Wikipedia:Reference desk/Mathematics

Welcome to the mathematics section
of the Wikipedia reference desk.

skip to bottom

Select a section:

• Computing
• Entertainment
• Humanities
• Language
• Mathematics
• Science
• Miscellaneous
• Archives

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

• Select the section of the desk that best fits the general topic of your question (see the navigation column to the right).
• Post your question to only one section, providing a short header that gives the topic of your question.
• Type '~~~~' (that is, four tilde characters) at the end – this signs and dates your contribution so we know who wrote what and when.
• Don't post personal contact information – it will be removed. Any answers will be provided here.
• Please be as specific as possible, and include all relevant context – the usefulness of answers may depend on the context.
• Note:
  • We don't answer (and may remove) questions that require medical diagnosis or legal advice.
  • We don't answer requests for opinions, predictions or debate.
  • We don't do your homework for you, though we'll help you past the stuck point.
  • We don't conduct original research or provide a free source of ideas, but we'll help you find information you need.

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

• The best answers address the question directly, and back up facts with wikilinks and links to sources. Do not edit others' comments and do not give any medical or legal advice.

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

August 28

Automorphisms

I want to prove the following result: Suppose I have an irreducible factor of the polynomial $x^{n}-1$, say h(x) in $\mathbb {Z} _{p}[x]$ (p prime). Now my first question is: If it is said that the Frobenius automorphism fixes h(x) does it mean that the coefficients of h(x) are permuted. Why does it happen? Secondly supposing we consider a splitting field in which the roots of h(x) exist. Will the Frobenius automorphism (over the splitting field Z_p(a,b,c...)) permute the roots a,b,c... and why? Finally why can't we define a splitting field for a polynomial defined over any general ring as opposed to over fields as the article Splitting field says? Thanks--Shahab (talk) 05:21, 28 August 2009 (UTC)

(I may be misunderstanding your questions, so if you do not find my answers helpful please ask again.) The Frobenius automorphism $\sigma$ on $K=\mathbb {Z} _{p}$ is the identity (there are no non-identity automorphisms of $\mathbb {Z} _{p}$). Therefore $\sigma$ fixes all polynomials over K, including h(x), no matter what h(x) is. Now let L be the splitting field of h over K, and let $\sigma$ be the Frobenius automorphism of L. Since $\sigma$ fixes K (why do we know this?), $\sigma h=h$, so the roots of h and $\sigma h$ in L are the same, and thus $\sigma$ permutes the roots of h; that is, $\sigma$ is in the Galois group of L over K.

Your last question is somewhat subtler. Recall that when we wish to adjoin a root of an irreducible polynomial f to a field K, we construct the field $L=K[X]/(f(X))$. (This is a field because (f(X)) is a prime ideal in the ring K[X], which is true because f is irreducible.) There is nothing really stopping us from doing the same operation to a ring R; fundamentally, the reason why we don't do this is because the results are uninteresting (I haven't verified this myself, but I assume that that is so, otherwise we would talk about splitting rings). However if we try to construct these "splitting rings" we can sometimes lose nice properties. For example, let $R=\mathbb {Z} /4\mathbb {Z}$, the cyclic ring of 4 elements, and let $f(x)=2x+1$. Let $S=R[X]/(f(X))$; then S has only one element, so there is no embedding of R into S anymore. Eric. 98.207.86.2 (talk) 07:48, 28 August 2009 (UTC)

Thanks Eric. I understand that the Frobeinus map over $\mathbb {Z} _{p}$ is the identity due to Fermat's little theorem. My main problem was the second question which is still not clear as it was incompletely asked. What I really want to do is this: Given that $g(x)\in \mathbb {Z} _{p^{r}}[x]$ is such that $g\equiv h(\mod p)$ (which I take to mean that the coefficients of g(x) and h(x) agree mod p) and g(x) is the irreducible factor of $x^{n}-1$ then the Frobenius map permutes the roots of g(x). Now $\mathbb {Z} _{p^{r}}$ is the ring in question and the roots must lie in the splitting field hence my last question. Assuming I have some splitting field F of g(x) I reason like this: (All computations are mod p and $\sigma$ is the Frobenius automorphism) g=h and so $\sigma (g)=\sigma (h)$ (where $\sigma$ is really acting upon the coefficients). But $\sigma (h)=h$ and so $\sigma (g)=h=g$. Now I want to use the fact(?) that automorphisms permute the roots of irreducible polynomials when the coefficients are fixed by the automorphism (I'll apply the Frobenius map on F). My second question is that I want a proof of this. Also please excuse me if some of what I have written is obvious or redundant. Since I am mostly self taught so there are quite a few gaps in my knowledge. Thanks again--Shahab (talk) 09:17, 28 August 2009 (UTC)

Are you trying to understand how to lift polynomial factorizations from Z/pZ to the p-adic integers, as in the algorithm to factor integer polynomials described in the original LLL paper? You can probably do that more easily than by developing a Galois theory for arbitrary commutative rings. JackSchmidt (talk) 14:41, 28 August 2009 (UTC)

Unfortunately I'm confused. I see that are you given an irreducible polynomial g (by "the irreducible factor" of $X^{n}-1\in \mathbb {Z} [X]$ I suppose you mean the one with degree $\phi (n)$, which you then project into $\mathbb {Z} _{p^{r}}[X]$ by taking the coefficients mod $p^{r}$) over $\mathbb {Z} _{p^{r}}$. However you introduce two new polynomials h and f and I don't know what those are. You also mention the Frobenius map on $\mathbb {Z} _{p^{r}}$; however, in general, this map (of raising to the power p) is not a ring homomorphism when $r>1$, so it would be unlikely to permute the roots of g or do anything nice.

Your last fact, that field automorphisms permute the roots of irreducible polynomials when the coefficients are fixed by the automorphism is correct. Eric. 216.27.191.178 (talk) 20:09, 28 August 2009 (UTC)

I made the mistake of writing f when I really meant h. I'll try to explain myself more clearly now. (BTW Theorem 1 of this paper online is the closest thing I could find on the internet to what I am trying to prove). I have some fixed irreducible factor of x^n-1 in Z_p[x] which is h. In Z_p^r[x] the same h goes by the name g. The roots of g lie in some extension field F (I don't know how such a field can be constructed. Perhaps we take the quotient field of Z_p^r and then find the splitting field) of Z_p^r. I want to raise all elements of F to pth power and then show that roots are only permuted. Thanks--Shahab (talk) 08:02, 29 August 2009 (UTC)

How do I prove the fact that that field automorphisms permute the roots of irreducible polynomials when the coefficients are fixed by the automorphism is correct.--Shahab (talk) 08:02, 29 August 2009 (UTC)

(dedent) Ah, that is clear. First I'll address the easy part (namely, the part I know how to prove). Suppose $\sigma$ is an automorphism of the field L, the fixed field of $\sigma$ is K, and f is a polynomial over K that splits completely in L; we wish to prove that $\sigma$ permutes the roots of f. For any polynomial g over L and $\alpha \in L$, we have $\sigma (g(\alpha ))=(\sigma g)(\sigma \alpha )$ (because $\sigma$ is an automorphism of L), so $\sigma$ maps the roots of g bijectively to the roots of $\sigma g$. In the case of f, we know $f=\sigma f$ (since the coefficients of f are fixed by $\sigma$); thus $\sigma$ bijectively maps the roots of f to the roots of f, i.e., permutes the roots of f.

As for what you're trying to prove, we run into difficulties. (The ring $R=\mathbb {Z} _{p^{r}}$ does not have a quotient field for r > 1 because R is not an integral domain; in particular $p\cdot p^{r-1}=0$.) I ran through an example to see if I could find the result you were looking for: take p = 2, r = 2, and $f(x)=x^{2}-x+1\in \mathbb {Z} _{p}[x],g(x)=x^{2}-x+1\in R[x]$. f and g are irreducible, and divide $x^{6}-1$. If we square the coefficients of g, we get another polynomial $x^{2}+x+1\in R[x]$, which being unequal to g, cannot have the same roots (even were we working in a ring where they both had all their roots). Eric. 98.207.86.2 (talk) 11:32, 29 August 2009 (UTC)

The best bet for understanding this is to just work with the p-adics. Use Hensel's lift to get the factorization over the p-adic integers, find the ring of integers A of the splitting field of the p-adic polynomial, the galois group is abelian so you can take an arbitrary factor of (p) as P and then look at A/P^r as the "extension field" of Z/p^rZ. Make sure that n and p are coprime (6 and 2 is naughty) otherwise the factorization can change. The paper you mentioned has a nice reference [2] that explains the LLL Z[x] factorization algorithm. The permutation of the roots happens in A/P^r the same as it does in A and the same as it does in A/P. If you are naughty and take n and p to have common factors, then the number of roots is not stable, and so the permutation actions cannot possibly be equivalent. JackSchmidt (talk) 13:06, 29 August 2009 (UTC)

That sounds quite reasonable to me. I am wondering if you could explain a few details for me. We have $L=\mathbb {Q} _{p}(\zeta _{n})$ is the splitting field of $x^{n}-1$ over $K=\mathbb {Q} _{p}$. I forget why L/K is abelian (cyclic, right?) but I can look that up myself. If n and p are coprime then L/K is unramified, but I don't see what that gets us. If f is a polynomial over $R=\mathbb {Z} _{p}$ we can project it to $R/p=\mathbb {Z} /p$, and that projection should split completely in A/P, where we have a Frobenius automorphism that permutes its roots in A/P, However I don't see how that helps us in A and A/P^r. Also, what do you mean by "the number of roots is not stable"? Eric. 98.207.86.2 (talk) 22:31, 29 August 2009 (UTC)

The Galois action on the field of fractions of A, L, restricts to a nice action on A itself. That action permutes the n roots of x^n-1 in A, and that action is permutation isomorphic to the action of the Galois group of A/P over Z/pZ on the roots of x^n-1. However, if n and p are not coprime, then there is a different number of n'th roots of unity in L versus A/P, so the actions cannot be permutation isomorphic. For example, if n=2 and p=2, then one has that both L=Q_2 and A=Z_2 have 2 square roots of unity, but Z/2Z=A/P has only one, Z/4Z has two, and Z/8Z has four. JackSchmidt (talk) 00:35, 30 August 2009 (UTC)

Thanks. I see my trouble now, I had gotten myself confused about the definition of Frobenius map#Frobenius_for_local_fields. Eric. 98.207.86.2 (talk) 01:35, 30 August 2009 (UTC)

Nonconvex polyhedra

I have a question: why is it the polyhedra {5/2, 3} and {5/2, 5} both exist but {5/2, 4} doesn't? Professor M. Fiendish, Esq. 12:38, 28 August 2009 (UTC)

The article on the four Kepler-Poinsot polyhedra states that "Augustin Cauchy proved the list complete by stellating the Platonic solids, and almost half a century after that, in 1858, Bertrand provided a more elegant proof by facetting them." Therefore, I suppose that from the definition of a regular star-polyhedron it follows directly that its vertices span a Platonic polyhedron as convex hull. If this supposition is correct, making a complete list is just a matter of checking a finite number of cases. (I wrote "I suppose" because I did not look for the exact definition. Of course, one can give apparently weaker definitions, but equivalent a posteriori, that make the proof more difficult and more general.) --pma (talk) 18:05, 28 August 2009 (UTC)

{5/2, 4} would have axes of 5-fold and 4-fold rotational symmetry; but these can coexist only in a hyperbolic geometry. —Tamfang (talk) 23:55, 9 September 2009 (UTC)

nested gaussian integrals.

First off: This is not for a homework problem

I am doing some work which requires calculation of nested time integrals (which come up in perturbation theory). I was able to work out a nice formula for a second order equation of the form

${\begin{aligned}&\int _{-\infty }^{\infty }dt_{2}\int _{-\infty }^{t_{2}}dt_{1}\exp \left(-{\frac {t_{2}^{2}}{2\sigma _{2}^{2}}}-i\Delta _{2}t_{2}\right)\exp \left(-{\frac {t_{1}^{2}}{2\sigma _{1}^{2}}}-i\Delta _{1}t_{1}\right)\\&\ \ \ \ \ \ =\pi \sigma _{1}\sigma _{2}\exp \left(-{\frac {1}{2}}(\Delta _{1}^{2}\sigma _{1}^{2}+\Delta _{2}^{2}\sigma _{2}^{2})\right)\operatorname {erfc} \left(i{\frac {\Delta _{2}\sigma _{2}^{2}-\Delta _{1}\sigma _{1}^{2}}{\sqrt {2\sigma _{1}^{2}+2\sigma _{2}^{2}}}}\right)\end{aligned}}$ .

Since there are nice efficient ways to calculate the error function, I don't need to do any time-consuming numerical integration for the second-order nested integral. But I am stuck coming up with a similar solution to a third-order nested integral.

