Wikipedia:Reference desk/Archives/Mathematics/2008 February 28

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February 28

Lebesgue measure of a complement?

No doubt this is trivial, but I don't have handy access to a measure theory text and web searching has availed me nothing. Is there a tidy expression for the Lebesgue measure μ(Rⁿ \ S) of the complement of an arbitrary measurable subset S of Rⁿ in terms of the measure μ(S) of S? My guess is that if μ(S) = c is finite, then μ(Rⁿ \ S) = ∞, and if not then there is nothing in general we can say about μ(Rⁿ \ S).—PaulTanenbaum (talk) 01:03, 28 February 2008 (UTC)

That's correct. The addivity formula for a measure continues to hold for sets of infinite measure, so you get µ(S)+µ(Rⁿ\S) = µ(Rⁿ) = ∞, which amounts to what you guessed (two elements of [0,∞] sum to ∞ iff one of them is ∞). You can also derive the same result using finite measures only by writing the Lebesgue measure as a limit of finite measures (e.g. measures supported on larger and larger closed balls). Bikasuishin (talk) 01:32, 28 February 2008 (UTC)

Finding the length of a vector

here's my question:

suppose u=[1, -2]. determine 3u.

so i do, i get [3, -6].

but then the question asks

find the length of the vector.

huh? i have no idea how to do this; none of my notes tell me how. i thought about finding its magnitude √(x²+y²), but you end up with √45. i checked the answer at the back of my textbook and it says 3√5. can anyone help me here? --24.109.218.172 (talk) 01:51, 28 February 2008 (UTC)

oh wait... i think i figured it out. √45 equals about 6.708; 3√5 equals about 6.708. does anyone know how they got 3√5?--24.109.218.172 (talk) 01:52, 28 February 2008 (UTC)

3u is three times the length of u. So |3u|=3|u|=3√(1²+(-2)1²)=3√5. Notice that 3√5=√(3^2*5)=√45. Taemyr (talk) 01:57, 28 February 2008 (UTC)

ahh i see thanks man.24.109.218.172 (talk) 02:59, 28 February 2008 (UTC)

A calculator is generally unhelpful in learning math. http://upload.wikimedia.org/math/5/e/5/5e5ea894cc317ec4bb22f46beec9d791.png —Preceding unsigned comment added by Imagine Reason (talk • contribs) 03:01, 28 February 2008 (UTC)

If you wanted to write ${\sqrt {xy}}={\sqrt {x}}{\sqrt {y}}$, all you needed was to take a look at Help:Formula. And no, I don't think I agree. A calculator can be extremely helpful if one uses it properly. -- Meni Rosenfeld (talk) 09:21, 28 February 2008 (UTC)

√45=√(9x5)=√9√5=3√5 should make it totally clear.. (another √2√2=√(2x2)=2) or √72=√2x36=√2√36=6√2 .. it helps if you spot that the value in the root has a factor that is a square eg 4,9,16 etc is a factor87.102.84.112 (talk) 10:27, 28 February 2008 (UTC)

point to point reflection formula

What is the formula for generating the reflection curve for a point to point reflection (not formula for ellipse)?

I don't understand. The curve in question is exactly (one of) the ellipse(s) with those two points as foci. -- Meni Rosenfeld (talk) 09:30, 28 February 2008 (UTC)

My error. I was thinking of 2 parabolas facing each other rather than an ellipse. The formula for an ellipse is the correct formula.

Are you asking how to prove/find that the 'foci' reflect onto each other for an ellipse? If so you're going to need the slope of the ellipse at a given point, which is given by the differential of the equation of the ellipse, and then be able to work out what is the reflected ray - if so ellipse, reflection, Ellipse/Proofs, dot product might be useful - ask for more if I was along the right lines —Preceding unsigned comment added by 87.102.84.112 (talk) 10:21, 28 February 2008 (UTC)

Linear equations Part II

Question 1

I think Ive managed to do the previous question:
${\frac {3a}{2}}-4={\frac {1}{2}}$
${\frac {3a}{2+1}}-4+4={\frac {1}{2}}$ (<---Plus 4 to the top of the fraction and 1 to the bottom)
${\frac {3a}{3}}={\frac {9}{3}}$
$a=3$

