one? The zero vector is frequently written ${\displaystyle {\vec {0}}}$, so I'm partial to writing the all-ones vector as ${\displaystyle {\vec {1}}}$, but I don't know how popular this is, and I don't know if a reader might confuse it with the identity matrix. I'm writing for a graph theory audience, if that helps pick a notation. --Bkkbrad (talk) 20:36, 27 December 2009 (UTC)[reply]

(1, 1, 1, ..., 1) ? I just had need of one in sixth dimension and actually wrote (1, 1, 1, 1, 1, 1). They don't come up very often, depending on you having made a choice of basis whereas most vector theory is designed to be independent of basis. I've not done any graph theory for a long time though. checking Adjacency matrix#Properties there's one like my first example, so maybe graph theory is not that different.--JohnBlackburne (talk) 20:53, 27 December 2009 (UTC)[reply]

I've mostly seen the notation ${\displaystyle {\underline {1}}}$, ${\displaystyle {\vec {1}}}$ or ${\displaystyle \mathbf {1} }$ (depending on your style for vectors). I believe I've seen ${\displaystyle {\boldsymbol {\iota }}}$ (bolded iota) used once or twice as well. Another alternative notation may simply be ${\displaystyle \mathbf {e} }$ (depending on your preferred vector style of course) - the rationale being that ${\displaystyle \mathbf {e_{1}} =(1,0,\dots ,0)}$ and ${\displaystyle \mathbf {e_{2}} =(0,1,0,\dots ,0)}$ and so on - although I've seldom encountered it and would avoid it as ${\displaystyle \mathbf {e} }$ is not of unit length by this definition. I do not think any of these will be confused with the identity matrix, which is usually ${\displaystyle I}$. All this is mostly from a statistical background. x42bn6 Talk Mess 02:46, 29 December 2009 (UTC)[reply]

December 28

Hi. I was searching the web recently for an example of a set of real numbers that is Lebesgue measurable but not Borel measurable. Such sets must exist, because I know that the cardinality of the family of Lebesgue measurable sets is greater than the cardinality of Borel measurable sets (${\displaystyle 2^{\mathfrak {c}}>{\mathfrak {c}}}$). However, this argument does not furnish an example.

So, I found a website claiming that the following set is Lebesgue measurable but not Borel measurable: The set of real numbers with continued fraction expansions of the form ${\displaystyle \left[a_{0};a_{1},a_{2},\ldots \right]}$ such that there is some sequence of integers ${\displaystyle i_{0}<i_{1}<\cdots }$ with ${\displaystyle a_{i_{n-1}}}$ dividing ${\displaystyle a_{i_{n}}}$ for all ${\displaystyle n\geq 1}$.

Evidently, this set can be seen to be Lebesgue measurable because it is an analytic set, which I don't really understand. I'm not sure how to see that it's not a Borel set, although there is some talk about it here, which I again don't really understand. On to my question: Can someone either help me better understand the example I've presented, or else demonstrate another example where we can see why it's Lebesgue-but-not-Borel measurable? Thanks in advance for any illumination anyone can provide. -GTBacchus^(talk) 00:50, 28 December 2009 (UTC)[reply]

So if all you want to know is that there is some Lebesgue measurable set that's not Borel, here's a cheapo way of doing it: Look at the almost-bijection between the Cantor set (thought of, let's say, as the set of infinite sequences of zeroes and ones) and the interval [0,1] given by putting a binary point before the infinite sequence and reading it off in binary notation. I say almost bijection because there are countably many points in [0,1] that have two distinct binary representations and therefore are mapped to by two elements of the Cantor set.

Now do the Vitali set construction inside [0,1], and pull it back to the Cantor set. The resulting set cannot be Borel, because, if it were, the Vitali set would also be Borel, and therefore Lebesgue measurable (there are a couple of details here that I'm confident you can check for yourself).

But if you now think of the Cantor set as the usual middle-thirds construction inside the reals, well, that's a set of measure zero, so the pullback of the Vitali set is a subset of a set of measure zero, and is therefore Lebesgue measurable (specifically, measure zero).

This is almost cheating. I'd encourage you to pursue the more interesting stuff about an analytic set that's not Borel. Unfortunately this requires much more theory. It's theory well worth learning, though. --Trovatore (talk) 02:13, 28 December 2009 (UTC)[reply]

That's very good; thank you. I can understand that construction, and relate it to my study partner, who was asking me about this. As for the analytic set business... yeah. It looks like a lot of point-set topology. I know that stuff's good for the soul, but I'm a number theorist! One of these days. :) -GTBacchus^(talk) 02:29, 28 December 2009 (UTC)[reply]

(e/c) Since the Cantor space 2^ω is homeomorphic to a closed measure-zero subset of the real line (namely, the middle-thirds Cantor set K), you can use any nonmeasurable set in Cantor space to achieve the same goal. This is because a homeomorphism between two spaces gives an isomorphism of the Borel algebras of the spaces. So any standard example of a nonmeasurable set inside Cantor space – an uncountable set with no perfect subset, a non-determined set, etc. – will correspond to a non-Borel, Lebesgue measurable subset of K. This avoids the "almost-injective" argument.

The point of the "continued fraction" thing in the original post is to use this same trick: the map that sends a real number to its continued fraction expansion is a homeomorphism from the irrationals to ω^ω.

The original question seems to be how to show that the set in question is analytic, and how to show that it is not Borel. To show that a subset of Cantor space or Baire space is analytic, you just check that the set is definable by a logical formula of the correct complexity, namely ${\displaystyle \Sigma _{1}^{1}}$ in the analytical hierarchy. This is more complicated for the real line, which is one reason descriptive set theorists avoid working with the real line directly. But the desired result here can be obtained pretty easily using the isomorphism fact I just mentioned and the standard fact that the continuous image of an analytic set is still analytic.

To show that the set described in the original post is not Borel is harder, and I don't know what the original proof could have been. The way I would prove it is by showing that some standard ${\displaystyle \Sigma _{1}^{1}}$-universal set reduces to the set in question. That would just be a standard textbook exercise, but as Trovatore says you would have to invest some time in the textbook to do it. The two places this sort of thing comes up are in descriptive set theory and in generalized recursion theory. Kechris' book Classical descriptive set theory has detailed information about universal ${\displaystyle \Sigma _{1}^{1}}$ sets. — Carl (CBM · talk) 02:57, 28 December 2009 (UTC)[reply]

I've seen this "divisibility" argument many times, and it always seemed like it should be trivial to check that it's just encoding either the problem of whether a linear order is a wellordering, or whether a tree is wellfounded. When I actually started thinking about it, I realized it's not quite as easy as maybe I expected.