$\int _{-\infty }^{\infty }dt_{3}\int _{-\infty }^{t_{3}}dt_{2}\int _{-\infty }^{t_{2}}dt_{1}\exp \left(-{\frac {t_{3}^{2}}{2\sigma _{3}^{2}}}-i\Delta _{3}t_{3}\right)\exp \left(-{\frac {t_{2}^{2}}{2\sigma _{2}^{2}}}-i\Delta _{2}t_{2}\right)\exp \left(-{\frac {t_{1}^{2}}{2\sigma _{1}^{2}}}-i\Delta _{1}t_{1}\right)$

Any help would be greatly appreciated. mislih 20:32, 28 August 2009 (UTC)

I think you can express the triple integral in terms of the erf function [or antiderivatives of it] (and, possibly, even the analog multiple integral). I'll give you some hints here below.

• You want an integral of a function of the form exp(-Ax²-By²-Cz²-2ax-2by-2cz) on the set {x<y<z}. Here A,B,C are positive constant, and a,b,c are actually imaginary numbers. Notice that {x<y<z} is the intersection of two half-spaces, with an angle of π/3.
• Since the integral is clearly an entire function of a,b,c, you can do the computation just for all real a,b,c and then extend by analytic continuation the result to all complex a,b,c; the extension will be automatic (recall that erf(z) is an entire function).
• Now complete the square, that is, here, use the identity exp(-(u+k)²)=exp(-u²-2ku)exp(-k²), and then change variable. This way you transform the initial into the integral of exp(-x²-y²-z²) on a suitable domain which is the intersection of two affine half-spaces, whose angle and position wrto the origin, of course, depend on all the initial coefficients (there is also a factor coming from the complete-the-square thing and from the Jacobian of the affine transformation of the variables).
• Thanks to the rotation invariance of exp(-x²-y²-z²), you can rotate the domain of integration so that its edge (i.e. the singular line of the boundary) is a line parallel to the z axis. A partial integration in the z variable gives you a factor sqrt(π) in front of the double integral of exp(-x²-y²) on an angular domain on the x,y plane (that is, the orthogonal section of the previous three-dimensional domain).
• Again by rotation invariance, you may assume that the angular domain is bounded by a half-line parallel to the y-axis, and you are left with the integral of exp(-x²-y²) on this domain. A first integration produces a partial integral of the form erf(αx+β)exp(-x²), that you still have to integrate in x>0. (I don't see yet a formula for the latter antiderivative of erf(αx+β)exp(-x²); but if the matter is just about numerics, it certainly has a very fastly convergent power series expansion).

I hope it's clear enough; of course to complete the computation you have some work to do in order to write everything in terms of the initial data. In case ask here again and we'll try to answer.--pma (talk) 21:47, 29 August 2009 (UTC)

August 29

Why use Poisson distribution as an approximation to binomial?

I understand that the Poisson distribution can be used as an approximation to a binomial distribution (if n is sufficiently large and p sufficiently small). I've seen it demonstrated, and I think I get the math involved. But why would you want to use this approximation instead of just using the binomial? What is the advantage of using the Poisson? —Preceding unsigned comment added by 118.241.57.8 (talk) 02:03, 29 August 2009 (UTC)

One could just consider it an out-of-date notion, relevant for a time when calculation was more difficult.Julzes (talk) 02:16, 29 August 2009 (UTC)

One reason is you might have no clue how big n is, or p, but you may still have a good estimate of λ = np. Michael Hardy (talk) 02:32, 29 August 2009 (UTC) ...and besides, obviously it's simpler. Michael Hardy (talk) 02:33, 29 August 2009 (UTC)

The last answer is better than mine.Julzes (talk) 03:56, 29 August 2009 (UTC)

• As Michael said, the Poisson distribution is much simpler. Why should you (the mathematician, not the computer) do calculations involving two parameters, and think about the problem in terms of two parameters, when all the relevant information is contained in one?
• The faster computers become, the more difficult the calculations people use them for. Calculating just one value of a distribution is of course easy, but it's not unthinkable a problem would require billions of such calculations. In such cases there may be a significant difference between evaluating a binomial distribution, which requires 3 factorials, and Poisson which only requires one.

-- Meni Rosenfeld (talk) 21:43, 29 August 2009 (UTC)

The binomial distribution is (a limiting case of) the hypergeometric distribution for large populations, and the poisson distribution is (a limiting case of) the binomial distribution for large samples. You may like to study Cumulant#Cumulants_of_some_discrete_probability_distributions to see that the relationship between binomial and poisson distributions are like the relationship between ellipses and parabolas. The parabola can be used as an approximation to an ellipse. But why would you want to use this approximation instead of just using the ellipse? Bo Jacoby (talk) 22:47, 29 August 2009 (UTC).

The prospective user of a quick approximation to something where an exact calculation is possible but may take more time needs to keep in mind that the trade-off may not be a good one because of the need to keep track of errors. If the approximation really is good enough on its face, that's one thing; but where there is uncertainty it may be necessary in very complicated cases to do some sort of prior study of the nature of the trade-off before one decides whether to use the quick approximation.Julzes (talk) 04:00, 30 August 2009 (UTC)

Some folk here might be interested to check out Stein's method, which is one way of generalizing the binomial -> poisson limit. HTH, Robinh (talk) 08:29, 30 August 2009 (UTC)

The Tychonoff theorem

There's every reason to believe that the product space X is (in terms of sets) different from all the spaces X(alpha) in the family. Now take the product space X of the collection of all compact spaces. X is compact hence it belongs to the collection, and so X is different from X (in terms of sets). Is this where proper class come in? Standard Oil (talk) 12:35, 29 August 2009 (UTC)

Right. Within the category of (compact) topological spaces you are only allowed to do products of families of spaces indicized by sets. Indeed what you proposed is just a topological version of a paradox that you may repeat in all versions: the sup of all ordinals is not an ordinal; the union of all sets is not a set.. &c. This does not mean that you can't endow a proper category with a structure that mimes the usual "small" structures (a group structure, a topological structure and so on).--pma (talk) 13:10, 29 August 2009 (UTC)

Question about sets

Hey, I've got a logic question relating to sets that I havn't been able to quite figure out. Maybe someone here could help me. OK, here goes:

There are 6 hypothetical sets (some of them are real sets, but some of them are non-standard). These are N, W, Z, Q, I, and R.

${\sqrt {17}}$ belongs to sets R and I,

$0.86$ belongs to sets R and Q,

${\sqrt {64}}$ belongs to sets R, Q, Z, W, and N,

$-10/2$ belongs to sets R, Q, and Z.

I've gathered that set R are all real numbers, set I are irrational numbers, set Q are rational numbers, and set N are natural numbers. I am have some difficulty with sets W and Z, however. Can anyone help? Thanks in advance, --Anonymous —Preceding unsigned comment added by 做工器 (talk • contribs) 19:59, 29 August 2009 (UTC)

$\mathbb {Z}$ is standard notation for the integers (it comes from German). I've never come across "W" as a standard name for a set before and can't guess what it would be from the information provided. --Tango (talk) 20:07, 29 August 2009 (UTC)

Ah, Blackboard bold says it could be "whole numbers". I guess whoever is using this notation uses W to refer to whichever of {0,1,2,...} or {1,2,3,...} isn't being called the natural numbers (definitions vary, you see - some people consider 0 natural, some don't). --Tango (talk) 20:09, 29 August 2009 (UTC)

The sets in the order presented are contained within the next one - so N would be {1,2,...} and W is {0,1,2,...}. Acb314 (talk) 10:55, 30 August 2009 (UTC)

... maybe, but Q (whatever it is) cannot be contained in I (whatever it is) since Q contains at least three given members that are not in I. Logically, there is no way to infer the entire contents of these hypothetical sets from a few members. We can, however, say that a minimal solution is R={√17, 0.86, 8, -5}; I={√17}; Q={0.86, 8, -5}; Z={8, -5}; W=N={8}. Gandalf61 (talk) 12:06, 30 August 2009 (UTC)

I think "the order presented" means the order in the statement "${\sqrt {64}}$ belongs to sets R, Q, Z, W, and N". That makes Acb314's argument as good as any, but it's of course true that we can't know much about these sets. —JAO • T • C 12:17, 30 August 2009 (UTC)

That was lack of reading comprehension on my part -indeed Q will not be contained in I! Assuming N, Z, Q, R have the usual meanings, which they seem to, N is inside Z inside Q inside R, so my logic was that N would also be inside W from the order they were written in - not mathematically rigorous I know. I throws a spanner in the works unfortunately, although I is "bigger" than Q and contained within R. I agree that there isn't enough information in the examples to tell what N and W are just from those. Acb314 (talk) 14:31, 30 August 2009 (UTC)

Average Value of a Function

Hello, someone asked me a question about this and I have been thinking about it ever since. Working only in the reals, the question is to find the average value of a function over a given set. Obviously, if the set has a finite number of points, then using the definition of average, I can just evaluate at each point and then take the average. If the set is a finite interval, then we just integrate over the interval and then divide by the length of the interval. Extending this, if the set is a bounded region in R^n, then I just integrate over the set and then divide by the volume of that region. Here are the questions:

1.To find the average, it is okay to divide by the Lebesgue measure of the underlying set over which we integrate, right? Is it correct to say, for a bounded set, that to find the average value of a function over that set, just integrate over that set and divide by its Lebesgue measure, assuming that its Lebesgue measure is nonzero?

2.If this is correct, then how do we deal with sets that have measure zero? If the set is finite, we are good. But what if the set has an infinite number of points with measure zero, like the Cantor set or the rationals in [0,1]? I don't want to be dividing by zero but I can't simply evaluate at each point and take the average.

3.In addition, how do we deal with sets with infinite Lebesgue measure? I am sure I can't also "divide" by infinity after the integration. For example, what is the average value of $f(x)=e^{(-x^{2})}$ over $[0,\infty ]$?

4.Furthermore, what is meant when we are asked to find the average value of a function without specifying the region? Is it safe to assume that the entire domain of the function is meant or is there some other standard? Thanks!-Looking for Wisdom and Insight! (talk) 20:52, 29 August 2009 (UTC)

#1 is correct not only for bounded sets, but for any set whose measure is finite. In #3, the way it's usually done involves conditional convergence, so you might get different answers depending on how the region of infinite volume is approached. The idea is: take an average of a subset of finite measure, then take a limit as as that subset in some way approaches the whole domain. There's no non-uniqueness problem if the function is everywhere nonnegative, nor if the positvie and negative parts both have finite integrals; it's only when both are infinite that there is such a problem. Simple example:

$\lim _{a\to \infty ,\ b\to -\infty }\int _{b}^{a}{\frac {x\,dx}{1+x^{2}}}.$

If b = −2a, then the limit is (1/2)log(1/4), but if b = −a, then the limit is 0. See also Cauchy principal value. Michael Hardy (talk) 21:19, 29 August 2009 (UTC)

[ec] I don't think the term "average value of a function" is generalizable to the kind of cases you are thinking about. For these cases it may be more useful to consider the expected value of a function of a random variable, where the latter follows a given distribution.

1. This seems correct.
2. There is probably no generic solution. For a countable set you may choose to assign a weight for each point, or choose an enumeration and take $\lim _{n\to \infty }{\frac {1}{n}}\sum _{i=1}^{n}f(q_{i})$. For a non-countable set you may want to choose a countable subset of "key points" (for the cantor set, a natural choice is the set of interval endpoints).
3. You can choose a weighting function with a finite integral. In some cases (e.g. periodic functions) it is appropriate to consider $\lim _{t\to \infty }{\frac {1}{t}}\int _{0}^{t}f(t)\;dt$ or $\lim _{t\to \infty }{\frac {1}{2t}}\int _{-t}^{t}f(t)\;dt$.
4. It seems safe to assume the entire domain.

