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September 12

how to predict the next number with 4 digits

what is the percentages to predict a totally exact number each time?

-if someone of you know anything details aboutthe prediction or formula applied to find a exact number with 4 digits each time,please let me know .thanks for sharing.

If I understand your question correctly, you are asking what the probability of guessing one correct number in 10000 numbers (0000 to 9999). Assorted conditions apply - the main one being that the next number is independent of the previous number. If that is the case then the answer is 1/10000 as a probability and 100 * 1/10000 as a percentage. -- SGBailey (talk) 11:26, 12 September 2008 (UTC)

What did they use before graphing calculators?

Pre-calc seems to depend on graphing calculators. But obviously they weren't always around. What did they use before the graphic calculator for pre-calc? I'm assuming the abacus and slide rule weren't sufficient. ScienceApe (talk) 21:43, 12 September 2008 (UTC)

What part of precalc requires a graphing calculator? I'm not quite sure what precalc consists of (I was educated in England), but I think you can do it all with pencil and paper. Nothing at school relies on complex calculations. Of course, there were people in pre-calculator days who had to perform complex calculations, beyond the abacus and slide rule. They used pen, paper and a lot of time and effort. Algebraist 21:48, 12 September 2008 (UTC)

I don't know. The professor just said that there are parts of the class that required a calculator, and that such problems were impossible to do without the use of a calculator (even for him). It just made me wonder how they were able to do such calculations before the graphing calc was in common usage for highschool/college students. ScienceApe (talk) 03:32, 13 September 2008 (UTC)

See Human computer for one alternative. A friend of mine asked a class when they thought the industrial revolution was, wide variations, one answer was 1960 :) Dmcq (talk) 21:51, 12 September 2008 (UTC)

True, but I don't think professional human computers were ever made available to students in high schools. Algebraist 21:57, 12 September 2008 (UTC)

Oh okay then, books of tables. Students used thin booklets of four or five figure tables, professionals used books with 7 figure tables and you learn rules to interpolate for the last couple of digits. You could get huge great big thick books with all sorts of tables including even tables of random numbers. Dmcq (talk) 22:08, 12 September 2008 (UTC)

Here's a classic professional mathematicians book of tables Handbook of Mathematical Functions: with Formulas, Graphs, and Mathematical Tables Dmcq (talk) 22:16, 12 September 2008 (UTC)

I have to say that in my time doing maths (junior school to degree level), I only ever used a graphics calculator once or twice, and that was when we were being taught how to use them. They're banned in most exams, and I can't think of any good reason to use one really. Most graphs you don't need to actually know the exact curve, just a sketch and a few quick calculations are enough. -mattbuck (Talk) 23:10, 12 September 2008 (UTC)

That's weird. I've taken pre-calc three times so far (nothing to do with my intelligence, just school politics), and all of them require a graphing calculator. I'm not talking about Math in general, it seems only pre-calc uses it really. ScienceApe (talk) 03:37, 13 September 2008 (UTC)

Could you give us an example of a question from your textbook, or maybe from an exam, that is allegedly impossible to answer without the assistance of a graphing calculator? One of the old-timers will probably be able explain how it would be done with more primitive tools. --tcsetattr (talk / contribs) 06:16, 13 September 2008 (UTC)

Graphing calculators are great for getting a quick rough idea of what a function does and finding out roughly where the axis intercepts are. Once you've got that, it's much easier to know what you need to do work things out more precisely (for example, you know how many intercepts there are so you know when to switch from finding more to proving there are no more). I can't think of anything that absolutely requires a calculator, but it is much quicker that working it out analytically from scratch, of plotting it by hand. --Tango (talk) 08:23, 13 September 2008 (UTC)

Reading this thread, I was also puzzled by which part of pre-calculus actually requires a graphing calculator, as opposed to it being a handy convenience tool. Digging around a bit, I found this question paper from this pre-calculus course outline, which includes the following question:

“ Use your graphing calculator to sketch the graph and label the vertex, y-intercept, and x-intercepts:

1. y = -2(x + 3)² + 7
2. 4x + y – 2x² = 9

”

IMHO, the student who algebraically determines the co-ordinates of the vertex and intercepts and confirms these by sketching the graphs by hand has actually learned something about parabolas, quadratic equations and how graphs change under certain linear transformations of co-ordinates. But the student who follows the instructions, copies a graph from the screen of their graphing calculator, and reads off and writes down the approximate locations of the vertex and intercepts has only learned how to use a graphing calculator and has learned little or no mathematics. It is a perfect example of a problem where the route to the answer is more important than the answer itself. Gandalf61 (talk) 11:41, 13 September 2008 (UTC)

That's a question intended to test your ability to use a graphing calculator. If that's specifically on the sylabus, presuming they expect you to need to know it later. Judging by the name, I expect pre-calc is meant to prepare you for calculus - you don't need a graphing calculator for any calculus course I've done... --Tango (talk) 13:47, 13 September 2008 (UTC)

