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March 16

grasping math to an intermediate level

hi, I like to get to an intermediate level of understanding of different subjects and I want to get a decent grasp of mathematics, but I'm having a hard time with it. I have a BS in math and an MS in statistics. I feel as if I'm wandering blindly in a mathematical universe, feeling the obvious, stumbling over the less-obvious, but without confidence in my travels. When I started to learn typography, say, I had that blind feeling at first but with study soon found the fundamentals and understood the basic ideas and questions in that field.

I want to get to that level of confidence with math, but I'm not sure how to get there. Obviously it's a big world out there, but is there a field that I should concentrate study on (analysis maybe?) to get an intermediate grasp (i.e., not expert and not blind beginner) ? thanks 72.150.136.11 (talk) 14:24, 16 March 2008 (UTC)

In all honesty, it's pretty much impossible to get an intermediate grasp of everything, since there is so much. I mean, I am doing an MMath, and I'd say that by doing so I was intermediate in the subjects I do, but there are others I simply have no clue about - topology, fluid mechanics, relativity, anything related to statistics... maths is so vast, and a lot of it is so remote from anything you do at school as to be an entirely different subject. Group theory is probably a useful thing to try and understand, as is basic vector calculus and mathematical analysis. I personally like complex analysis and combinatorics. Of course, combinatorial game theory is also fun to try. -mattbuck (Talk) 15:00, 16 March 2008 (UTC)

Linear algebra, while rather boring, is also a useful subject to have a working knowledge of - it seems to pop up all over the place. However, if you've got a BS in Maths I would expect you've already covered the basics of all these areas. If you want to do more Maths, you need to specialise, there's no way to have a better than undergraduate understanding of more than one or two areas. --Tango (talk) 15:19, 16 March 2008 (UTC)

I disagree with the last statement, there are many researchers who are active in several quite distinct areas. It's not easy, of course.

Mathematical logic is very important to know the basics of (say, up to the level of understanding Godel's theorems). Likewise for topology, which at the very least one needs to know to the level of feeling the difference between metric features and topological ones. I recall that when I have been applying for a math MSc, virtually all schools frowned upon me not having taken a topology course in my BA. Fortunately I had learnt the material independently - my peers who hadn't had some trouble with many of the courses.

Of course, there's the obvious stuff, like analysis (real, complex, functional, differential geometry...), algebra (groups, rings, fields, Galois theory...), discrete (combinatorics, graphs, ...), and of course computer science (computational complexity, information theory, machine learning, ...) and applied math (numerical methods, mathematical physics, game theory...). Disclaimer: This categorization is based on my own personal view and the subjects are not equally divided. -- Meni Rosenfeld (talk) 17:37, 16 March 2008 (UTC)

Well, I guess you run into trouble defining an "area" of maths. There are plenty of people that do research in what would usually be considered distinct areas, but they're usually working in some kind of overlap between them - pretty much all branches of maths overlap somewhere. Are there many people that have a solid understanding of the whole of several areas? I would say most understand just the parts of those areas that are relevant to their research. It does depend very much on how large you consider an area to be (is algebra a single area or is ring theory a distinct area to Galois theory? - obviously, you can understand both ring theory and Galois theory to a high level, but understanding the whole of algebra to a high level is pretty challenging without even trying to start on any non-algebraic areas). Put simply: I think we're both right for appropriate definitions of "area". --Tango (talk) 18:12, 16 March 2008 (UTC)

Thanks for these answers. I thought it might be impossible for a not-particularly-gifted person to walk with confidence in the math spaces. I just hate that feeling of flailing around blindly, holding only to those proofs I really really understand. Sometimes I feel like I'm missing a sense that some others have; which I suppose is true for everyone on some level. I'll explore mathematical logic and topology and see how that goes. thanks everyone 72.150.136.11 (talk) 19:00, 16 March 2008 (UTC)

While we all strive for perfection, nobody gets there. However far you go, there's always going to be more you don't understand, so you'll probably always be walking blindly, you'll just be walking in more advanced parts of the world of maths. I think the important thing is to enjoy it - study the areas of maths that interest you and that you find fun. There's little point studying anything else. --Tango (talk) 19:31, 16 March 2008 (UTC)

From what you say, I'm not sure that what you're looking for is so much a subject as a teaching style. You need some reintroductions to subjects you already know, from a different viewpoint. At the moment, I don't know of any better way to find good sources than to read things randomly and hope to hit gold (as I have occasionally), but someone else on the board might. If you want to grasp the material, try some popularizers, avoiding as best you can those who dumb it down instead of actually making it clearer. I was just reading Indra's Pearls, for instance, which gives some solid visual meaning to some pretty abstract techniques from several fields (analysis, group theory, linear algebra, complex analysis, maybe some others). If you want to understand the motivation, that is the goals, behind something, look into its history. Even the original papers on the subject, if you can get them. For instance, I've heard over and over again this same fable about complex numbers being the solution to the equation x^2+1=0, which ties in nicely to some other equations, like x^2-2=0 and x+1=0. It wasn't until fairly recently that I found out that complex numbers were motivated for the first time, after thousands of years of people intentionally ignoring their effects on quadratic equations, by the solution to the cubic equation. There are a lot of nice books that give the story behind that, not least the Ars Magna itself. Also, something that will always help you grasp a subject is to use it. Set your sights on a number of small, random goals, and see what you can do with them. Abstraction is bad on an empty stomach - you need a lot of bulk to soak it up. Once you can navigate real situations with confidence, the patterns you find among them will motivate and support abstraction. The exercizes in textbooks are a start, but don't rely on them for originality, follow your own curiosity. I'm trying to keep my studies broad-based, since I like them that way, but if you want to focus on a single subject for awhile, there's certainly plenty to choose from. It doesn't sound like you plan to use this study immediately, so focus your decision on how much you can learn about learning. It may be best to relearn something concrete as an introduction. Euclid's geometry, for instance, moving through some historical works like the Arithmetica Infinitorum or Newton's papers, then up through the crisis a century ago with Cauchy and Weierstrass et al, to get to analysis. An interesting stop you might take on the way is in Algebraland - most of calculus, as it applies to polynomials, can be derived without hardly touching limits. Black Carrot (talk) 02:28, 17 March 2008 (UTC)

I would have to agree with Black Carrot there, for me, my confidence in my mathematical ability comes from my thorough understanding of the basics (that is HS level and some college level) mathematics, I'm so comfortable with these rules, that I have little trouble knowing whether or not they apply in a situation I've never encountered in a text. Memorization of the basic rules to a very exacting degree helps with that a lot. And curiosity is definitely the best motivator, I have learned more from my own curiosity than from those who have taught math to me. (Mainly do to the fact that I wasn't challenged in HS math class even AP) A math-wiki (talk) 23:49, 21 March 2008 (UTC)

I wouldn't recommend memorising the basic rules, it's much better to make sure you fully understand them and why they work. Once you've done that, you'll find you never forget them without having to make any actually effort to memorise them. (I frequently make mistakes in basic integration [as you may have seen lower down!], and that's because I didn't understand it when I first learnt it so tried to just memorise the methods, and ended up memorising them incorrectly and now get confused between my incorrect memory and my correct understanding and end up making mistakes - had I just taken the time to work through it and understand it, I wouldn't have the problems I do now.) --Tango (talk) 00:37, 19 March 2008 (UTC)

Just out of curiosity (which we have already established to be important), care to elaborate what is it that you have incorrectly memorized? -- Meni Rosenfeld (talk) 16:55, 19 March 2008 (UTC)

Well yes tango, of course you should understand it. I had intended to mean that you should not only know the rule but thoroughly understand it as well. I would argue one should never use a theorem they don't fully understand, and I try to observe that whenever reasonably possible. A math-wiki (talk) 23:25, 21 March 2008 (UTC)

Local Sidereal Time Conversion

How do I convert local sidereal time to something that is more commonly used? Is this actually a measure of time? I read the article about sidereal time and it appeared to be an angle measurement, a measure of place. So when someone tells me "do this action at 13:20" local sidereal time, is this something that makes sense, or is it just astrological mumbo jumbo? And if it makes sense, how do I convert it to Eastern Standard Time or Hawaii Standard Time or GMT or anything else that is more commonly used? —Preceding unsigned comment added by RastaNancy (talk • contribs) 20:45, 16 March 2008 (UTC)

Sidereal time is measure of time, but it's not easy to convert to normal time. Regular time is determined by watching the Sun move across the celestial sphere, sidereal time is determined by watching the stars move - the stars move only due to Earth's rotation on it's axis, the Sun moves because of that and because of Earth orbiting it. Therefore, a solar day differs from a sidereal day (by about 4 minutes). This means that to convert from solar to sidereal time you need to know the date in order to work out how much they differ by. The easiest way to convert would be to find a converter online, I'm sure there are plenty. --Tango (talk) 21:42, 16 March 2008 (UTC)

Possibility of solving this equation

I have the following equation:

${\it {\lambda _{i}}}=c\int _{0}^{L}\!\left|-1+R\theta \ '\left(s\right)\cos \left({\it {\phi _{i}}}\right)\right|{ds}$

I can choose any values for ${\it {\phi _{i}}}$, and get the corresponding values of $\lambda _{i}$ through an experiment. The aim is to find the function $\theta \left(s\right)$ as accurately as possible. I am wondering if it is possible at all to do this. I was trying to expand $\theta \left(s\right)$ in a Fourier series with n terms (and n undetermined coefficients) and then using n different values of ${\it {\phi _{i}}}$ to get a system of n equations. However, it turns out that all these equations are linearly dependent. Note that ${\it {\phi _{i}}}$ is not a function of s. Any ideas on how best we can find $\theta \left(s\right)$ will be highly appreciated. Regards, deeptrivia (talk) 21:40, 16 March 2008 (UTC)

Do you have any additional knowledge about $\theta '$? If it is known to be monotonic then I have an idea which might work in some cases. The essence is to start with $\phi \approx \pi /2$, for which the absolute value can be solved; then decrease $\phi$ until you see a deviation from normal. The rate of deviation should give you the ratio $\theta '/\theta ''$ at the x-intercept. This might then be solvable. -- Meni Rosenfeld (talk) 22:56, 16 March 2008 (UTC)

Thanks, Meni. Although I can choose any values for ${\it {\phi _{i}}}$, I can't choose too many. Practically, n is two or three, at most six. $\theta '$ is smooth but not monotonous. Also, I don't need to find $\theta$ exactly, but just as best as I can. How best do you think can I find it by finding $\lambda _{i}$ for, say, n wisely chosen values of ${\it {\phi _{i}}}$ ? Regards, deeptrivia (talk) 23:13, 16 March 2008 (UTC)

PS: I think in my case it is safe to assume that

${\it {\lambda _{i}}}=c\int _{0}^{L}\!\left(1-R\theta \ '\left(s\right)\cos \left({\it {\phi _{i}}}\right)\right){ds}$

Then, since R and ${\it {\phi _{i}}}$ are not functions of s, we can directly integrate and get:

${\it {\lambda _{i}}}=c\left(L-R\cos \left({\it {\phi _{i}}}\right)\left(\theta \left(L\right)-\theta \left(0\right)\right)\right)$

Consequently, it appears that no matter how we vary φ, all we can find about θ is the difference between its values at 0 and L. Now, if we remove the assumption I just made regarding the absolute value, we can still break the integral up into a countable number of parts that can be directly integrated, and using a similar logic, we can argue that we cannot know anything more than θ(L) - θ(0) by varying φ. Am I right? Thanks, deeptrivia (talk) 23:33, 16 March 2008 (UTC)

It's certainly very hard to deduce information about $\theta$ in the general case, but it's not theoretically impossible. If you break up the interval, in some segments you will have $\theta (a)-\theta (b)$ and in some $\theta (b)-\theta (a)$, and when you add them up they don't cancel. Since changing $\theta$ can change the integral, knowledge of the integral can at least eliminate some possibilities. -- Meni Rosenfeld (talk) 00:15, 17 March 2008 (UTC)

Thanks. Am I at least right for the case where there is no sign change, becaause $R\theta \ '\left(s\right)\cos \left({\it {\phi _{i}}}\right)\ll 1$ and no need to break the interval? Regards, deeptrivia (talk) 00:47, 17 March 2008 (UTC)

Yes, indeed. That is what I called the "normal" case, where the integral is reduced to a multiple of $\theta (L)-\theta (0)$. Deviations from this tells you something about $\theta$, but how to extract anything useful from it, especially with limited observations, is a mystery. -- Meni Rosenfeld (talk) 06:32, 17 March 2008 (UTC)

Am I missing something? If $\theta \left(s\right)$ solves the equation, then so does $\theta \left(s\right)+C$, so you can only determine solutions up to an additive constant.  --Lambiam 17:17, 17 March 2008 (UTC)

Certainly, but it appears that it's rather difficult, if not impossible, to even determine it that far. --Tango (talk) 17:28, 17 March 2008 (UTC)

March 17

is mathematics subjective?

So my understanding is that math isn't falsifiable, because it doesn't make physical predictions. Is it then subjective?

For example, whether "something has been proved", is that a question of viewpoint, a subjective thing, or is it objective? —Preceding unsigned comment added by 79.122.42.52 (talk) 01:41, 17 March 2008 (UTC)

I can't speak for all of mathematics but probability is subjective.