-- Meni Rosenfeld (talk) 21:25, 29 August 2009 (UTC)

With periodic functions, I'd just take the average value over one period. Michael Hardy (talk) 02:20, 30 August 2009 (UTC)

Fun little problem

Here's a fun little problem I saw yesterday. Which is bigger $e^{\pi }\,$ or $\pi ^{e}\,$? No calculator, computer, or search engine allowed; just your brain and some paper. Try to find a proof, i.e. no estimations allowed. It took me a little while to think of a proof, but in the end it's not that difficult. I would like to see if anyone can come up with a proof that uses a different method to mine. If this is too easy for you, then after you've posted a proof (which must use a different method to any others that yours follows) then please turn your attention to my serious problem above. ~~ Dr Dec (Talk) ~~ 23:13, 29 August 2009 (UTC)

Consider the inequality $\scriptstyle e^{x}>x^{e}$, that is $\scriptstyle x>e\ln x$. So we're interested in the sign of $\scriptstyle f(x):=x-e\ln x$, which is nicely continuous and differentiable for $\scriptstyle x>0$. $\scriptstyle f'(x)=1-e/x$ so the only solution to $\scriptstyle f'(x)=0$ is $\scriptstyle x=e$. $\scriptstyle f(e)=0$, of course, and it's readily seen that this is a minimum. Hence, $\scriptstyle e^{x}>x^{e}$ for all positive $\scriptstyle x$ (except $\scriptstyle x=e$). —JAO • T • C 23:40, 29 August 2009 (UTC)

Please. It's $\scriptstyle e\ln x$, not $\scriptstyle e\,lnx$. The backslash BOTH prevents italicization and provides proper spacing. Michael Hardy (talk) 06:24, 30 August 2009 (UTC)

Oops, of course. I thought my spacing problems were somehow due to \scriptstyle, so I just thought it weird instead of thinking myself stupid. Thanks for setting me straight on that; I fixed those uglies now. —JAO • T • C 08:59, 30 August 2009 (UTC)

Essentially the same as the above. Taking logs we see we are comparing $\pi$ to $e\log \pi$. Now, differentiate $x/\log x$ to compare it to $e$. Phils 23:46, 29 August 2009 (UTC)

But I'd like to see a non-log proof, if at all possible. ~~ Dr Dec (Talk) ~~ 23:59, 29 August 2009 (UTC)

It is easy to re-write Jao's proof to not use logs. Let $f(x)=e^{x}x^{-e}$ for x > 0. We have $f'(x)=e^{x}x^{-e}(1-e/x)$. This is zero exactly when x = e, so f has a unique minimum at e. Proceed as before. Eric. 98.207.86.2 (talk) 01:18, 30 August 2009 (UTC)

Declan, you've been posting some interesting topics, but they really aren't Reference Desk questions. Have you thought about starting a blog? 67.122.211.205 (talk) 06:16, 30 August 2009 (UTC)

Really? I know that this one is boarderline, but all of my other threads have been perfectly valid reference desk question. My latest questions have been The real numbers and subsets of the real numbers, Commutivity of addition, Associativity of composition and Line bundles and cohomology. I've just read them all again, and they seem perfectly valid reference desk question. I don't know, what do other people think of these five questions? ~~ Dr Dec (Talk) ~~ 09:54, 30 August 2009 (UTC)

I think Dr Dec's questions are entirely appropriate for the Mathematics Reference Desk, and I welcome him as a new regular contributor here. But this is off-topic so if any further discussion is required. let's take it to Wikipedia talk:Reference desk. Gandalf61 (talk) 11:55, 30 August 2009 (UTC)

I agree. I'll admit that this thread is a very unusual RD thread, posing a problem the OP has already solved. But as the goal was not to see if others can solve it, but rather to see if there are other ways to solve it, I took it as a sincere fragment of a pursuit of deeper understanding. Which, after all, is what the RD is intended for, isn't it? (Dec's questions have also certainly been more interesting than the typical homework questions, but I don't think that is disputed.) —JAO • T • C 12:04, 30 August 2009 (UTC)

Fun little macro with a (math) issue...

I posted this here because this is more a math question than a computer question, though it involves both...

Im making a little macro for fun with AutoIt v3 which opens up MsPaint and draws. There are currently values called xMomentum and yMomentum which guide the pen. Basically the program starts with xMomentum = 0, yMomentum = 0, then each step it generates a random number between -2 and 2 and adds that to xMomentum , and does it again seperately to yMomentum, then the mouse is moved that many pixels.

What i am seeming to get is momentums that build up, so the pen starts to fly across the screen in a (non straight) line. Although a few straight bits are cool, id like more changes to happen. My thought is this:

I could have variables xTop and xBottom, where the random number is no longer between -2 and 2 its between xTop and xBottom. I would adjust these based on xMomentum so that the "faster" the pen goes in either X direction, the more likely it is to change direction slightly. (of course, i would do the same for corresponding y values.)

I dont know how to do this mathematically though, any ideas? All thoughts are welcome! Thanks! :)

97.112.117.236 (talk) 23:38, 29 August 2009 (UTC)

There are a lot of ways you could do this. One way would be to subtract k times the momentum from the momentum at each step where k is a small positive constant. This is basically like introducing a frictional force proportional to the velocity.

Another way would be to just cap the momentum. If at any step it exceeds the cap, reduce it to the cap. Rckrone (talk) 00:19, 30 August 2009 (UTC)

I spoke shortly with my father, and he unearthed a problem. Lets say i add a random integer in the set [-2, 2] as i did before. this is added to a momentum, so there are 3 integers which do not decrease the speed, and 2 that do. Thus 0 needs to be removed from my random integers.

I sort of like the frictional thing.... but can i use that and still get the possibility of both increase or decrease? Maybe i should add my random number (-2, -1, 1, or 2) and then do this subtraction? this would have the possibility of speeding up or slowing down, but add the friction as well. am i correct?

97.112.117.236 (talk) 00:26, 30 August 2009 (UTC)

Yeah I meant add the friction part in addition to what you're already doing. With the friction alone it would be pretty dull: just slowing to a stop and then sitting still. Using the set {-2, -1, 0, 1, 2} doesn't cause any problems, although {-2, -1, 1, 2} works fine too. As long as the average is 0 the momentum won't favor one direction or the other. Rckrone (talk) 00:43, 30 August 2009 (UTC)

Perhaps so. I have opted to eliminate 0 just in case :) 97.112.117.236 (talk) 00:49, 30 August 2009 (UTC)

Resolved

August 30

Principle of mathematical induction

Please solve this problem of mathematical induction

sinx + sin3x + ... +sin(2n-1)x =sin^2nx/sinx —Preceding unsigned comment added by Tipusultan11 (talk • contribs) 07:08, 30 August 2009 (UTC)

If you can tell us where you are getting stuck in solving this problem maybe we can help you with it. What have you tried to do so far? Do you understand induction? Eric. 98.207.86.2 (talk) 07:47, 30 August 2009 (UTC)

Have you to use trig identities? You can do it easily using Euler's formula‎‎ Dmcq (talk) 11:37, 30 August 2009 (UTC)

...and powers thereof. ~~ Dr Dec (Talk) ~~ 11:43, 30 August 2009 (UTC)

As an induction problem it seems simpler than taking powers of the Euler formula, expanding, then comparing terms in the binomial expansion. The base case would be when n = 1, i.e.

$P(1):\ \sin(x)={\frac {\sin ^{2}(x)}{\sin(x)}}\ ,$

and this is clearly true. Next, assume that P(k) is true and prove that this implies that P(k+1) is also true. Well for the case P(k) we have

$P(k):\ \sin(x)+\sin(3x)+\cdots +\sin((2k-1)x)={\frac {\sin ^{2}(kx)}{\sin(x)}}\ .$

We assume that this equality holds. To prove that P(k) implies P(k+1) we simply need to show that

${\frac {\sin ^{2}(kx)}{\sin(x)}}+\sin((2k+1)x)={\frac {\sin ^{2}((k+1)x)}{\sin(x)}}\ ,$

which I won't do because I'm not going to do all of your homework for you ;oP ~~ Dr Dec (Talk) ~~ 12:30, 30 August 2009 (UTC)

...and this last step isn't too hard if you use the formula $\sin(x+y)=\sin(x)\cos(y)+\sin(y)\cos(x).$ ~~ Dr Dec (Talk) ~~ 12:42, 30 August 2009 (UTC)

Maybe I'm a stickler for detail, but the conjecture, as written, is false for x=kπ, k integer. --Stephan Schulz (talk) 12:44, 30 August 2009 (UTC)

It's not false, it's true for all x. I assume your problem is that the RHS has a zero denominator for x = mπ (m an integer). Well there's no problem: it's perfectly well defined. The limit of the RHS exists, and is equal to zero. See the article on L'Hôpital's rule. (In fact you don't even need that: just completing the inductive steps that I have outlined above show you that the singularities at x = mπ are removable.) ~~ Dr Dec (Talk) ~~ 12:50, 30 August 2009 (UTC)

Sure, the function can be steadily continued at the undefined points, as limits from both sides agree. But as written, it is not so continued. Even your base case is, strictly, wrong. As I wrote, stickler for details ;-) --Stephan Schulz (talk) 13:15, 30 August 2009 (UTC)

Why is the base case wrong? It has a removable singularity - there is no problem! The base case doesn't require steadily continuing anything. We are canceling functions. I've not tried to evaluate the numerator and denominator before canceling. Would you argue that x/x = 1? I hope not! Well, you're trying to say that there's a problem with the statment that x/x = 1 by saying that the base case isn't well defined. In fact the expression x/x is perfectly well defined; its value at x = 0 is indeterminate, but can be calculated as a limit. Where did I evaluate anything above? ~~ Dr Dec (Talk) ~~ 13:18, 30 August 2009 (UTC)

To make Stephan happy, let's rephrase the problem. Show, using mathematical induction, that for any positive integer n, and any real number y:

$\lim _{x\to y}(\sin(x)+\sin(3x)+\cdots +\sin((2n-1)x))=\lim _{x\to y}{\frac {\sin ^{2}(nx)}{\sin(x)}}\ .$

I always tacitly assume the idea of a limit when I evaluate a function. I'm sure most other mathematicians do too. ~~ Dr Dec (Talk) ~~ 13:30, 30 August 2009 (UTC)

I suspect you're missing a negation somewhere above. But yes, for me (the function defined over the reals by the formula) x/x is different from (the function defined over the reals by the constant formula) 1. In set notation, 1 is $(R\times \{1\})$, and x/x is $(R\backslash \{0\}\times {1})$. The one is a proper subset of the other (and so can of course be extended), and hence they are not the same. --Stephan Schulz (talk) 13:54, 30 August 2009 (UTC)

...and this helps to answer Tipusultan11's question, how? (p.s. So what if I haven't not missed no negatives?) ~~ Dr Dec (Talk) ~~ 14:02, 30 August 2009 (UTC)

Stephan, let's stop this now. You can carry on worrying about the validity of expressions like x/x, so that I can try to help answer people's questions here on the RD. Maybe Tipusultan11 should just hand his homework in with the solution: Question not well defined? ~~ Dr Dec (Talk) ~~ 14:08, 30 August 2009 (UTC)

For OP's interest, it will probably help him get full marks if he puts sin x != 0 somewhere. --91.145.89.58 (talk) 16:28, 30 August 2009 (UTC)

...meaning? ~~ Dr Dec (Talk) ~~ 16:39, 30 August 2009 (UTC)

In the case OP has to compose math test answers, it will steer him away from all trouble if he mentions that there's no equality when sin x = 0; x = pi * n. Forgive if my english is bad. --91.145.89.58 (talk) 16:52, 30 August 2009 (UTC)

Dr Dec. You write: "I always tacitly assume the idea of a limit when I evaluate a function. I'm sure most other mathematicians do too". How can you be that sure after meeting a mathematician who do not tacitly assume the idea of a limit? Your tacit assumption may lead you into trouble. Stephan's objection is perfectly valid. Bo Jacoby (talk) 17:29, 30 August 2009 (UTC).

Even if Stephan were a mathematician, the statement "most other" does not mean "all". Besides that, Stephan is a computer scientist and not a mathematician and as such does not count when it comes to qualifying the valididity of statements involving mathematicians (exhaustive or not). One is not going to go very far with one's mathematical thinking if one stops at the first problem. If one were to be presented with a problem that involved indeterminate forms, then one may stop if one chose. However, I might prefer to proceed with common sense. ~~ Dr Dec (Talk) ~~ 19:01, 30 August 2009 (UTC)

When composing functions with restricted domains, the domain of the composition is the largest set on which all the intermediate functions are defined. For example, working over the real numbers, the function given by f(x) = (√x)² is only defined for x≥0, even though it can be extended to a function defined for all x. (Such an extension must of course be non-unique, which is why we cannot ask what f(-2) is.) If a student claimed that (sin x)/x was defined at 0, I would mark it as incorrect. And I don't mean this pedantically; I don't believe I've ever said "And since (sin x)/x is 1 at 0..."; instead I would only say "And since (sin x)/x goes to 1 at 0..." The distinction between a function and its continuation, completion, or closure is an important one which becomes more important the deeper you go in mathematics; in my experience a practicing mathematician would be unlikely to blur this line, even in informal discussions. Tesseran (talk) 05:53, 31 August 2009 (UTC) [Edit: fixed typo, thanks Pt.] Tesseran (talk) 18:51, 31 August 2009 (UTC)

As functions on R, sin²x/sinx is definitely not equal to sinx. The OP's teacher may or may not care if he/she attends to that kind of detail, but if for instance this were an analysis class that would not fly. 67.100.146.151 (talk) 08:16, 31 August 2009 (UTC)

By the way, (sin x)/x goes to 1 and not 0, as x→0. — Pt _(T) 10:35, 31 August 2009 (UTC)

Program the function f by f(x) = x/x. Then f(x) evaluates to one when x is not equal to zero, and gives either an "zero divide" error message, or the value 0 (because of the zero in the numerator) when x = 0. Which programming languages return f(0) = 1 by evaluating lim_x→0f(x) ? Bo Jacoby (talk) 22:07, 1 September 2009 (UTC).