But it's still not hard. I think this works: Assume given a transitive relation R on the naturals, that we want to check whether it's wellfounded. Code R as the following element of Baire space: ${\displaystyle f(n)=p_{n}\Pi _{k<n,nRk}f(k)}$, where p_n is the nth prime. Then for k<n, f(k) divides f(n) just in case nRk. So the continued fraction given by f is in the set if and only if R is illfounded. --Trovatore (talk) 08:23, 28 December 2009 (UTC)[reply]

Yes, that will work. The difficulty I was referring to is in showing

(*): the set of (codes of) non-well-founded trees is ${\displaystyle \Sigma _{1}^{1}}$ complete

in the first place, which you would need to do to really believe that you had solved the original problem. In jargon, (*) is just the normal form theorem for ${\displaystyle \Sigma _{1}^{1}}$ formulas plus the normal form theorem for ${\displaystyle \Pi _{1}^{0}}$ formulas. None of this is truly hard, but it takes a little time to go through it all. I think it could be done very well to a general mathematical audience in two 50-minute seminars. — Carl (CBM · talk) 13:36, 28 December 2009 (UTC)[reply]

Oh, well, I'm used to seeing ${\displaystyle \Sigma _{1}^{1}}$ defined as projections of trees, which makes your (*) pretty trivial. If you have some other characterization in mind, sure, there's a bit of work in showing all the characterizations are equivalent. --Trovatore (talk) 19:11, 28 December 2009 (UTC)[reply]

Is every (associative) K-algebra a universal enveloping algebra of some Lie algebra (over K)? -- Taku (talk) 05:23, 28 December 2009 (UTC)[reply]

No. Universal enveloping algebras are infinite-dimensional (see Poincaré–Birkhoff–Witt theorem). 86.15.141.42 (talk) 16:09, 28 December 2009 (UTC)[reply]

Right. Assume associative algebras are infinite-dimensional too. Or how about a quotient of a universal enveloping algebra? (By the way, I tend to think the answer is yes (if the question is formulated correctly)) -- Taku (talk) 22:07, 28 December 2009 (UTC)[reply]

You are probably going to have to work hard to make the answer yes. Universal enveloping algebras have PBW-type bases, this is very special. The question about quotients is easy because the UEA of a free Lie algebra is a free algebra. 86.15.141.42 (talk) 01:59, 29 December 2009 (UTC)[reply]

Good point. So, the question really boils down to which associative algebra admits a PWB-type basis; that's certainly not the case in general. (I guess a quotient one is trivial.) Anyway, thanks. -- Taku (talk) 06:52, 29 December 2009 (UTC)[reply]

Many thanks for all responses! My high school teacher said though that there isn't math after 12th grade; I read the Lockhart's lament suggested, but my math teacher has a math degree and he's always right. He also says that he knows all math like all math teachers. He won many math competitions and came first place, and he says math is about problem solving. He also says that all problems have been solved, and that's why it matters to show working in math. Who's the world best mathematician? Who's won the most math competitions? I understand that math is not about calculating but how else do you solve math problems like x + 2 = 5? Thanks but please don't insult me though. —Preceding unsigned comment added by 122.109.239.199 (talk) 09:06, 28 December 2009 (UTC)[reply]

If there's no math after 12th grade, how did your teacher get a math degree? -- Meni Rosenfeld (talk) 10:33, 28 December 2009 (UTC)[reply]

By doing high school math, and solving complex math problems? I don't know. That's sort of what I'm asking. Why's there math after 12th grade and what math's there? My high school teacher says there isn't and he's always right about math (he's a math nerd as my friends at school used to say). I'm trying to understand what sort of math is done if calculating isn't there. My math teacher is an expert on calculus, and isn't calculus the highest math? He said that calculus was taught at uni. So does he have a degree in calculus or something? Please explain? Many of youy say my high school teacher's wrong, but how's that? He has a math degree. Many thanks. —Preceding unsigned comment added by 122.109.239.199 (talk) 11:18, 28 December 2009 (UTC)[reply]

Look, we've already answered your questions. We told you that mathematics does involve calculations, but there's much more to it than only that.

Your teacher may have a math degree, but the people who have responded to your query have about a dozen math degrees between them. If you trust him so much, and since he appears eager to discuss these matters with you, and you also have an algebraist friend, then what are you asking us for?

Personally I don't believe you are asking these questions seriously. But if I'm wrong and you are, please read our replies carefully and don't ask us to repeat ourselves. -- Meni Rosenfeld (talk) 12:25, 28 December 2009 (UTC)[reply]

Yeah, so what math's there higher than 12th grade if it isn't calculating. Please give an example of such math. Give a mind-boggling math problem to me please. I'm dead serious about these questions. I just wnat to understand the purpose of math, and what really math is after 12th grade math. My high school teacher says that teachers know most about math since that's the main profesional math job. Applied math's there also. That's why I'm confused why my math teacher's wrong. —Preceding unsigned comment added by 122.109.239.199 (talk) 13:05, 28 December 2009 (UTC)[reply]

Ask your teacher to define "math." I've heard it said the exact opposite way at the beginning of calculus. Some calc teachers say that everything before calc is not math at all. Without having a discrete definition of what math means, you are just having a stupid semantic argument that gets nowhere. Further, nobody here appears to care what your teacher thinks. So, attempting to form an argument between your teacher and everyone else is a trollish waste of time. As for math beyond 12th grade - there is a lot of it. For example, given two vectors of unequal size, how to you find the optimal alignment of the shorter one against the longer one? That is nothing but a bunch of calculations. Have you done vectors or vector alignment? It is very important if you get into health informatics (which is a blend of math and computer science and health). -- kainaw™ 13:18, 28 December 2009 (UTC)[reply]

[ec] I think it's safe to say Mathematical logic is higher than 12th grade and isn't calculating. For mind-boggling math problems, you needn't look further than the Millennium Prize Problems, which involve varying amounts of calculations.