OK, not entirely sure where you got ${\frac {3a}{2+1}}$ from - the easiest way to do this would be to multiply through by 2, thus eliminating the fraction, then move all the constants to the right hand side and divide through by the coefficient of a (3). You got the right answer though. -mattbuck (Talk) 13:14, 28 February 2008 (UTC)

The next one proved a little more difficult:
${\frac {2}{3}}-{\frac {5b}{6}}={\frac {3}{2}}$
${\frac {2\times 2}{3\times 2}}-{\frac {5b}{6}}={\frac {3\times 3}{2\times 3}}$ (lowest common denominator)
${\frac {4}{6}}-{\frac {5b}{6}}={\frac {9}{6}}$
And Something Goes next.. im not sure what

You are correct so far. Multiply through by 6 and it should become obvious. -mattbuck (Talk) 13:14, 28 February 2008 (UTC)

Question 2

Im not sure how to do the questions like these because the numbers are different on both sides. $7x+4=4x-5$

What you want to do is get all the terms in x onto one side, and all the constant terms onto the other. If you do that, it should be obvious. -mattbuck (Talk) 13:14, 28 February 2008 (UTC)

(edit conflict) I don't understand your method for the first question. It gets the right answer, but I can't see how... There is a standard method for solving linear equations:

1. Use addition and subtraction to get all constant terms on one side of the equals sign and all terms with an x (or whatever your variable is) in on the other.
2. Combine the terms so you have one term on each side (this is just basic addition and subtraction).
3. Divide both sides by the coefficient of x
4. Read off the answer

That method will work for all the questions you've mentioned. --Tango (talk) 13:18, 28 February 2008 (UTC)

Recursively defined sets, Part II

I'm having trouble trying to determine some of the language in the following problem:

Consider the following set C, a subset of the set of all bitstrings:

Basis step:

$\lambda \in C$ where $\lambda$ is the empty string.

Recursive step:

$\omega \in C\rightarrow S_{1}(\omega )\in C$

$\omega \in C\rightarrow S_{2}(\omega )\in C$

$\omega \in C\rightarrow S_{3}(\omega )\in C$

Where...

$S_{1}:C\rightarrow C$ given by $S_{1}(\omega )=1\cdot (\omega \cdot 0)$

$S_{2}:C\rightarrow C$ given by $S_{2}(\omega )=1\cdot (\omega \cdot 1)$

$S_{3}:C\rightarrow C$ given by $S_{3}(\omega )=0\cdot (\omega :\omega )$

(Note $\cdot$ and $:$ are character and string concatentation respectively.)

My question is as follows:

Evaluate $\omega _{1}=(S_{3}\circ S_{3}\circ S_{3})(0).$

Now what does this mean? Do I just generate the set using just $S_{3}$ on three seperate steps and concatenate?:

$S_{3}=\lbrace 0\rbrace \,$

$S_{3}=\lbrace 0,000\rbrace \,$

$S_{3}=\lbrace 0,000,0000000\rbrace \,$

$w_{1}=0\cdot 000\cdot 0000000\cdot 0$

Then I suppose that's wrong. I'm fairly certain I have to concat the (0) at the end. Trouble is there is no mention of the $\cdot$ operator and there is no indication of what $\circ$ means. Last time I checked it was the function composition operator. Help start me off with this one! Damien Karras (talk) 13:16, 28 February 2008 (UTC)

It is function composition. S₃ is a function, you just apply that function three times. If it's easier, think of it as $\omega _{1}=S_{3}(S_{3}(S_{3}(0)))$, it means the same thing. --Tango (talk) 13:21, 28 February 2008 (UTC)

Apologies, my brain is useless - the bracketed function notation confused me. Am I right I declaring, therfore that

$w_{1}=0000~0000~0000~000?\,$

Damien Karras (talk) 15:19, 28 February 2008 (UTC)

That's what I got. --Tango (talk) 16:43, 28 February 2008 (UTC)

Second order, linear ODE with non-constant coefficients

${\frac {d^{2}x}{dt^{2}}}-at{\frac {dx}{dt}}-ax=0$

$a$ is a constant. Is there a closed-form solution for $x(t)$? —Keenan Pepper 16:34, 28 February 2008 (UTC)

Depending on what you're willing to accept as "closed-form", yes, there is. $x_{0}(t)=\exp(at^{2}/2)$ is a solution, so you can reduce the problem to a first order ODE by looking for solutions of the form $x(t)=x_{0}(t)z(t))$. The solutions turn out to be:

$x(t)=e^{at^{2}/2}\left(\lambda +\mu \cdot \mathop {\rm {erf}} {\Big (}{\sqrt {\frac {a}{2}}}t{\Big )}\right)$

where erf is the error function. Bikasuishin (talk) 16:54, 28 February 2008 (UTC)

That's not solved for $x(t)$, because $x$ appears in the left-hand right-hand side. Thanks for that one solution though; now I can apply reduction of order myself. —Keenan Pepper 17:51, 28 February 2008 (UTC)

Sorry, the x on the right-hand side was a mistake on my part. Fixed. Bikasuishin (talk) 18:13, 28 February 2008 (UTC)

Sinusoidal tourism model

I'm looking for a website that would contain information on how the amount of tourists in Japan (or another country) varies periodically throughout the year.Tuesday42 (talk) 16:46, 28 February 2008 (UTC)

mmmh some data here http://www.tra.australia.com/content/documents/Visitor%20Profile/Japan_visitor_profile_07.pdf page 7 which you can read and find out that they come from "Source: Australian Bureau of Statistics, Overseas Arrivals and Departures Data [Cat. 3401.0]" - so this might be a good source. I suppose (that was visitors from japan if that matters). Found using search for "tourism figures <country> month" or similar.87.102.84.112 (talk) 17:48, 28 February 2008 (UTC)

May as well get the actual publication, from here. Unfortunately a quick look on Japan's Statistics Bureau site didn't reveal any tourism data (the ABS gets their figures as administrative data from the Department of Immigration and Citizenship, so perhaps the Japanese equivalent of that could be of use). Confusing Manifestation(Say hi!) 22:25, 28 February 2008 (UTC)

That doesn't seem to be a sinusoidal model, which is what I really need. I was hoping that amount of tourists in a given area would vary sinusoidally due to seasonal weather. In retrospect, I should have suggested a tropical area.Tuesday42 (talk) 02:00, 1 March 2008 (UTC)

Maybe not tropical - if it's sunny all the year round - suggest somewhere that only has good weather in the summer - how about Cleethorpes?87.102.79.228 (talk) 09:42, 1 March 2008 (UTC)

Why do they teach complicated mathematics to middle and highschool students?

There's no doubt that mathematics is important to humanity. Engineering depends on it, along with statistics, and other fields. But why do they teach complicated math to the average highschool student who may or may not intend to go into those fields? If such a person decides to become a writer, historian, or go into business, he would never need something like geometry, calculus or pre-calc. So why? 64.236.121.129 (talk) 19:18, 28 February 2008 (UTC)

Full-on calculus is pretty rare for the average high school student. Geometry, algebra, and the like, however, are inextricably tied to everyday life. Those who understand the subject are better equipped, then, to intelligently handle situations where math arises. Additionally, mathematical topics tend to reinforce concepts like critical thinking and interpretation of data which arise in even more wide-ranging fields. Personally, the branch of mathematics I'd like to see get more of a push at the high school level is probability (and to a lesser-though-related degree, statistics). The lotteries wouldn't like me very much if I ever pulled that off. Finally, it helps develop the student as a well-rounded capable individual. Similar ideas could be put forth that engineers don't need writing classes, but having met engineers who hold to such an opinion, I can firmly state that there's a world of difference in the career potentials of engineers who express themselves well and those who can't. — Lomn 19:40, 28 February 2008 (UTC)

I can't really agree that Geometry is used in everyday life. Geo isn't exactly about knowing that a triangle has three sides, it's complicated proofs, which no average person ever uses. And yea, I guess the average HS student doesn't need to take Calc, but my point was, HS math teaches convoluted maths to students who typically won't ever use it. In theory I guess math should push critical thinking, but you don't actually have to know what you are doing in math to do well. You can get by just by following patterns and memorizing formulas without knowing what the point is. 64.236.121.129 (talk) 20:42, 28 February 2008 (UTC)

If you can pass the exam just be memorising things (which is often the case), it's a bad exam. That's true (and quite common) in many subjects, not just maths. It's a serious problem with modern education. We didn't do many (if any) complicated proofs in Geometry before A-level - it was mostly calculation based, which is what's useful in everyday life. --Tango (talk) 20:48, 28 February 2008 (UTC)