No no no, pre-calc has absolutely nothing to do with Calculus nor does it prepare you for Calc (don't ask me why they call it pre-calc, I don't know). I'm guessing you don't have pre-calc in UK or where ever you are from, but it's here in America. Pre-calc is basically the study of functions, graphs, and advanced algebra. ScienceApe (talk) 17:38, 13 September 2008 (UTC)

...and how to use graphing calculators, apparently ;) Gandalf61 (talk) 21:08, 13 September 2008 (UTC)

Early pocket calculators were expected to be used for hard stuff like 34201·4837, but actually they are used for computing 3+18, which is no longer done by hand. Teachers expect students to use graphing calculators for sketching y = −2(x + 3)² + 7, but probably it is used more for sketching y = 2 and y = x and the like. Bo Jacoby (talk) 13:36, 13 September 2008 (UTC).

In the past before learning calculus a student would just have to plot each y point for different x values. They would then learn as part of the calculus course how to find the minimum and maximum points and possibly inflexion points which would make sketching the graph and finding the zeros much easier. Dmcq (talk) 14:42, 13 September 2008 (UTC)

I certainly did that during my A-levels, 3 or 4 years ago - maybe it's just the US that teaches using graphics calculators rather than actually plotting graphs. (I did have a graphics calc for my A-levels, but I don't remember using it very often.) --Tango (talk) 14:45, 13 September 2008 (UTC)

Here in Ontario, Canada, we learn to graph lines, parabolas, and high-order functions by hand in grades 9, 10, and 12, respectively. Graphing calculators are sometimes used for advanced algebra because the other steps involved are more important than solving the zeros of an equation like y=2x^5+3x^4-23x^3+9x^2-9x+200.

I'm assuming that pre-calculus corresponds to Ontario's advanced functions course, which is about high-order polynomials, continuity, zeros, asymptotes, and the like. Advanced functions is a pre-requisite for the calculus & vectors course, so in this sense it can be termed "pre-calculus". --Bowlhover (talk) 03:13, 14 September 2008 (UTC)

(American, currently undergraduate.) I can't really speak for the effectiveness of a particular Pre-Calc course, but the purpose of precalculus is to get you ready for calculus by introducing or reviewing things you'll need to know about, like limits, the transcendental functions, and basic use of cartesian coordinates. If it's been decided you'll need to be able to use a calculator in calculus, that's a reasonable tool to teach the use of in precalculus. It can't replace any of the other knowledge you'll need, other than the use of older tools like log tables, but it's not an unreasonable thing to put a little time into, just to make sure everyone's on the same page. Most of the things I've used a calculator for in school involved precise calculations that couldn't be conveniently done by hand, in answer to questions like, "What is the measure of angle A to three significant digits?" or "What is the determinant of this matrix?" Neither of those makes use of an advanced calculator's graphing, approximation, or symbol manipulation capabilities, and no question a national test will ask does either. If you can't do it with a scientific calculator and some thought, it shouldn't come up. I remember teachers using the displays of graphing calculators to teach theory, since it's nice to have a fast and accurate picture of what you're learning about, but that doesn't come up on a test. That still doesn't explain the ubiquity of the TI-83 in our curriculum, which I blame on secret kickbacks from calculator distributors. Black Carrot (talk) 06:16, 14 September 2008 (UTC)

One other thought, on rereading the second Original Post. It's not entirely outside the realm of possibility that your teacher isn't actually good at math. Just something to think about. Black Carrot (talk) 06:19, 14 September 2008 (UTC)

So we seem to have reached the conclusion that the parts of a pre-calculus course for which you absolutely require a graphing calculator (rather than just a numerical scientific calculator) are just those parts that teach you how to use a graphing calculator. Yes ? Gandalf61 (talk) 08:20, 14 September 2008 (UTC)

No, I don't think so. The professor said parts of the exam require the use of a graphic calculator. ScienceApe (talk) 14:17, 14 September 2008 (UTC)

Since you failed to provide an example of a question that can't be answered without the graphing calculator (and the only example given by anyone else was "use your graphing calculator to do something that's already easy enough to do with pencil and paper alone") you forfeit the right to object to the conclusion. --tcsetattr (talk / contribs) 20:17, 14 September 2008 (UTC)

No, the professor really did say that. If you don't believe me, well that's your problem I guess. I didn't provide an example because we haven't gotten to those kinds of problems yet. The rest of your post is silly and absurd. Saying you forfeit the right to do anything is childish at best, rude at worst especially on the reference desk. I suggest you keep that kind of talk off of here, it's not civil. ScienceApe (talk) 03:19, 21 September 2008 (UTC)

Before graphing calculators they used common sense (except for people who didn't use common sense). Look: If you don't understand graphs of polynomials, rational functions, exponential and logarithmic functions, etc. without a graphing calculator then you don't understand, and a calculator won't change that. Calculators are not supposed to be a substitute for using your head. Nor are they supposed to be an anesthetic. Michael Hardy (talk) 23:13, 14 September 2008 (UTC)

Excuse me? I never made a claim that calculators replaced "using your head". You're arguing a strawman here it seems. My professor said there were functions that would be impossible to graph without a calculator, even for people with complete knowledge of functions, polynomials, logs, etc. I was curious to know how they graphed such things before calculators. From the responses here, I presume they simply didn't graph such complex problems. ScienceApe (talk) 03:28, 21 September 2008 (UTC)

Also graphing a real function on a desktop computer was available like a decade before graphing calculators were. If you're asking a question on this refdesk, then you likely have access to a computer powerful enough to graph at least simple functions.