Big O of sin x

$\sin(x)={\mathcal {O}}(f(x)){\mbox{ as }}x\to \infty$

Can anything meaningful be said about f(x)? SpinningSpark 13:13, 30 August 2009 (UTC)

Could you please qualify your notation. I know ${\mathcal {O}}$ as a function ring. Do you mean O(f)? ~~ Dr Dec (Talk) ~~ 13:19, 30 August 2009 (UTC)

Sorry, I meant Big O notation. I copied the style from sorting algorithm so someone should take a look at that article if it's wrong. SpinningSpark 13:34, 30 August 2009 (UTC)

Well the notation means that, for sufficiently large x, $|\sin(x)|\leq k|f(x)|$ for some fixed positive constant k. This isn't very illuminating. The Big O notation is usually used to describe growth, but since the sine function doesn't grow it's a bit of a non-starter. For example, if $g(x)=O(f(x))$ then we know that the infimum of $|f(x)|$, for very large x, grows like the infimum of $|g(x)|$. But the infimum of $|\sin(x)|$ is zero: as small as it gets; so this doesn't tell us anything. Why did you ask the question? Can you give us more information and context? ~~ Dr Dec (Talk) ~~ 13:58, 30 August 2009 (UTC)

(ec) Saying sin(x) = O(f(x)) means that there is some k such that for x sufficiently large we have |sin(x)| ≤ k|f(x)|. This will always be satisfied provided that 1 ≤ k|f(x)|, which is the same as 1/k ≤ |f(x)|. So sin(x) = O(f(x)) means that the values of f(x) will eventually stay outside of some interval around 0. That is, there is some ε such that ε ≤ |f(x)| for x sufficiently large. It's not hard to see that this condition is equivalent to sin(x) = O(f(x)). Another way to say this is that the lim inf of f(x) as x goes to infinity is not zero. Staecker (talk) 14:05, 30 August 2009 (UTC)

Is this true? $|\sin(x)|$ will become zero infinitly many times, not matter how large x is. For example, try $f(x)=2\sin(x).\,$ Clearly $|\sin(x)|\leq |2\sin(x)|$ for all x (where k = 1), but f is zero infinitly many times and does not have a well-defined limit. I guess that the statement might be that If a limit exists then it will not be zero. ~~ Dr Dec (Talk) ~~ 14:18, 30 August 2009 (UTC)

While $\scriptstyle \lim _{x\rightarrow \infty }2\sin x$ does not exist, $\scriptstyle \liminf _{x\rightarrow \infty }2\sin x$ certainly does; it is –2. But shouldn't it be $\scriptstyle \limsup _{x\rightarrow \infty }|f(x)|$ (which is 2 in your example) that should not be 0? —JAO • T • C 14:30, 30 August 2009 (UTC)

Good point! I missed the inf. In that case then

$\lim _{x\to \infty }\inf |f(x)|\neq 0$

as Staecker quite rightly says. ~~ Dr Dec (Talk) ~~ 14:42, 30 August 2009 (UTC)

I think Dec's objection is valid- |f(x)| = |2sinx| does have lim inf equal to zero, but it is certainly the case that sin(x) = O(f(x)). So there are more possibilities for f. Perhaps the correct weaker condition is that: $\liminf _{x\to \infty }|f(x)|\neq 0$ or $\liminf _{x\to a}|f(x)|=0$ for all a being an integer multiple of π. (Or something like that.) Staecker (talk) 17:38, 30 August 2009 (UTC)

Sorry Dr. Dec, there is no context, it just arose out of some idle scribblings on scrap paper. SpinningSpark 14:38, 30 August 2009 (UTC)

Just to check that I am following this, would it be right to say,

$\sin(x)=\mathrm {O} (1)){\mbox{ for }}k\geq 1$

$\because |\sin(x)|\leq k|1|$

SpinningSpark 15:29, 30 August 2009 (UTC)

That's correct. To say that a function f is such that f(x) = O(1) means that, for sufficiently large x, f is bounded, i.e. |f(x)| ≤ k for all sufficiently large x. Like I said: the big O notation is used to describe the growth of a function. Say that f(x) = O(g(x)) then what does that mean? Well consider the cone given by y = ±g(x). Then f(x) will, for sufficiently large x, always stay inside a cone of that shape. So, e.g., if f(x) = O(x) then, for sufficiently large x, f(x) will always lie inside a cone given by the two straight lines y = ±kx, for some k (i.e. f(x) ≤ k|x|). ~~ Dr Dec (Talk) ~~ 15:50, 30 August 2009 (UTC)

The usefulness of sine and cosine

So, let's assume that sine and cosine haven't been invented. We want to try to express, in big O notation, the statement that the limit of the infimum of some smooth function f is non-zero (whilst also, theoretically, being able to take any other real value). How could that be done if we didn't know about sine or cosine? (If we don't know about sine or cosine then we couldn't use a power series for them either!) ~~ Dr Dec (Talk) ~~ 14:51, 30 August 2009 (UTC)

That's $\liminf f>0$. The big-Oh notation says something about absolute size. You cannot separate statements about the liminf and limsup. Besides, $O(\cdot )$ only implies an upper bound. There is a corresponding notation for lower bounds ($\Omega$, see Big Oh notation), although it's not as commonly seen. The definition also has absolute values. I don't think it is worthwhile trying to express non-quantitative statements in big-Oh notation. Phils 16:16, 30 August 2009 (UTC)

After reading the above more carefully, I realize that you are referring to the fact that $\sin x=O(f(x))$ implies $\liminf |f|>0$, i.e. the function to tested is in the $O(\cdot )$. In that case $1=O(f(x))$ works just as well. My remark about absolute value still holds. Phils 16:26, 30 August 2009 (UTC)

In computer science when we want to be precise, for a given function f, we define O(f) as the set of functions h for which there exists a constant K such that for sufficiently large x, $h(x)/f(x)<K$. So we would then say $\sin \in O(1)$ where we are abusing notation and writing "1" to denote the constant function $x\mapsto 1$. Re Declan: big O notation in the treatment I'm accustomed to is insensitive to additive constants. I don't think it would be used in the way you are asking. 67.122.211.205 (talk) 01:52, 31 August 2009 (UTC)

What mathematics things are vs. what mathematical things do

Is only the latter important? It seems to me like defining 'dog' as something that barks, bites, etc. But not telling what a 'dog' is. On the other hand, the only way of defining 'division' I can thing of is over its output.Quest09 (talk) 16:50, 30 August 2009 (UTC)

...so are you suggesting we define a dog by its output? YUK! :oP ~~ Dr Dec (Talk) ~~ 17:01, 30 August 2009 (UTC)

But surely all objects are defined by their interaction with obsevers anyway? The "are" is only assessed by interaction with the world. --Leon (talk) 19:18, 30 August 2009 (UTC)

Here's an interesting example of this problem (borrowed from a book, but I've forgotten which one EDIT maybe the one in the OUP very short intro series by Gowers). What "is" the king in chess? It's not a particular piece in your chess set, because if you've lost your king you can use any old bit of junk lying around. Is a king anything more than "something which can move only a single square in one direction except for castling, and...."? 87.194.213.98 (talk) 19:25, 30 August 2009 (UTC)

If you can find a difference between the two then perhaps you can find a scientific explanation of Transubstantiation ;-) Dmcq (talk) 19:53, 30 August 2009 (UTC)

Mathematicians are concerned with the fact that the reals form a complete ordered field (what they do), and often impatient with things like the fact that the reals can be regarded as equivalence classes of Cauchy sequences or as Dedekind cuts, etc. (what they (supposedly?) are).

The analogy to the king in chess is on the mark.

But this might be argued about philosophically. Michael Hardy (talk) 01:35, 31 August 2009 (UTC)

The fact that you have mentioned two completely different constructions of the reals that are both very popular shows how unimportant constructions are to the vast majority of mathematics. "The reals are the completion of the field of fractions of a set with at least one element and a successor function." is the most useful definition (albeit lacking a few key details for the sake of conciseness). --Tango (talk) 01:43, 31 August 2009 (UTC)

That's exactly my point in mentioning the two examples. Michael Hardy (talk) 21:05, 31 August 2009 (UTC)

The issue you're up against is the endless quest for syntactic proofs of theorems, since who can really say anything about the semantics of uncountable entities like real numbers, to say nothing of (e.g.) Hilbert spaces (the substrate of quantum theory)? Jean-Yves Girard has some pretty good rants along these lines, e.g. [1] 67.122.211.205 (talk) 03:11, 31 August 2009 (UTC)

PUZZLING PUZZLE

Guys I have been trying to solve this for a long time but I am not able to..please help me with this puzzle.(Thot it would fit in here better rather than the misc. desk)

The diagram shows a 4 x 5 grid with some filled cells. Find the numbers in the remaining cells according to the following rules: a) Each number can only take values from 1 to 5. There are 4 such full sets(1-5). b) Sum of each row is same. c) Sum of each column is same. d) Adjacent numbers cannot be the same.

-	-	-	1	5
-	-	-	-	1
2	4	-	-	-
4	-	-	-	-

thanks —Preceding unsigned comment added by 117.193.136.45 (talk) 16:53, 30 August 2009 (UTC)

An impractical but possible way to solve any such problem is to formulate and solve it as an Integer Programming Problem. There are softwares which automatically solve IPP's and you could use those.--Shahab (talk) 17:46, 30 August 2009 (UTC)

But, I would hope, 117.193.136.45 wants some kind of mathematical algorithm to solve the problem. I know that computers use algorithms, but that's just no fun! I was trying to find a connexion with sudoku. ~~ Dr Dec (Talk) ~~ 18:28, 30 August 2009 (UTC)

1	3	5	1	5
5	2	4	3	1
2	4	2	5	2
4	3	1	3	4

Just used good old fashioned trial and error. It's clear that the row and column sums must be 15 and 12 respectively.--RDBury (talk) 19:33, 30 August 2009 (UTC)

My solution was

2	4	3	1	5
4	2	3	5	1
2	4	3	1	5
4	2	3	5	1

which has a more obvious pattern. I can't see how it could have taken any length of time at all. Dmcq (talk) 19:41, 30 August 2009 (UTC)

Umm Dmcq, what about the constraint: d) Adjacent numbers cannot be the same? -hydnjo (talk) 19:49, 30 August 2009 (UTC)

Oops sorry, I should have read through to the end. Silly me Dmcq (talk) 20:09, 30 August 2009 (UTC)

I found the same solution as RDBury, also by trial and error. Note that since the column sums must be 12, there are only three ways to complete column 1 and three ways to complete column 5. That gives you 9 starting points, most of which quickly lead to dead ends. Gandalf61 (talk) 19:58, 30 August 2009 (UTC)

Damn. I followed this from Science to Miscellaneous, but didn't realize (didn't read) that is had also been moved here. However, under the premise that it is better the teach a man to fish than to give him a fish, I'll copy my answer here. -- Tcncv (talk) 23:59, 30 August 2009 (UTC)

There is a solution (and only one), but it would spoil the fun for me to give it outright. One way to find it is to use a backtracking approach, which is a way to solve many logic problems like this. (See also the Ariadne's thread article.) First, make a choice and see where that takes you. Fill in other cells that can be derived from that choice. When you cannot fill in any more, make another choice. If you come to a dead end, backtrack to your most recent choice and select another. If you run out of options at that point back up further and select another option for the previous choice.

This puzzle has a couple of good places to start. For ease of reference, I'll refer to the cells by row and column – (1,1) through (4,5). First you need to figure out the needed row and column sums. Since the sum of all of the available numbers is 4 × (1 + 2 + 3 + 4 + 5) = 60, the rows must each total 15 and the columns 12. Take a look at column 1. You have two numbers already whose sum is 6, so the remaining two numbers in cells (1,1) and (2,1) must also equal 6 (giving a column total of 12). Your options are 1 & 5, 2 & 4, and 5 & 1. The combination 3 & 3 would be an immediate adjacency violation; so would 4 & 2, given the existing value in cell (3,1). Tentatively select one pair and fill them in, keeping track of the other options in case you have to backtrack.

Now take a look at row 1. You already have three of five values, so you can apply the same logic. After that, look at column 2, then row 2, then column 3 etc. As you consider numbers, you can immediately eliminate any that cause adjacency violations. You can also watch for cases where the same number is used more than the allotted four times. That would also prompt you to back track.