In the UK, calculus isn't taught until A-level, so you have to explicitly choose to take A-level maths to learn it. Geometry and basic algebra are very useful in the real world - being able to do basic right-angled trig could come in handy when working out how to lay out a room, for example. Algebra can be used to work out if a special offer is really a good deal. I agree with Lomn that there should be more probability (and, as he says, to a lesser degree, stats) - probability is very counter intuitive and without a proper understanding of it people make quite serious mistakes. --Tango (talk) 20:18, 28 February 2008 (UTC)

Of all the mathematics which I had learn in high school, I hated geometry and statistics the most. It's not that I cannot do geometry but that I just can't see myself solving real world problem in geometry and that real world engineers uses vectors to solve problems rather than using geometry.

As for statistics, it's useful but just not taught right to me in highschool. Furthermore as I didn't have computers in highschool, doing a statistics problem which I made up myself is excruciating painful (because adding lots of numbers together is very painful without a computer). Hence my dislike for statistics. 202.168.50.40 (talk) 21:19, 28 February 2008 (UTC)

Vectors are part of geometry, so I'm not quite sure what you mean. The difference between using vectors and doing geometry in the way it's taught at lower levels is pretty much just one of notation - the actual maths is the same. I've also never enjoyed studying statistics... I'm not sure why not. I enjoy playing around in Excel analysing data, but studying it formally has always seemed boring to me - probably because stats at school generally involved calculating means, etc. of tables of random numbers, which, as you say, is excruciating. --Tango (talk) 21:49, 28 February 2008 (UTC)

To be fair, when I say that I'd like to see the average student work more with statistics, it's really that I'd like the average student to better understand the general theory and limitations of statistics -- I hate doing the work, too. But I think it'd be great if more people understood why a random poll of 1000 Americans is meaningful (within America) as well as the caveats that accompany it. — Lomn 22:21, 28 February 2008 (UTC)

I'm rather confirming your statement here, but a random poll of 1000 Americans is meaningful? Don't you have to choose people from appropriate demographics to get statistically significant results from such a small sample? --Tango (talk) 22:28, 28 February 2008 (UTC)

Yes and no. I don't recall the exact number, but a fully random sample of approximately 1000 people is sufficient to give around ±5% results with 95% confidence for the US population. A table for populations up to 100,000 provides a pretty good benchmark for the estimate. Theoretically, one shouldn't adjust for demographics. In practice, as Black Carrot notes, some sort of adjustment has to be made, though I'm no expert on how this is actually applied. — Lomn 00:00, 29 February 2008 (UTC)

In terms of policy, I think the "may or may not" bit of the original post is also critical. Public school is not a trade school. It's supposed to prepare you for any occupation or hobby you could want. (Side effect of freedom of choice.) And, given the dropout and change-of-major rates at most colleges, you don't know what you want yet, even if you think you do. My sister is a software engineer right now, at a huge company. Until she took a mandatory computer programming class in high school, she wanted to be a vet. You have be given a taste of everything, by force if necessary, or you'll lose a lot of golden opportunities. I'm not saying you've been introduced to the subject particularly well, but you should at least recognize that you need to be introduced to the subject. Also, be careful what you assume doesn't use math. You mention business. Statistics and finance are the lifeblood of business. Not everyone in business knows that, but that's their problem. Black Carrot (talk) 22:38, 28 February 2008 (UTC)

On the sample thing - for a "random" sample, 1000 would be plenty. Since it's impossible to get a random sample, you have to counter the inherent biases by breaking it into demographics. Black Carrot (talk) 22:41, 28 February 2008 (UTC)

There's no doubt that maths gets complex. But complicated is a different matter. To me, that word suggests making it more complex than it needs to be. Surely the job of any teacher, and a maths teacher in particular, is to explain concepts in ways the students can understand, by de-complexifying them and expressing them in simpler ways appropriate for their age group. Once they've grasped that idea, then the teacher might proceed to more complex stuff, on the assumption that the students are up to speed. If that's actually not the case, then sure, what comes next will seem mystifying and incomprehensible. That's why depending on just what's presented in the class room is generally not enough, and doing homework (and not necessarily limiting yourself to the exercises that are set by the teacher) is important. A music student does most of their learning not at the weekly lesson with the teacher, but while at home practising by themselves, finding out what works and what doesn't, and trying out different approaches in an effort to find some of their own solutions. If they haven't mastered playing a scale error-free, they won't master a harder piece. Same with maths and other disciplines. -- JackofOz (talk) 23:01, 28 February 2008 (UTC)