The amount of mathematical knowledge currently in existence is too great for even a brilliant individual to learn in a lifetime. So no, neither your teacher nor any other math teacher knows most of what there is to know in math. -- Meni Rosenfeld (talk) 13:23, 28 December 2009 (UTC)[reply]

Whoever claims there is no math beyond highschool level is like a kid that believes that the universe ends just out his little garden. There are thousands of streets instead, towns, mountains, seas and oceans, woods and deserts as large as a state, and rivers and lakes.... and this is just what we know; then there are planets stars and galaxies... The only thing that gives me a comparable feeling of immensity is the stupidity of your teacher's claim! :-) --pma (talk) 15:16, 28 December 2009 (UTC)[reply]

Have you seen Pleasantville? --Trovatore (talk) 21:52, 28 December 2009 (UTC)[reply]

I haven't; after reading the plot summary it sounds a nice movie --pma (talk) 15:26, 29 December 2009 (UTC)[reply]

Is it not obvious that our esteemed original poster is amusing himself by seeing how much we try to enlighten a fool? Michael Hardy (talk) 01:55, 29 December 2009 (UTC)[reply]

Yes, an obvious timewaster. As the masochism poster, I feel that those who've replied since are exhibiting that very thing.→→86.152.78.244 (talk) 11:27, 29 December 2009 (UTC)[reply]

You people shouldn't insult the OP so bad. When I was in high school I thought just like him "I know all the math, I'm the best in the world". Things started to change after I finished year 12 and now I feel like my knowledge is a needle in the ocean (and in reality it's less than a bacterium). It takes time for people to realize and appreciate the real stuff. And for a mind boggling problem, how about theory building? It's all so easy when you read it but trying to develop the tools for the first time is, to me, only a genius can do. Money is tight (talk) 12:39, 29 December 2009 (UTC)[reply]

Actually I do not think that the OP's antics surprise anyone, since it is well known that a majority hold the belief that mathematics is "arithmetical computations" (educated people and high school teachers inclusive). What is somewhat awkward is that the OP expresses his ignorance openly. Most people are so confident that they are "right about mathematics", that they believe it is not necessary to explicitly note its purpose. Rather thay feel they may implicitly assume its "arithmetical nature" in their assertions. That said, I agree with you, and especially with your last statement (Galois being an excellent example of this). --PST 13:21, 29 December 2009 (UTC)[reply]

You are missing the point. Of course most people are confused to some extent about mathematics. It's the specifics of what the OP said and how that triggered our troll-detectors. Things like

• Asking us what his algebraist friend does, instead of asking him.
• Telling us he wants to be a mathematician although he knows little about it and hates what little he knows.
• Explaining that his teacher is always right. Virtually nobody believes that some other individual is always right.
• Reporting that his math teacher said there is no math after 12th grade and that math teachers know all math, when of course anyone with a math degree would not say that.
• Not updating his questions based on our answers - e.g. "[Without calculating, how] do you solve math problems like x + 2 = 5" after we explained that there are calculations in math, and asking for examples of higher math after we've given them.
• And of course other subtleties in his style.

-- Meni Rosenfeld (talk) 15:27, 29 December 2009 (UTC)[reply]

Why feed the trolls (per Don't feed the trolls)? If a majority feels that the OP's posts were not made in full seriousness, the posts should simply be ignored. In fact, a better alternative would be to remove the posts altogether. Although I certainly agree with your judgement of the OP's questions, insulting him is counterproductive; whatever the circumstance (anyway, I think he is enjoying our insults, as Michael points out). --PST 02:34, 30 December 2009 (UTC)[reply]

Well I guess he is somewhat trolling. I was just trying to "assume good faith" but Meni made his point. Money is tight (talk) 03:30, 30 December 2009 (UTC)[reply]

Assuming good faith is something I completely agree with - so much so that I have the userbox on my page. But the point of WP:AGF is to assume good faith when lacking evidence to the contrary. In this case there is certainly no lack of evidence.

In my view, the other point of AGF is to always consider, if possible, the distant possibility that the posts are genuine, give the OP a chance to explain himself, and not do anything that would be incredibly offensive (which is why I disagree with PST's suggestion of removal). Of course, in particularly disruptive cases this is not an option.

Anyway, posts like those of 86 and Michael Hardy are not meant to insult the OP, only to signal to others that they should not waste their time. -- Meni Rosenfeld (talk) 08:55, 30 December 2009 (UTC)[reply]

And avoid the fate that nearly befell Matilda's aunt in Matilda Who told Lies - and was Burned to Death. :) Dmcq (talk) 09:19, 30 December 2009 (UTC)[reply]

After all I think is OK being polite with maths trolls, especially at christmas times... Go and look at a troll's pub... all the other trolls treat them like scum you know, I mean real trolls working in categories such as history politics sexuality social sciences women studies...--pma (talk) 12:07, 30 December 2009 (UTC)[reply]

Considering that two interesting queries (Wikipedia:Reference desk/Mathematics#transcendental_functions and Wikipedia:Reference desk/Mathematics#No_math_after_12th_grade.3F) have resulted in the troll's silly questions, I would say that he has done more good than he intended... (of course, it would have been more productive if his posts were ignored sooner, but I think that he has left permanently by now). --PST 12:32, 30 December 2009 (UTC)[reply]

Hello. Can a sum or difference of two tangents be expressed as a product or quotient? For example,

${\displaystyle \sin x-\sin y=2\cos \left({\frac {x+y}{2}}\right)\sin \left({\frac {x-y}{2}}\right).}$

Thanks in advance. --Mayfare (talk) 21:03, 28 December 2009 (UTC)[reply]

Just from the definition of tan, we have

${\displaystyle \tan x\pm \tan y=\left({\frac {\sin x}{\cos x}}\right)\pm \left({\frac {\sin y}{\cos y}}\right)={\frac {\sin x\cos y\pm \sin y\cos x}{\cos x\cdot \cos y}}={\frac {\sin(x\pm y)}{\cos x\cdot \cos y}}.}$

Is this the kind of thing you had in mind? ~~ Dr Dec (Talk) ~~ 23:37, 28 December 2009 (UTC)[reply]

Yes. Thank you very much Dr Dec. Have a happy new year. --Mayfare (talk) 18:34, 29 December 2009 (UTC)[reply]

December 29

Does anyone know the formula for the hypervolume of an 8-sphere? This is so I can remove all those question marks from Nine-dimensional space... 4 = 2 + 2 04:23, 29 December 2009 (UTC)[reply]

See n-sphere. PrimeHunter (talk) 04:32, 29 December 2009 (UTC)[reply]

The article Lockhart's Lament, linked above has the sentence "The measurement of triangles will be discussed without mention of the transcendental nature of the trigonometric functions." Can someone please explain to me what the transcendental nature of trigonometric functions has to do with measurement of triangles. Thanks-Shahab (talk) 07:09, 29 December 2009 (UTC)[reply]

Could you please provide the context for that remark (the exact number of the page on which it was made)? I read that article earlier this year and do not recall such a statement being made. But it is quite possible that I have forgotten its presence. --PST 07:36, 29 December 2009 (UTC)[reply]

It's on the last page (Pg 25) of the article. The section deals with an "honest catalog" of the curriculum(The trigonometry course). The complete paragraph runs as follows:

Two weeks of content are stretched to semester length by masturbatory

definitional runarounds. Truly interesting and beautiful phenomena, such as the way the sides of a triangle depend on its angles, will be given the same emphasis as irrelevant abbreviations and obsolete notational conventions, in order to prevent students from forming any clear idea as to what the subject is about. Students will learn such mnemonic devices as “SohCahToa” and “All Students Take Calculus” in lieu of developing a natural intuitive feeling for orientation and symmetry. The measurement of triangles will be discussed without mention of the transcendental nature of the trigonometric functions, or the consequent linguistic and philosophical problems inherent in making such measurements. Calculator required, so as to

further blur these issues.