The probability of an event depends on what the observer/questioner knows. So two different observers can assign two different probability to the same event if they have different knowledge. 202.168.50.40 (talk) 02:26, 17 March 2008 (UTC)

Mathematics is objective in the sense that all proofs are derived from a set of axioms, so unlike the physical sciences where different experiments can lead to different results, whenever you start from the same axioms you'll get the same result for your proof. That said, the choice of axioms is somewhat subjective, since there is nothing to say that axiom set A is any "better" than axiom set B, although set A may result in something more familiar such as the natural numbers. Confusing Manifestation(Say hi!) 03:26, 17 March 2008 (UTC)

Sets of axioms are necessarily completely different, according to Gödel's incompleteness theorems. Because there are an infinite number of sets, subjectivity can not exist. Mac Davis (talk) 05:06, 17 March 2008 (UTC)

Mathematics is falsifiable, in that guesses can be checked. That's what falsifiability, as a doctrine, is for in the sciences - make sure your clever guesses aren't dead wrong. In mathematics, what's called proof serves roughly the same purpose. We can check our guesses. Whether we can check the system we use to check our guesses is open to philosophical debate, as the system of empirical validation used in the sciences is. Maybe it works, but maybe we live in god's big toe and he's playing games with us. That's a different level of fact-checking, and one that even falsifiable theories can't pass. So, to answer your question, mathematics is subjective to roughly the same extent that science is. Black Carrot (talk) 06:14, 17 March 2008 (UTC)

Mathematics differs in subjectivity from other sciences in a quantitative, not qualitative, way. Mathematics certainly has a higher standard of derivation - while in science, repeating an experiment several times passes as "proof" for its result, in mathematics we prefer to formulate our statements in an abstract way and derive them in a way we consider "logical". But don't be fooled into thinking that mathematics is completely rigorous, objective and devoid of human weaknesses. Not so much because of the choice of axioms (since it is understood that establishing a theorem doesn't mean that it's absolute, but rather that it follows from the specified axioms), but because of the choice of phrasing and logical inference. We have rules like "if we know A and we know A->B then we can deduce B", but there is nothing objectively "true" about this rule, other than that humanity's experience over the eons seems to support it. To this extent, mathematics relies on empirical evidence the same way as other sciences, the only difference being the amount of expected evidence. -- Meni Rosenfeld (talk) 06:49, 17 March 2008 (UTC)

The axioms of the logic in use are, to me, just as much part of the "choice of axioms" as the more mathematical axioms. It turns out that these are deeply intertwined anyway; a lovely result I saw once was that DeMorgan's Laws as logical axioms are equivalent to axiomatically stating that maximal ideals of commutative rings are necessarily prime -- one statement appears purely logical, the other most definitely in the realm of math not logic, yet in practice they are equivalent axioms. What I'm getting at with this little digression is that I don't really see the distinction you make between "choice of axioms" and axioms of logic that we assume. -- Leland McInnes (talk) 14:19, 17 March 2008 (UTC)

As I understand it - something that is 'subjective' depends on who is examining it. So a good maths proof should not be subjective.

Really though 'subjective' and 'objective' have no meaning here ie all mathematical knowledge should be 'objective'. Subjective examples of mathematical things include: imperfect models of phyical behaviour, best estimates of statisical behaviour/probabilities - eg things that are (educated) guesse.

I suggest reading about 'subjective and objective' and forming your own opinion. Picked these definitions 'at random' from the net:

Subjective: characteristic of or belonging to reality as perceived rather than as independent of mind : phenomenal

Objective: of, relating to, or being an object, phenomenon, or condition in the realm of sensible experience independent of individual thought and perceptible by all observers : having reality independent of the minds objective reality

Emphasis mine.87.102.13.144 (talk) 11:59, 17 March 2008 (UTC)

Answer - when something has been proved mathematically that should be objective, and not subjective knowledge. (It's work noting that peoples perception of what the words 'subjective' and 'objective' mean can be a bit subjective...)87.102.13.144 (talk) 14:07, 17 March 2008 (UTC)

Yes, something that has been proved should be objective, in the sense that we would want it to be. Unfortunately, it is not, per the responses above. -- Meni Rosenfeld (talk) 15:06, 17 March 2008 (UTC)

Our choice of what to try and prove is very subjective, but the actual proof is objective. We don't prove XYZ, we prove that our choice of axioms implies XYZ. That choice of axioms is very subjective, but the implication is entirely objective (give or take Godel, anyway - even then, I don't think it's subjective, it's just not wholly reliable). So, I guess the answer to the question "Is maths subjective?" is that maths is a subjective collection of objective facts (there is more to maths than a collection of facts, of course, but for the sake of this discussion I think that definition will do). --Tango (talk) 15:59, 17 March 2008 (UTC)

I disagree. First, nobody proves theorems in a completely formal language. People use natural language to describe their proofs, and hope that others will interpret it as they meant and agree that each of their implications is indeed valid. Sometimes it works, but there is still a lot variance in people's interpretation of arguments in a language such as used in articles. Second, even if the proof was written in a completely formal language, it still relies on the person's ability to validate each step. This validation is also composed of steps that can only be described in a natural language ("look at this line, see what's written here, compare it to what you see there..."). Especially for a long and complicated proof (like most proofs will be when written in a formal language), the possibility of human error is huge. It's also not unthinkable for the same person to make the same error repeatedly because of a certain mindset he is in. Thus different people will report different views on the correctness of a proof. In fact, when given a certain proof, there is no way for us to tell if it is truly valid (whatever that means) or is it just that every person who has ever examined it happened to mistakenly accept it. Thus I reiterate my claim that everything in mathematics is subjective, but to a much lesser extent than some other sciences. -- Meni Rosenfeld (talk) 16:15, 17 March 2008 (UTC)

What you say is certainly true, but I'm not sure if that's really a form of subjectivity - if it is, then nothing is truly objective and the question is moot. --Tango (talk) 17:16, 17 March 2008 (UTC)

Personally, I do believe there is an "objective truth", but that anything a human can ever discover is only a crude approximation of it, and thus subjective. The original question is not moot, as a priori one might think that modern mathematics has, at least in some parts, achieved the holy grail of true objectiveness - my argument is that this is not the case. It's probably as close as we'll be for a while, though. -- Meni Rosenfeld (talk) 18:38, 17 March 2008 (UTC)

Simultaneous Equation

how to solve this simultaneous equation 4x+y+8=x²+x-y=2 —Preceding unsigned comment added by 60.48.198.225 (talk) 07:32, 17 March 2008 (UTC)

You have the two equations $4x+y+8=2\;\!$ and $x^{2}+x-y=2\;\!$. There is a simple operation you can take to get rid of the y's and have an equation with only x. Can you see it? -- Meni Rosenfeld (talk) 07:35, 17 March 2008 (UTC)

Infinitely Many Factors

Consider the ring of formal power series over something like the integers. Is there any power series that can be split into infinitely many (non-unit) factors? That is, is there one where the process of pulling off factors will never stop with some number of irreducible factors? Black Carrot (talk) 10:02, 17 March 2008 (UTC)

Maybe I'm missing some subtlety in the question, but isn't

$\sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots =x\prod _{n=-\infty }^{\infty }\left(1-{\frac {x}{\pi n}}\right)$

an example of a power series with an infinite number of irreducible factors ? Gandalf61 (talk) 10:44, 17 March 2008 (UTC)

Another example (unless I, too, misunderstand the question) is $\prod _{n=0}^{\infty }\left(1+2^{-n}x\right)=\sum _{n=0}^{\infty }{\frac {2^{n}}{\prod _{k=1}^{n}(2^{k}-1)}}x^{n}$. -- Meni Rosenfeld (talk) 10:50, 17 March 2008 (UTC)

Those are units: (1-ax)(1+ax+a^2*x^2+a^3*x^3...) = 1. You can always pull off infinitely many unit factors, even from something irreducible. Black Carrot (talk) 11:12, 17 March 2008 (UTC)

Oh, I see. No idea then. -- Meni Rosenfeld (talk) 11:22, 17 March 2008 (UTC)

Let's see if I have got this straight. You want an infinite number of factors each of which has, say, integer coefficients, is irreducible and is not a unit when considered as a formal power series over the integers ? Let's take out the largest power of x that we can, so we have an initial factor of x^k for some k (possibly 0). Doesn't this mean that the constant term of every other factor cannot be 0 (otherwise it is reducible) or +/-1 (otherwise it is a unit when considered as a formal power series) ? So you have x^k times an infinite number of factors each of which has constant term with absolute value greater than or equal to 2 - which seems to imply that the coefficient of the lowest power of x in the product is not finite, so the product is not a power series. Gandalf61 (talk) 11:29, 17 March 2008 (UTC)

What is meant by "non unit factor"?87.102.13.144 (talk) 11:47, 17 March 2008 (UTC)

In this context, a unit is something you can divide by without leaving the number system. For instance, in the whole numbers 1 is a unit, but anything bigger isn't. That's why 1 isn't considered prime - every number would have infinitely many prime factorizations, eg 5 = 5*1 = 5*1*1 = 5*1*1*1... You're right Gandalf, I hadn't thought of looking at it like that. So, does that mean that power series over the rationals also have finitely many factors? Since everything with a constant term would be a unit, and everything without could be divided by x. Black Carrot (talk) 12:04, 17 March 2008 (UTC)

Meni, should that be 2^n at the top of your last formula? Black Carrot (talk) 12:09, 17 March 2008 (UTC)

Yes, indeed. Fixed. -- Meni Rosenfeld (talk) 14:55, 17 March 2008 (UTC)

Was the first example given not an example? eg a power series over rational fractions expressed, but made of factors which are irrational (contain 1/pi)?87.102.13.144 (talk) 14:12, 17 March 2008 (UTC)

The original question needs to be made more precise. What do you mean by "something like the integers"? For example, are algebraic integers something like the integers? And wat about the complex numbers? Should the factors be over the same ring as the power series? Can the factors again be formal power series, or should they be polynomials?  --Lambiam 14:49, 17 March 2008 (UTC)

Gandalf has solved one form of the question: I believe his proof shows that if R is a Noetherian integral domain, then any element of R[[X]] can be expressed as a product of finitely many irreducibles in R[[X]]. Algebraist 16:09, 17 March 2008 (UTC)

math: polynomial as the difference of two squares

How do you determine if a polynomial is the difference of two squares? —Preceding unsigned comment added by 72.94.230.129 (talk) 14:08, 17 March 2008 (UTC)

Assuming that by "square" you mean the square of a polynomial, and that the coefficients are from a number system that is closed under division by 2, such as the rationals and the reals, any polynomial P can be rewritten as:

$\left({\frac {P+1}{2}}\right)^{2}-\left({\frac {P-1}{2}}\right)^{2}.$

 --Lambiam 14:33, 17 March 2008 (UTC)

I'm not sure that's what the OP means. We studied polynomials last year in algebra I. We learned the difference of two squares as $(x+S)(x-S)\,\!$, in which S is any number. This then takes the form $x^{2}-S^{2}\,\!$. So, to determine if a polynomial is the difference of two squares, just look for two square terms (such as $x^{2}-64$). The solution to a polynomial of the form $(x+S)(x-S)$ is $x=\pm {S}$. So the solution to $x^{2}-64$ would be $x=\pm {8}$.

Disclaimer: I learned these things last year so they are subject to possible error.

Zrs 12 (talk) 19:48, 17 March 2008 (UTC)

That's right, but the principle can be used for more complicated polynomials, as well. For example $x^{2}-y^{2}+8x+4y+12=x^{2}+8x+16-y^{2}+4y-4=(x+4)^{2}-(y-2)^{2}=(x+4+y-2)(x+4-y+2)=(x+y+2)(x-y+6)$. If you want everything to have integer coefficients, then I'm not sure there is a general algorithm to find out if you can write a polynomial as the difference of two squares, you just have to fiddle around with it and see what happens (you can try completing the square as a starting point). --Tango (talk) 20:08, 17 March 2008 (UTC)

We also have an article on this. See difference of two squares. Zrs 12 (talk) 20:06, 17 March 2008 (UTC)

Tango: A polynomial (over Z, say) is the difference of two squares iff it can be factorised with the sum of the (two) factors being divisible by 2 (i.e. corresponding coefficients in the factors have to be of the same parity). Thus a poly is a difference of two squares iff its reduction mod 2 is a square. Algebraist 23:25, 17 March 2008 (UTC)

Or rather not. We get an algorithm, anyway. There's an algorithm that factors a polynomial over Z into its irreducible factors, so we can just test all possible factorizations. I've no idea if this can be improved on. Algebraist 23:28, 17 March 2008 (UTC)

need a more indepth determination

Bold textHow do you determine if a polynomial is the difference of two roots? Thats how the question was brought to me.I don't quite get it either. —Preceding unsigned comment added by 72.94.241.142 (talk) 19:36, 17 March 2008 (UTC)

The difference of two roots? Well, assuming you mean square root, any number is a square root because it is able to be squared. Just go down a number line and mulitiply each term by itself. For example $S$ is the square root of $S^{2}$; 4 is the square root of 16. However, every number is not a square because not every number has an integer square root. I think the answer to your previous question above may be what you seek. Zrs 12 (talk) 19:55, 17 March 2008 (UTC)

Solve for y: x = y - 1/y

How can you solve the equation $x=y-{\frac {1}{y}}$ for y in terms of x? I know that the solution is $y={\frac {x\pm {\sqrt {x^{2}+4}}}{2}}$, and this works when you substitute it back into the original equation, but how can this be found analytically? (I used a calculator). --BrainInAVat (talk) 20:59, 17 March 2008 (UTC)

• Try multiplying both sides by $y$... you should then have a quadratic in terms of $y$ which can be solved using the "usual method". --Kinu ^t/_c 21:14, 17 March 2008 (UTC)
  • I'm at a loss, sorry. You get $xy=y^{2}-1$, but then what?--BrainInAVat (talk) 01:29, 18 March 2008 (UTC)

Take it all over to one side and you have a quadratic in y (just treat x like a number), you can then solve it using the quadratic formula and you'll get the answer you gave above. --Tango (talk) 01:53, 18 March 2008 (UTC)

If you are not familiar with it, you can take a look at out Quadratic equation article. -- Meni Rosenfeld (talk) 07:19, 18 March 2008 (UTC)

This can be a little confusing the first time you see it - here:

x=y - 1/y

xy=y² - 1 (equation 1)

y² - xy - 1 = 0

Now complete the square

(y-x/2)² - x²/4 - 1 = 0

As you should be able to see multilpling by y gives a quadratic in y (equation 1)

The solve the quadratic by completing the square - you can see that x is a parameter of the quadratic. Hopefully from the last equation you should be able to get the result.