Good luck. -- Tcncv (talk) 23:20, 30 August 2009 (UTC)

Sudoku boards and groups with 9 elements

Think about a completed sudoku board: it's a 9 × 9 board where we must fill the board with the numbers {1,…,9} in such a way that the same number can't appear in the same colomn, the same row, or in any of the 3 × 3 sub-squares that make up the 9 × 9 total board. Now, for me, the first two conditions (i.e. the same number can't appear in the same colomn or the same row) remind me of a group table for a group, say G, with nine elements. Is there a correspondence between completed Sudoku boards and groups with nine elements? It's been about five years since I did any finite group theory or number theory, so please forgive my ignorance. But I seem to rememeber subgroups forming little blocks in the table like the 3 × 3 blocks we have on sudoku boards. Is there a 1-1 correspondence between group tables and sudoku boards (I doubt it, but I'm so rusty that I forget). Although, I seem to remember some enumeration results from representation theory of finite groups, but again; it's so long ago that they're nothing more than quiet little bells ringing ~~ Dr Dec (Talk) ~~ 18:51, 30 August 2009 (UTC)

There are only two groups of order 9: the cyclic group and the direct product of two groups of order 3, so no such correspondence exists. If you draw the Cayley table of a group, that is, a grid whose rows and cols correspond to the group elements and whose ij entry is the product of the corresponding elements, a subgroup doesn't look like a "sub-square" in which every element (of the big group) occurs once, so this is also not like the small squares in a sudoku grid. You might be interested in Latin square, a correct sudoku grid is a special kind of Latin square. As for rep thy, you might have character tables in mind perhaps.87.194.213.98 (talk) 19:21, 30 August 2009 (UTC)

I already know about Latin squares, but I thought a group theoretical approach might be interesting. I know, from Lagrange's theorem, that the order of any subgroup must divide the order of the group, so subgroups of order 1, 3 and 9 are all possible. Could you please say a few words as to why there aren't any order 9 groups with subgroups of order 3? ~~ Dr Dec (Talk) ~~ 19:29, 30 August 2009 (UTC)

There are, all groups of order 9 have subgroups of order 3. If <g> is cyclic of order 9 then g^3 generates a subgroup of order 3, and if a group G is the direct product of two groups of order 3 then it has several subgroups of order 3. Have a look at the Mathematics_of_sudoku article, they mention that not-Burnside's lemma can be used to enumerate solutions up to a notion of equivalence. That article also talks about Cayley tables and sudoku grids.87.194.213.98 (talk) 19:40, 30 August 2009 (UTC)

Excellent! I'd seen the maths of sudoku section in the sudoku article, but had some how managed to miss the fact that there was an entire article. I know what I'll be doing for the next half hour... ~~ Dr Dec (Talk) ~~ 21:15, 30 August 2009 (UTC)

There are 2 non-isomorphic groups of order 9. The actual number of Cayley tables that form groups is a lot more than that. The problem is that groups require associativity which puts way more structure on the array than is applicable for Sudoko squares.--RDBury (talk) 19:51, 30 August 2009 (UTC)

If we restrict ourselves to the set {1,...,9} then the only isomorphisms are permutations of the symbols and you can permute the symbols in a completed Sudoko board without breaking it, so I don't think that makes much difference (although rigour requires we mention it). What if we consider quasigroups instead? According to the article there is (modulo permutation of symbols) a 1-1 correspondence between quasigroups and Latin squares. Is there anything interesting to say about those quasigroups that correspond to a Suduko board? --Tango (talk) 01:35, 31 August 2009 (UTC)

As an aside note that all groups of order p² (p prime) are abelian and the converse of Lagrange's theorem holds for all abelian groups.--Shahab (talk) 19:55, 30 August 2009 (UTC)

If G is a group of order p² where p is prime, it must equal to its center. To see this, observe that the center of G, Z(G), is either trivial or equals G, since G/Z(G) cannot be cyclic (here we are using the hypothesis that p is prime). Now note the fact that the center of a p-group (a group of prime-power order) can never be trivial, and therefore G is abelian (the fact that the center of a p-group is non-trivial is often proven using a simple argument with orbits and group actions). --PST 12:13, 31 August 2009 (UTC)

Random trig question

I know that it is possible to perform inverse trigonometric functions with a calculator. I have been told that it is not possible by hand. However, trigonometry existed centuries before calculators were invented. So, is it possible to perform these function by hand? If not, how did the ancient mathematicians do it? Intelligentsium 23:21, 30 August 2009 (UTC)

It is possible, it just depends how much time you have on your hands. For example, in the olden days people used log tables to evaluate logarithms. These were basic grids that once you knew the base of the log (say n) and the value of the variable (say x), would give you the answer (say log_n(x)). Well, in that case, finding a number in the middle of the table that was close to the number you wanted to find the inverse of would tell you to which base, and of which value, it was the inverse of. I assume that trig' functions were the same. It's just like a group table: find the number in the group table and you know what's the inverse of what. ~~ Dr Dec (Talk) ~~ 00:13, 31 August 2009 (UTC)

Well, obviously, when you see a table of values of sine and cosine in a book published in 1908, or 1650, that tells you something. Maybe what was meant was that the methods needed in order to do it efficiently by hand cannot be taught at the very elementary level at which instruction in trigonometry takes place today.

As for log tables, where you looked up log_n(x) if you knew n and x, that's not how it was done because it's not an efficient way to do it. Base-10 log tables were commonplace. If you wanted log₂(3), you used a base-10 log table and found (log₁₀(3))/log₁₀(2). Tables of base-e logarithms also existed, but were not in an appendix to every book. Michael Hardy (talk) 01:28, 31 August 2009 (UTC)

Of course you can compute inverse trig functions by hand, for example with Padé approximants. Stupendous amounts of research went into developing methods to minimize the amount of labor required by those computations. In practice you would look your number up in a table and use linear or quadratic interpolation on the nearest values you could find to the one you wanted. But somebody had to prepare the tables, and they did it without computers... 67.122.211.205 (talk) 01:42, 31 August 2009 (UTC)

And the books of tables all had little errors in them which is why Charles Babbage designed his Difference engine. Dmcq (talk) 07:47, 31 August 2009 (UTC)

Maybe we should add that if it were impossible to compute values of inverse trigonometric functions by hand, then it would be impossible to do so via computers as well. Michael Hardy (talk) 21:07, 31 August 2009 (UTC)

I wouldn't say that. I mean, sure, they are Turing equivalent, but some amounts of hand calculation are just impractical. You can calculate pi to 100's of digits by hand, maybe even 1000's of digits, but not millions of digits. But it can be done with computers. 70.90.174.101 (talk) 00:59, 1 September 2009 (UTC)

August 31

Empty set and phi

So, in my topology class last year, a student referred to the empty set as phi. Then, this past week, one of my professors did the same thing. It seems to me that $\emptyset$ is more like a 0 with a line through it than $\phi$, and the 0 would match up symbolically as well. Is it related at all to $\phi$ or are these two people incorrect in calling it $\phi$? StatisticsMan (talk) 01:04, 31 August 2009 (UTC)

I've not heard "phi" for the empty set, and calling it that is incorrect or at least not standard. However $\varnothing \$ is a common variant on $\emptyset \$ that looks a good bit like $\phi \$. Eric. 98.207.86.2 (talk) 01:16, 31 August 2009 (UTC)

Nor have I. People mis-reading the symbol me thinks. ~~ Dr Dec (Talk) ~~ 01:21, 31 August 2009 (UTC)

Ditto. It is an easy mistake to make, but I've always heard it pronounced as "the empty set" or just "empty". --Tango (talk) 01:26, 31 August 2009 (UTC)

This is my thought when it was just a student in the class saying it last year. When a professor says it, then it makes me wonder, though it still seems wrong. Thanks for your thoughts. StatisticsMan (talk) 02:42, 31 August 2009 (UTC)

From the page Earliest Uses of Various Mathmatical Symbols:

The null set symbol (Ø) first appeared in N. Bourbaki Éléments de mathématique Fasc.1: Les structures fondamentales de l'analyse; Liv.1: Theorie de ensembles. (Fascicule de resultants) (1939): "certaines propriétés... ne sont vraies pour aucun élément de E... la partie qu’elles définissent est appelée la partie vide de E, et designée par la notation Ø." (p. 4.)

André Weil (1906-1998) says in his autobiography that he was responsible for the symbol:

Wisely, we had decided to publish an installment establishing the system of notation for set theory, rather than wait for the detailed treatment that was to follow: it was high time to fix these notations once and for all, and indeed the ones we proposed, which introduced a number of modifications to the notations previously in use, met with general approval. Much later, my own part in these discussions earned me the respect of my daughter Nicolette, when she learned the symbol Ø for the empty set at school and I told her that I had been personally responsible for its adoption. The symbol came from the Norwegian alphabet, with which I alone among the Bourbaki group was familiar.

So no relation to phi. Tesseran (talk) 03:35, 31 August 2009 (UTC)

Indeed. The symbol originated as an Ø and is completely unrelated to phi. Gandalf61 (talk) 07:50, 31 August 2009 (UTC)

Wow. This is the first time I've heard of the symbol for the empty set not being phi. For the past 10, maybe 15 years I was sure it was. Our article really should state explicitly that it's not. -- Meni Rosenfeld (talk) 20:37, 31 August 2009 (UTC)

Done - it now states that explicitly. Gandalf61 (talk) 21:56, 31 August 2009 (UTC)

So, how about the way some computer scientists write zero, that is again Ø? Why do they need it? Did it follow the notation of the empty set ?(via von Neumann construction?) or where did it originate? And is it still used? How they distinguish zero from the empty set? :( --pma (talk) 22:22, 31 August 2009 (UTC)

The slash through a zero is to distinguish it from the letter O. You don't usually need to distinguish between zero and the empty set - it should be clear from context. --Tango (talk) 22:43, 31 August 2009 (UTC)

Method of Characteristics to solve PDEs

First, I do admit that this is a homework problem. The question is to solve
$2tu_{t}+xu_{x}=0$
using the method of characteristics. I have done that and the solution is
$u(x,t)=g\left({\frac {x}{\sqrt {2t}}}\right)$. Then the initial condition $u(x,0)=f(x)$ is given but how does this work? I can't just plug in t=0 in my solution and figure out what g(z) is. How can I resolve this? Any ideas/hints? Did I do something wrong? I checked my solution by plugging it into the PDE and it works fine. Thanks!97.118.56.41 (talk) 03:52, 31 August 2009 (UTC)

The method of characteristics requires the characteristics to be transverse to the initial manifold (here it's $\scriptstyle \mathbb {R} \times \{0\}$, the x-axis). On the contrary here the initial manifold is itself a characteristic line (the support of a ch.). As a consequence, the initial values of a solution can't be prescribed on it, and, indeed, the equation already tells you that any solution defined on $\scriptstyle \mathbb {R} \times \{0\}$ has to be constant there. --pma (talk) 06:57, 31 August 2009 (UTC)

Sequence limit

Resolved

--Shahab (talk) 17:22, 31 August 2009 (UTC)

I want to find the limit of the sequence $(an+b)^{1/n}$ where a and b are constants? I know that $n^{1/n}\rightarrow 1$ but how should I use this result. Thanks--Shahab (talk) 10:10, 31 August 2009 (UTC)

Lim x->0 (1+x)^1/x is e. Use this result, and come back if you have any problems. Rkr1991 ^{(Wanna chat?)} 10:24, 31 August 2009 (UTC)

That limit cannot be used, since we are taking a limit as n goes to infinity here.

However, notice that:

$an\leq an+b\leq (a+b)n$

$(an)^{1/n}\leq (an+b)^{1/n}\leq [(a+b)n]^{1/n}$

for n≥1. That should get you your answer. --COVIZAPIBETEFOKY (talk) 13:08, 31 August 2009 (UTC)

(of course, I assumed that b≥0, here; if it's not, simply reverse the signs) --COVIZAPIBETEFOKY (talk) 13:09, 31 August 2009 (UTC)

Thanks. However I can't resolve all cases still. What will happen if $a+b<0$ or $a<0$? Then I can't use the fact that $a^{1/n}\rightarrow 1$. Is the sequence divergent in those cases?--Shahab (talk) 14:49, 31 August 2009 (UTC)

The expression an + b is just a finite scalling and finite translation of n. Since x⁰ = 1 for any real number x, and you already know that n^1/n → 1 as n → ∞. I would say that (an + b)^1/n → 1 as n → ∞. Of course, depending on your choices of a and b you may have complex values for your sequence. This doesn't matter though because you will always have a real limit. For a complex number z we see that |z^1/n| → 1 as n → ∞ and (more importantly) arg(z^1/n) → 0 as n → ∞, i.e. z^1/n tends to a number with modulus 1 and argument 0, i.e. z^1/n → 1. ~~ Dr Dec (Talk) ~~ 16:10, 31 August 2009 (UTC)

I don't understand how you conclude (an + b)^1/n → 1 from n^1/n → 1. COVIZAPIBETEFOKY's approach made sense to me because I used the results related to algebra of sequences and the sandwitch theorem. Perhaps you can quote which result you are using. Thanks--Shahab (talk) 16:24, 31 August 2009 (UTC)

Indeed, COVIZAPIBETEFOKY's response gives a water-tight argument, and will lead you to a solution. I was trying to give some intuative comments as to why the limit is what it is. It is simple to see that the limit is 1, and I have shown that above. But intuition is not a proof, and of course you should use something along the lines of COVIZAPIBETEFOKY's rigorous analytical argument. If you can't see why my comments might lead you to conclude (and then try to prove) that the limit is one then maybe spend some time thinking it over. It's very natural and very intuative. The bottom line is that x⁰ = 1 for all fixed real numbers x. Basically, the only problem you get is from n^1/n. Once you know that it doesn't pose any problems then you're in the clear. ~~ Dr Dec (Talk) ~~ 17:21, 31 August 2009 (UTC)

My first solution didn't have accurate bounds in the general case; this one works (as long as a≥0, which has to be, since you can't always take an n^th root of a negative number), by taking the absolute value of b.

$a\leq an+b\leq (a+|b|)n$

$a^{1/n}\leq (an+b)^{1/n}\leq [(a+|b|)n]^{1/n}$

Where both equations are true for sufficiently large n.