I believe that teaching different ways of thought is helpful to any person. Understanding mathematics, functions, and programming bettered my thought significantly, and I can think of things in more than one perspective, kind of. Mac Davis (talk) 23:04, 28 February 2008 (UTC)

You could say..why teach history if they are going to be a plumber with no interest in the past? We teach for many reasons: To increase the intellect of the individuals that are being taught, to potentially inspire them into an avenue of work/development that will lead them to have a fulfilling life, to give people a basic knowledge of things that are 'important' to society, to provide a good entry to a subject that the individual has never looked at in detail, to teach frameworks/ideas, to promote thinking and asking questions, to challenge the brain/mind of individuals. How would you ever know you want to do maths without being given a branch into it? Yes the stuff isn't everyday-useful but really what is? Do i need to be able to understand verbs, adjectives and nouns? No. In my entire life post-schooling I have never had anybody ask me to identify what they are, and I doubt they are 'hugely important' to my understanding of sentences/books/whatever. You specialise more the further into education you go, but up to the 'opt out' level of schooling we really should try to give students a starting-block knowledge (at various levels of depth) for the main subject areas that people value. As the tv show Numb3rs introduction says "We all use maths everyday, to predict weather, to tell time. Maths is more than formulas or equations; it's logic, it's rationality, it's using your mind to solve the biggest mysteries we know" - A good grounding in maths/math-theories can help people be logical/approach things in a more structure way. Much like the formal-sciences can help with things like that - it's not so much the fine-grain detail of what you learn, but the frameworks, structures and boundaries that you learn to operate in and around that are important. ny156uk (talk) 23:23, 28 February 2008 (UTC)

The world is getting complicated, and you can't trust what you don't understand. That's why America is so behind in math and science - they don't care and they don't trust. Would anyone complain about English classics being taught? I don't think so. The anti-intellectualism in America is horrifying. Imagine Reason (talk) 00:44, 29 February 2008 (UTC)

Why learn math? Because solving problems is fun and joy and satisfaction. Most math courses teach some theory first and then presents you with exercises to check that you understand the theory. In my opinion that is a mistake. It would be better first to present some problems, which you can understand but not solve, and then teach the tools which enables you to solve the problems. Most problems have two solutions, a elementary brute force method, and an elegant shortcut. You do not appreciate the shortcut unless you know the elementary method. Bo Jacoby (talk) 01:52, 29 February 2008 (UTC).

New York City's new elementary math curricula for the past few years starts off each lesson with some very interesting problems. I don't think it goes as far as unsolvable questions, but it's awfully good. Imagine Reason (talk) 05:56, 29 February 2008 (UTC)

I could also turn this question backwards: why do they teach history in high schools? – b_jonas 10:43, 29 February 2008 (UTC)

All human knowledge is interesting. In high school I learnt, for example, how to identify a glaciated landscape; how to say "blood sucking, bandy-legged vulture" in German; and several stanzas of The Rime of the Ancient Mariner. I don't think that information will ever give me any pratical benefit, yet I don't regret learning those things. Gandalf61 (talk) 12:55, 29 February 2008 (UTC)

I agree, it's always best to motivate a theory before teaching it. Sometimes, it's fun to learn a bit of maths just for the sake of it, but usually you want to learn it in order to do something useful, so you should be told that useful thing before you start. --Tango (talk) 13:42, 29 February 2008 (UTC)

One thing I wish highschool math teachers would do is tell students what the maths they are teaching is actually used for. In my entire time in HS, or even middleschool, they never told us what these maths they were teaching us, were for. I remember a student asked a teacher what are we going to use these complicated algebra theories she was teaching us for, and she said "absolutely nothing". I had to learn on my own, what maths was for. Physics, engineering, etc.

I think the comment the ideology that, "we have to force everything on students whether they like it or not, just so they have a choice of what they want to do later in life" is not quite "right", in the sense that it's a good idea. There are basics to math that everyone should learn, sure. But the more complicated maths should be left to the future engineers, and such. Of course it isn't up to me, so things remain as they are. But yes, I agree that seems to be one of the reasons why they teach maths to students.