Regards-Shahab (talk) 07:52, 29 December 2009 (UTC)[reply]

Not sure what it's about. Trigonometric functions of any rational number besides 0 are always transcendental when you use radians, perhaps he was just referring to the problems of incommensurables the Greeks had with things like the square root of 2? Dmcq (talk) 08:35, 29 December 2009 (UTC)[reply]

I think he means that in many high school problems given, the lengths of the sides of the triangles involved are usually assumed to be simple values such as 1 or ${\displaystyle {\sqrt {2}}}$, and this convolutes one's understanding of more general triangles. --PST 09:49, 29 December 2009 (UTC)[reply]

I think he means that learning how to calculate the length of a certain side given an angle and the length of another side is not interesting maths (it's just computation) whereas learning (or, even better, deriving) that the trigonometric functions are transcendental is interesting maths. --Tango (talk) 14:56, 29 December 2009 (UTC)[reply]

A question about let me thinking. Is it too awkward to say that there is no math before 12th grade? Until 12th grade you are only performing calculations, you are more a human calculator than a real mathematician. ProteanEd (talk) 12:52, 29 December 2009 (UTC)[reply]

The word "mathematics" refers to a wide spectrum of activities. It includes things such as addition and multiplication, learned in grade school, and also includes esoteric facts about more complicated objects that are learned in graduate school. These are not fundamentally different topics, but are simply different expressions of the same subject. You have to crawl before you can walk, and later you learn to drive; all of these are forms of locomotion. — Carl (CBM · talk) 12:56, 29 December 2009 (UTC)[reply]

I disagree (in fact, I believe that there is not mathematics even in the 12th grade; at least in the current American education system). Mathematics is, after all, creativity. It is not about attaining tools such as computational ability, but rather about using the tools for a particular purpose. Thus mindless computation, on its own, is not mathematics, unless it is done in a creative manner. Even calculus, in the mechanical sense of solving problems, is not mathematical unless one does it creatively. Thus, while the routine technique of finding the maximum values of polynomials of degree two on a compact interval loses its mathematical value at a certain stage, inventing new techniques in numerical analysis can be safely referred to as mathematics. In another context, one could refer to painting blue dots on a white sheet of paper as "painting", but such would not be considered to be of true artistic value. Combining a wide range of colors with different tone, depth and emotion to create a painting is true art. --PST 13:10, 29 December 2009 (UTC)[reply]

You're free to argue that the English language should be changed. At the present, "mathematics" clearly refers both to the things we learn in kindergarten and the things we learn in graduate school. Similarly, finger-painting is indeed painting. Maybe you're talking about real mathematics, but that sort of discussion will always devolve into polemics. — Carl (CBM · talk) 13:16, 29 December 2009 (UTC)[reply]

I guess you are right. --PST 13:22, 29 December 2009 (UTC)[reply]

That definition is entirely subjective. In particular, at which point mathematics becomes creative is not properly ascertainable. If "creative" means "original", many mathematicians never end up doing mathematics. In the same way that the composition of a short letter and the composition of a classic piece of fiction are both termed "writing", everything from basic addition to the discovery of new theorems in esoteric fields is mathematics. So, yes, mathematics is done in high school and before. —Anonymous Dissident^Talk 13:30, 29 December 2009 (UTC)[reply]

As I mean it, doing mathematics is not necessarily creating original maths, nor one needs to discover original results to enjoy it greatly, in the same way that doing music is not necessarily composing. When I was 6 I had a wonderful teacher, which was about 19, so sweet. She made us solve nice addition problems drawing colored balls... when she would come to me and lean over my little desk, how happy I was! I scarcely listened towhat she would tell me, but only hear the sound of her sweet voice and sniff the perfume of her black hairs. Maybe that was no original maths, but how exciting! ;)--pma (talk) 15:39, 29 December 2009 (UTC)[reply]

What do you think about this statement:

Suppose f is a continuous function and h is a (continuous) symmetry for f i.e.

${\displaystyle f\circ h=h\circ f}$

Then for any ε there exists δ such that if d(f,f ')<δ then there exists h ' such that h ' is a simmetry for f ' and d(h,h ')<ε.

assuming d is the standard distance given by the uniform norm.

Is it true or false?--Pokipsy76 (talk) 13:57, 29 December 2009 (UTC)[reply]

I see it more false than true in general, though I don't have a counterexample. Could you specify the domain? If f₀ is an Anosov diffeomorphism and f₁ is (sufficiently) C¹ close to f₀ then f₁ is Anosov and conjugated to f₀ with a conjugation H which is C⁰ close to the identity, together with its inverse H^-1. So by conjugation with the same H you do have a h₁ C⁰ close to h₀ doing the job. --pma (talk) 14:36, 29 December 2009 (UTC)[reply]

I didn't want to be specific about the domain, I wanted to ask the question in general, however I'm curious in particolar for functions in R and R². I supopose that a first step to find a counterexample could be to find a function which doesn't have any nontrivial simmetry... are there such functions?--Pokipsy76 (talk) 15:26, 29 December 2009 (UTC)[reply]

Consider the following counter-example on ${\displaystyle \mathbb {R} .}$ Let ${\displaystyle f_{0}:=\mathrm {id} }$ and ${\displaystyle h_{0}\in C^{0}(\mathbb {R} ,\mathbb {R} ),}$ any function, with ${\displaystyle h_{0}\neq \mathrm {id} .}$ Say ${\displaystyle \|h_{0}-\mathrm {id} \|_{\infty }>r>0.}$ Then ${\displaystyle f_{0}}$ and ${\displaystyle h_{0},}$ of course, do commute, and I claim that there exists a function ${\displaystyle f}$ as close in the uniform distance to ${\displaystyle f_{0}}$ as we want (in fact, even in the ${\displaystyle C^{\infty }}$ distance if you wish), with the property that any ${\displaystyle h\in C^{0}(\mathbb {R} ,\mathbb {R} )}$ that commutes with ${\displaystyle f}$ necessarily has a uniform distance ${\displaystyle \|h_{0}-h\|_{\infty }\geq r.}$