Once you've got it, keep an eye out for equations of this type as it's a useful thing to remember.87.102.74.53 (talk) 11:35, 18 March 2008 (UTC)

March 18

Proof of Chain Rule

Let

$y=f(u)$ (1)

$u=g(x)$ (2)

where $f(u)$ and $g(x)$ are both differentiable functions. Then

${\frac {dy}{dx}}$ (3)

$=\lim _{\Delta x\rightarrow 0}{\frac {f(g(x+\Delta x))-f(g(x))}{\Delta x}}$ (4)

$=\lim _{\Delta x\rightarrow 0}\left[{\frac {f(g(x+\Delta x))-f(g(x))}{g(x+\Delta x)-g(x)}}\cdot {\frac {g(x+\Delta x)-g(x)}{\Delta x}}\right]$ (5)

$=\lim _{\Delta x\rightarrow 0}{\frac {f(g(x+\Delta x))-f(g(x))}{g(x+\Delta x)-g(x)}}\cdot \lim _{\Delta x\rightarrow 0}{\frac {g(x+\Delta x)-g(x)}{\Delta x}}$ (6)

Treat the arrow $\rightarrow$ as equal sign $=$. We can do the same operation on both sides of the arrow without changing the relationship

$\Delta x\rightarrow 0$ (7)

$x+\Delta x\rightarrow x$ (8)

Function $g(x)$ is continuous since it is differentiable. Apply $g(\cdot )$ to both sides of (8)

$g(x+\Delta x)\rightarrow g(x)$ (9)

$g(x+\Delta x)-g(x)\rightarrow 0$ (10)

Let

$\Delta u=g(x+\Delta x)-g(x)$ (11)

Replace (11) into (10)

$\Delta u\rightarrow 0$ (12)

Therefore

$\lim _{\Delta x\rightarrow 0}\cdot$  implies $\lim _{\Delta u\rightarrow 0}\cdot$ (13)

Replace (2) into (11)

$g(x+\Delta x)=u+\Delta u$ (14)

Replace (2), (11), (13) and (14) into (6)

${\frac {dy}{dx}}=\lim _{\Delta u\rightarrow 0}{\frac {f(u+\Delta u)-f(u)}{\Delta u}}\cdot \lim _{\Delta x\rightarrow 0}{\frac {g(x+\Delta x)-g(x)}{\Delta x}}={\frac {dy}{du}}{\frac {du}{dx}}$ (15)

Q.E.D.

Is the proof of chain rule above correct and rigorous? - Justin545 (talk) 06:25, 18 March 2008 (UTC)

There are some questionable details. First, if we want a proof we can consider "rigorous", we would want to avoid treating functions as quantities (e.g., u instead of $g(x)$) and using Leibniz notation (${\tfrac {du}{dx}}$). So as a first step you should try formulating the proof without using u or y, only f, g and their composition $h=f\circ g$ (equivalently, $h(x)=f(g(x))$. Second, the limit notation, $\lim _{x\to x_{0}}f(x)=L$, is one unit. You shouldn't take out the $x\to x_{0}$ and treat it as something that stands on its own. This would be acceptable for a handwaving proof, but not for a rigorous one. -- Meni Rosenfeld (talk) 07:37, 18 March 2008 (UTC)

>> "the limit notation, $\lim _{x\to x_{0}}f(x)=L$, is one unit. You shouldn't take out the $x\to x_{0}$ and treat it as something that stands on its own."

I think you mean the result of (13) is incorrect or not rigorous. Does it mean the whole proof should be re-derived in a completely different way or we can somehow fix the problem so that we don't have to re-derive the whole proof? If (13) is not rigorous, is there any example which opposes it? Thanks! - Justin545 (talk) 09:00, 18 March 2008 (UTC)

(13) and the derivations that lead to it are "not even wrong" in the sense that in the standard framework of calculus they are pretty much meaningless - if you look at the standard rigorous definitions of limits, you will see that they do not allow a function to be used as a variable. It is "correct" in the sense that intuitively, the limit of a function "when" the variable approaches some value is equal to the limit when some othee function approaches its appropriate limit value. However, this "when" business lacks a rigorous interpretation and is haunted by Bishop Berkeley's ghosts.

I have thought about how one might amend the proof, and realized that you also have a mistake much earlier. Step (5), dividing and multiplying by $g(x+\Delta x)-g(x)$, is only valid if $g(x+\Delta x)-g(x)\neq 0$, but there is no reason to assume that should be the case. Take, for example

$g(x)=\left\{{\begin{array}{ll}x^{2}\sin {\tfrac {1}{x}}&x\neq 0\\0&x=0\end{array}}\right.$

- a perfectly differentiable function at 0, and yet $g(x)=0=g(0)$ infinitely many times in any neighborhood of 0. Thus your proof will not work for it. Those kinds of pathological counterexamples are one of the things that separates rigorous proofs from not-so-rigorous ones. -- Meni Rosenfeld (talk) 10:58, 18 March 2008 (UTC)

>> "Step (5), dividing and multiplying by $g(x+\Delta x)-g(x)$, is only valid if $g(x+\Delta x)-g(x)\neq 0$, but there is no reason to assume that should be the case. Take, for example..."

I think $g(x+\Delta x)-g(x)$ will never be zero since $\Delta x$ "is not zero", $\Delta x$ is just a value that "very close to zero". Thus, $g(x+\Delta x)-g(x)$ will only close to zero but $g(x+\Delta x)-g(x)$ will not be zero, and I believe the step (5) would be still correct. As for your example, we may first need to evaluate

$\lim _{\Delta x\rightarrow 0}g(0+\Delta x)$ (16)

$=\lim _{\Delta x\rightarrow 0}(0+\Delta x)^{2}\sin {\frac {1}{0+\Delta x}}$ (17)

$=\lim _{\Delta x\rightarrow 0}{\Delta x}^{2}\sin {\frac {1}{\Delta x}}$ (18)

$=\lim _{\Delta x\rightarrow 0}{\Delta x}^{2}\cdot \lim _{\Delta x\rightarrow 0}\sin {\frac {1}{\Delta x}}$ (19)

But what will $\lim _{\Delta x\rightarrow 0}\sin {\frac {1}{\Delta x}}$ evalute to? I'm not sure... - Justin545 (talk) 01:41, 19 March 2008 (UTC)

You've made two mistakes here. First, $g(x+\Delta x)-g(x)$ can be zero for arbitrarily small values of $\Delta x$. That's what Meni's example shows. Your (18)=(19) is also mistaken: it would be valid if both limits in (19) existed, but as it happens the second one doesn't. Btw, your error at step (5) is a reasonably common one: IIRC, it occurs in the first few editions of G H Hardy's A Course of Pure Mathematics. Though there are other ways round it, perhaps the best is to avoid division at all in the proof. This has the advantage that your proof immediately generalises to the multi-dimensional case. Algebraist 02:36, 19 March 2008 (UTC)

>> "First, $g(x+\Delta x)-g(x)$ can be zero for arbitrarily small values of $\Delta x$. That's what Meni's example shows."

Meni's example is not so obvious to me why $g(x+\Delta x)-g(x)=0$ where

$g(x)=\left\{{\begin{array}{ll}x^{2}\sin {\tfrac {1}{x}}&x\neq 0\\0&x=0\end{array}}\right.$ (20)

Could you provide more explanation for it? Or could you tell what theorem supports that $g(x+\Delta x)-g(x)$ could be exactly zero?

>> "perhaps the best is to avoid division at all in the proof."

Division could be avoided at all, but it is "intuitive" since the definition of derivative involves division. Besides, even this proof involves division I think. If it does involve division, the proof would be considered non-rigorous. - Justin545 (talk) 03:08, 19 March 2008 (UTC)

>> "(13) and the derivations that lead to it are "not even wrong" in the sense that in the standard framework of calculus they are pretty much meaningless - if you look at the standard rigorous definitions of limits, you will see that they do not allow a function to be used as a variable."

I'm afraid I don't get it that "rigorous definitions of limits do not allow a function to be used as a variable" and why the derivations lead to (13) is meaningless. - Justin545 (talk) 03:37, 19 March 2008 (UTC)

A better question is how are they not meaningless. Where in your textbook did anyone mention taking the $\Delta x\to 0$ notation, treating it as a formula on its own, and doing manipulations on it? -- Meni Rosenfeld (talk) 16:48, 19 March 2008 (UTC)

This proof is really from my textbook except the steps from (7) to (14) are missing. The missing steps is my creation since I have no idea how does step (6) become step (15). I want to know, in detail, how does step (6) become step (15) so I added those steps and make discussion here to see if it's correct or not. - Justin545 (talk) 05:30, 20 March 2008 (UTC)

In this case, the proof in your book is wrong (that happens too). Step 5 cannot be justified without more assumptions on g. Your steps 7-12 describe intuitively correct ideas but are far from being rigorous. If g is "ordinary" enough for step 5 to hold, it is possible to justify the leap from (6) to (15), but if you want it to be rigorous you need to rely only on the definition of limits, not on your intuitive ideas of what they mean. -- Meni Rosenfeld (talk) 12:03, 20 March 2008 (UTC)

If you want a similar proof that really works, one way would be to apply the mean value theorem to f at (4). This allows you to replace $f(g(x+h))-f(g(x))$ 163.1.148.158 (talk) 12:54, 18 March 2008 (UTC)

The mean value theorem I found is

$f'(c)={\frac {f(b)-f(a)}{b-a}}$

where $c\in (a,b)$. But I have no idea how to apply it to f at (4) and why it's needed to replace $f(g(x+h))-f(g(x))$? Thanks! - Justin545 (talk) 02:38, 19 March 2008 (UTC)

To the OP: it is not necessary to avoid division to make the proof rigorous, but it is one way of doing it. I meant division specifically by values of the domain or codomain of f and g (since these are the things that become vectors when you generalise), but I see I failed to say it. Apologies. The definition of the derivative need not involve such division (the one lectured to me didn't, for example), and one could argue that it shouldn't. Not sure if one would be right, mind. To your specific question, Meni's function is zero whenever x is 1/(nπ) (n a non-zero integer). Thus we have g(x)=0 for arbitrarily small x. Algebraist 03:34, 19 March 2008 (UTC)

>> "The definition of the derivative need not involve such division (the one lectured to me didn't, for example), and one could argue that it shouldn't."

The familiar definition of derivative is

${\frac {dy}{dx}}=f'(x)=\lim _{\Delta x\rightarrow 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}$ (21)

It seems you was were saying that (21) is not a "rigorous" definition. It sounds pretty odd to me. I thought (21) is the only way of defining derivative. There are many lemmas or theorems about derivative in my textbook are originated from (21). It's not easy to imagine there other there are other definition without division. - Justin545 (talk) 05:13, 19 March 2008 (UTC)

No, that's not what he was saying. He said that you can define the derivative without division, not that you should. Definition (21) (at least the $f'(x)=\lim _{\Delta x\rightarrow 0}{\frac {f(x+\Delta x)-f(x)}{\Delta x}}$ part) is rigorous and is indeed the standard definition. There is nothing wrong with division, except for division by zero. The main flaw in your proof is dividing by $g(x+\Delta x)-g(x)$ which may be zero. Just because $\Delta x\neq 0$ doesn't mean that $g(x+\Delta x)\neq g(x)$. This is just common sense, you don't need my complicated example for that. -- Meni Rosenfeld (talk) 16:48, 19 March 2008 (UTC)

>> "To your specific question, Meni's function is zero whenever x is 1/(nπ) (n a non-zero integer). Thus we have g(x)=0 for arbitrarily small x."

I'm afraid I'm not able to proof (20) is zero when $x\in \left\{{\frac {1}{n\pi }}{\Bigg |}n\in \mathbb {Z} \land n\neq 0\right\}$. But I think $g(x+\Delta x)-g(x)$ will be zero when $g(x)=a$ where $a$ is any fixed constant. (Edit: which means I was ridiculously wrong. Apologies.) - Justin545 (talk) 05:42, 19 March 2008 (UTC)

x²sin(1/x) is zero whenever sin(1/x) is zero, which happens whenever 1/x is a multiple of pi, which happens whenever x = 1/npi for some integer n. You know, you're not really all that wrong. You have the right idea, you just don't have the tools to implement it. Here's roughly how my analysis textbook solves the problem. First, you define a new function h(y). I'll skip the details about intervals and mappings, and just say that it's focused on f and ignoring g, and assumes some interesting value c has been chosen. Let h(y) = (f(y)-f(g(c)))/(y-g(c)) if y does not equal g(c), and let h(y) = f'(y) if y=g(c). All that should be possible by assumption. Since g is differentiable at c, g is continuous at c, so h of g is continuous at c, so lim x->c (hog)(x)=h(g(c))=f'(g(c)). By the definition of h, f(y)-f(g(c))=h(y)(y-g(c)) for all y, so ((fog)(x)-(fog)(c)) = (hog(x))(g(x)-g(c)), so for x not equal to c we have ((fog)(x)-(fog)(c))/(x-c) = (hog(x))(g(x)-g(c))/(x-c). Taking the limit of both sides as x->c, then (fog)'(c)=lim x->c ((fog)(x)-(fog)(c))/(x-c) = (lim x->c hog(x))(lim x->c (g(x)-g(c))/(x-c)) = f'(g(c))g'(c). Black Carrot (talk) 06:36, 19 March 2008 (UTC)

>> "x²sin(1/x) is zero whenever sin(1/x) is zero, which happens whenever 1/x is a multiple of pi, which happens whenever x = 1/npi for some integer n."