Let me also mention that apart from graphing calculators, we have (surprise) calculators with text-only or numeric-only screen. These are much cheaper and were also available a decade earlier, and you can use these to do calculations as well. You can even use them to graph a function with the help of a graphing paper and a pencil. – b_jonas 16:32, 19 September 2008 (UTC)

September 13

Calculus Question

Integral of 1 over radical (6x-x^2) —Preceding unsigned comment added by 169.229.75.140 (talk) 01:10, 13 September 2008 (UTC)

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misevaluation, but it is our policy here to not do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn how to solve such problems. Please attempt to solve the problem yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. Thank you. Algebraist 01:13, 13 September 2008 (UTC)

I agree with Algebraist that this looks like a homework question, but I'm feeling generous so I'll give you a hint: My first step would be to use substitution to get rid of the 6x term, it should be easy from there. Good luck! --Tango (talk) 08:19, 13 September 2008 (UTC)

In my schooldays, many years ago, the integral of 1/(2ax-x^2)^0.5 was a standard form. Times change.—81.132.236.163 (talk) 14:25, 13 September 2008 (UTC)

You say that as if memorising a standard form over understanding the method is a good thing. 91.143.188.103 (talk) 18:25, 15 September 2008 (UTC)

It may well be in the back of a standard calculus text, I don't honestly remember - I'm pretty sure it is once you do the substitution, though (that's the easy bit I referred to!). It may depend on the textbook - if I were writing it, I'd keep the list short and expect readers to do the simple conversions to turn their problem into a standard one! --Tango (talk) 14:37, 13 September 2008 (UTC)

I was curious, so I dug out my calculus textbook from the bottom of the box of notes I brought it home for the summer in (Salas, Hille and Etgen if you're interested) and what you get after the substitution is on the back cover, the OP's question isn't. It seems Salas and I are of like mind! --Tango (talk) 14:42, 13 September 2008 (UTC)

Now the OP has had time to think. First substituting x=3+3t and later substituting t=sinv gives

(6x−x²)^−1/2dx = (6(3+3t)−(3+3t)²)^−1/2d(3+3t) = (18+18t−(9+18t+9t²))^−1/23dt = (1−t²)^−1/2dt

= (1−sin²v)^−1/2d(sinv) = (cos²v)^−1/2cosvdv = dv

So

(6x−x²)^−1/2dx = d(arcsin(x/3−1))

Bo Jacoby (talk) 01:24, 14 September 2008 (UTC).

OK, at least one nonsensical answer appears above and at least one that is needlessly complicated. Let's try being straightforward: if you're not instantly thinking completing the square then there's a gap in your understanding that you badly need to fill. So:

${\begin{aligned}6x-x^{2}&=-(x^{2}-6x)\\&=-(x^{2}-6x+9)+9{\text{ (add }}-9{\text{ to complete the square and}}\\&{}\qquad \qquad \qquad \qquad \qquad \qquad {\text{compensate by adding }}9)\\&=-(x-3)^{2}+9\\&=9-(x-3)^{2}\\&=9\left(1-{\frac {(x-3)^{2}}{9}}\right)\\&=9\left(1-\left({\frac {x-3}{3}}\right)^{2}\right)\\&=9(1-u^{2}).\end{aligned}}$

If

$u={\frac {x-3}{3}}$

then

$du={\frac {dx}{3}},\quad 3\,du=dx$

so

$\int {\frac {dx}{\sqrt {6x-x^{2}}}}=\int {\frac {3\,du}{3{\sqrt {1-u^{2}}}}}=\arcsin u+C=\arcsin \left({\frac {x-3}{3}}\right)+C.$

Pretty routine once you've practiced a bit. But if you don't recognize this immediately as a completing-the-square problem, you need to learn this: the purpose of completing the square is always to reduce a quadratic polynomial with a first-degree term to a quadratic polynomial with no first-degree term. Michael Hardy (talk) 03:27, 14 September 2008 (UTC)

What are the chances

My wife went to work the other day and my son joked and said "We may never see her again." We laughed a bit but I got to wondering, what are the chances that we may never see here again? How does one go about establishing the probability that any one particular event like that is canoccur? I think that you would start let's say with the number of highway fatalities in the area in any one week or year, and then perhaps calculate the number or vehicals that travel on the roads that make it home safe every day and so on. I never studied statistics or probability - so I am curious how the though process in this sort of event is calcutate. Just so we are clear - I do wish my wife to come every day.142.68.147.80 (talk) 17:43, 13 September 2008 (UTC)