Indeed, let ${\displaystyle a\in \mathbb {R} }$ such that ${\displaystyle |a-h_{0}(a)|\geq r.}$ There exists a function ${\displaystyle f}$ such that ${\displaystyle a}$ is a globally attractive fixed point of ${\displaystyle f}$, meaning that ${\displaystyle f(a)=a}$ and ${\displaystyle f^{\,n}(x)\to a}$ as ${\displaystyle n\to \infty }$ for all ${\displaystyle x\in \mathbb {R} ;}$ such an ${\displaystyle f}$ may be chosen arbitrarily close to ${\displaystyle f_{0}}$ (take e.g. ${\displaystyle f(x):=x-\epsilon \,\arctan(x-a)}$ for a small ${\displaystyle \epsilon >0}$). Let ${\displaystyle h\in C^{0}(\mathbb {R} ,\mathbb {R} )}$ a function that commutes with ${\displaystyle f}$. Then we have ${\displaystyle h(a)=h(f^{n}(a))=f^{n}(h(a))\to a}$ as ${\displaystyle n\to \infty }$ so ${\displaystyle h(a)=a}$ and ${\displaystyle \|h-h_{0}\|_{\infty }\geq |h(a)-h_{0}(a)|=|a-h_{0}(a)|\geq r.}$ Is that clear? Actually this argument could be adapted in order to show that the original property fails to hold with ${\displaystyle f_{0}:=\mathrm {id} }$ and ${\displaystyle h_{0}\neq \mathrm {id} }$ with any topological manifold as a domain. --pma (talk) 20:28, 29 December 2009 (UTC)[reply]

Very nice proof!--Pokipsy76 (talk) 19:37, 5 January 2010 (UTC)[reply]

December 30

When using master theorem, f(n) and n^log_ba must have a polynomial difference. The example shown is f(n)=n/log n. With a=2 and b=2, n^log_ba=n. So, the claim is that n/log n and just n have a non-polynomial difference. Another example I saw changes a to 4 so n^log_ba=n². The claim is that n/log n and n² have a polynomial difference. I'm left wondering exactly what the "polynomial difference" is. Is it taking (n/log n)-(n) and claiming that is non-polynomial? -- kainaw™ 02:30, 30 December 2009 (UTC)[reply]

"Difference" here is being used multiplicatively -- the quotient of n and ${\displaystyle n/\log n}$, which is ${\displaystyle \log n}$, is not polynomial. Formally, the master theorem requires ${\displaystyle f(n)\in O(n^{\log _{a}(b)-\epsilon })}$ for some positive ${\displaystyle \epsilon }$; or equivalently, requires ${\displaystyle {\frac {f(n)}{n^{\log _{a}(b)}}}\in O(n^{-\epsilon })}$ or ${\displaystyle {\frac {n^{\log _{a}(b)}}{f(n)}}\in \Omega (n^{\epsilon })}$. Eric. 131.215.159.171 (talk) 03:21, 30 December 2009 (UTC)[reply]

In any case, what everybody would do as a first step with this ${\displaystyle T(n)=aT(n/b)+f(n)}$ is writing ${\displaystyle n=b^{k}}$ and turning the recurrence into the form ${\displaystyle \tau (k+1)=a\tau (k)+\phi (k);}$ solutions of the latter have an immediate representation in terms of discrete convolutions, and a whole machinery for growth estimates is available, to bound the solution in terms of ${\displaystyle \phi (k).}$ As I see it, in these cases it should be better not to make everything into a theorem (especially with such a name), that makes things more rigid. --pma (talk) 10:09, 30 December 2009 (UTC)[reply]

I was born on September 21, 1944; the USA was born on July 4, 1776. How do I calculate on what date I will become exacty 25% as old as the USA? 206.54.145.254 (talk) 18:00, 30 December 2009 (UTC)[reply]

here then here. use the first link to work out the number of days old the US was when you were born, the date you want is 1/3 of that number of days later (as for you to be 1/4 the age of the US on a date the US was 3/4 that age when you were born, and 1/4 is 1/3 of 3/4. Even if you could do the calculations yourself the site's a useful check.--JohnBlackburne (talk) 18:24, 30 December 2009 (UTC)[reply]

I have this function:

${\displaystyle y=\arcsin {\frac {{\sqrt {3}}{\sqrt {n^{2}-sin^{2}\theta }}-sin\theta }{2}}+\theta }$

If I know "n", I can easily find the minimum value of y: just use a graphing calculator to graph y against theta ad ask it to find the minimum. If I know the minimum and the standard deviation on the minimum, how do I find "n" and the standard deviation on "n"? --99.237.234.104 (talk) 20:54, 30 December 2009 (UTC)[reply]

You'll have to explain a bit better. The way I understand the function, it is unbounded for values of n for which it is defined, so you can't speak of its (global) minimum. Also, I don't understand what standard deviation means in this context. -- Meni Rosenfeld (talk) 06:53, 31 December 2009 (UTC)[reply]

OK, here's a short explanation. If you're interested, also read the long explanation (which is pretty cool, IMHO).

Short explanation: "n" is a constant. If I know the numerical value of "n", I can plot a y vs. theta graph and compute the graph's minimum y value between theta=0 and theta=pi/2.

However, I don't know the numerical value of "n"; that's what I'm trying to calculate. I experimentally measured the minimum y value, and as with any experiment, there is an error margin associated with the measured value. How do I calculate "n" from this data? Also, how do I calculate the margin of error on n?

Long explanation:

I'm doing an experiment to determine the refractive index of ice (this is the "n"). To do this, I'm taking a photograph of a 22 degree halo and measuring its radius. I've worked out, using some physics, that ${\displaystyle y=\arcsin {\frac {{\sqrt {3}}{\sqrt {n^{2}-sin^{2}\theta }}-sin\theta }{2}}+\theta -pi/3}$ gives the angle of deflection (the y value) in terms of the angle of incidence of light on an ice crystal. The minimum possible angle of deflection is equal to the radius of the halo. It follows that if I measure the radius of the halo, I can calculate the refractive index of ice. It turns out that the radius of the halo depends VERY sensitively on "n": a difference of 0.01 in "n" corresponds to a difference of 0.8 degrees in the radius. Since I can measure radius to an accuracy of 0.02 degrees, I should get a very precise fix on "n". --99.237.234.104 (talk) 10:43, 31 December 2009 (UTC)[reply]

That's much better - in particular ${\displaystyle 0<\theta <\pi /2}$ was an important piece of information. Is it also true that ${\displaystyle 1\leq n\leq 2}$? Otherwise there's a problem.