Thanks! Now I understand it.

>> "You know, you're not really all that wrong. You have the right idea, ..."

Excuse my rewiring of your response for readability:

Here's roughly how my analysis textbook solves the problem. First, you define a new function $h(y)$. I'll skip the details about intervals and mappings, and just say that it's focused on $f$ and ignoring $g$, and assumes some interesting value $c$ has been chosen. Let

$h(y)={\begin{cases}{\frac {f(y)-f(g(c))}{y-g(c)}},&y\neq g(c)\\f'(y),&y=g(c)\end{cases}}$

All that should be possible by assumption. Since $g$ is differentiable at $c$, $g$ is continuous at $c$, so $h$ of $g$ is continuous at $c$, so

$\lim _{x\rightarrow c}h(g(x))=h(g(c))=f'(g(c))$.

By the definition of $h$,

$f(y)-f(g(c))=h(y)[y-g(c)]$, $\forall y\neq g(c)$,

so

$f(g(x))-f(g(c))=h(g(x))[g(x)-g(c)]$,

so for $x$ not equal to $c$ we have

${\frac {f(g(x))-f(g(c))}{x-c}}={\frac {h(g(x))[g(x)-g(c)]}{x-c}}$.

Taking the limit of both sides as $x\rightarrow c$, then

$(f\circ g)'(c)=\lim _{x\rightarrow c}{\frac {f(g(x))-f(g(c))}{x-c}}=\left[\lim _{x\rightarrow c}h(g(x))\right]\left[\lim _{x\rightarrow c}{\frac {g(x)-g(c)}{x-c}}\right]=f'(g(c))g'(c)$.

Did I misunderstand your response? Thanks! - Justin545 (talk) 09:01, 19 March 2008 (UTC)

After "By the definition of h", it should be for all y. If y=g(c), both sides are equal to zero, and the equality still holds. That one line is pretty much the goal of the whole thing, finding a way to get that conclusion without dividing by zero anywhere. Black Carrot (talk) 16:02, 19 March 2008 (UTC)

But I think it should be

${\begin{cases}f(y)-f(g(c))=h(y)[y-g(c)]&\forall y\neq g(c)\\f'(y)=h(y)&\forall y=g(c)\end{cases}}$ (22)

by definition of $h$.

By the way, I think derivatives of composition functions should be able to rewritten rewrite to be rewritten in Leibniz notation as below

$(f\circ g)'(c)={\frac {df(g(c))}{dc}}={\frac {d}{dc}}f(g(c))$ (23)

$f'(g(c))={\frac {df(g(c))}{dg(c)}}={\frac {d}{dg(c)}}f(g(c))$ (24)

Justin545 (talk) 07:02, 20 March 2008 (UTC)

Calculating inflation by consumer price index statistics

Hello,

I'm currently working on HMAS Melbourne (R21), which is currently at Featured Article Candidates. One of the reviewers has requested that more recent financial figures be provided for the various sums of money (listed at User talk:Saberwyn/HMAS Melbourne (R21)) mentioned in the history of the ship. I've been pointed to Australian consumer price index statistics (available at [1] from 1969 to 2007, and [2] from 1949 to 1997 - Both open directly as Excel spreadsheets).

I've looked at them, and realise I don't have the first idea as to using these statistics to convert, for example, a AU$1.4 million figure from 1985 to a AU$ figure in 2007. Any help, at least a point in the right direction to an online tutorial or something, would be muchly appreciated. -- saberwyn 10:32, 18 March 2008 (UTC)

Well, the table says that AU$69.7 in 06/1985 are equivalent to AU$157.5 in 06/2007 in terms of overall purchasing power, so 1.4m in 1985 AU$ would roughly amount to 1.4m*157.5/69.7 = 3.2m in 2007 AU$. Bikasuishin (talk) 12:21, 18 March 2008 (UTC)

You calculate percentage of inflation from Year 1 to Year 2 by using the formula $Inflation={\frac {(CPI\ in\ Year\ 2)-(CPI\ in\ Year\ 1)}{(CPI\ in\ Year\ 1)}}\times 100$

For example, according to the table CPI in March 1988 was 87 and in March 1998 it was 120.3. Thus, by using the above formula, we find that prices in this period have inflated by ${\frac {120.3-87}{87}}\times 100\approx 38.28\%$. Therefore, AU$ 1,000 in March 1988 was equivalent to AU$ $1000+1000\times 0.3828=1382.8$ in March 1998. Cheers,  ARTYOM  17:14, 18 March 2008 (UTC)

Thank you for your assistance, it has been a great help!! -- saberwyn 00:12, 19 March 2008 (UTC)

Area Between Curves

I'm having a little trouble with an area between curves problem. The functions are f(x)=x^3-x and g(x)=3x. What I'm trying to figure out is if I should add the two areas (the one bellow and the one above the curve) or if i should subtract the area from underneath the X axis from the one above it. Thanks RedStateV (talk) 14:28, 18 March 2008 (UTC)

The distance between two points is the absolute value of one minus the other. So long as you keep your signs straight, it doesn't matter where either curve is in relation to the x-axis or to the other. — Lomn 14:42, 18 March 2008 (UTC)

So should I be getting a answer of 8 then? RedStateV (talk) 14:52, 18 March 2008 (UTC)

That's not what I get, but either one of us could have got the arithmetic wrong. I've done it twice and got the same answer both times, do you want to try doing it again and see if you still get 8? --Tango (talk) 15:17, 18 March 2008 (UTC)

My silicon master confirms that the answer is indeed 8. -- Meni Rosenfeld (talk) 15:35, 18 March 2008 (UTC)

Yeah, I can't integrate, it's definitely 8... sorry. --Tango (talk) 17:31, 18 March 2008 (UTC)

Are you going from x=-2 to x=2 ? If so, you can take advantage of the symmetry of the curves - take the area between the curves from x=0 to x=2 and then double it. I also get 8. Gandalf61 (talk) 15:49, 18 March 2008 (UTC)

Ok, so the value of the area bellow the X axis is not negative then. Good, Thanks. RedStateV (talk) 16:03, 18 March 2008 (UTC)

Area is, in fact, always positive (at least in the usual settings). What would negative area (or length, or volume) even mean? — Lomn 16:11, 18 March 2008 (UTC)

To answer this question we need to go back to the source. Lengths, areas etc. are measures given to sets in, say, an Euclidean space. To this extent they cannot be negative. But we can generalize to a sort of multiset in which elements can appear several times or even a negative number of times, and define measures on these in the natural way. Then the area of a set which only has negative points will be negative. In analytic geometry, there are formulae for area that can give negative results if an absolute value is not taken explicitly, such as $S={\frac {1}{2}}\det {\begin{pmatrix}x_{A}&x_{B}&x_{C}\\y_{A}&y_{B}&y_{C}\\1&1&1\end{pmatrix}}$ for a triangle. The sign is traditionally taken to represent the orientation, with only the absolute value representing the "true" area. But the signed number can be taken to be the area, if a negatively oriented triangle is seen as containing its points negatively. The same goes for calculating length by simply subtracting two coordinates, or area under a function by evaluating its integral. -- Meni Rosenfeld (talk) 17:21, 18 March 2008 (UTC)

If in doubt - plot the graph. Always check for places where the line crosses y=0 because if you don't notice this can mess up your areas if you are using integration.87.102.47.176 (talk) 16:56, 18 March 2008 (UTC)

If you're calculating the area between a curve and the x-axis, then the curve crossing the axis is significant, however this question is about the area between 2 curves, so it's where the curves cross each other that's significant, not where they cross the x-axis. --Tango (talk) 15:45, 19 March 2008 (UTC)

Rational function conjecture

If $f:\mathbb {R} \rightarrow \mathbb {R}$ is an analytic function such that $x\in \mathbb {Q}$ implies $f(x)\in \mathbb {Q}$, must $f$ be a rational function? I expect it's false, so what's a counterexample? —Keenan Pepper 19:02, 18 March 2008 (UTC)

How about f(x)=2^(lg(x+1))?, where lg is log base 2. GromXXVII (talk) 20:08, 18 March 2008 (UTC)

Silly me...bad domain. GromXXVII (talk) 20:10, 18 March 2008 (UTC)

And it equals x+1, which is rational. Nice try, though! I can't think of a counterexample. It wouldn't surprise me if it were true, but I can't prove it. --Tango (talk) 20:13, 18 March 2008 (UTC)

Any ideas? Or search keywords? —Keenan Pepper 15:26, 19 March 2008 (UTC)

I believe it's true, based on the following thinking. The question here is really, can we construct an irrational number with a finite number of rational parts? If we can then your conjecture is false, but if not then it is true. And since by definition an irrational number cannot be constructed by a finite number of rational components the conjecture is true. A math-wiki (talk) 03:15, 20 March 2008 (UTC)

I don't follow this at all. Are you sure you're talking about my question and not its converse? It's easy to show that every rational function whose denominator has no real zeros is analytic on the real line and maps rationals to rationals. My question was about going the other direction: does being analytic and mapping rationals to rationals guarantee that something is a rational function? —Keenan Pepper 05:19, 20 March 2008 (UTC)

Lovely question. I don't know the answer. Here's one approach that occurs to me: Let's try and figure out whether, if f is represented by a power series that converges everywhere on R, and f takes rationals to rationals, then f must be a polynomial. If that can be proved then it doesn't finish off the conjecture but might be a large step towards a general proof; a counterexample disposes of the question.

So consider all sequences of reals $\langle a_{0},a_{1},\ldots \rangle \in \mathbb {R} ^{\omega }$ such that the corresponding power series converges everywhere (if "converges everywhere" is too messy, we could simplify it a bit by putting bounds on the $a_{i}$ that guarantee the power series converges via the ratio test or something). That's a nice topological space (a Polish space) in which we might be able to work some magic. Now enumerate the rationals as $q_{0},q_{1},\ldots$. For what values of $\langle a_{0},a_{1},\ldots \rangle$ is $a_{0}+a_{1}q_{0}+a_{2}q_{0}^{2}+\ldots$ a rational number? It'll be a reasonably simple set, maybe F-sigma or something (have to think about this) -- can we find a big chunk of this set that's compact? Because then we pull the same trick with $q_{1}\,$ and try to get a nonempty compact set that works for both $q_{0}\,$ and $q_{1}\,$ and keep going. If we can do this forever then we have a descending sequence of compact sets, so it has a nonempty intersection, and we've got out counterexample. But there's a lot of ifs; I don't know whether it can be made to work. --Trovatore (talk) 03:42, 20 March 2008 (UTC)

Oh, I didn't make explicit why a counterexample to my version answers the original question. That's because the power series is assumed to converge everywhere. That means it extends to the whole complex plane giving an entire function -- no poles. Therefore, if it were a rational function, it would have to be a polynomial. --Trovatore (talk) 03:50, 20 March 2008 (UTC)

That's true if the function is analytic over the whole complex plane, but not if it only has to be analytic on the real line. Counterexample: $f(x)=1/(x^{2}+1)$ (it has two poles, but they're both imaginary). —Keenan Pepper 05:19, 20 March 2008 (UTC)

But that wasn't the hypothesis -- the hypothesis is that the function has a power series that converges everywhere on the real line. That isn't true for 1/(x^2+1). You have to use different power series with different centers and glue them together. --Trovatore (talk) 07:17, 20 March 2008 (UTC)

Ah yes, you're right. —Keenan Pepper 13:24, 20 March 2008 (UTC)

Analytic *doesn't* mean that there's a power series convergent everywhere --- see our article, where $1/(1+x^{2})$ is given as an example. Analicity is a local property: there must be a power series representation for f valid in a neighbourhood of every point.79.67.177.9 (talk) 10:30, 20 March 2008 (UTC)

I think Keenan requires the function ƒ to be meromorphic but not necessarily holomorphic. Gandalf61 (talk) 11:05, 20 March 2008 (UTC)

Trovatore asked a different question than Keenan. Keenan's hypothesis is that the function is analytic (by which I think he means holomorphic on the real line, but not necessarily elsewhere). Trovatore's hypothesis is that the function has a power series which converges everywhere. The two questions are related in that a counterexample to Trovatore's is also a counterexample to Keenan's. Read his argument more carefully to understand why. -- Meni Rosenfeld (talk) 12:11, 20 March 2008 (UTC)

One idea on different lines: for a counterexample it's enough to show there exist analytic f that map rationals to rationals and grow faster than any polynomial 163.1.148.158 (talk) 11:24, 20 March 2008 (UTC)

I'll clarify my argument a bit Keenan, your right in some ways, I'm talking about the converse of your hypothesis. But what I was pointing to was to try to show that if your hypothesis was false, then the converse is also false which as we know, would be a contradiction. The key to my line of thought is that f is really an operator on x and a rational operator will have $f:\mathbb {Q} \rightarrow \mathbb {Q}$ but an irrational operator must for some rational input, give a irrational output or for some irrational input, give a rational output for some x. (e.g. sin and it's inverse arcsin) Therefore f is a rational operator. A math-wiki (talk) 22:39, 20 March 2008 (UTC)

Both a statement and its converse can be false. Are you confusing converse with negation? --Tango (talk) 22:46, 20 March 2008 (UTC)

Tango, I think u slightly missread my last post, I was suggesting he try to show that IF his hypothesis is false, THEN the converse (which is true) would have to be false also. A math-wiki (talk) 23:03, 20 March 2008 (UTC)

March 19

Angle bisector proof

In triangle ABC, draw angle bisectors AD and CE, where D is on BC and E is on AB. If angle B is 60 degrees, show that AC=CD+AE.