This is pretty easy to generalize to any polynomial of arbitrary degree, by the way. --COVIZAPIBETEFOKY (talk) 17:16, 31 August 2009 (UTC)

Great. Thanks COVIZAPIBETEFOKY and Dr Dec!--Shahab (talk) 17:22, 31 August 2009 (UTC)

COVIZAPIBETEFOKY, there does seem to be one problem this the inequality

$a^{1/n}\leq (an+b)^{1/n}\leq [(a+|b|)n]^{1/n}.$

The sequence can take complex values, for example a = –1 and b = 0. When n is even we get imaginary numbers; so the inequality signs don't make such sense. Besides (–1)^1/4 has four possible values! Surely there should be some modulus signs? ~~ Dr Dec (Talk) ~~ 17:29, 31 August 2009 (UTC)

COVIZAPIBETEFOKY specified that his solution is only valid for a≥0 (although when a=0 b must be positive). In those cases we can assume we're talking about the positive real root. Otherwise the sequence diverges. And if we allow complex values then it's not particularly well defined and definitely doesn't converge in general although I guess you could set up a scheme to choose roots so that it would converge to any value on the unit circle you wanted. However in all cases (except a=b=0) it can be said that |(an + b)^1/n| goes to 1. Rckrone (talk) 18:09, 31 August 2009 (UTC)

I know, I was just prompting him to leave a revised inequality that involved muduli. ~~ Dr Dec (Talk) ~~ 18:13, 31 August 2009 (UTC)

p.s. the sequence does not diverge. It's limit is 1 for all real numbers a and b. As my post above showed: for any complex number z we have that |z^1/n| → 1 and arg(z^1/n) → 0 as n → ∞. This implies that the modulus of the sequence converges to 1 for all real number a and b. Find some a and b so that the sequence diverges. ~~ Dr Dec (Talk) ~~ 18:16, 31 August 2009 (UTC)

arg(z^1/n) doesn't go to 0 in general since z^1/n is ambiguous. The nth roots of z are evenly spaced around the circle. You could pick a value of arg(z^1/n) for each n to make a sequence that went to 0. You could also choose a sequence that goes to any value you like, or one that diverges. Just as an example, if for real a and b with a<0 we take the "obvious" root for each n odd, I would think that would be the real root (which is negative), in which case the odd terms go to -1. Rckrone (talk) 18:32, 31 August 2009 (UTC)

Really? Well, if $z=Re^{i(\theta +2\pi k)}\,$ where k is an integer then $z^{1/n}=R^{1/n}e^{i(\theta +2\pi k)/n}\,$ and clearly, for a fixed choice of k, $i(\theta +2\pi k)/n\to 0\,$ as $n\to \infty \,$. ~~ Dr Dec (Talk) ~~ 19:31, 31 August 2009 (UTC)

k is not fixed for all n. I can choose k_n to be any sequence of integers I want and it'll still produce a sequence of nth roots of z. Going to my example before: Let z = |z|e^iπ and choose k_n = n/2 for even n and k_n = (n-1)/2 for odd n. The nth term of my sequence is |z|^1/ne^iπ(n+1)/n for even n and |z|^1/ne^iπ for odd n. This sequence goes to -1 (the arguements go to π). It's not hard to see how you can create sequences that have lots of different behaviors depending on how you pick the nth root, and in the general complex case there's not an obvious "preferred" scheme for picking roots. Rckrone (talk) 19:58, 31 August 2009 (UTC)

(@dec)I would not worry about complex numbers; that exercise clearly refers to real numbers; it's quite implicit in the notation that a>0 and b is real. One could have asked the same question in a more general setting (complex numbers/matrices/linear operators/Banach algebras or whatever), but since nobody did, COVIZ's is, I think, exactly the right answer to the OP. --pma (talk) 18:18, 31 August 2009 (UTC)

The original question simply says that a and b are constant. It doesn't mention the positivity of a or b. COVIZ has the right solution. With a few moduli signs it would work even when a and b were taken to be complex. ~~ Dr Dec (Talk) ~~ 18:22, 31 August 2009 (UTC)

So, in my opinion the additional information on complex numbers may be of some interest, indeed; but in this and similar situations, it should be better to add it as a further remark or side remark, not as a correction to a totally satisfying answer. Otherwise, the OP will get confused, the first person who answerred will give you an explanation, you'll reply, I'll no help adding my boring remarks, a useless debeate will start, and this desk becames a mess again. It's not to criticize; I'm just looking with great admiration at those fellows here that are able to give short, concise, yet complete and illuminating explanations; let's try to learn how they do. --pma (talk) 21:38, 31 August 2009 (UTC)

Agreed! ~~ Dr Dec (Talk) ~~ 16:54, 1 September 2009 (UTC)

Millimeters

I speak of the probe on the left.

After reading the metre article, I still have a question regarding measurements -- as a dentist, I measure periodontal pockets (the sulci of gum tissue surrounding teeth in patients with periodontal disease) with a periodontal probe to the nearest mm. Because we (dentists in general) round to the nearest mm, and because there is some inherent intra- and interexaminer discrepancy due to probing force, velocity, angulation and the degree of inflammation present at the base of the sulci, any measurement is expected to be within 1 mm of what another competent dentist would record. My question -- since a mm is 1/1000 of a meter and a meter is arbitrary (in that there is no reason to have chosen the half-meridian for the original calculation), had a mm been 0.9 or 0.8 of our actual mm, wouldn't our measurements be more accurate? DRosenbach ^{(Talk | Contribs)} 12:53, 31 August 2009 (UTC)

With the smaller units, if you were still able to reliably measure within one "short" mm, then your measure would be more accurate. But I assume there would be a tradeoff between the this gain in accuracy and the loss in accuracy due to the examiner discrepancies. If the millimeter were smaller, it would be even more difficult for examiners to measure accurately. Staecker (talk) 13:06, 31 August 2009 (UTC)

True, but it's perfectly possible that the optimal trade-off between the stated precision, and the ability to reliably make measurements with that precision, is at an increment less than a mm. However, in other applications the optimum may be higher, so it's not universally preferable to choose a smaller meter. If this is significant you (where "you" is a guild of some sort) can always define your own optimal unit of measurement for your application. -- Meni Rosenfeld (talk) 20:25, 31 August 2009 (UTC)

There are two sources of error: the various sorts of individual variation that you mentioned, and the rounding to the nearest mm. If the definition of mm were changed, then the second source of error would be reduced (because you're rounding less) but the first would not be changed. If the first source of error is the dominant source of error (e.g., dentists can't be more accurate than 2 or 3mm because of measurement error), then changing the definition of mm will have negligible effect, whereas if the second source of error is the dominant source of error (e.g., dentists can measure down to 0.01 mm but this precision is lost because of rounding to the nearest 1mm), then rounding to a smaller unit would significantly reduce error. Eric. 216.27.191.178 (talk) 20:34, 31 August 2009 (UTC)

Octahemioctahedron

This is the octahemioctahedron, with tetrahedral symmetry... 3/2 3 | 3	...but what about this? Can it have octahedral symmetry?
Cantellated tetrahedron 3 3 | 2	Cuboctahedron 2 | 3 4

Does the octahemioctahedron have tetrahedral or octahedral symmetry? Since it's a facetting of the cuboctahedron with octahedral symmetry. Thanks. Professor M. Fiendish, Esq. 12:58, 31 August 2009 (UTC)

The article you linked - infobox - symmetry group. The answer is in there. It seems correct.83.100.250.79 (talk) 15:06, 31 August 2009 (UTC)

If you consider the eight triangular faces as being identical and indistinguishable (right hand diagrams) then it has octahedral symmetry. If you consider the triangular faces as separated into two families of four, distinguished by whether you are seeing the "top" of "bottom" surface of the triangle (red and blue triangles in the left hand diagrams) then the octahedral symmetry is broken and what remains is tetrahedral symmetry. Gandalf61 (talk) 18:21, 31 August 2009 (UTC)

Usually by "symmetries of a polyedron", or more generally "symmetries of a metric space", people mean its group of isometries... so I'd say the former you said is the natural choice. I suddenly had a suspect, and googoled "colored polyhedra" and "colored symmetries": they are there. Everything that may be thought by a human mind is already there. --pma (talk) 21:59, 31 August 2009 (UTC)

I contributed to the article and images. The Wythoff symbol 3/2 3 | 3 has tetrahedral symmetry, two triangle colors. There may be a secondary reflective construction with octahedral symmetryLooking at Schwarz triangle options, I don't see a simple form, maybe as a composite form?! I'm asking around by email too. Tom Ruen (talk) 22:33, 31 August 2009 (UTC)

OK, what about the tetrahemihexahedron? It's a facetting of the octahedron, which has octahedral symmetry.

Tetrahemihexahedron.

Especially since the dual, the tetrahemihexacron, is made up of 3 intersecting infinite square prisms (octahedral symmetry! There's a cube in the central intersection.)

Tetrahemihexacron.

Professor M. Fiendish, Esq. 04:42, 2 September 2009 (UTC)

Oh, maybe we should check the symmetries of the octahemioctacron.

Octahemioctacron.

Professor M. Fiendish, Esq. 04:42, 2 September 2009 (UTC)

Thank you, math reference desk

Hi. I'm just posting here to say that I passed my Complex Analysis qualifying exam, and you guys helped me a lot while I was studying for that. This page is one heck of a resource, and that wouldn't be true if not for the patient and friendly mathematicians who answer questions here. You guys rock, individually and as a group, and I wish a thousand blessings upon all your houses. :) -GTBacchus^(talk) 19:19, 31 August 2009 (UTC)

I second the above totally.--Shahab (talk) 05:09, 1 September 2009 (UTC)

Weighted sum of distance from a curvy line

Consider a busy road which is a curvy line on a flat plane. The sound intensity at every point on the curvy line is the same - call it a sound intensity level of 1, say. The sound intensity at an X,Y point on the plane is I think the sum of the noise from each point on the curvy line, except that the noise from each point is weighted by the inverse of its distance squared, since sound energy decreases with distance.

My question is - what would be the best practical way to calculate the sound intensity at an X,Y point on the plane?

All I have is a paper map which includes the road, and my maths is sub-calculus. Ideally I would like to have contour lines of the sound intensity over the plane, but just being able to calculate the sound intensity at two different points will do. I realise that this is an abstraction, but its a start. The calculations could be done manually or is there any suitable mapping software that could be used? Thanks 92.27.79.62 (talk) 20:30, 31 August 2009 (UTC)

Well, if $\textstyle \gamma :I\to \mathbb {R} ^{2};$ (or $\textstyle \mathbb {R} ^{3}$ if you like) is the arc-length parametrization of your road, the intensity at the point x it is proportional to $\textstyle \int _{I}|x-\gamma (s)|^{-2}ds$. here |v| denote the euclidean norm or v. Then, depending on the curve, it may be possible to compute the integral, or just evaluate it numerically. In any case I hope it is just an ideal calculus, and you do not have to face a real traffic noise! --pma (talk) 21:08, 31 August 2009 (UTC)Warning: In fact, the above integral was a mathematical representation of a quantity induced with a inverse square law by a source uniformly disributed on the curve γ, provided one assume a superposition principle. But I do not see why it should be true here. --pma (talk) 11:51, 2 September 2009 (UTC)

If it's a real road map, one way you could get an estimate manually for a specific point is to draw concentric circles around your point on the map with a compass. Then for each ring between consecutive circles estimate the length of road in the ring and take the distance away for all that road to be the average of the radii of the two circles. So if you have length d of road between circles of radius r and s, then the contribution of that part of the road is d/((r+s)/2)². Then sum up the contributions of all the rings. This is going to be least accurate for road that's very close to your point, so you would probably want to draw your circles closer together as you get closer to your point. Depending on how accurate you want to be, this could be a lot of work. Rckrone (talk) 21:11, 31 August 2009 (UTC)

Yes, but the sample points should be equally spaced on the road, to avoid having to weight each individually, (in what is already a time consuming process..) 83.100.250.79 (talk) 22:09, 31 August 2009 (UTC) Ignore this - had a better idea..:

I have a better idea - if you can get the x,y coordinates of a set of points along the road (choosing points so that the 'zig-zag' straight line is fairly close to the original road) - then enter the points into a spreadsheet (or specially prepared program) - the program could then calculate the contour lines for the sound levels - using the distance between adjacent points as a weighting factor, and the inverse square rule as well.

This way you only need to measure points once - but you can get an entire contour map made.

Would you be able to make a program to do that (any programming experience?) - it would be quite easy to write - someone would probably volunteer to do that on the computer desk, as it is not too difficult (I could write that if necessary).83.100.250.79 (talk) 22:30, 31 August 2009 (UTC)

Yeah, this method is much better than mine since you can get a computer to do most of the work without having to feed it too much data, and it's more accurate. Rckrone (talk) 22:41, 31 August 2009 (UTC)

Assuming there is no equation for the road - one solution would be to scan the map - and then use software to convert the road into a line of points (tricky bit - but possible) - then the sums of intensities to give contours would be actually trivial for the computer.

If you want to follow this route (pun!) then the hardest step is finding software which will take an image (assume you have scanned or photographed the map, and removed any stuff that is not the road) and convert it to a set of points describing the line/curve of the road. You should ask for this if this sounds useful to you. I would guess the maths desk would be better for the algorthym, but the computer desk would be more likely to know of software.