So you have a function ${\displaystyle f(n,\theta )=\arcsin {\frac {{\sqrt {3}}{\sqrt {n^{2}-sin^{2}\theta }}-sin\theta }{2}}+\theta }$. For any n, we let ${\displaystyle a(n)}$ be the value of ${\displaystyle \theta }$ for which f is minimal, and ${\displaystyle b(n)=f(n,a(n))}$ the value of the minimum. What you want is to find the inverse function of b.

Since b is computed using a, it is natural to first find an expression for a. This requires finding where the derivative of f is 0, but the resulting equation seems unsolvable algebraically. It can still be found numerically.

I've done some numeric calculations; the first interesting thing to note is that ${\displaystyle b(n)=2a(n)}$ (though this is irrelevant for the solution). The second is that b can be approximated fairly well with a polynomial - for example, ${\displaystyle b(n)\approx -4.51+11.41n-7.91n^{2}+2.04n^{3}}$. Given b you can solve for n numerically, for example by graphing.

If it happens to be known that ${\displaystyle 1\leq n\leq 1.5}$, then a much better, and simpler, approximation is ${\displaystyle b(n)\approx 0.2654+0.4413n+0.3412n^{2}}$.

The error in n is simply the error in b divided by ${\displaystyle b'(n)}$. -- Meni Rosenfeld (talk) 11:39, 31 December 2009 (UTC)[reply]

Thanks for the response! Yes, n is between 1 and 1.5 (in fact, it's around 1.31). How accurate are your approximations? I'm expecting this experiment to be capable of giving 5 significant digits for "n", and I don't want rounding errors to worsen the accuracy of the result. --99.237.234.104 (talk) 22:20, 31 December 2009 (UTC)[reply]

A better approximation can be found by optimizing for ${\displaystyle 1.30\leq n\leq 1.32}$, resulting in ${\displaystyle b(n)\approx 0.3460547+0.3289593n+0.3795664n^{2}}$. This one can easily give you 5 significant figures for n (the difference is less than ${\displaystyle 2\cdot 10^{-7}}$ in the specified range). -- Meni Rosenfeld (talk) 08:27, 1 January 2010 (UTC)[reply]

I'm working through a problem concerning the size of a population after n generations, assuming that no members of the population die, and that the number of members of each generation is given by 2^n-1, where n is the generation number. For example, in the first generation there is 1 member, the second generation has 2, the third 4, the fourth 8, etc.

I know that I require a total population of roughly 1.5 x 10²⁵ and want to know how many generations I require. So far, all I have is:
${\displaystyle \sum _{k=1}^{n}2^{k}\approx 1.5\times 10^{25}}$
Any body know the formula for the sum of 2 to the power n? Searching on Google just seems to bring up things like the sum of n to the power 2, which I already know and doesn't appear to be of much use to me in this instance. Logarithms tell me it's a bit less than 84 generations, which is confirmed by a quick Excel spreadsheet, but I was hoping for something a bit more 'mathematical'.--80.229.152.246 (talk) 20:59, 30 December 2009 (UTC)[reply]

I think the formula ${\displaystyle \sum _{k=0}^{n}2^{k}=2^{n+1}-1}$ may be of some use. At the scale of your application, the "-1" really doesn't matter, of course, but the formula itself does provide justification for your answer. --Kinu ^t/_c 22:29, 30 December 2009 (UTC)[reply]

See also Geometric series. -- Meni Rosenfeld (talk) 06:47, 31 December 2009 (UTC)[reply]

December 31

I've read milenium problems and wow! The math's mindboggling! My math teacher's a math degree and he sais he couldn't solve Ryan's hypo-whatsisis (I don't know what a hypothises is?). He's tried many calculators to solve it failed (he sais he'll solve it though since he's a math degree, but Wiki sais its to hard to solve). Now I know why math's there after 12th grade. Many thanx! Though there's another milenium problem Hodge conjecture? My math teacher sais it's fake math because there isn't numbers and formulas. Is Hodge real math? Please explain Hodge conjecture though! I know many of youy think I'm not serious but I'm serious. It's my English (i'm not a native speaker), so may be it sounds I'm not serious. But I now understand math's more than calculators, that's math is more than 12th grade (my high school teacher's wrong then). Sorry and many thanx. —Preceding unsigned comment added by 122.109.239.199 (talk) 04:09, 31 December 2009 (UTC)[reply]

Is Gauss–Jordan elimination the same as Reduced Row Echelon Form? --33rogers (talk) 07:46, 31 December 2009 (UTC)[reply]

Gauss Jordan elimination involves taking the coefficient matrix and reducing it to reduced row echelon form, with corresponding changes on the right hand side. Then the system is solved by back substitution. This whole procedure is called Gauss Jordan elimination.[1] Reduced row echelon form is the type which the coefficent matrix ultimately becomes.-Shahab (talk) 07:51, 31 December 2009 (UTC)[reply]

I want to show that the torus, described as "the surface of revolution obtained by rotating a circle about an axis in its plane and disjoint from it", is homeomorphic to ${\displaystyle S^{1}\times S^{1}}$ (part of Problem 4 here). I suppose this is meant to be easy but my analytic geometry is sufficiently weak that I'm having a devil of a time turning this description into something concrete enough to write down an explicit map and formally verify all the conditions. Here's my attempt so far:

It's not too hard to see that ${\displaystyle S^{1}\approx \mathbf {R} /\mathbf {Z} }$. Define ${\displaystyle f:\mathbf {R} /\mathbf {Z} \to \mathbf {R} ^{3}}$ by ${\displaystyle f(s)=(\cos 2\pi s,0,\sin 2\pi s)+(2,0,0)}$ so that the image of f is the circle of radius 1 in the xz-plane centred at (2,0,0). Let R(t) be rotation around the z-axis by the angle 2πt. Finally define ${\displaystyle h:\mathbf {R} /\mathbf {Z} \times \mathbf {R} /\mathbf {Z} \to \mathbf {R} ^{3}}$ by ${\displaystyle h(s,t)=R(t)f(s)}$. It is clear that h is injective, and it's continuous because each component of ${\displaystyle R(t)f(s)}$ is a linear polynomial in sines and cosines of s and t (as R(t) is just a rotation matrix). Since the domain of h is compact, this implies that h is a homeomorphism onto its image.