I've figure out that if the intersection of AD and CE is F, then <CFA and <EFD are 120 degrees and <DFC and <EFA are 60 degrees, but I'm not sure what's next. Thanks, 76.248.244.196 (talk) 01:06, 19 March 2008 (UTC)

Your four angles around F are all the angles there are around F and you have them adding up to 60+60+30+30 = 180. Surely this should be 360, so some of your figuring has gone astray. Present what working you have done so far and maybe it will become clearer... -- SGBailey (talk) 07:24, 19 March 2008 (UTC)

I've correct my numbers above, but I'm still not sure where to go next. 76.248.244.196 (talk) 01:36, 20 March 2008 (UTC)

OK, so you know ALL the angles in the diagram (<ABC=60, <BAD=a, <BCA=c=120-a, <BAD=a/2, <BDA=120-a/2, <BEC=120-c/2=60+a/2 <EXD=360-60-(120-a/2)-(60+a/2)=120). Now you can use the law of cosines. -- SGBailey (talk) 10:25, 20 March 2008 (UTC)

Algebra question

Given any three natural numbers, show that there are two of them, a and b, that a^3b-b^3a is divisible by thirty.

I have no clue how to prove it, please help. Thanks, 76.248.244.196 (talk) 01:09, 19 March 2008 (UTC)

Note that "divisible by thirty" is equivalent to "divisible by two, three, and five". I would first factor the polynomial as $a^{3}b-b^{3}a=ab(a+b)(a-b)$. If either a or b is even, then this product is even, and if they are both odd, then (a+b) is even so the product is still even. Therefore $a^{3}b-b^{3}a$ is even for every pair of natural numbers. A similar argument can be used to show that it is also a multiple of three (work modulo three). So the polynomials for all three of the pairs are multiples of six, and you can use the pigeonhole principle to show that at the polynomial for at least one of the pairs is also a multiple of five (work modulo guess-what). —Keenan Pepper 01:29, 19 March 2008 (UTC)

I don't understand the modulo part. Where do I go with this? 76.248.244.196 (talk) 01:52, 20 March 2008 (UTC)

If you haven't learned modular arithmetic yet, it'll be quite challenging. It's really not that complicated, though. "Working modulo n" just means considering the remainder you get when you divide by n. For n=2, you split numbers up into even and odd. For n=3, you split them up into three categories: multiples of 3, numbers that are one more than a multiple of 3, and numbers that are two more than a multiple of 3. Just consider all the possibilities, and use the fact that if you have n+1 numbers, then at least two of them must have the same remainder when you divide by n (because there are only n possibilities; that's the pigeonhole principle). Good luck! —Keenan Pepper 05:26, 20 March 2008 (UTC)

Divisibility curiosity

If n is a positive integer that is not divisible by 6, (n^2-1) seems to be always a multiple of 24. Why? Imagine Reason (talk) 02:54, 19 March 2008 (UTC)

It only works if neither 2 nor 3 divides n. Difference of two squares says a^2-b^2 = (a-b)*(a+b). Can you use that to rewrite n^2-1 and make arguments about where 2 and 3 must divide? PrimeHunter (talk) 03:18, 19 March 2008 (UTC)

You're right, it's about 2 and 3. So n^2-1 = (n-1)(n+1), which when n is not a multiple of 2 equals a multiple of 4. And since n isn't a multiple of 3, (n-1)(n+1) must be, but that's as far as I got. Imagine Reason (talk) 19:05, 19 March 2008 (UTC)

You need to think slightly harder to show n²-1 is a multiple of 8, not just of 4. If n is odd, then it is either one more or one less than a multiple of 4. Algebraist 19:34, 19 March 2008 (UTC)

Oh, two consecutive even numbers must include one that is a multiple of 4. Imagine Reason (talk) 02:10, 20 March 2008 (UTC)

Gödel's Incompleteness Theorems: Is The Math Reliable?

Many sciences depend on the math to prove something and use it for rigorous study. But Gödel's incompleteness theorems states:

For any consistent formal, computably enumerable theory that proves basic arithmetical truths, an arithmetical statement that is true, but not provable in the theory, can be constructed.1 That is, any effectively generated theory capable of expressing elementary arithmetic cannot be both consistent and complete.

Therefore, I would like to know are all the theories we use (for biology, chemistry, physics, medicine, computer science, etc.) considered to be consistent theories themself? And are all of maths we learn from elementary school to university considered to be reliable and don't contradict each other? - Justin545 (talk) 07:00, 19 March 2008 (UTC)

What do you mean by "reliable"? I would say the mathematics underlying biology, chemistry, etc is far less likely to be in error than the biology and chemistry themselves. But if you're looking for apodeictic certainty -- the sort of thing that, by its nature, cannot be wrong -- well, sorry, we don't have any of that. In my humble opinion, anyway. We'll settle for being right; we don't have to be completely certain.

Or as the Eagles put it -- "I could be wrong, but I'm not". --Trovatore (talk) 07:18, 19 March 2008 (UTC)

Math is used as a tool for studying many sciences. If the tool itself is "problematic" or "questionable", the consequences of employing it are very likey to be wrong! "reliable" means "consistent" and "don't contradict". Incompleteness theorems, in other words, states: if every arithmetical statement that is true and is provable in the theory, the theory is inconsistent but it is complete. So what I want to know is: the math we use is either

(1) consistent but not complete, or

(2) complete but not consistent - Justin545 (talk) 07:48, 19 March 2008 (UTC)

Well, we don't know for certain, but the general view is that we are in the consistent but not complete case, which is really not as bad as it sounds at first. If you know any group theory, consider that there are plenty of facts about groups that cannot be deduced from the axioms for a group alone -- the theory of groups, as given by the most basic group axioms, is not complete. In some sense this is because there are different models, different groups, that all meet those basic axioms, and thus have truths that are not derivable from just those axioms. You can think of arithmetic as being similar, with different models, just with the proviso that, unlike groups, we haven't found models that disagree on any arithmetic facts you or I would generally care about. -- Leland McInnes (talk) 12:11, 19 March 2008 (UTC)

Could you give an example of what you mean, there? I've done quite a bit of group theory and have never come across something that's true but can't be proven from the axioms of a group (together with ZF). --Tango (talk) 13:39, 19 March 2008 (UTC)

>> "which is really not as bad as it sounds at first"

It sounds bad to me... since we are not able to justify our math.

>> "there are plenty of facts about groups that cannot be deduced from the axioms for a group alone...the theory of groups is not complete."

I don't know any of group theory, but: Could those set of un-deducible facts themself be considered as axioms? Will group theory be complete if we make those facts axioms? - Justin545 (talk) 02:50, 20 March 2008 (UTC)

What I mean is that the group axioms don't uniquely define the group, but rather a whole slew of possible objects each of which satisfies the axioms of being a group. Thus there isn't a unique model of "group" specified by the axioms, but rather each and every different group is a different model that satisfies the basic group axioms. There are things that are true of particular groups that you can't deduce from just the group axioms -- you need more information (more axioms in essence) to pin down which group (or class of groups) you are talking about. Thus there are truths that occur in systems that fulfill the group axioms that are not provable from the group axioms alone. Does that make more sense? -- Leland McInnes (talk) 17:26, 19 March 2008 (UTC)

Right, but arithmetic (and set theory) are quite a different case from group theory. Arithmetic is not the study of models of arithmetic; it's the study of numbers. All models of arithmetic have (copies of) all the true natural numbers, but some of them also have fake natural numbers. The one true Platonic intended model of arithmetic has only the true ones, and none of the fake ones, and is unique up to a canonical isomorphism. There's a limit to what we can find out about the behavior of the true natural numbers from a fixed set of axioms and first-order logic alone. That doesn't mean we have to stop there. --Trovatore (talk) 17:49, 19 March 2008 (UTC)

Let me insert a response here: essentially, yes. I was going for a loose analogy suggesting that incompleteness isn't really a horrible thing. As to models of arithmetic, there is the question of what the intended model is, and, for the sufficiently messy cases where we can't practically distinguish is from some fake model, whether it even matters. I would liken it (again, an analogy, so don't take it too literally) to science trying to model (in a different sens of the word) some objective reality -- we can't know the objective reality, only our model of it, but as long as we can't tell the difference between our model and the reality (i.e. where our model hasn't been falsified) we may as well consider our model as true. -- Leland McInnes (talk) 20:55, 19 March 2008 (UTC)

Sure, there are plenty of things that can't be proven using just the axioms of a group, but those things aren't true. $gh=hg\quad \forall g,h\in G$ can't be proven just from the group axioms, because it isn't true in general. That's not incompleteness, it's just a false statement. If you want it to be true, you have to add an additional assumption (that the group G be cyclic, say). If the statement can be stated in terms of only the group axioms, and is true, then it can be proven using only the group axioms. If it can't be stated using only those axioms, then it being impossible to prove isn't a case of incompleteness. A framework is incomplete if there are unprovable true statements within that framework. --Tango (talk) 18:06, 19 March 2008 (UTC)

Tango, I think you have not thought these things through terribly well. At least it isn't clear to me what you mean by a framework, or unprovable but true within a framework. Is a framework a first-order theory, or a model, or what exactly? --Trovatore (talk) 18:19, 19 March 2008 (UTC)

Let me be a little less Socratic and hopefully more constructive (took me more time to figure out how to say this than it did to ask a question). Let's take a specific example. Peano arithmetic neither proves nor (we suppose) refutes the claim "Peano arithmetic is consistent" (the claim is usually abbreviated Con(PA). Therefore there are models of PA in which Con(PA) is true, and there are models of PA in which Con(PA) is false. So we can make an analogy with your example statement "multiplication is commutative": There are models of group theory (that is, groups) in which "multiplication is commutative" is true, and there are other models of group theory in which it's false.

Here's the big difference: There's no such thing as "the intended group", the group that defines the truth value of "multiplication is commutative in group theory". We're interested in Abelian groups, and we're also interested in non-Abelian groups, and you just have to specify which ones you're talking about.

But Peano arithmetic (we suppose) really is consistent. The models of PA that think otherwise are wrong about that. That's not to say they're not interesting (people devote whole careers to them), but merely by their opinion on this one issue, they prove that they are not the intended model. --Trovatore (talk) 18:35, 19 March 2008 (UTC)

Ok, I think I understand what you're saying now. I'm not sure I agree, though. Group theory is defined in terms of set theory. Once you've determined a model of set theory, your model of group theory is completely determined (a group is simply a set together with a function - both concepts defined outside of group theory). Is there a (reasonable) model of set theory in which all groups are abelian? --Tango (talk) 18:56, 19 March 2008 (UTC)

Whoah, we have to be careful here -- the phrase "group theory" is being used in two different ways (my fault, probably). When I say "model of group theory==group", I'm using "group theory" to mean the first-order theory defined by the three axioms (identity existence, existence of two-sided inverses, associativity). That's different of course from "group theory" as in "the study of groups", which is not a formal first-order theory at all. Please re-read my remarks keeping this clarification in mind -- they won't have made any sense at all if you were thinking of "model of group theory" as meaning "model of the study of groups(?)". --Trovatore (talk) 19:02, 19 March 2008 (UTC)

Ok, but I think my point still stands. Group theory, in that sense, is still built on set theory. Any model of group theory must be a model of set theory, since it has to satisfy ZF plus the 3 axioms of a group. Can you have such model of set theory in which all groups are abelian? For example, set theory provides all kinds of methods of combining sets to produce sets - those method can be used to combine groups and produce other groups. Is there a model in which all such possible combinations are abelian? --Tango (talk) 19:22, 19 March 2008 (UTC)

No, of course not. It's a theorem of ZF that there exist non-Abelian groups. But you're still mixing things in a confusing way -- whatever a "model of group theory" is, it's certainly not something that "satisfies ZF plus the three axioms of a group"; that doesn't even make sense; the ZF axioms are in a different language from the group axioms. If by "group theory" we mean the three axioms, then "model of group theory" means precisely "group", and does not imply that the model satisfies the ZF axioms. That's the sense in which I was using the phrase "model of group theory". --Trovatore (talk) 19:47, 19 March 2008 (UTC)

[edit conflict] I think you're still misunderstanding. Sure, it's possible to define groups as a special kind of set in ZF set theory. But that is not what we are talking about here. You are probably confused by the fact that ZF is an immensely more complex system then the meager 3 axioms of groups (to which I will refer as GP). But they are the same thing for this discussion. Each of them is a collection of rules governing a world of objects. A bag of objects can either satisfy these rules, in which case it is called a model of the theory, or not. In the case of ZF, the models are very complicated and hard to point out, but I think Godel's constructible universe is an example of one. For GP, every simple little group is a model, and the elements of the group are the basic objects. In ZF, you can have models that satisfy choice, and models that don't; in GP, you can have models that satisfy commutativity, and models that don't. -- Meni Rosenfeld (talk) 19:55, 19 March 2008 (UTC)