Yes, you would need to looks at how many fatalities there are doing a particular activity and divide it by how much that activity is done. You could improve the accuracy by restricting it to people like your wife (does she have a good no claims bonus? If so, you should look at how many people with good no claims bonuses die on the roads divided by how much they drive). You should probably base it on miles driven, rather than number of cars, although the risk may well not be proportional to the length of the journey. You would need to gather lots of data to get accurate statistics. Either that, or you could try googling - someone has probably done such a study already for most common activities (well, they certainly will have studied it in order to work out insurance premiums, but they may not have published the results). --Tango (talk) 18:05, 13 September 2008 (UTC)

And then double this number, because you yourself could die too. Sentriclecub (talk) 09:39, 14 September 2008 (UTC)

You can't necessarily double it, since you'll be doing different things and are a different person, so have different risks. Also, you can't just add the probabilities, you have to minus off the chance of your both dying otherwise it will be counted twice (this becomes obvious if the change of each of you dying is more than 50% since adding them together would get a probability of over 100%). --Tango (talk) 12:55, 15 September 2008 (UTC)

I originally thought about just writing as my first response and then double this number and leaving it at that. Allowing either of you two to figure out the subtle oversight you both made. Then I thought about the response that I eventually went with. But I feared that it may be misinterpreted that I literally mean to do all the suggested research, calculate a number, then multiply it by 2.000 and assume uncertainty of .001 but I gambled that my response would not dissected as literal advice of how to exactly measure the chance. I only wished to elucidate that the question implicitly assumes that the husband is immortal during the time period. For the elaborateness of the first response, nowhere is mentioned that there are two ways to meet "the condition". To be 100% accurate--here it goes-- the chance that "we may never see her again" is approximately double the chance of "she may die". Okay, but this statement can be refuted a million different ways, for starters you could say that the radius of the probability of the husband's death is .3% different than the probability of the wife's death. However, in the end, questioner gets a better and more accurate answer if he reads our first responce each, only. I guess its just that I am obsessed with probability puzzles. My favorite probability enigma is one that says 13% of people with 160 IQ couldn't pass a mensa entrance exam, if they are already a member of Prometheus socity (having truthfully passed an IQ test which they scored 160+). I merely believe that doubling the answer, is more accurate than to 1x it. If you thought I meant 2.000 and forbade 1.7 or 2.3, then its just misunderstanding on your part caused by failure of communication on my part. I only elaborate on this in great detail because I think Zain made a good point about your answer of the APR question, where your answer carried a broad generalization and you converted it into a special case. Nominal is only opposite of real, in a very limited number of cases. Inflation, sure, but APR no. I'm only sporadically involved ref desk contributor, so hope you can tolerate me until the end of the day. I'm obsessive, and its counterproductive for me (and for other contributors) to stray off from the original questioner's topic. So I apologize for excessiveness. Sentriclecub (talk) 21:50, 15 September 2008 (UTC) Forgot placeholder, oops. Made my point, from an hour ago, thanks.

You also need to take into account that both father and son could lose their eye sight. "Thus never seeing her again." —Preceding unsigned comment added by 203.202.144.223 (talk) 04:27, 16 September 2008 (UTC)

To make it really complex and assuming that god exists, the events that you said 'you may never see her again' and that 'you really don't see her again', may be dependent and in that case just saying that 'you won't see her again' may increase the probability of the event!

Topology Expert (talk) 13:28, 17 September 2008 (UTC)

September 14

Validity of proof

Is this proof valid? If so, can you help me find a reference for it on the Internet? --Bowlhover (talk) 02:00, 14 September 2008 (UTC)

What's highlighted in green is accurate, and is a straightforward enough application of basic principles that it doesn't need a reference. As a formatting consideration you might try taking the fractions out of their math brackets and just writing them normally, p²/q². Black Carrot (talk) 05:50, 14 September 2008 (UTC)

"Since all numbers which can be represented by p/q are by definition rational, the above implies that all rational square roots are roots of a ratio of two perfect squares," should be rephrased somewhat, the implication is in the wrong direction. Maybe, "Since all rational numbers can by definition be represented in the form p/q, the above implies that all rational square roots are roots of a ratio of two perfect squares." Black Carrot (talk) 05:54, 14 September 2008 (UTC)

The proof does depend on the unstated assumption that gcd(p²,q²)=1 if gcd(p,q)=1. This is true in a unique factorization domain such as Z, but is not true in all integral domains. Other proofs make this dependence clear by using unique factorisation directly. Gandalf61 (talk) 08:36, 14 September 2008 (UTC)

Yep I agree with both contributors above, the proof is flawed. The sentence "p²/q² can be assumed to be in lowest terms" is where it all falls down but it was heading the wrong way from the beginning. Dmcq (talk) 08:50, 14 September 2008 (UTC)