I'm not sure how satisfactory a solution to the original problem this is. Any advice appreciated! — merge 13:37, 31 December 2009 (UTC)[reply]

This is not much of an answer, but can you intuitively feel why the result is true? Can you imagine the torus as a cartesian product in the manner described? In my eyes, the most important aspect of topology is attaining an intuitive feel for the topology of a space; although many textbooks encourage defining explicit homeomorphisms between topological spaces, it is not necessary as long as you can intuitively visualize the homeomorphism. However, when confronted with a first example, it is sometimes instructive to construct an explicit map. In this case, consider using your intuitive feel for the torus as a cartesian product when constructing a homeomorphism. Does this help? --PST 14:59, 31 December 2009 (UTC)[reply]

Thanks for your response! Yes, the basic intuition is clear enough to me for all the spaces mentioned in the problem. What I'm specifically concerned about at this stage is grounding the intuition in correct proof. The above was exactly my attempt to do the explicit construction you suggested (modulo writing out all the rotation matrix details, which don't seem all that relevant). — merge 15:28, 31 December 2009 (UTC)[reply]

How about this:

${\displaystyle (x,y,z)=(\cos(2\pi t),\sin(2\pi t),0)(2+\cos(2\pi s))+(0,0,\sin(2\pi s)).}$

Michael Hardy (talk) 01:48, 1 January 2010 (UTC)[reply]

Thanks! I'm less concerned about the actual formula than whether the argument is sound, though (and whether there's a better way to do it). — merge 11:20, 1 January 2010 (UTC)[reply]

What you did seems fine. Demonstrating a homeomorphism seems like a good way to go about proving that they're homeomorphic. 67.100.146.151 (talk) 05:01, 2 January 2010 (UTC)[reply]

Yes, after looking at it for a while it seems all right to me. I think I've also managed to show the other spaces in the problem are equivalent. Thank you! — merge 12:21, 2 January 2010 (UTC)[reply]

I'm looking to see if an algorithm for a certain problem (described below) exists. I've run into this problem a number of times in different contexts, so suspect it might be solved, but don't know what it might be named. For the time being, I've coined the name "random shopping algorithm".

The problem begins with presented with a collection of "goods" with different costs and a given sum of money. The algorithm must then choose a random selection of goods to purchase whose total is equal to the given sum. It does not work to simply choose items at random until the sum is reached as that would select one high cost item more often than a lot of small cost items.

--jwanders^Talk 22:25, 31 December 2009 (UTC)[reply]

I'm not sure what your intended criteria for selection is but see subset sum problem. If you want all subsets with an equal sum to have the same probability then you have to know the total number of solutions and then pick one of the solutions. Or do you want each individual item which is both inside and outside at least one solution to have the same probability or as close to the same as possible? Then we would also have to define a closeness measure when there are items with different probabilities. PrimeHunter (talk) 23:33, 31 December 2009 (UTC)[reply]

Please: "This criterion is..." or "These criteria are....". Michael Hardy (talk) 01:15, 1 January 2010 (UTC)[reply]

Hmm, no, I don't think the subset sum problem quite matches. As an example, say you have $20 and the items available are A for $10 or B for $1. The three possible outcomes here are (2As), (1A + 10Bs), or (20Bs). The algorithm I'm trying to find would chose one of these three with an equal probability. You right that I need some way of A) counting the number of possibilities N, randomly selecting n<N, and B) creating possibility n directly. As I said above, I'm hoping this is a solved problem that someone will recognise and know the name of. --jwanders^Talk 00:16, 1 January 2010 (UTC)[reply]

So your algorithm is:

• generate the N possibilities by the subset sum algorithm, and remember them in a list
• choose a random number between 1 and N, call it x
• output the x-th element of the list in step 1

Doesn't that do it? Staecker (talk) 01:19, 1 January 2010 (UTC)[reply]

January 1

To me its amazing that inferances from the rules of mathematics can predict what happens in the real world. As a second question, have there ever been empirical studies that have led to the rules of mathematics being altered? 78.146.210.81 (talk) 11:43, 1 January 2010 (UTC)[reply]

Isn't it amazing the way the real world acts as a model for mathematics ;-) Naive set theory is a good example of some maths that was found to be wrong after a bit of study and had to be amended, see the section on paradoxes at the end. Dmcq (talk) 11:59, 1 January 2010 (UTC)[reply]

An important aspect of mathematics lies within the relationships between different branches. For instance, one could use algebra in topology, without delving deep into the subject (but of course, a deeper form of algebra may shed new light on topology, if applied). Likewise, although contradictions were found in naïve set theory approximately 100 years ago, it continues to be applied to this day. In fact, many branches of mathematics could function without reference to Russel's paradox or any deep form of axiomatic set theory. On the other hand, as I pointed out, fields such as set-theoretic topology require a very deep form of set theory; naïve set theory alone does not suffice. --PST 12:29, 1 January 2010 (UTC)[reply]

The OP might not be that interested, and there is an overlap between who helps at the two help desks, but he or she might ask something close to what is being asked here at the Science desk and shuttle back and forth when it seems interesting if a more complex answer is desired. Part of the issue is what the 'real world' is from a scientific standpoint. There are some who will argue that it is a mathematical object when all is said and done. I wouldn't hold out too much hope for a serious reply at the Computer Science desk, but you might actually get the most interesting one there. Ultimately, the world consists of objects with rules for their behavior (it seems), and the fact that an abstraction from the real world (mathematics) would be able to feed back predictions on the behavior of the real world isn't what surprises anyone. It's that purposeless questions internal to the abstractions can be taken far from their natural sources and still find applications that is often surprising.Julzes (talk) 12:56, 1 January 2010 (UTC)[reply]

Of course there are religious/philosophical questions like whether you support the Axiom of choice or the Axiom of determinacy or go for a third way like projective determinacy. I think I read a science fiction story once where the laws of nature kept getting more complicated to cover up problems as more and more flaws were shown up in the previous laws. So relativity and quantum mechanics only became true recently as problems and paradoxes were shown in the previous simpler laws. Eventually it'll all become so complicated we'll never be able to show something is actually wrong or it might all disappear as just too paradoxical. :) Dmcq (talk) 14:09, 1 January 2010 (UTC)[reply]

Nobody has answered the first question. Does that mean the answer is "Nobody knows"? 92.24.69.222 (talk) 18:54, 1 January 2010 (UTC)[reply]

What was the first question. All I see there is "To me it is amazing that..." etc. That's not a question. Did you mean "Why does that work"? Or maybe "Am I the only one who's amazed?" Or did you maybe want an answer that would leave you unamazed? Michael Hardy (talk) 00:30, 2 January 2010 (UTC)[reply]