Ok, I get you. So, if I'm understanding your definition of completeness correctly, a theory being complete is basically equivalent to there only being one model satisfying it? Since, if there are two models of the theory, they must differ in some way and that way gives rise to a statement which is true in one model and not true in the other. --Tango (talk) 20:05, 19 March 2008 (UTC)

Not quite. It's possible for two models to satisfy all the same first-order statements, but to be nonisomorphic. For example the theory of torsion-free abelian groups is complete, but there are nonisomorphic torsion-free abelian groups. --Trovatore (talk) 20:38, 19 March 2008 (UTC)

If memory serves, all torsion-free abelian groups are of the form ${\mathbb {Z}}\times \cdots \times {\mathbb {Z}}$. I sometimes get a little confused with the orders of logical statements, but is $\exists g\in G,\forall h\in G,\exists n\in {\mathbb {N} },h=g^{n}$ not a first order statement satisfied by only one of those groups? --Tango (talk) 21:32, 19 March 2008 (UTC)

It's not a first-order statement in the language of groups. The language of groups has no function symbol for nth power and no symbol for the set of all natural numbers. --Trovatore (talk) 21:40, 19 March 2008 (UTC)

Ah, good point. I think we've got there, I have no more questions! Thank you. (Well, I'm sure one will come to me at 3am, but that can wait until tomorrow. ;)) You know... I really do wish my Maths dept. had a proper course on logic, it seems a really major topic to miss out (we did a bit in 1st year, but it was really just half a module on set theory in rather vague terms - the phrase "first order logic" did not appear once). I've done some reading on the subject, I should do some more... --Tango (talk) 21:47, 19 March 2008 (UTC)

For the record, the rationals Q, the reals R, and the p-adic integers Z_p are all torsion-free abelian groups. Your structure theorem holds for finitely generated abelian groups. Tesseran (talk) 16:03, 21 March 2008 (UTC)

Excellent point. That doesn't change my (nevertheless incorrect) point, though. --Tango (talk) 16:11, 21 March 2008 (UTC)

Theories in physics (thought to be the trunk of the science "tree") are not necessarily consistent. Helpfulness of established theories begin and end with orders of magnitude. This is why we have semiclassical physics, and the mesoscopic scale, and why we differentiate "Physics in the Classical Limit," Relativity, and quantum theory. Mac Davis (talk) 08:01, 19 March 2008 (UTC)

Did I misunderstand Gödel's Incompleteness Theorems or the Incompleteness Theorems is really about distinguishing between classical physics and modern physics? I thought Incompleteness Theorems is just all about the math but not the physics. And Incompleteness Theorems should be able to be applied to all kinds of science, not just physics. I'm not offending, just hope someone can clarify the concept. - Justin545 (talk) 09:23, 19 March 2008 (UTC)

The incompleteness theorems don't apply particularly well to the kind of math you're probably familiar with. That is, they aren't relevant. They claim that a specific very sensible, very general way of justifying things doesn't work very well in certain contexts. That doesn't mean that what we were trying to justify is wrong, just that we'll have to look somewhere else for confidence in it. It also throws essentially no doubt on actual arithmetic, which deals only with fairly small numbers and can be justified by direct experience and some common sense. Black Carrot (talk) 08:10, 19 March 2008 (UTC)

Theorems are proved based on axioms. Experience in proving theorems made mathematicians conjecture that every true statement could eventually be proved. This conjecture turned out to be naive. The incompletenes theorem states that the conjecture is not true: the fact that some statement cannot be proved does not imply that the statement is false. The incompleteness theorem does not threaten the reliabilty of mathematics. Bo Jacoby (talk) 11:06, 19 March 2008 (UTC).

>> "The incompleteness theorem does not threaten the reliabilty of mathematics."

I think you mean the mathematics we use is consistent but not complete since there are still some true statements can not be proven by mathematics, and also you said the mathematics is reliable. But your opinion sounds a bit different with the other. For example, some one said "mathematicians believe that mathematics is consistent". Which means mathematicians "can not prove" mathematics is consistent. - Justin545 (talk) 01:20, 20 March 2008 (UTC)

Before the incompleteness theorem mathematics was supposed to be consistent and complete. After the incompleteness theorem mathematics is known to be incomplete. The incompleteness theorem does not clarify whether mathematics is consistent or not. So I do not say that mathematics is reliable as a consequence of the incompleteness theorem, nor do I say that mathematics is unreliable as a consequence of the incompleteness theorem. Bo Jacoby (talk) 05:03, 22 March 2008 (UTC).

In general, mathematicians believe that mathematics (however we may choose to define that term) is consistent. This is mainly because we have not found an inconsistency (a statement P such that both P and not-P can be proved). We can even express this "conjecture" as a (humungously complex) arithmetical statement. Problem is that we also know, thanks to Gödel, that we cannot prove this statement - at least, not without stepping up to some more powerful axiom system, which then leads a "turtles all the way down" type of regression. Bottom line is, most mathematicians say "that's interesting and slightly weird" but they don't lose sleep worrying that mathematics might be inconsistent. On a scale of rational evidence-based confidence, you can put the consistency of mathematics right up at the 99.99% mark. Gandalf61 (talk) 12:21, 19 March 2008 (UTC)

>> "(a statement P such that both P and not-P can be proved) ... thanks to Gödel, that we cannot prove this statement"

I believe Gödel used "logic" to build his Incompleteness Theorems. But isn't logic a kind of mathematics? If logic is a kind of mathematics, Gödel was using a tool, about which its consistency can not be sure, to prove his Incompleteness Theorems. In other words, Incompleteness Theorems is questionable since the logic itself is questionable. - Justin545 (talk) 01:57, 20 March 2008 (UTC)

It sounds like we can not use logic to justify the logic itself. It's meaningless! If we are doubt of the logic, we should also doubt of the natural language, such as English, Chinese,... etc., we use, since the logic is just a symbolization of our natural language. We can do inference by the logic and we can also do inference by our language. - Justin545 (talk) 02:15, 20 March 2008 (UTC)

Learn to live with uncertainty. (Like you have a choice....) --Trovatore (talk) 02:23, 20 March 2008 (UTC)

Maybe, learn to live with confidence in the logic and the language. - Justin545 (talk) 02:28, 20 March 2008 (UTC)

Confidence is one thing; fully justified certainty is quite a different thing. You seem to be looking for the latter. You're not going to find it. --Trovatore (talk) 02:33, 20 March 2008 (UTC)

Knowing the incompleteness theorems somewhat shakes my confidence in the logic and math. I was just trying to find my confidence in them by this discussion. I'm not pursuing the absolute certainty. Imperfection is allowed. - Justin545 (talk) 03:01, 20 March 2008 (UTC)

Ah, I see. Well, I suppose maybe they should shake your confidence. Just not very much. The take-away message is that mathematics is not really different in kind from the experimental sciences -- you can have confidence in it because it's observed to work, not because it's built up from an unassailable foundation via unassailable steps. The latter idea never really did make sense, even before Gödel -- there was always an infinite regress built into it, as you've noticed. But Gödel does seem to have made people come to terms with this more. --Trovatore (talk) 03:13, 20 March 2008 (UTC)

Literally, your prior response "I would say the mathematics underlying biology, chemistry, etc is far less likely to be in error than the biology and chemistry themselves." seems to contradict "mathematics is not really different in kind from the experimental sciences". Well, just my picking hobby, I'm not trying to "offend" you again. And excuse my English, I don't know why you use "kind" in italic.

>> "not because it's built up from an unassailable foundation via unassailable steps."

The foundation may not be unassailable, but the stpes is unassailable I think. That's why I like deduction more than induction.

>> "The latter idea never really did make sense"

The latter idea? - Justin545 (talk) 03:40, 20 March 2008 (UTC)

I said the mathematics was "far less likely" to be in error. That's a difference in degree, not a difference in kind. Your English seems to be pretty good, but I see on your user page that you're not a native speaker -- are you familiar with the phrases "different in degree" and "different in kind"? --Trovatore (talk) 04:15, 20 March 2008 (UTC)

Sure, I know what are "different in degree" and "different in kind". Maybe I understand your point now. Well, thanks for the compliment. But actually I can not write articles without a dictionary. Besides, my English grammar is questionable. - Justin545 (talk) 05:03, 20 March 2008 (UTC)

Poincare Video

I tried to access the video at [3], but it doesn't work. Does anyone know where I can find a good copy of it? Black Carrot (talk) 08:14, 19 March 2008 (UTC)

Diophantine equation

I've got the equation n(2n+1)=a(a-1), both n and a to be positive integers. The solutions (10,15) and (348,493) came by searching, but nothing higher has been found. I'm wondering if the absence of further solutions can be shown from the fact that as a and n increase, their ratio will tend to the irrational root 2?—81.132.237.15 (talk) 13:16, 19 March 2008 (UTC)

Next two solutions are (11830, 16731) and (401880, 568345). As you conjecture, there is indeed a connection with rational approximations to the square root of 2. Gandalf61 (talk) 15:50, 19 March 2008 (UTC)

So there will be an infinite number of solutions?—81.132.237.15 (talk) 19:26, 19 March 2008 (UTC)

Yes, there are an infinite number of solutions. Gandalf61 (talk) 20:03, 19 March 2008 (UTC)

You can prove this by solving your equation using the quadratic equation (using n or a as the bound variable doesn't matter). Then solve the limit for n/a as a -> infinity:

$n={\frac {{\sqrt {8a^{2}-8a+1}}-1}{4}}$ (solve for n. this solution is used because they must be positive integers)

$\lim _{a\rightarrow \infty }{\frac {{\sqrt {8a^{2}-8a+1}}-1}{4a}}$ (solve the limit for n/a)

$=\lim _{a\rightarrow \infty }{\frac {\sqrt {8a^{2}-8a+1}}{4a}}$ (lim_a->inf (1/4a) is just 0 so we can leave it out)

$={\sqrt {\lim _{a\rightarrow \infty }{\frac {8a^{2}-8a+1}{16a^{2}}}}}$ (put everything inside sqrt and apply power rule in limits)

$={\sqrt {\lim _{a\rightarrow \infty }{\frac {8a^{2}}{16a^{2}}}}}$ (do the same thing, limit of (-8a+1)/(16a^2) is 0 as well so we can leave it out)

$={\sqrt {\frac {1}{2}}}$ (answer)

I think that's right...

--wj32 ^t/c 05:48, 20 March 2008 (UTC)

Yes, that proves that if there are integer solutions then the ratio n/a approaches 1/sqrt(2). It does not prove that there are infinitely many integer solutions, or indeed that there are any integer solutions at all. You could apply the exactly the same analysis to the equation 2n² = a², which has no integer solutions. The integer solutions to the original equation n(2n+1)=a(a-1) can be derived from the convergents of the continued fraction expansion of sqrt(2). Gandalf61 (talk) 09:41, 20 March 2008 (UTC)

Complex Analysis

Determine, with motivation, all funtions that is analytic on {z/ |z| < 4}, with f( 0 ) = i and |f( z ) <= 1 on {z/ |z| < 4} —Preceding unsigned comment added by 163.187.240.51 (talk) 13:33, 19 March 2008 (UTC)

Simple tip: When trying to get other people to do your homework, don't include phrases like with motivation. Also, you'd have better chances if you explained specifically what you've tried so far, and what you're having trouble with. —Keenan Pepper 15:23, 19 March 2008 (UTC)

Oh, what the heck. I'm feeling nice, so I'll practically give you the answer: Maximum modulus principle. —Keenan Pepper 15:28, 19 March 2008 (UTC)

>> Thanks very much!!!! It is actually not my homework but anyway thanks! And I will remember all your advise for the future.......

How do you explain the fractional root of a number?

I can't find an answer to this because I'm not sure if I'm using the right terminology. What I mean is, how does one explain what the "1/2"th root of a number is? Or the "2/3"rd root? I understand that the nth root of A is that number x which when multiplied by itself n times yields A, but how do I explain a fractional root in similar terms? Thanks in advance, Narxysus (talk) 17:22, 19 March 2008 (UTC)

the n th root of a is in fact a^1/n eg sqrt(10) = 10^1/2

Therefor the a/b th root of n is n^b/a. That's a consistent intepretation. So the 1/2 root of x is in fact x squared.83.100.183.180 (talk) 18:24, 19 March 2008 (UTC)

It just occured to me that you will be able to work this out if you have knowledge of logarithms. Or at least they offer a method of understanding the problem - suggest reading about them.83.100.183.180 (talk) 20:28, 19 March 2008 (UTC)

It looks like you are trying to do two conceptual steps at once here. First you should be asking what is the fractional power of a number, such as power of 1/2, and we can try to answer that question if you do. Only then should you ask about fractional roots, and the answer is that the 1/2th root of A is that number x such that $x^{1/2}=A$. -- Meni Rosenfeld (talk) 18:29, 19 March 2008 (UTC)

The (A/B)th root of a number C is what, multiplied by itself A times, gives C^B. Black Carrot (talk) 22:16, 19 March 2008 (UTC)

Ah, I understand it now - if I want ½th root of A, what I really want is A^(1/(1/2)) which is really A^2. It's a little tricker to explain, but I understand it, anyway. Thanks everyone! Narxysus (talk) 04:49, 20 March 2008 (UTC)

There's one little catch, it's not always correct to assume that a radical like a 4th root can be expressed as a 1/4th power, this has to do with even roots of negative values, for which the answers are not real numbers, but complex numbers. A math-wiki (talk) 22:51, 20 March 2008 (UTC)

It's not just even roots of negative values - except for square roots of positive real numbers, some of the roots are always going to be complex. ${\sqrt[{n}]{x}}$ is generally taken to mean the positive real root when x is a positive real number. If x isn't a positive real number, the notation doesn't have an obvious meaning, it can mean any of the n roots. $x^{1/n}$ is usually taken to be multivalued, with no attempt to define which root you're talking about. --Tango (talk) 02:41, 21 March 2008 (UTC)

problem setting up the equation

1. In this problem, we will analyze the profit found for sales of decorative tiles. A DEMAND EQUATION shows how much money people would pay for a product depending on how much of that product is available on the open market.