Is it necessary to deal with any numbers except integers? A perfect square and its square root must be integers, and p and q can be assumed to be. --Bowlhover (talk) 14:27, 14 September 2008 (UTC)

If I may interject two points. First of all, I find the section too detailed for inclusion in the main article. Perhaps there is enough material scattered about to start an article on the square root of rational numbers. There are a number of proofs indicated in the square root of 2 article, and I would rather see the section in irrational number kept brief and in summary style. Second, the lengthy proof is not necessary. One can prove consider]ably more in much less space by appealing to Gauss's lemma (for which there already is an article): Any rational number which is an algebraic integer is a (rational) integer. siℓℓy rabbit (talk) 14:51, 14 September 2008 (UTC)

Richard Dedekind's proof in [1] seems a good one to me. It doesn't depend on proving the Fundamental Theorem first and is fairly short, it only depends on numbers being ordered. Dmcq (talk) 08:16, 16 September 2008 (UTC)

Would like a math expert to look over my experimentation with a simple physics equation

$f_{k}={\frac {joules}{meter}}={\frac {d}{dx}}E_{th}$

Frictional Force $=f_{k}=Newtons=Joules/meter=Force$ shows that the kinetic frictional force is the derivative wrt x of internal kinetic energy of an isolated system as the block undergoes a displacement, x. Since $f_{k}$ is a constant, and is equal to $\mu _{k}\,$*N (where N = the normal force) then $\mu _{k}\,$ can be thought of as...

Help me from here, please. I want to continue the line of thinking that $f_{k}\,$ is a derivative of internal kinetic energy.

A month ago, I realized that $\mu _{k}\,$ is a percent relative the kinetic frictional force to the Normal Force, and this insight really helped me understand everything in physics much better. i.e. $\mu _{k}\,$=.40 means that of the Normal force, the kinetic frictional force is about 40% the Normal force.

I also really liked the algebraic proof that $\mu _{s}\,$ is the solution to the equation $\mu _{s}=\tan \alpha \,$ where alpha is the angle of incline which a block begins to slide, from rest position.

So today, I'm trying to round out my collection and would like an insightful way to represent an intuitive relationship between $\mu _{k}\,$ to work or internal-kinetic-energy.

So, ignoring angles of incline, and ignoring mass and gravity (let them be 0, m₁, and g all respectively) is there any way for me to algebraically rearrange variables or shuffle equations to get a neat relation between \mu_k and work or internal kinetic energy? I haven't found one, so if one doesn't exist, are there any other outside the box ways to look at these relationships between variables?

When I usually try to creatively play with formulas, if the term is ${\mu _{k}}^{2}$ or something like a difference of terms, such as-- ${\mu _{k}}$ initial minus ${\mu _{k}}$ final, terms like these have more flexibility and you can apply more algebra to get more new equations from old ones.

Thanks in advance for any input. My goal is to learn physics slowly and very analytically, always looking for ways to inject redundant mathematical thinking into my notes that I keep. Sentriclecub (talk) 12:23, 14 September 2008 (UTC)

addendum from the kinetic friction page...

When an object is pushed along a surface, the energy converted to heat is given by:

$E=\mu _{\mathrm {k} }\int F_{\mathrm {n} }(x)dx\,$

where

F_n is the normal force,

μ_k is the coefficient of kinetic friction,

x is the coordinate along which the object transverses.

I may have wrongly presumed that calculus leads to nowhere. (I'm stronger in algebra than calculus) Sentriclecub (talk) 12:29, 14 September 2008 (UTC)

numerical analysis

plz help me solve this example with detailed solution. Find a root of the equation x sin x + cos x = 0, using Newton-Raphson method. Miral b (talk) 12:42, 14 September 2008 (UTC) Retrieved from "http://en.wikipedia.org/wiki/Wikipedia:Reference_desk/Mathematics" Hidden category: Non-talk pages that are automatically signed

Rearranging the equation gives x = −cot(x). Looking at the graphs of y = x and y = −cot(x) you can see that there are an infinite number of solutions. As you only need to find one solution, you could find the solution that lies between 3π/4 (where x sin(x) + cos(x) = 0.9588...) and π (where x sin(x) + cos(x) = -1) - I am assuming x is in radians here. Starting from either of these initial values, Newton-Raphson converges quickly to a solution, giving at least 8 decimal places in just 4 iterations. I set it up in Excel - alternatively you could use your favourite programming language, or do it with a calculator in less than 10 minutes. Gandalf61 (talk) 13:25, 14 September 2008 (UTC)

The function $f(x)=x\sin x+\cos x$ is an even function: $f(-x)=f(x)$. The Taylor series

$x\sin x+\cos x=1-\sum _{k=1}^{\infty }{\frac {(-x^{2})^{k}}{2(k+1)(2k)!}}=1+{\frac {1}{2}}x^{2}-{\frac {1}{8}}x^{4}+{\frac {1}{144}}x^{6}-{\frac {1}{5760}}x^{8}+\cdots$

contains only even powers of x. So it is simpler to set $t=x^{2}$ and solve the nth degree equation

$1-\sum _{k=1}^{n}{\frac {(-t)^{k}}{2(k+1)(2k)!}}=0$

for some sufficiently big value of n.

