Wikipedia:Reference desk/Archives/Mathematics/2008 April 21

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

April 21

Prime ideals of a ring structure defined on a power set

Suppose Ω is a countably infinite set. Let R be the power set of Ω. Define a ring structure on R with operations $a\times b=a\cap b\ ,\ \ a+b=(a\cup b)\setminus (a\cap b)$. (Equivalently, consider R to be the direct product of copies of Z₂ indexed by Ω, then interpret each R element as a "truth table" for the indexed elements of Ω - each element r of R is identified with the subset of Ω for which the correponding components of r are 1. Then consider how addition and multiplication work). Note that the ring is commutative with unity.

Now, for any subset S of Ω, let P(S) be the power set of S - the set of all subsets. I've proven that P(S) is an ideal of R for any S. I'm now trying to show that any prime ideal has the form P( Ω \ {s} ) for some s in Ω. I'm finding it hard to prove that every prime ideal is of the form P(S) for some S, but if I can get that then the last bit is fairly easy.

So, if I is a prime ideal, then let S be the union of all sets in I (remember I is a collection of subsets of Ω, so S is the set containing every point that lies in an element of I). What I am actually struggling with is proving P(S) is a subset of I.

The three important rules I have are:

• If A is in I and X is in C, then their intersection is in I. (since I an ideal)
• If A and B are in I, then $\textstyle (A\cup B)\setminus (A\cap B)$ is in I. (since I an ideal)
• If the intersection of A and B is in I, then one of A or B is itself in I. (since I prime)

So, if X is a subset of S, I need to show one of these three things:

• X is the intersection of some subset K of S with any subset Y of Ω
• There's a subset Y that ISN'T in I, but the intersection of Y and X is in I
• X is the union, excluding the intersection, of two subsets K and L in I

But I'm damned if I can work it out. If I knew there were points outside of S I can do it (I will know this eventually, but not until after), or if I could construct a set not in I whose intersection with X was in I, but I can only seem to do these things if I make extra assumptions about things being nice. Can anybody give me a suggestion? Thanks, Maelin (Talk | Contribs) 10:13, 21 April 2008 (UTC)

After a bit of thought, I've realised the problem: the result is not true. Consider the collection of all finite subsets of Ω. This is clearly an ideal, so by a well-known application of Zorn's lemma it extends to a maximal (and hence prime) ideal. This ideal is not of the form given. In a different language, you have been asked to prove that every ultrafilter on a countable set is principal. Algebraist 11:23, 21 April 2008 (UTC)

Is Zorn's Lemma applicable? It doesn't look like it is to me, since you could have a chain of finite subsets (start with a unit set, adjoin a new point each time) with no upper bound in the ideal (the only upper bound is Ω - not finite, so not in the poset). I'm not sure if you weren't actually doing something more clever than just a direct application, but I don't think we meet the requirements. Maelin (Talk | Contribs) 23:39, 21 April 2008 (UTC)

Zorn's lemma is applicable to show that every ideal in a unital ring is contained in a maximal ideal and to show that every filter is contained in an ultrafilter. The union of a chain of ideals is an ideal. The poset involved is the poset of ideals containing the ideal of finite sets (or the poset of filters containing the filter of cofinite sets). It might seem confusing since the ring in this particular case is also a poset, but in general you are interested in the poset of ideals, not the set forming the ring. JackSchmidt (talk) 00:33, 22 April 2008 (UTC)

So, it turns out that it ought to have been the direct sum, and hence only finite subsets in the ring, which I've managed to do. Thanks for the help! Maelin (Talk | Contribs) 11:29, 23 April 2008 (UTC)

DAYS DETERMINATION

Could there be a formulae for determining a particular day of the week, in which a particular date and month fell or will fall? e.g. by applying that formulae how would you determine the day of the week in which this dates apply?.......tenth of march of the year 1768? without necessarily looking In the calendar?41.220.120.202 (talk) 14:31, 21 April 2008 (UTC)DAVIS

It's not easy for most people to do in the head. See Calculating the day of the week. PrimeHunter (talk) 14:37, 21 April 2008 (UTC)

Yes, there is. And it is pretty funny that you should ask that. I was looking for the same thing and I found it. Looks up the article about Zeller's congruence here. If you want something which you can do in your head to determine the date, that would be Conways's doomsday rule.A Real Kaiser (talk) 00:20, 22 April 2008 (UTC)

See also the Calendar FAQ. – b_jonas 09:39, 24 April 2008 (UTC)

Partial Derivatives

This is the question i've been set to answer: If $f$ is a twice differentiable function of a single variable, find $f=z{\sqrt {x^{2}+y^{2}}}$ that satisfies ${\frac {\delta ^{2}z}{\delta x^{2}}}+{\frac {\delta ^{2}z}{\delta y^{2}}}=x^{2}+y^{2}$.