A. Suppose that a market research company finds that s price of p=$20, they would sell x=42 tiles each month. If they lower the price to p=$10, then more people would purchase the tiles, and they can expect to sell x=52 tiles in a month's time. Find the equation of the line for demand equation. Write your answer in the form p=mx+b (hint: write an equation using two points in the form (x,p)) A company's revenue is the amount of money that comes in fromsales, before business costs are subtracted. For a single product, you can find the revenue by multiplying the quantity of the product sold, x, by the demand equation, p.

B Sustitute the result you found from part a into the equation R=xp to find the revenue equation. simplified answer. —Preceding unsigned comment added by Lighteyes22003 (talk • contribs) 19:52, 19 March 2008 (UTC)

Answer effectively here you need to find the line that connects (20,42) and (10,52) ie sales (x) is a function of cost (p) - you can assume the relationship is a straight-line. —Preceding unsigned comment added by 83.100.183.180 (talk) 20:20, 19 March 2008 (UTC)

To make things a bit more obvious, your looking to start with Point-slope form of a line A math-wiki (talk) 23:42, 21 March 2008 (UTC)

Graphing calculator peculiarity

Why does the graph of sin(e^x) appear to be a sine wave when graphing on the interval (19.9999, 20}? Zooming in furthur on the function, the graph returns to the chaos I expected to see. I'm wondering why this occurs at all. I'm assuming that my calculator's low resolution (I'm using a TI-84 Plus, which has a screen about 94 pixels wide and 62 pixels high) is partly to blame, But I'm wonding if there is a better, more mathematical reason.

First you should remember that it is (approximately) a sine wave in this interval, only a very high frequency one. If it looks like a low frequency sine wave, the effect is known as aliasing and results from the signal frequency being close to a multiple of the sampling frequency. -- Meni Rosenfeld (talk) 22:51, 19 March 2008 (UTC)

(ec) It should look like a sine wave on a small enough scale - e^x is a differentiable function, which basically means it looks roughly like a straight line on a small enough scale, so sin(e^x) is going to look roughly like sin(ax+b), which is a sine wave. The chaos you see on a large scale is caused by the low resolution - the calculator just picks a few points to calculate and joins them with straight lines, which doesn't work if the function is changing direction very frequently, which this one does (for large values of x). --Tango (talk) 22:53, 19 March 2008 (UTC)

It should be emphasized that (19.9999, 20) is not a small enough scale - (19.9999999, 20) is. -- Meni Rosenfeld (talk) 23:02, 19 March 2008 (UTC)

True, if I've done the sum right (which I didn't think to do before answering), there should be around 8000 periods in that interval, so I doubt 94 pixels is going to be enough to resolve them! --Tango (talk) 23:19, 19 March 2008 (UTC)

March 20

What is the summation formula for a negative power?

What's the solution for this?: $\sum _{i=1}^{n}i^{-1}$

That's the Harmonic series (or, at least, a partial sum of it). The article should tell you quite a lot. If you have any specific questions after reading it, ask away! --Tango (talk) 00:15, 20 March 2008 (UTC)

It is called a harmonic number. Bo Jacoby (talk) 01:03, 20 March 2008 (UTC).

Proving Trigonometric Identities

Normally I'm okay at this, but I've been stuck on this one for a while. Prove the identity: (1+Tan^2x)/tan^2x=Csc^2x

I've got the basic trig identities down, but I just can't get this problem... Any help?

• Try simplifying the left side by separating the numerator... things just might cancel out relatively nicely. :) --Kinu ^t/_c 01:56, 20 March 2008 (UTC)

${\frac {1+\tan ^{2}x}{\tan ^{2}x}}$

Use the identity for tan(x)

$={\frac {1+{\frac {\sin ^{2}x}{\cos ^{2}x}}}{\frac {\sin ^{2}x}{\cos ^{2}x}}}$

Simplify.

$=\left(1+{\frac {\sin ^{2}x}{\cos ^{2}x}}\right)\left({\frac {\cos ^{2}x}{\sin ^{2}x}}\right)$

Change the fraction.

$=\left({\frac {\sin ^{2}x+\cos ^{2}x}{\cos ^{2}x}}\right)\left({\frac {\cos ^{2}x}{\sin ^{2}x}}\right)$

Use the identity sin^2(x)+cos^2(x)=1

$=\left({\frac {1}{\cos ^{2}x}}\right)\left({\frac {\cos ^{2}x}{\sin ^{2}x}}\right)$

Cancel.

$={\frac {1}{\sin ^{2}x}}$

Use the identity for csc(x).

$=\csc ^{2}x$

--wj32 ^t/c 05:57, 20 March 2008 (UTC)

Using Kinu's clue, here's another proof:

${\frac {1+\tan ^{2}x}{\tan ^{2}x}}$

Simplify the fraction.

$={\frac {1}{\tan ^{2}x}}+1$

Since cot(x)=1/tan(x):

$=\cot ^{2}x+1$

Because $\csc x={\sqrt {\cot ^{2}x+1}}$ (see List of trigonometric identities):

$=\csc ^{2}x$

--wj32 ^t/c 06:08, 20 March 2008 (UTC)

• Yes, that.... but I was somewhat hesitant to put the complete solution up, since this is the "help" desk, not the "answer" desk... --Kinu ^t/_c 16:34, 20 March 2008 (UTC)

Fundamental group and de Rham cohomology

Is there a simple relationship between the fundamental group of a manifold and its first de Rham cohomology group? The concepts seem very similar, like the de Rham cohomology group is a "continuous version" of the fundamental group, but I don't know how to quantify that. The Hurewicz theorem sounds very close, but I think I'm missing something (probably because I don't yet understand any kind of cohomology other than de Rham, and that only vaguely). —Keenan Pepper 05:44, 20 March 2008 (UTC)

The answer is indeed given by the Hurewicz theorem, as well as any suitable comparison isomorphism. What you get is that the first de Rham cohomology of X is dual to the abelianization of π₁(X) with scalars extended from Z to R (if you take de Rham cohomology with real coefficients). In other words, consider the mapping:

${\begin{aligned}H_{\textrm {dR}}^{1}(X,\mathbb {R} )\times \pi _{1}(X,x_{0})&\to \mathbb {R} \\(\omega ,\gamma )&\mapsto \langle \omega |\gamma \rangle =\int _{\gamma }\omega \end{aligned}}$

taking a 1-form and a loop to the integral of the 1-form along the loop. This is well-defined (it only depends on the cohomology class of $\omega$ and on the homotopy class of $\gamma$), bilinear in the obvious sense, and the only loops such that $\langle \cdot |\gamma \rangle =0$ are the commutators in the fundamental group. It becomes a non-degenerate pairing on $H_{\textrm {dR}}^{1}(X,\mathbb {R} )\times \pi _{1}(X,x_{0})^{\textrm {ab}}$. Bikasuishin (talk) 18:12, 20 March 2008 (UTC)

Wow, it seems so simple now! One more question: What's the simplest example you can think of of a manifold with a nonabelian fundamental group? I want something a little more concrete to think about. —Keenan Pepper 18:32, 20 March 2008 (UTC)

Think of the twice-punctured plane. The fundamental group is the (non-abelian) free group on two generators (loops around the two punctures), whereas the cohomology is just R². An example of a loop that is non-trivial in the fundamental group but homologically trivial is the "figure 8" loop around the two punctures. Bikasuishin (talk) 23:38, 20 March 2008 (UTC)

Isn't the figure eight (once round each hole) non-trivial in homology? For a homologically trivial curve, you want to go round each puncture twice in opposite directions, corresponding to the word aba^-1b^-1. Algebraist 11:08, 21 March 2008 (UTC)

Yes, sorry, you're correct of course. Bikasuishin (talk) 12:09, 21 March 2008 (UTC)

Discrete cosine transform

I'm trying to understand the "why" of DCT. (It's part of a larger goal to understand what's going on inside JPEG/MPEG.) In the Discrete cosine transform article these formulas are given for DCT-II and DCT-III:

$X_{k}=\sum _{n=0}^{N-1}x_{n}\cos \left[{\frac {\pi }{N}}\left(n+{\frac {1}{2}}\right)k\right]\quad \quad k=0,\dots ,N-1.$

$X_{k}={\frac {1}{2}}x_{0}+\sum _{n=1}^{N-1}x_{n}\cos \left[{\frac {\pi }{N}}n\left(k+{\frac {1}{2}}\right)\right]\quad \quad k=0,\dots ,N-1.$

It's then stated, but not proven, that they are inverses (with a constant multiplier). Every other source I've found does the same thing. It seems like they expect the inverse relationship to be so obvious it doesn't need proving, but to me it's quite surprising. Intuitively I expect a function full of cosines to be inverted by a function full of arccosines.

Of course I can numerically verify the results (which shows that I didn't misread the formulas) and I even proved N=1, N=2, and N=3 by expanding the sums and substituting the result of one formula into the other. But that gets really tedious by the time you get to N=4, and I'm not seeing any way to generalize it.

Can someone provide the missing proof? --tcsetattr (talk / contribs) 08:03, 20 March 2008 (UTC)

It should be relatively easy to find proofs of the Discrete Fourier transform. The DCT is just the real part of it: apply the operator Re to both sides of the identity, and use .  --Lambiam 11:29, 20 March 2008 (UTC)

That's the kind of hand-waving that has left the question unresolved after a long time of trying to find the answer. I need details. And doing it without a detour through complex numbers would be a bonus. Here's what happens when I try to confirm what you said:

The Discrete Fourier transform article says

$X_{k}=\sum _{n=0}^{N-1}x_{n}e^{-{\frac {2\pi i}{N}}kn}\quad \quad k=0,\dots ,N-1$

The real part is:

$X_{k}=\sum _{n=0}^{N-1}x_{n}\cos -{\frac {2\pi }{N}}kn\quad \quad k=0,\dots ,N-1$

Because cos(-x) = cos(x) the minus sign can be dropped:

$X_{k}=\sum _{n=0}^{N-1}x_{n}\cos {\frac {2\pi }{N}}kn\quad \quad k=0,\dots ,N-1$

How is that equivalent to the DCT-II formula above? It has 2pi instead of pi, and n instead of n+1/2. They aren't the same thing at all! I've barely got started on this and I'm lost already. Is there no one who will actually write everything out so it can be understood? --tcsetattr (talk / contribs) 23:33, 20 March 2008 (UTC)

You shouldn't expect arccosines in this case because you're not taking the cosine of the input, but rather multiplying it by the cosine of something else. So there might be secants in the inverse, but not arccosines.

Let me add the right normalization factor and rename some variables so that I can substitute one equation in the other:

$X_{n}=\sum _{j=0}^{N-1}x_{j}\cos \left[{\frac {\pi }{N}}\left(j+{\frac {1}{2}}\right)n\right]$

$x_{k}={\frac {2}{N}}\left[{\frac {1}{2}}X_{0}+\sum _{n=1}^{N-1}X_{n}\cos \left[{\frac {\pi }{N}}\left(k+{\frac {1}{2}}\right)n\right]\right]$

Now substituting the first in the second I get:

$x_{k}={\frac {2}{N}}\left[{\frac {1}{2}}\sum _{j=0}^{N-1}x_{j}+\sum _{n=1}^{N-1}\sum _{j=0}^{N-1}x_{j}\cos \left[{\frac {\pi }{N}}\left(j+{\frac {1}{2}}\right)n\right]\cos \left[{\frac {\pi }{N}}\left(k+{\frac {1}{2}}\right)n\right]\right]$

$x_{k}={\frac {1}{N}}\sum _{j=0}^{N-1}x_{j}\left(1+\sum _{n=1}^{N-1}2\cos \left[{\frac {\pi }{N}}\left(j+{\frac {1}{2}}\right)n\right]\cos \left[{\frac {\pi }{N}}\left(k+{\frac {1}{2}}\right)n\right]\right)$

$x_{k}={\frac {1}{N}}\sum _{j=0}^{N-1}x_{j}\left(1+\sum _{n=1}^{N-1}\cos \left[{\frac {\pi }{N}}(j+k+1)n\right]+\sum _{n=1}^{N-1}\cos \left[{\frac {\pi }{N}}(j-k)n\right]\right)$

So it all hinges on whether the parenthesized part, $1+\sum _{n=1}^{N-1}\cos \left[{\frac {\pi }{N}}(j+k+1)n\right]+\sum _{n=1}^{N-1}\cos \left[{\frac {\pi }{N}}(j-k)n\right]$, equals N when j=k and 0 otherwise. And it does, but I admit to not understanding in a deep way why it does. It follows from the weird identity

$\sum _{q=0}^{N-1}\cos \left({\frac {\pi pq}{N}}\right)={\begin{cases}N&p=0\\1&p{\mbox{ odd}}\\0&{\mbox{otherwise}}\end{cases}}$

which I can prove, but again without all that much insight. Let me know if you want the details. This does seem much less elegant than the complex case, which is straightforward to prove and easy to understand geometrically. I don't see any obvious way to adapt a proof of the complex case to the real case. -- BenRG (talk) 02:13, 21 March 2008 (UTC)

I have to disagree with complex numbers being easy to understand geometrically. They double the number of spatial dimensions you have to visualize. That's only making things harder. Thanks for the other ideas though: splitting the product of cosines into a sum was the big step I wasn't coming up with on my own. The last identity is new to me too. --tcsetattr (talk / contribs) 00:10, 22 March 2008 (UTC)

Incidence matrix

Hi. I was playing this video on the ocw.mit.edu site related to graphs, networks and incidence matrices and the teacher said that $A^{T}y=0$ is one of the most fundamental relationship in applied maths (A is the incidence matrix of a network). He then went on to show if y is the current vector then $A^{T}y=0$ is Kirchoff's circuit law. What I want to know is in what other sense is this relationship important and why should we consider it fundamental. Thanks.--Shahab (talk) 09:50, 20 March 2008 (UTC)

I don't know about applied math, but it closely resembles homology, which is fundamental in abstract math. I think, though, that he meant to imply a connection to the topics he mentioned at the beginning, like fluid flow in a hydraulic system and balancing of forces in a weight-bearing structure. Black Carrot (talk) 09:42, 22 March 2008 (UTC)

Distribution Theory

If Xi ~ N(0,1), derive the probability density function of Xi^2. Write down the probability density function of SigmaXi^2 ..Jacques

The probability that Xi^2 is within an interval dx of some number x is equal to the probability that Xi is within an interval ${\frac {d{\sqrt {x}}}{dx}}dx$ of either ${\sqrt {x}}$, or $-{\sqrt {x}}$ (because those are the only solutions of Xi^2 = x). Use that together with some calculus, and come back if you're stuck and having trouble with something specific. —Keenan Pepper 18:41, 20 March 2008 (UTC)

Chemistry question!