The two smallest solutions to x sin x + cos x = 0 are

$x\approx \pm 1.19968i$

The smallest real solutions are

$x\approx \pm 2.79839$

In order to reach this accuracy you need n ≥ 9. The J-code used is %:_1 _2{1{>p.((*1&o.)+2&o.)t.2*i.9 , and the result produced is 0j1.19968 2.79839. Bo Jacoby (talk) 22:14, 14 September 2008 (UTC)

OK, you need to find when x*sin(x) + cos(x) = 0. Find when cos is negative and when sine is positive. For instance in the second quadrant. Take a value in the second quadrant reasonably close to pi but also reasonably close to 3pi/2 as the first approximation (since the 'x' in x*sin(x) increases sin(x) considerably). Then apply the formulae.

Topology Expert (talk) 13:23, 17 September 2008 (UTC)

Cos and sine are real-valued functions so you cannot speak of an imaginary solution. They aren't (for complex arguments), and sure you can. Please don't propagate these kinds of misconceptions. Fredrik Johansson 08:02, 20 September 2008 (UTC)

Is impulse related to work?

When an object is pushed along a surface, $\mu _{\mathrm {k} }\,$ is the ratio of the energy converted into heat vs this denominator:

$\mu _{\mathrm {k} }={\frac {E_{th}}{\int F_{\mathrm {n} }(x)dx\,}}$

How do i verbally say the denominator? Is it the indefinite integral of the Normal Force (as a function of x) with respect to x?

I am picturing an example. A constant force is applied to block A which slides 2 meters on flat surface, then up a 60 degree incline for the final 8 meters.

I see a lot of new ways to further explore uses of $\mu _{\mathrm {k} }\,$ now, but needed to know what the denominator is in spoken words.

If this block is pulled by a string in such a way that the block has constant velocity of 1 meters per second (and the string always pulls parallel to the block's momentum), then I could integrate the normal force as a function of time, then convert it into Joules. This would involve solving for the impulse, then converting an impulse into work. I never covered subjects from Calc_II. Is it straightforward to convert an impulse to an amount of work, given this example? I have read the articles on these subjects, but they get too complicated too fast, and I get overwhelmed. I should be able to better attempt my first question if I am not afraid of making a calculus goof. Sentriclecub (talk) 13:28, 14 September 2008 (UTC)

For the denominator, you can just say "the integral of the normal force with respect to x" in most cases. Saying "integrating with respect to x" usually implies that every non-constant term in the integrand is somehow a function of x. You shouldn't really say "indefinite integral" in this context since, strictly speaking, the integral should have lower and upper limits (respectively, the initial and final positions of the object being moved). In the specific example you gave, it is very easy to convert the impulse into work since you know everything about the applied force -- in fact, if you use all SI units, the two quantities are numerically equal since you set the velocity to be 1 m/s. With Calc_1 knowledge, you should be able to quickly verify this using the equations $J=\int Fdt$ and $W=\int Fdx$ where J is the impulse and W is work. The complete answer to your question of whether there is a general relationship between impulse and work is more complicated. You definitely need to know the actual masses of the objects in question. I might explain this more in depth later, but for now I'll give you some helpful information that may lead you to work out the complete answer yourself. First, note that $K={\frac {p^{2}}{2m}}$ where K is the kinetic energy and p is the momentum. Second, review the work-energy theorem, which states that the total work done on an object is equal to the object's change in kinetic energy. 97.90.132.94 (talk) 07:08, 15 September 2008 (UTC)

Thanks, and I hope you'll explain more, after a few days (link now available from my user page ). I have a huge love for mathematics. I spend additional time, when learning physics, to try under understand the equations from a extremely mathematical point of view. I love working through proofs, and I will start working immediately on the part you gave me. That term with momentum^2, I've never seen before, so I'll jump right on it. I'm self taught calculus, so if I'll need more calculus knowledge to understand it, then thats a helluva motivation! Thanks very much! Sentriclecub (talk) 14:56, 15 September 2008 (UTC)

The $K=p^{2}/2m$ equation is just an extrapolation from p = mv. Anyway, I should probably revise what I said earlier. As far as I can tell, there aren't any particuarly useful relationships between impulse and work (and I stress the word useful). Nevertheless, here is a helpful tip in case you haven't discovered this yet: sometimes you will need to use the equations $J=mv-mv_{0}$ and $W_{net}={\frac {1}{2}}mv^{2}-{\frac {1}{2}}mv_{0}^{2}$ and notice that $v^{2}-v_{0}^{2}=(v+v_{0})(v-v_{0})$. Sorry if I might have wasted your time with my earlier statement. You would still profit from learning more about the work-energy theorem, though. 97.90.132.94 (talk) 19:16, 16 September 2008 (UTC)