If you don't do this I think the solution will be a lot of hand measuring, and calculations.83.100.250.79 (talk) 22:06, 31 August 2009 (UTC)

Thanks. Remember as I said my maths is beneath calculus, so the first responce means nothing to me. I remember I have a computer program somewhere that is designed to digitise lines from images of graphs, so that would probably help. Or I could do it manually. This is a practical real problem, by the way, not an exercise. You can assume as a starting point that I have coordinates for points along the road. I have not done any amateur computer programming for a long time - the only language I was fluent with was GWBasic. I will try and find the digitisation program and get back to you later. 78.147.28.17 (talk) 10:20, 1 September 2009 (UTC)

There's a side-issue which has occurred to me - since there are an infinite number of points in a line, won't the sum of the sound from these points along the road be in theory infinate? Or at least, wont the calculated intensity of the sound at X,Y compared with that at the road be heavily dependant on how many points along the road I decide to use? If I use for example twice the number of road-points, then the total sound intensity is approximately doubled. 78.144.246.12 (talk) 14:44, 1 September 2009 (UTC)

If you sample the curve with 2x points, you divide the result with 2. You need to sample relatively evenly in space, eg with a rectangular grid. --91.145.89.58 (talk) 15:21, 1 September 2009 (UTC)

Yes using N points divide the total by N, also bear in mind to weight each point by the approximate length it represents. - using the weights - you then divide by the sum of the weighting lengths at the end, instead of N.83.100.250.79 (talk) 18:38, 1 September 2009 (UTC)

I'm not sure about the two comments above - if for example the road formed a hairpin bend, and you went and stood so that you had road noise on either side of you, then you would expect the intensity of the noise to add and be greater where you stood than on the road itself. Dividing the sum of the noise by the number of points, or the total of the inverse squared distance would not give this effect I think. 78.146.3.82 (talk) 20:38, 3 September 2009 (UTC)

September 1

Integration

Earlier this year our lecturer gave us an integral to solve, then realised she had made a mistake, and gave us an easier one.

The Integral was ʃ { 2 / [ 3X Ln ( 2x + 1 )]}dx, which she changed to ʃ { 2 / [ 3X Ln ( 2x)]} dx, which equals ⅓ Ln[ Ln ( 2x ) ] + C

This was relatively easy, but what I would like to know, is how to solve ʃ { 2 / [ 3X Ln ( 2x + 1 )]}dx, and why is it that simply adding a 1 to the term in the parentheses makes it so much harder ? I am only a first year Canterbury ( New Zealand ) University student. Thanks. —Preceding unsigned comment added by 202.36.179.66 (talk) 04:08, 1 September 2009 (UTC)

Probably because if you use the u substitution method (which I'm guessing was the suggested one) to try to evaluate the second ("mistake") integral, it doesn't work out too well, whereas with $\int {\frac {2}{3x\ln {(2x)}}}\,dx$, if you let u = ln(2x) then you end up getting du = 1/x dx which substitutes in nicely. However if you use the same method with $\int {\frac {2}{3x\ln {(2x+1)}}}\,dx$ and set u = ln(2x+1) then du = 2/2x+1 dx, which unlike the first problem, does not substitute in nicely. Ginogrz (talk) 04:44, 1 September 2009 (UTC)

Just to clarify, in the "easier" integral, the solution should be ${\dfrac {2}{3}}\left(\ln {(\ln 2x)}\right)$ (as the u-substitution yields ${\dfrac {2}{3}}\int {\dfrac {du}{u}}$). --Kinu ^t/_c 04:54, 1 September 2009 (UTC)

As far as I can tell, the more difficult integrand doesn't have an indefinite integral in terms of elementary functions. ~~ Dr Dec (Talk) ~~ 07:42, 1 September 2009 (UTC)

How do you determine that? Black Carrot (talk) 22:22, 1 September 2009 (UTC)

Risch algorithm has some details, although only in that superficial hand-wavy kind of way that encyclopedia maths articles always seem to gravitate to. Zunaid 09:57, 2 September 2009 (UTC)

Also found Differential Galois theory, according to which the problem of solving indefinite integrals using elementary functions is analogous to the problem of finding roots of polynomial equations using radicals. Read the article for more details. Zunaid 10:08, 2 September 2009 (UTC)

Hyperbolic Functions

Given the value of, say, $\cos \theta$ one can determine $\sin \theta$ and $\tan \theta$ by drawing a right angled triangle with the appropriate side lengths. Is there an equivalent for the hyperbolic functions? I know you can use relations such as $\cosh ^{2}\theta -\sinh ^{2}\theta =1$ but this isn't what I'm looking for. Thanks. 92.4.122.142 (talk) 17:39, 1 September 2009 (UTC)

If you know $\cos \theta$ then you only know the angle. You would also need to know the length of one of the sides to determine the triangle unically. If we knew the length of the hypotenuse, say h, then we would know that the adjacent side has length $h\cos \theta$ and the opposite side has length $h\sin \theta .$ Knowing one of the non-right-angle angles only gives us the triangle up to a scalling. After all: cosine gives the ratio of the length of two sides, and that's invariant under scalling. ~~ Dr Dec (Talk) ~~ 17:53, 1 September 2009 (UTC)

You have picked up on a minor slip up when writing my initial post that has no bearing on either the statement "Given the value of, say, $\cos \theta$ one can determine $\sin \theta$ and $\tan \theta$" or my question. 92.4.122.142 (talk) 18:12, 1 September 2009 (UTC)

Ay, ay, ay. No need to be so touchy. I was trying to help! You don't need to use any triangles. If you know $\cos \theta$ then you know $\sin \theta$ up to sign: $\sin \theta =\pm {\sqrt {1-\cos ^{2}\theta }}.$ If you know $\cos \theta$ and you know $\sin \theta$ up to sign then you know $\tan \theta$ up to sign:

$\tan \theta =\pm {\frac {\sqrt {1-\cos ^{2}\theta }}{\cos \theta }}\ .$

If you're trying to fit these around a triangle then you'll want 0 < θ < π/2, and so all the ±'s just become +. The same goes for the hyperbolic trig functions: we know that $\cosh ^{2}\theta -\sinh ^{2}\theta =1$, so once we know either $\cosh \theta$ or $\sinh \theta$ we can work out the rest with some algebra. Although there won't be a nice circle picture, you'd get branches of a hyperbola. Please, 92.4.122.142, assume good faith; I was honestly trying to help you. ~~ Dr Dec (Talk) ~~ 18:44, 1 September 2009 (UTC)

My apologies. I thought you were deliberately being pernickety about my question just for the sake of it. I see now that was a wrong assumption to make. Thank you for your help and again, I'm sorry. 92.4.122.142 (talk) 18:47, 1 September 2009 (UTC)

It does not appear that you have to apologize. It does not appear that your initial post was uncorrect. You were talking of determining sin(θ) and tan(θ), not a triangle. --84.221.68.104 (talk) 22:57, 2 September 2009 (UTC)

For trig functions, we can draw the unit circle (x² + y² = 1) and then any point on the circle has coordinates (cosθ, sinθ) for some value of θ. The value of θ corresponds to twice the area of the circle swept to our point, which is also equal to the angle from the x-axis. This let's us draw a right triangle with unit hypotenuse, and one angle θ that has legs cosθ and sinθ. For the hyperbolic functions we can do a similar thing with the unit hyperbola (x² - y² = 1). Any point on the right branch of the hyperbola has coordinates (coshθ, sinhθ) for some value of θ. This θ is also twice the area swept out along the hyperbola to that point, but unlike with the circle that doesn't correspond to any angle that I know of, so I think that's pretty much where the analogy ends. Rckrone (talk) 18:17, 1 September 2009 (UTC)

Actually I guess you could draw a right triangle with one leg equal to 1, and then the hypotenuse would be coshθ and the other leg would be sinhθ for some θ. θ wouldn't correspond to any obvious property of the triangle though. Rckrone (talk) 18:24, 1 September 2009 (UTC)

September 2

Variables confusing me

The "Circular references in computer programming" section of Circular reference gives an example of how spreadsheets can be confused by circular references, using the following wording: "This [the previous example] leads to what is technically known as a livelock. It also appears in spreadsheets when two cells require each others' result. For example, when the value in Cell A1 is equal to the value in Cell B1 plus 5, and the value in Cell B1 is equal to the value in Cell A1 plus 5." Is this possible? I can't imagine a situation in which x = y + 5 AND y = x + 5, unless our spreadsheet putter-togetherer is making more basic errors than circular references. Unless I misunderstanding something here, would someone please revise this statement so that it's far more useful? Nyttend (talk) 02:38, 2 September 2009 (UTC)

It isn't possible that x = y + 5 and y = x + 5 and that's not what the article is implying. It is possible to enter these formulas into a spreadsheet though. The spreadsheet won't be able to compute anything from them and that's why you get an error message. You can enter all sorts of mathematical nonsense into a spreadsheets and a lot of other computer programs, and they give you error messages back because they are nonsense.--RDBury (talk) 06:04, 2 September 2009 (UTC)

That sentence in the article is written very badly. It implies that each cell contains a specific value, and that each of these values is five more than the other. This is of course simply impossible. What is meant is that A1 contains the formula "B1+5" and B1 contains the formula "A1+5", which is possible but leads to an error when the spreadsheet tries to evaluate the formulae. Algebraist 10:46, 3 September 2009 (UTC)

Thanks for the explanations; I wasn't altogether sure that I wasn't missing something. Algebraist, your comment gets at it exactly. Nyttend (talk) 13:31, 3 September 2009 (UTC)

Continuation of the Method of Characteristics to solve PDEs

pma, thanks for the answer! I understand some things but not all so I have a couple of conceptual questions about the method itself if you (or anyone) be kind enough to help me. I state the PDE here just for completeness. $2tu_{t}+xu_{x}=0,\,u(x,0)=f(x)$

• First of all, you say that the PDE tells you that any solution defined on the initial manifold (the x-axis) will be a constant. Is it because the PDE is homogeneous? So that when we compare it to the complete derivative of u(x,t) we obtain

$u'(x,t)=u_{x}x'+u_{t}t'\Rightarrow u'=0,x'=x,t'=2t$

so u must be a constant? Is this how you concluded because u'=0?

• Second, what does "transverse" mean? I looked at the article but does it mean that the characteristics have to be at exactly right angles or they just have to be NOT tangent to the initial manifold?
• Third, how do you know that here the initial manifold is itself a characteristic line (the support of a characteristic)?
• Fourth, why does this method fail if the initial manifold is itself a characteristic line? Is there like a test (or something you look at) in the very beginning to determine if this method will work on this PDE or not so that one doesn't do all the work only find out at the very end that the initial condition cannot be satisfied?
• Fifth, what is the resolution in a case like this? Do I just say that the method of characteristics just cannot be used to solve this PDE? Can another method be used or does this (somehow) guarantee that a solution does not exists or if it does exists, it may not be unique? There a lot of confusion (in my head) regarding the vocabulary and what we really are "doing" when we use this method. So thank everyone!97.118.56.41 (talk) 02:56, 2 September 2009 (UTC)

1. Correct. Indeed, the computation you did proves that any solution is constant along any characteristic line. For the x-axis in particular, the argument could be somehow simplified this way: let $u(x,t)$ be a solution of class $C^{1}$ defined on $\textstyle \{(x,0):\,a<x<b\}$. Then the equation in these points becames $u_{x}(x,0)=0$, so $u(x,0)$ is constant.

2. Correct: as a particular case of a more general definition, a vector is transverse to a submanifold iff the submanifold has codimension 1 and the vector is not tangent to it. Notice that (answerring a subsequent question of yours) it is simple to test whether the characteristic lines are transverse to submanifold M. Here, it just means that at any point (x,t) of M, the vector (x,2t) is not tangent to M. You can check e.g. that $\mathbb {R} \times \{1\}$, and also the unit circle $\{(x,t):\;x^{2}+t^{2}=1\}$ enjoy the transversality condition.

3. Is it clear to you how to write the system of ODE for the characteristics lines? It is:

$\xi '(s)=\xi (s)$,

$\tau '(s)=2\tau (s)$.

You can check that these curves (more precisely, their supports) are a family of (half) parabolas with vertices in the origin, and the 4 semiaxes. Any solution $\textstyle u:\Omega \subset \mathbb {R} ^{2}\to \mathbb {R}$ has to be constant along each of them : precisely, along any arc interval of a characteristic line contained in its domain $\scriptstyle \Omega$. Note that since each characteristic springs off from the origin, it also follows that any solution defined on $\mathbb {R} ^{2}$ is necessarily a constant.

Also note that if the equation is not homogeneous, e.g., if you have 1 on the RHS, then for each solution of it you'll have $u(\xi (s),\tau (s))=s+c$ along a characteristic (the constant c depending on the characteristic). More generally, if you consider the semilinear version of your equation, that is

$\textstyle xu_{x}+2tu_{t}=a(x,t,u)$,

again the solutions are determined along any of the characteristics line above, as soon as you know the value of the solution in one of its points. The reason is that now the PDE restricted to the characteristic lines gives you an ODE for $v(s):=u(\xi (s),\tau (s))$, precisely

$v'(s)=a(\xi (s),\tau (s),v(s))$,

which is again enough to determine it completely from the value in one point. (Let's say that $\textstyle a(t,x,u)$ is at least Lipschitz, to have unicity for the Cauchy problem).