Mathematicians discover properties of natural numbers, shapes etc. and then invent notation, theories and structures to describe, investigate and explain what they have discovered, and go on to make more discoveries about what they invented. The answer has to be "both". Dbfirs 19:36, 1 January 2010 (UTC)[reply]

"Shape" can be formulated in the language of natural numbers, after employing various other mathematical (algebraic and analytical) ideas. --PST 00:58, 2 January 2010 (UTC)[reply]

That's true, but the discoveries were made thousands of years before this formulation was invented. Dbfirs 09:57, 2 January 2010 (UTC)[reply]

Um, I'm not quite sure what the claim is supposed to mean. On its face, to talk about shape using only natural numbers, you need second-order logic; I'm not sure why you wouldn't just bring in the real numbers from the start, which are more natural to use when discussing shapes. --Trovatore (talk) 10:12, 2 January 2010 (UTC)[reply]

Not sure what kind of answer is wanted. If we had an answer that could be codified then it would be mathematical and then we'd simply have something saying it itself is applicable using methods which work only because it is applicable. Something like that which doesn't sound very sensible when applied to the real world but worked okay for Gödel in maths. That we are able to make sense of the world is only because it is reasonable. Dmcq (talk) 22:15, 1 January 2010 (UTC)[reply]

This is a classic philosophical question about the nature of mathematics. See mathematical realism for the "discovered" viewpoint, and see several alternative views on that page. Staecker (talk) 22:31, 1 January 2010 (UTC)[reply]

January 2

I have been wondering if the following factorization has a history, if there is any reasonably simple way to answer this. Oddly, if it is new, it was discovered to be factorable by one person (me) and actually factored by someone else (User:PrimeHunter) who knew the quick way to do it. My own reaction is to be agnostic on the question of its originality. It's not far from the subject of cyclotomic polynomials, but I can't see any particular way the question is likely to have come up other than by simply experimenting as I was (as a new user of PARI/GP). I also haven't a clue as to whether its factorability has any more than educational value (in regards to small-number coincidences?), though I am looking into it on paper. Here is the result:

${\displaystyle x^{8}+x^{7}+x^{6}+x^{5}+x^{4}+x^{3}+x^{2}+x+20=(x^{4}-x^{3}+x^{2}-3x+4)(x^{4}+2x^{3}+2x^{2}+4x+5)\,}$

I'm virtually positive that the following is a re-discovery, but I'd be interested in its history as well:

${\displaystyle x^{4}+x^{3}+x^{2}+x+12=(x^{2}-2x+3)(x^{2}+3x+4)\,}$

Julzes (talk) 00:26, 2 January 2010 (UTC)[reply]

These sorts of factorizations are often discovered (in more general situations), so it is unlikely that it has not been stumbled upon before, as you point out. However, I have not thought about this one, or an analogue (recently) in particular, so I cannot guarantee anything about it. You might like to read the books "Cyclotomic fields" and "Cyclotomic fields II", by Serge Lang, in the GTM series. --PST 07:41, 2 January 2010 (UTC)[reply]

That's a good suggestion. I think of the subject as more fundamental and simple than it surely is and another couple of Lang's books is not going to hurt me.Julzes (talk) 09:49, 2 January 2010 (UTC)[reply]

It seems by an off chance ive forgotten how to factorise the ${\displaystyle x^{3}-27}$ in ${\displaystyle {\lim _{x\to 3}{\frac {x^{3}-27}{(x-3)}}}}$, its been a while since ive did this so any help would be appreciated. —Preceding unsigned comment added by 121.214.27.148 (talk) 04:16, 2 January 2010 (UTC)[reply]

Two things you should know:

${\displaystyle a^{3}-b^{3}=(a-b)(a^{2}+ab+b^{2})\,}$

and a basic fact from algebra: If you plug the number 3 into a polynomial and get 0, then (x − 3) is a factor of that polynomial; if you plug in 42 and get 0, then (x − 42) is a factor, etc.

This looks like another typical case of someone learning calculus without really knowing the algebra prerequisites, which means you'll probably pass but you'll find it unpleasant. Michael Hardy (talk) 04:43, 2 January 2010 (UTC)[reply]

I think that one cannot generalize such situations without sufficient evidence regarding the circumstances. For instance, I was self-taught in calculus; theoretical ideas such as "maxima" and "minima" evoked interest in me when I was first presented a calculus book. But I never "liked" elementary algebra at the time, and thus never really thought about these sorts of factorizations (for instance) when doing calculus, even when I saw them arise in the theory of limits. But once I actually noticed that I was taking these (algebraic) ideas for granted in my intuitive visualization of limits, I thought it worthwhile to develop some intuition about them, and in doing so, generalize them to higher degree polynomials (and other sorts of factorizations). Thus it was not a question of me not being able to understand basic algebra; rather, I did not see its purpose (or interest) prior to studying calculus and did not attempt to understand it. And this was perfectly appropriate (at least in my view) because I saw its need in consolidating my intuition when doing calculus, and subsequently sought to think about it in greater depth. That said, I was not "dumb" at algebra prior to studying calculus; I was reasonably competent, though there were some minor aspects which although I knew, I did not examine in close depth. Succinctly, my point is that it may be more effective for someone to realize the importance of algebra in calculus with time, rather than be forced to do it (algebra) without having a clue about where they will use it in mathematics (not every mathematician would have enjoyed doing elementary algebra (and by this I do not mean original discoveries; just mindless application of identities) without reference to calculus (for instance), for a long period of time, although I could be wrong). But of course, I am assuming one is also not being forced to do calculus; unless calculus is found interesting at the time of its study, it is not worthwhile to do anything earlier in depth anyhow. --PST 07:34, 2 January 2010 (UTC)[reply]

....and notice that if you get something other than 0 in the numerator, then the question of what to do with the limit becomes trivial. Michael Hardy (talk) 04:44, 2 January 2010 (UTC)[reply]

What are "the" (or "some") geometric interpretations of the three main geometric algebra involutions (reversion, grade involution, and Clifford conjugation)? I believe that grade involution corresponds to reflection through the origin, but I don't know about the other two.

ALSO, when representing a geometric algebra using a faithful matrix representation, where ordinary matrix multiplication corresponds to a geometric product, are there analogous matrix involutions to these geometric algebra involutions. When using a sensible basis, the transpose is equal to the reverse, but I've found nothing for the other two.--Leon (talk) 13:09, 2 January 2010 (UTC)[reply]

How to prove the continuity of f(x)=ln(x)? --84.62.205.233 (talk) 13:12, 2 January 2010 (UTC)[reply]