Thanks, actually I did solve my question. J * (v_avg) = W Sentriclecub (talk) 11:58, 18 September 2008 (UTC)

earliest date of use or description

I began using computers to verify results of logical equations I solved previously by hand in about 1963. Then in about 1978 I found a computer method for reducing logical equations to minimum form in “Digital/Logic Electronics Handbook” by William L Hunter (pages 112-113, Tab Books, Blue Ridge summit, PA) ISBN 0830657740 ISBN 0830657746 ISBN 9780830657742 called the Harvard chart Method of logical equation reduction. In 1981, I published a modification of the method to reduce multi-valued equations to minimum form as a computer program here. In doing the modification I may have become unconsciously aware of the ability to count and to sort sets and multisets using an indexed array as demonstrated here.

Nonetheless, I did not become consciously aware of this method until at least 1995, followed by its copyright and online publication in 1996 here and again in 2006 here. Consequently, I am searching for any publication prior to my own which might describe or demonstrate this method to count and to sort sets and multisets or to find the date the Counting sort was first described or published with its original definition found here and the date the Pigeonhole sort was first described or published with it's original definition found here. 71.100.10.11 (talk) 14:04, 14 September 2008 (UTC)

I looked at your BASIC program and am puzzled by the statement 20 a(a)=a(a)+1 . a cannot be an array and an index at the same time. The claim that your sorting routine is the fastest in the world is not proved. Usually the speed of sorting routines is measured by the asymptotical behavior of the time consumption as a function of file size when the file size is big. Bo Jacoby (talk) 11:11, 15 September 2008 (UTC).

I think the original program was written in Zbasic. Yes, I know that speed claim is based on assumption since for one thing I did not create Instant sort, the hardware version, until Nov. 1996. ~ WP:IAR 71.100.4.227 (talk) 16:49, 15 September 2008 (UTC)

In fact I just pulled out an old copy of zbasic and while it has no problem with an index of the same name as an array is does require that you dimension the array, unlike earlier versions of interpreter Basic which allowed up to ten dimensions before requiring a dim statement. 71.100.4.227 (talk) 18:13, 15 September 2008 (UTC)

I seem to half remember that in some early variants of BASIC (BBC Basic?) ordinary variables, typed variables functions and names all had separate address spaces, so you could do something like that (and have the string A$, integer A% and so on to do something else with). With the source in memory and 16k to play with people would actually use all the single-character variable names they could in a large program. Edit: Yes, I found a reference:

Note that in BBC BASIC the different types of variable are completely independent. For example, the integer variable list%, the numeric variable list, the string variable list$ and the arrays list%(), list() and list$() are all entirely separate.

From [2] -- Q Chris (talk) 12:57, 15 September 2008 (UTC)

Please note that 71.100.*.* has a history of creating original research articles such as Articles for deletion/Rapid sort and Articles for deletion/Optimal classification. See also Wikipedia talk:Reference desk/Archive 26#Money/Golden Calf/Anti Semmitic postings for repeated trolling of reference desks. --Jiuguang (talk) 17:20, 15 September 2008 (UTC)

Although both articles were moved to here and here and the latter is now listed by the National Institute of Standards and Technology here and are quite happy at their new homes, who would ever have guessed that user Jiuguang Wang is a stalker? Other users be warned. 71.100.4.227 (talk) 18:17, 15 September 2008 (UTC)

September 15

having problems with conversion of measurements

somethings i can do but this is not one of them. here are some examples.15 meters to millimters?3.5 tons to pounds?6800 seconds to hours?could someone please help me understand how you change these? —Preceding unsigned comment added by 71.185.146.144 (talk) 09:14, 15 September 2008 (UTC)

I put them in google and it does them for me...

http://www.google.co.uk/search?hl=en&q=15+meters+in+millimeters&meta= http://www.google.co.uk/search?hl=en&q=3.5+tonnes+in+pounds&meta= http://www.google.co.uk/search?hl=en&q=6800+seconds+in+hours&meta= Simple as that. Of course without google things get a bit harder... 194.221.133.226 (talk) 10:35, 15 September 2008 (UTC)

To change between units you need to know how many of the one unit are in the other unit. Then you multiply by how many of the first unit you have. Say that there are 5 "apples" in a "bag" and you have 2.4 bags, then you have 2.4 * 5 = 12 apples. To go the other way, there is 1/5 = 0.2 bags per apple, so 9 apples is 9 * 0.2 = 1.8 bags. So all you need to know is how many millimetres in one metre, how many pounds in one ton, seconds in an hour etc. If you don't know these from memory then the relevant wikipedia articles will help you. -- SGBailey (talk) 11:28, 15 September 2008 (UTC)

Annual Percentage Rate

So I am in a confusion about what “APR” is. In particular, in the simplest case (ignoring everything but the principal) whether it is a nominal or effective rate.