4. In particular, clearly, you can't prescribe arbitrarily the value of a solution at more than one point in each characteristic line; or, to be precise: in each arc interval of characteristic contained in the domain of the solution. On the contrary, for instance, $\mathbb {R} \times \{1\}$ instead of $\mathbb {R} \times \{0\}$ is OK, and in fact you can check that you can solve uniquely in $\mathbb {R} \times \mathbb {R} _{+}$ the PDE with the condition u(x,1)=f(x). Also, you can solve uniquely the PDE on $\mathbb {R} ^{2}\setminus \{(0,0)\}$ if you prescribe the values of a solution on the unit circle as initial manifold. Indeed, each arc of parabola in the domain meets (trasversally) this initial manifold exactly once. In general, if the transversality condition holds on the (codimension 1) submanifold M, thanks to a result of general topology, there exists a nbd U of M such that each arc of a characteristic line in U meets M exactly once (transversally). In this situation, each function $f$ of class C¹ on M extends uniquely to a solution of the PDE on U.

5. It should be clear that the original PDE can't be solved prescribing e.g. u(x,0)=f(x) for all x>0 if f is not a constant: it's not a limit of the method, but just a matter of fact. BTW, the method of characteristics can be generalized to fully nonlinear first order equations, that is $F(x,u(x),\nabla u(x))=0$ with $x\in \mathbb {R} ^{n}$.--pma (talk) 19:23, 2 September 2009 (UTC)

PS: I took the liberty of compressing a bit your post, to make it more readable.

Period

Can someone please help me find the period if cos[n²], where n can take only integral values ? I know the answer is 8, but I can't get the method. Rkr1991 ^{(Wanna chat?)} 13:04, 2 September 2009 (UTC)

cos(0) = cos(64) certainly does not hold, so the answer can't be 8... or don't you mean period as in Periodic function? —JAO • T • C 13:13, 2 September 2009 (UTC)

My mistake, the question should read cos[πn²]. Rkr1991 ^{(Wanna chat?)} 13:33, 2 September 2009 (UTC)

Let's see ... cos(0) = 1 ... cos(π) = −1 ... cos(4π) = 1 ... cos(9π) = −1 ... cos(16π) = 1 ... is that beginning to look periodic ? Gandalf61 (talk) 14:07, 2 September 2009 (UTC)

Is that comment supposed to be mocking me ? cos(0)=cos(64π)=1, cos(π)=cos(81π)=-1, and so on... The answer is 8, my question is how ? Rkr1991 ^{(Wanna chat?)} 09:25, 3 September 2009 (UTC)

No, the answer is not 8, as Gandalf's helpful comment shows. (Hmm - just looked at Periodic function - the definition of period there is somewhat defective.) AndrewWTaylor (talk) 09:39, 3 September 2009 (UTC)

No mocking intended - it was meant to be a hint. The correct answer to the question as you posed it is definitely not 8. Let's start again.

You have a function f(n) defined on the integers by f(n) = cos(πn²). A quick check shows that f(0) = f(2) = f(4) = 1, and f(1) = f(3) = f(5) = −1. This can be generalised to f(2m) = 1 and f(2m+1) = −1 - or, more concisely, f(n) = (−1)ⁿ. So we have f(n + 2) = f(n) for all n, and f(n + 1) ≠ f(n). Therefore the period of the function is 2.

You are correct to say that f(n + 8) = f(n), but this only shows that 8 is a multiple of the period. To conclude that the period was 8 you would have to eliminate all possible smaller periods. Gandalf61 (talk) 09:46, 3 September 2009 (UTC)

I'm sorry if I was too hasty, but still, how do you prove that the period is 2, I mean, apart from the method of observation ? Rkr1991 ^{(Wanna chat?)} 09:56, 3 September 2009 (UTC)

Assuming that you can use the fact that the period of cos is 2π, just consider what is the parity of n² for integer n. — Emil J. 11:17, 3 September 2009 (UTC)

Just as a random question, are all non-constant periodic functions essentially trig functions? 92.4.122.142 (talk) 10:15, 3 September 2009 (UTC)

At least pretty much all periodic functions that occur in reality (or engineering) can be written as an infinite sum of trig functions; see Fourier series. So in that sense, pretty much yes (but not strictly all periodic functions). —JAO • T • C 10:34, 3 September 2009 (UTC)

How about all "single valued continuous periodic funtions are essentiall trig functions"83.100.250.79 (talk) 18:50, 3 September 2009 (UTC)

Could you give an example of a periodic function that cannot be expressed as the sum of trig functions? 92.4.150.32 (talk) 22:19, 3 September 2009 (UTC)

f(x)=0 if x is rational, f(x)=1 if x is irrational. So for example, f(x+1)=f(x) for all x, satisfying the definition of a periodic function. 67.122.211.205 (talk) 22:38, 3 September 2009 (UTC)

Non-Archimedean fields

We haven't got much on this topic, I'm noticing.

I'm studying non-Archimedean fields, and I'm stuck on what I think is an embarrassingly simple question. The notes I'm working from begin by defining a non-Archimedean absolute value, which is a map from a commutative ring to the non-negative reals with the following 3 properties:

1. $\left|a\right|=0\Leftrightarrow a=0$
2. $\left|ab\right|=\left|a\right|\cdot \left|b\right|$
3. $\left|a+b\right|\leq \operatorname {max} \left\{\left|a\right|,\left|b\right|\right\}$

Here's my issue. To do some of the very basic first proofs, I find myself wanting to say that $\left|-a\right|=\left|a\right|$. I can prove that quite easily... if the ring has unity. Otherwise, how do I know that a ring element will be the same size as its additive inverse? I think I must be missing something quite obvious. Thanks in advance for any hints. -GTBacchus^(talk) 21:19, 2 September 2009 (UTC)

Yes sir: since $(-a)^{2}=a^{2}$ from 2 it follows $|-a|^{2}=|(-a)^{2}|=|a^{2}|=|a|^{2}$ thus $|-a|=|a|$ because they are both non-negative. --84.221.68.104 (talk) 21:52, 2 September 2009 (UTC)

Ah, duh! Thank you. I knew it had to be something simple. -GTBacchus^(talk) 07:32, 3 September 2009 (UTC)

Actually, thinking about it, I'm not sure why $(-a)^{2}=a^{2}$ if the ring hasn't got unity. On the other hand, I spoke with the professor who prepared the notes I'm studying, and he said that by "ring" he means unital ring. I suppose most people do. -GTBacchus^(talk) 10:06, 3 September 2009 (UTC)

ab=(-a)(-b+b)+ab=((-a)(-b)+(-a)b)+ab=(-a)(-b)+((-a)b+ab)=(-a)(-b)+(-a+a)b=(-a)(-b) Algebraist 10:41, 3 September 2009 (UTC)

Mmm, nice... The first step is really all I had to see; it's clear from there. Thanks, though. -GTBacchus^(talk) 11:19, 3 September 2009 (UTC)

"He said that by "ring" he means unital ring. I suppose most people do." Note however that the theory of non-unital rings is quite different to the theory of unital rings (so that deciding upon whether the ring should have unity is not merely a convention). For instance, by Zorn's lemma, any unital ring has at least one maximal (proper) ideal - simply consider the set of all ideals which do not contain a unit (invertible element); this set is non-empty since (0) is a member. If the ring does not have a 1, however, this argument fails since one cannot define a "unit" and therefore non-unital rings need not possess maximal (proper) ideals. In particular, various constructions relating to maximal ideals have to be altered in the case of non-unital rings; notable is the concept of the Jacobson radical or the notion of quasiregularity.

There are also several interesting theorems which guarantee the existence of a 1 in a (not necessarily unital) ring. For instance, a right artinian "ring without 1" with no non-zero nilpotent ideals must have a 1. This is interesting because it shows how one can "locate" a 1 in a ring through almost "indirect arguments."

Similarly, the theory of non-unital modules also requires special distinctions when translating theorems from the theory of unital modules. For instance, if one defines a simple (not necessarily unital) module, as a module with proper annihilator as well as having no non-trivial proper submodules, one can show certain basic results. One example of this is that such a module is necessarily isomorphic to the quotient of R (the ring over which the module is defined) by a maximal right ideal, M. Additionally, one can show that there is an r in R, with x − r · x an element of M for all x. For unital modules, this is trivial since r = 1 works; for non-unital modules this requires a little more basic work.

I know that I have probably been slightly pedantic in this argument but nevertheless I feel that it is worthwile that one knows of this important distinction. Quite a few authors (such as I. N. Herstein) do not assume the existence of an unity in their rings and study this chosen distinction. Anyway, I think that you get the idea now. --PST 13:12, 3 September 2009 (UTC)

So, I was reading about non-unital rings, and I saw that there seems to be a canonical way to adjoin a multiplicative identity. Would this be another solution to the problem of showing that abs(a) = abs(-1); i.e. adjoin a "1" element, and then proceed as in a unital ring? Somehow that seems too easy... -GTBacchus^(talk) 13:19, 3 September 2009 (UTC)

How do you extend the absolute value from the nonunital ring to the adjunction? — Emil J. 14:01, 3 September 2009 (UTC)

Ah, good point. What I should have asked is whether one could use the extension with unity to more easily show that $(-a)^{2}=a^{2}$. I'm not sure if there's a natural way to extend the absolute value. -GTBacchus^(talk) 14:38, 3 September 2009 (UTC)

Yes, you can use the extension to prove that (−a)² = a². More generally, the same argument shows that the theory of unital rings is conservative over the theory of not-necessarily-unital rings wrt universal formulas in the language $\langle +,0,-,\cdot \rangle$. However, it's not easier than a direct proof: checking that the extension is a ring is a lot more work than Algebraist's proof above (though it is no less elementary, just longer and more tedious); furthermore, you need to show that (−a) = (−1)a in a unital ring, which is more or less the same proof as showing (−a)b = −(ab) in a nonunital ring, and that already implies your identity (it gives also a(−b) = −(ab) by symmetry, hence (−a)(−b) = −(a(−b)) = −−(ab) = ab). — Emil J. 15:28, 3 September 2009 (UTC)

Thank you for that explanation. -GTBacchus^(talk) 16:12, 3 September 2009 (UTC)

September 3

Econometrics

Does anyone have any suggestions for a good undergraduate level econometrics text book?--98.240.70.102 (talk) 00:02, 3 September 2009 (UTC)

I used Gujarati's Basic Econometrics, a very user-friendly introductory book, if somewhat lacking up-to-date treatment of recent topics such as the asymptotic (large-sample) approach.
A more modern choice can be Woolridge's Introductory Econometrics. Pallida  Mors 18:28, 3 September 2009 (UTC)

Binomial Expansions

Resolved

I have to determine the expansion in powers of x up to $x^{4}$ of $(1-x^{3})^{6}(1-x)^{-6}$. Now I can do this by some simple division and then expanding $(1+x+x^{2})^{6}$ but is there any way of reaching the same answer by expanding the two sets of brackets separately? Thanks 92.4.122.142 (talk) 12:47, 3 September 2009 (UTC)

Yes, expand the first factor as $1-6x^{3}+...$ and the second as $1+6x+(-6)(-7)x^{2}/2+...$ Ignore powers of x above 4, and then multiply out the resulting expressions. Tedious, but it gives the same answer as your "simple division" method.Caution, expressions may contain typos. AndrewWTaylor (talk) 13:55, 3 September 2009 (UTC)

Exactly! See Newton's generalised binomial theorem. (r = –6 is the case of the second bracket) ~~ Dr Dec (Talk) ~~ 17:56, 3 September 2009 (UTC)

Uniform convergence of $e^{-x^{2}}\,\sin({\frac {x}{n}})$ over $\mathbb {R}$

Hi there guys,

could anyone please suggest a test or approach to check whether convergence of $f_{n}(x)\,=\,e^{-x^{2}}\,sin({\frac {x}{n}})$ to $f(x)\,=\,0$ over $\mathbb {R}$ is uniform? I've tried everything I could think of (not much sadly) such as checking that obviously both $f_{n}(x)$ and $f(x)$ are continuous, and trying to find a value for the maximum of $e^{-x^{2}}\,sin({\frac {x}{n}})$, from which all I got was an ugly formula in xtan(x/n) (for x, not for f_n(x)), and I don't seem to be making any headway. I don't need to be walked through what to do, but if I could just get myself aimed in the right direction that'd be great (e.g. the name of a test or a property to look at) - thanks!

Spamalert101 (talk) 18:20, 3 September 2009 (UTC)

No special machinery is needed here. To make f_n bounded by epsilon, just choose M large enough that exp(-x²) is less than epsilon outside [-M,M] and then make n large enough that sin(x/n) is smaller than epsilon inside [-M,M]. Algebraist 18:28, 3 September 2009 (UTC)

Also, you may use the inequalities $|\sin(t)|\leq |t|$ for all $t\in \mathbb {R}$, and $\mathrm {e} ^{-x^{2}}|x|\leq 1$ for all $x\in \mathbb {R}$ that imply

$|f_{n}(x)|\leq {\frac {1}{n}}$ for all $x\in \mathbb {R}$.

--pma (talk) 22:09, 3 September 2009 (UTC)