If I have 12 coordinate sphere packing ie rhombic dodecahedral of spheres A

then there are 14 vertices : 6 (4line) vertices and 8 (3line vertices)

the (4line) vertices are surrounded by 6 A spheres (??)

the (3line) vertices are surrounded by 4 A spheres.

So if I place (smaller) spheres B at all the 14 vertices the formula is A1 B (6vertices/6coordinate)+(8verticles/4coodinate) = A1 B3

Is this correct?83.100.183.180 (talk) 14:57, 20 March 2008 (UTC)

duals (in a platonic sense) extended..

the vertices in rhombic dodecahedral packing appear to be the centres of tetrahedron or octahedron (taking the centres of surrounding) rhombic dodecahedra as vertices - is there a name for this type of (octahedron/tetrahedron) / (rhombic dodecahedron) dual relationship

article

Is there a space filling solid article - under another name, if not should there be one?

more

FCC is equivalent to connected rhombic dodecahedra, HCP packing forms a different shape "if the spheres of hexagonal close packing are expanded, they form a second irregular dodecahedron consisting of six rhombi and six trapezoids " from http://mathworld.wolfram.com/HexagonalClosePacking.html does this shape have no generic name?83.100.183.180 (talk) 15:13, 20 March 2008 (UTC) Sorry about the big list of little questions.83.100.183.180 (talk) 15:15, 20 March 2008 (UTC)

Our article on space filling tesellations is called Honeycomb (geometry). We also have an article on the rhombic dodecahedral honeycomb. The equivalent cell for hexagonal close packing is the trapezo-rhombic dodecahedron. Gandalf61 (talk) 15:35, 20 March 2008 (UTC)

Good thanks. Had no idea (or had forgotten) about 'honeycombs'83.100.183.180 (talk) 16:54, 20 March 2008 (UTC)

March 21

explicit rule

Can anyone give a step-by-step description on writing a rule in equation form of the following pattern?

2,6,24,120

• Well, without doing all of the work for you, check out factorial. Let us know if you're still stuck after that. :) --Kinu ^t/_c 00:40, 21 March 2008 (UTC)

lol I can't believe I didn't get that (the factorial part)! Still, I would appreciate it greatly if you can solve it for me as my brain is working a little more slowly than usual today.

ah (n+1)! is the answer...is there any other way to express this?

No, not really, that's the answer. You can write "(n+1)n(n-1)(n-2) ... 3.2.1", but that's just defining the factorial notation. --Tango (talk) 01:10, 21 March 2008 (UTC)

recursive rule

Speaking of rules and equations, can anyone help me write a recursive rule for the following equation?

2,5,14,41,122

Calculate the differences between the terms (5-2, 14-5, etc) and see if you spot a pattern. --Tango (talk) 01:11, 21 March 2008 (UTC)

yes i see a pattern...a_n=a_n-1+3^n-1

Yes. You can also search integer sequences in OEIS. "2,5,14,41,122" gives a non-recursive formula (and more complicated rules leading to other continuations). PrimeHunter (talk) 05:05, 21 March 2008 (UTC)

Yes, but that's not really very useful. What's useful is knowing how to determine the pattern yourself, having a website do your homework for you is pointless. --Tango (talk) 15:42, 21 March 2008 (UTC)

One issue with this question is that there are quite many patterns that have these first five numbers; how do you decide which is `the' pattern? SmaleDuffin (talk) 17:23, 21 March 2008 (UTC)

It's an issue with all such questions - I think to be sure you need to know the context. It's presumably a homework question and the class will have been studying certain types of sequence, so it's going to one of those types. --Tango (talk) 18:13, 21 March 2008 (UTC)

Exponential of an exponential

Can an equation of the form y = k(a + bc^x)^x be simplified? NeonMerlin 04:57, 21 March 2008 (UTC)

You could expand the xth power binomial, but I would say that's unsimplifying the above equation. A math-wiki (talk) 08:37, 21 March 2008 (UTC)

Peano Curve

Does anyone know where I could get hold of Peano's original paper on the curve, or a translation if it's in another language? Also, a question about the curve itself: Did Peano himself come up with these pictures of the curve, or did they come later when someone decided they wanted a concrete example? Black Carrot (talk) 05:50, 21 March 2008 (UTC)

Peano curve cites 'G. Peano, Sur une courbe, qui remplit toute une aire plane. Math. Ann 36 (1890), 157-160.', in case you missed that. The original text is available here. The article appears to contain a concrete example, but no pictures (at least in the pages I can get to work). I don't know about translations, but at least the article's in French, not a language of his own invention. Algebraist 11:04, 21 March 2008 (UTC)

Thanks! And apparently I can read it, which surprised the heck out of me. :) Black Carrot (talk) 07:37, 22 March 2008 (UTC)

Happy Easter!

Another pagan festival approaches and all those wishing to receive a 'happy easter' message get one.

My question is - what simple formulas give a good 'egg' shape. Links please, and it's just for curiousities sake, so don't exert yourselfs too much.87.102.16.238 (talk) 12:08, 21 March 2008 (UTC)

This doesn’t quite answer the question, but both Ellipsoid and Oval have a section on the egg shape. GromXXVII (talk) 12:35, 21 March 2008 (UTC)

Look here for all your oval curve needs. — Kieff | Talk 13:02, 21 March 2008 (UTC)

Super!87.102.16.238 (talk) 13:53, 21 March 2008 (UTC)

simple question

can we finde,or do we have afunction where,limit[f(x)] approches to infinity when ,x,approches to zero,while,f(0)=known value?it sounds no big deal but i think,yes there is such afunction exists.Husseinshimaljasimdini (talk) 12:23, 21 March 2008 (UTC)

Define f by f(0)=0, f(x)=1/x elsewhere. Bo Jacoby (talk) 12:35, 21 March 2008 (UTC).

Well 1/|x| elsewhere I believe, else you’d have to talk about one handed limits. GromXXVII (talk) 12:39, 21 March 2008 (UTC)

I wonder something like the differential of the dirac delta function?87.102.16.238 (talk) 15:17, 21 March 2008 (UTC)

as such take a look at Dirac_delta#Representations_of_the_delta_function the differential of the limits (n to infinity) of these functions.

eg y=d/dx( lim (e^{-x^2n} ))

when x=0 y=(1 error) =0, when x=0+delta x y=big

?87.102.16.238 (talk) 15:21, 21 March 2008 (UTC)

Also maybe consider the differential of the function y=e^{1/x^2} .At x=0 dy/dx = 0?87.102.16.238 (talk) 18:06, 21 March 2008 (UTC)

e^{1/x^2} isn't defined at 0, so can't have a derivative there. Even if you define y(0)=k though, for some k, it won't be continuous (because the limit at that point is positive infinity), so there'll be no derivative. Black Carrot (talk) 06:05, 22 March 2008 (UTC).Well, i was not expecting all these answers.As amatter

of fact i was thinking of the numbers of sectors we can get by dividing acircle.Now if the function ,F(θ)=2pi\θ,describes numbers of sectors we get by dividing acircle where θ is the centeral angle,then how many sectors we will get if,θ=0?is it zero?Husseinshimaljasimdini (talk) 11:46, 22 March 2008 (UTC)

Algebra with probabilistic events

I'm a college freshmen, and I'm no further into my mathematical education than vector calculus, but recently I picked up a book on probability theory. Even if it's mostly out of my league, it sure looks interesting. Anyway, I've encountered a statement that seems to me intuitively correct but algebraically wrong. Surely I'm mistaken; the question is, how?

Since I don't know to what degree the notation that the author uses is standard, I'll briefly recapitulate it. An event $A_{i}$ is a set of one or more of the elementary events within a sample space. $\Omega$ is the set of all elementary events, and $O$ is the empty set. $A_{1}+A_{2}$ is defined as the union of $A_{1}$ and $A_{2}$; $A_{1}A_{2}$ is the intersection of $A_{1}$ and $A_{2}$. As usual, addition and multiplication are associative and commutative, and addition is distributive over multiplication. Lastly, ${\overline {A_{1}}}$, the complement of $A_{1}$, is $\Omega -A_{1}$.

The statement that puzzles me is:

${\overline {A_{1}+A_{2}}}={\overline {A_{1}}}\;{\overline {A_{2}}}$.

That makes sense to me insofar as the event that neither $A_{1}$ nor $A_{2}$ happens is the same as the combined event that $A_{1}$ doesn't happen and $A_{2}$ doesn't happen. But:

${\begin{aligned}\Omega -A_{1}-A_{2}&=(\Omega -A_{1})(\Omega -A_{2})\\\Omega -A_{1}-A_{2}&=\Omega ^{2}-\Omega A_{1}-\Omega A_{2}+A_{1}A_{2}\\\Omega -A_{1}-A_{2}&=\Omega -A_{1}-A_{2}+A_{1}A_{2}\\O&=A_{1}A_{2}\\\end{aligned}}$

which clearly isn't necessarily true. —Saric (Talk) 17:26, 21 March 2008 (UTC)

Your deduction of the second line from the first is flawed, as is your deduction of the last from the third. You have been misled by the author's non-standard and ghastly notation into thinking that the things he has called +, - and * behave as you expect things with those names to behave: in the first case, you've assumed - interacts properly with * (it doesn't), and in the second you've used a cancellation law which doesn't hold in this setting. I recommend finding a better book. Algebraist 17:43, 21 March 2008 (UTC)

Oh, and FYI the statement in question is (one of) De Morgan's laws. Algebraist 17:47, 21 March 2008 (UTC)

(ec)I think the problem you're having is because set difference (ie. complement) and union aren't inverses of each other. $A=(A\cup B)-B$ is only true if $A\cap B=\emptyset$. You need to be careful when using "+" and "-" that you don't assume it works in the way as addition and subtraction of numbers. --Tango (talk) 17:51, 21 March 2008 (UTC)

The usual notation for this set subtraction, known as relative complement, is not but . Using that notation helps one to avoid the pit you fell in. The intersection of two such complements satisfies the identity

(P \ A) ∩ (Q \ B) = ((P ∩ Q) \ A) \ B.

There is no such thing here as "minus times minus equals plus".  --Lambiam 22:22, 21 March 2008 (UTC)

Solving "almost" superincreasing knapsacks problem

I know that it is really easy to solve the knapsack problem if all elements are super increasing, however, what if all of the elements were "almost super increasing". By almost super increasing, I mean that instead of having each successively larger integer being greater than the sum of all of the smaller integers. Each successively larger integer will be larger than, say, one-half of the sum all of the smaller integers. For example a valid set of "~1/2 almost super increasing integers" would be {1,2,3,4,5,8,12,18,27,40,60,90,135,205,305,460}; Would this instance of knapsack be as difficult to solve as the general knapsack problem? Or maybe somewhere between the difficulty of general knapsack and super increasing knapsack? 24.250.129.216 (talk) 19:16, 21 March 2008 (UTC)mathnoob

We don't know for sure whether the general knapsack problem is difficult to solve, so a proof that this is "somewhere between" easy and hard, but neither of the two, would imply a proof of P ≠ NP.

My gut feeling is that this is still NP-hard, but I don't readily see an applicable reduction like that of Exact cover to the knapsack problem.  --Lambiam 22:39, 21 March 2008 (UTC)

March 22

Uncountable Sets of Measure Zero

To a beginner (in measure theory that is) it seems that a set having (Lebesgue) measure zero is equivalent to that set being denumerable (countable or finite) but I just found out that the Cantor ternary set is both uncountable and it has measure zero. My question is, does anyone know of any other uncountable sets that have measure zero? What about an example of a real set which is not measurable at all? Thanks!
A Real Kaiser (talk) 03:37, 22 March 2008 (UTC)

For uncountable sets of measure 0, a bunch of copies of the Cantor set would work, of another set similarly constructed. As for non-measurable sets, see Vitali set. -GTBacchus^(talk) 03:48, 22 March 2008 (UTC)