The critical thinking part definitely is a reason why they teach maths, I just don't think it works. Even in college, they make computer science students learn complicated calculus theories that they will probably never use. You don't need calculus to program in C++ (I think. I didn't get very far in my Comp Sci studies, so I don't know what the advanced classes encompass). So the only reason to learn calc for a comp sci student is for the critical thinking skills. But like I said before, you can ace a calc class just by memorizing formulas and recognizing patterns. "Oh this is the equation that we learned in lesson 5. I'll just follow the pattern they taught in lesson 5 to solve this equation". No critical thinking involved.

I remember at the end of one of our calc classes, my prof asked one student what a Derivative was. And the student just kept saying the derivative was the derivative. He had no idea a derivative was just the slope at a given point on a function. He just knew how to find the derivative, and solve the equation. He did well in the class, but he had no idea what the point of it was though. Another time another professor came into our calc class and asked us what the point of a math degree was. The only answer the students could give was "to teach math to other people". Amazing. 64.236.121.129 (talk) 14:26, 29 February 2008 (UTC)

What you (OP) have observed is a symptom of a much larger and I think more dangerous problem with the education system in the United States of America. The reason math has become something the "people never use" is because it's taught in such a way to be unless to all but those who have the most aptitude for it. If it were taught correctly (as even I could do), I am quite certain most people would use at least Algebra 1 level math on a daily or near daily basis. The only real question is why has the problem been allowed to persist when it has surely been noticed by many people over the years. The answer to that question is very intriguing indeed. The way mathematics is currently taught (at the HS level) is in a purely utilitarian form. Mathematics is best learned by posing the question, and then developing an answer based on what one already knows. That requires one to be able to think freely and critically about a problem. There is a quote on my brother's myspace page that reads as follows; "The most dangerous man, to any government, is the man who is able to think things out for himself, without regard to the prevailing superstitions and taboos. Almost inevitably he comes to the conclusion that the government he lives under is dishonest, insane and intolerable, and so, if he is romantic, he tries to change it. And even if he is not romantic personally he is very apt to spread discontempt among those who are." --- H.L. Mencken. To actually understand the concepts in mathematics one must be able to think things out for them selves, but someone who can do that is dangerous (in the eyes of those in power). A math-wiki (talk) 12:33, 1 March 2008 (UTC)

Here is a though-provoking essay on mathematics education in the U.S.  --Lambiam 19:42, 2 March 2008 (UTC)

Thank you for that link, I owe you a pint! --Tango (talk) 00:59, 3 March 2008 (UTC)

Word. Excellent essay. Black Carrot (talk) 20:19, 3 March 2008 (UTC)

Mandelbrot set of Matrices

If you plot the Mandelbrot set with matrices instead of complex numbers, can you get interesting shapes with more than two dimensions? — DanielLC 23:18, 28 February 2008 (UTC)

I guess you don't get anything really different. Indeed, if you consider the iterates of 0 under a Mandelbrot-like transform $M\mapsto M^{2}+C$, all the matrices you get are polynomials in C. Assuming without loss of generality that C is upper-triangular (see Jordan normal form), you find that the asymptotic behaviour of that sequence is mostly given by the elements of the diagonal, i.e. the eigenvalues of C. The nilpotent part may matter at the border, but it is of little concern in general when looking at asymptotic dynamics. So I think it's mostly true that a matrix belongs to the "matrix Mandelbrot set" when all of its eigenvalues belong to the usual Mandelbrot set, and so you don't really get anything new.

On the other hand, Julia sets are a different matter altogether, since the iterates of a general matrix A under a transform like $M\mapsto M^{2}+C$ do not usually commute with one another, and thus cannot be reduced to a common upper-triangular form in general. I remember that there was a picture of a (3D slice of a) quaternion Julia set on the cover of one my math textbooks way back in high school, so people do draw that sort of things, and that's really a particular case of a "matrix Julia set". Bikasuishin (talk) 00:41, 29 February 2008 (UTC)

COMPUTE

COMPUTE: JUNE 18, 2008 + 31 WEEKS + 2 DAYS

THANK YOU.--Goon Noot (talk) 23:23, 28 February 2008 (UTC)

Here is a calendar, it shouldn't be hard to do. In the future, please avoid posting in all caps. It is the online equivalent of shouting. 134.173.92.241 (talk) 23:58, 28 February 2008 (UTC)

Here is a calculator. --hydnjo talk 01:48, 29 February 2008 (UTC)