I've started of by letting $u={\sqrt {x^{2}+y^{2}}}$, i know that ${\frac {\delta z}{\delta x}}={\frac {dz}{du}}{\frac {\delta u}{\delta x}}={\frac {dz}{du}}{\frac {x}{\sqrt {x^{2}+y^{2}}}}$, and a similar equation but with y rather than x. If i had ${\frac {\delta ^{2}z}{\delta ^{2}x}}$ i would be able to substitute it into the original equation and get a solvable ODE but i dont know how to get this, can someone help? Thanks. 212.140.139.225 (talk) 18:56, 21 April 2008 (UTC)

Can't you just do something simple: Set $z={\frac {1}{2}}x^{2}y^{2}$, which clearly satisfies the PDE, then set $x=y$ and you get $f=z{\sqrt {x^{2}+y^{2}}}={\frac {1}{2}}x^{2}y^{2}{\sqrt {x^{2}+y^{2}}}={\frac {1}{2}}x^{4}{\sqrt {2x^{2}}}={\frac {x^{5}}{\sqrt {2}}}$. I think I may be misunderstanding the question, because it seems really convoluted to have such a simple answer. --Tango (talk) 19:21, 21 April 2008 (UTC)

Given the setup of the problem, I would guess f=f(r) where $r={\sqrt {x^{2}+y^{2}}}$. This gives:

$f(r)=zr$

$z={\frac {f(r)}{r}}$

${\frac {\partial r}{\partial x}}={\frac {x}{\sqrt {x^{2}+y^{2}}}}={\frac {x}{r}}$

${\frac {\partial ^{2}r}{\partial x^{2}}}={\frac {1}{\sqrt {x^{2}+y^{2}}}}-{\frac {x^{2}}{(x^{2}+y^{2})^{\tfrac {3}{2}}}}={\frac {y^{2}}{(x^{2}+y^{2})^{\tfrac {3}{2}}}}={\frac {y^{2}}{r^{3}}}$

${\frac {\partial r}{\partial y}}={\frac {y}{r}};{\frac {\partial ^{2}r}{\partial y^{2}}}={\frac {x^{2}}{r^{3}}}$

${\frac {dz}{dr}}={\frac {f'(r)}{r}}-{\frac {f(r)}{r^{2}}}$

${\frac {d^{2}z}{dr^{2}}}={\frac {f''(r)}{r}}-2{\frac {f'(r)}{r^{2}}}+2{\frac {f(r)}{r^{3}}}$

${\frac {\partial z}{\partial x}}={\frac {dz}{dr}}{\frac {\partial r}{\partial x}}$

${\frac {\partial ^{2}z}{\partial x^{2}}}={\frac {d^{2}z}{dr^{2}}}\left({\frac {\partial r}{\partial x}}\right)^{2}+{\frac {dz}{dr}}{\frac {\partial ^{2}r}{\partial x^{2}}}=\left[{\frac {f''(r)}{r}}-2{\frac {f'(r)}{r^{2}}}+2{\frac {f(r)}{r^{3}}}\right]{\frac {x^{2}}{r^{2}}}+\left[{\frac {f'(r)}{r}}-{\frac {f(r)}{r^{2}}}\right]{\frac {y^{2}}{r^{3}}}$

and by symmetry:

${\frac {\partial ^{2}z}{\partial x^{2}}}+{\frac {\partial ^{2}z}{\partial y^{2}}}={\frac {f''(r)}{r}}-2{\frac {f'(r)}{r^{2}}}+2{\frac {f(r)}{r^{3}}}+{\frac {1}{r}}\left[{\frac {f'(r)}{r}}-{\frac {f(r)}{r^{2}}}\right]={\frac {1}{r}}f''(r)-{\frac {1}{r^{2}}}f'(r)+{\frac {1}{r^{3}}}f(r)$

by the problem statement:

${\frac {\partial ^{2}z}{\partial x^{2}}}+{\frac {\partial ^{2}z}{\partial y^{2}}}=x^{2}+y^{2}=r^{2}$

and equating these gives:

${\frac {1}{r}}f''(r)-{\frac {1}{r^{2}}}f'(r)+{\frac {1}{r^{3}}}f(r)=r^{2}$

$r^{2}f''(r)-rf'(r)+f(r)=r^{5}$

and now you have a simple differential equation to solve. --Prestidigitator (talk) 00:22, 22 April 2008 (UTC)

Improving mathematics marks

Hi. I am an grade 11 student and would like to improve my maths marks. Any hints? —Preceding unsigned comment added by 198.54.202.70 (talk) 19:54, 21 April 2008 (UTC)

My first advice: Don't just memorize formulas and routines. Contrary to popular belief, mathematics makes sense, and understanding where everything comes from will help whenever you are dealing with it. If there is anything you don't feel comfortable with, ask around about its meaning until it clicks.

That said, most ideas do need repetition and practice to process and completely swallow, and familiarity with the formulas will spare you the need to look them up. Thus, there's no avoiding spending more time in studies, reviewing material and solving exercises. Don't shy away from the problems marked as harder - being accustomed to such problems will make it a breeze to face the easier problems on the tests. -- Meni Rosenfeld (talk) 20:29, 21 April 2008 (UTC)

I agree completely with Meni. Also, remember to check your answers. It sounds like a boring and tedious waste of time at the end of a question, but it really can save you marks. If you find out that a parachute makes you fall faster than you would otherwise, you know you're wrong - so check it, just in case. --Tango (talk) 20:35, 21 April 2008 (UTC)

From my experience teaching, the students who did worst were the ones who relied on memorization rather than understanding. You want to know why things are, not just what they are. If you do that, you will ace every math class you take (and even be able to take on some questions that you haven't learned how to solve yet). Donald Hosek (talk) 23:43, 21 April 2008 (UTC)

While I fully agree with the advice above, what may be difficult is getting from here to there: if most of your math studies so far have involved memorizing formulas and conventions, it may not be obvious how to even begin to understand the logic behind them. A good and dedicated teacher can help a lot here, but those can be rare. The next best thing, IMO, would be a good book on elementary mathematics. Schoolbooks don't tend to be very good for this (although there may be exceptions); for various reasons most of what they contain tends to be exercises, exercises and exercises, with very little space left over to explain why things work the way they do and where the notation and other stuff comes from.

What you want, instead, is something from the "popular mathematics" shelf at the bookstore. I'm sure others here can offer their own suggestions on what books to choose, probably better than I can. If you asked me, though, one that I'd suggest would be Lancelot Hogben's Mathematics for the Million; I remember reading it myself (as a translation) when I was about your age, but it's much older than that, having been written in the 1930's and remaining in print continuously since then. Can't really argue with a track record like that. One thing it has going for it is that I'd be willing to bet a good sum that the way it presents mathematics will not be anything like the way your schoolbooks do it — and seeing that one can approach the same ideas in many different ways is an important part of learning to understand mathematics.

Pretty much anything by Martin Gardner is also likely to be good, and at least intellectually stimulating. It might not actually directly teach you anything that will be of any use in a test, but it will teach you things that will help you make sense of the stuff you need to learn for tests. —Ilmari Karonen (talk) 15:15, 22 April 2008 (UTC)

For those things you do need to memorize, I suggest flash cards. Take out cards as you learn the material, and focus on learning those which remain. StuRat (talk) 00:10, 23 April 2008 (UTC)

trinomials-algebra1

how do you factor trinomials, specificly, not square ones(ones that all # have square roots). —Preceding unsigned comment added by 66.44.248.145 (talk) 22:18, 21 April 2008 (UTC)

What kind of trinomials are you talking about? Quadratics? Could you give an example? --Tango (talk) 22:31, 21 April 2008 (UTC)

This is clearly a basic question. Note the subject indicating algebra 1. We're talking about factoring quadratics. My suggestion to the OP is to talk to your teacher about this, as you'll get the best and most appropriate feedback. High school math teachers are generally happy to provide help to their students. If not, try the teacher in the next classroom, who will be flattered that you've asked for help. Donald Hosek (talk) 23:41, 21 April 2008 (UTC)

Do you mean solving a cubic equation of the form $ax^{3}+bx^{2}+cx+d=0$? See Cubic function#Roots of a cubic function. It's not a very simple feat to do it symbolically. --Prestidigitator (talk) 00:43, 22 April 2008 (UTC)