The article annual percentage rate claims it is an effective rate. Through some online searching I’ve found conflicting sites that seem to say opposite things. Eventually I dug out my old theory of interest book (Theory of interest by Kellison) which said it was a nominal rate. Asking a friend, she pulled out hers (Mathematics of interest rate and finance by Guthrie and Lemon), which also says it is a nominal rate.

So I just wanted to check with anyone that might know more about this than me to see if the article is actually wrong, before changing anything (or telling the students in 101 my class something incorrect next week!) GromXXVII (talk) 12:53, 15 September 2008 (UTC)

I'm not going by the article, I have a Ba in finance, and graduated with high honors from the best college in the nation. The APR is not an effective rate. Here's two examples to clear things up.

A $100 bond pays a $3 coupon every 6 months.

The APR is 6.000%

The effective interest rate on the bond is 6.090%

A $100 bond pays a $4 coupon, every 12 months

The APR is 4.000%

The effective interest rate on the bond is 4.000%

I'll answer any further questions, and will provide relevant sources, at your request. The following example may be confusing, but here it goes anyway.

A credit card has a 29.9% APR

If I charged $200, and pay off $100 of it, then I'll have a $100 balance remaining. When my new bill comes in, I'll owe $102.492

The effective interest rate is 34.358%, and if you figure out how I computed this, you'll have came away with full understanding, as every assumption is reflected. Sentriclecub (talk) 13:11, 15 September 2008 (UTC)

Surely a $100 bond that pays $4 each 12 months is not a 6% rate? The credit card example is also flawed, in that many cards compute interest based on an average daily balance, thus not knowing when the partial payment was made makes it impossible to computer the next month's finance charge. Offering sources "upon request" is redundant, the request is implicit in the use of the reference desk. --LarryMac | Talk 13:24, 15 September 2008 (UTC)

Thanks fixed it, ask me how to derive a physics formula, no problem. Trying to multiply 4x1, that's a challenge. The credit card example isn't flawed, if you reread the stipulation that I started the month with a $100 remainder balance and made no new charges. The reason I don't want to offer sources, is to avoid the conundrum that its a real possibility that some sources may calculuate an EAR and APR in different ways. If I gave a source, I would be forced to say that the wikipedia article is wrong, which I don't believe. I would have given the source right away, but I want to compare my author's accuracy to the wikipedia article's accuracy. Secondly, I will point out that you are correct about when the payment is applied, however the example implies in my favor. Since I'm providing clearification, I don't want to give more information than necessary, but I also agree on your point that there is risk resulting with not specificating every detail such as was it februrary on or off leap year, or was it a month with 31 days, etc... Upon further inspection, the credit card example is not flawed. The rate per period is APR/n. LarryMac, may I respectfully ask if you have studied this stuff in school? Something is intriguing me, a subtle mistake you've made in your post. I take it you are a math expert in another area? I don't want to come across bad, I'm representing my school now, and I hope my behavior today is helpful and respectful. I'm 24, and one of the worst feelings I ever had on wikipedia was upsetting someone at the ref desk. So please be patient with me, and my only intention here today is to help answer this guy's question, while Primum non nocere to the other helpers. Sentriclecub (talk) 14:02, 15 September 2008 (UTC)

I understand the effective interest rate in your credit card example being 34.358%=(1+0.299/12)^12. But I’m still a little confused in that APR then appears to be a nominal interest rate (in that I could divide by 12 to get the monthly interest charged of 2.49%), but you said the Wikipedia article isn’t wrong? The first line is “Annual percentage rate (APR) is an expression of the effective interest rate” which seems to say the opposite. (as far as I know it’s not possible to be both an effective and nominal rate unless if the frequency of compounding is the period). GromXXVII (talk) 23:02, 15 September 2008 (UTC)

Yeah, I discussed that earlier over at a discussion page somewhere. Here is the quote--The first sentence is already prone to having people miseducated. Annual percentage rate (APR) is an expression of the effective interest rate the borrower will pay on a loan which is accurate, but completely misleading, because the APR is not an effective interest rate. It is an expression for one, but not one itself. -/quote I spoke to some of the editors in finance articles, and they're okay with me saying that the line is wrong. I just didn't want to offend anyone over there. APR is a nominal rate, (and is an effective rate if and only if n=1). I am going to work with wiki:FINANCE to help fix inaccurate statements in existing prominent articles. The finance articles on wikipedia are in poor shape (maybe 5 or so major ones are rife with errors) so I'm going to volunteer some time over there, even though I'm going to med-school and leaving finance behind me. It's where I'm needed the most, so I'll try and make myself compare articles to my textbooks and rule in favor of my book. At least that way, APR and the articles which link to APR will all be consistent. If someone else wants to use a source that APR is an effective interest rate, they can come behind me--I dont care. But as it stands now nominal interest rate, effective interest rate, real interest rate and annual percentage rate contradict, so I already have the permission from the wikiproject over there, to do a mass clean-thru of the pages. My 5 finance professors and all of their textbooks (5+5=10) are all consistent with each other, and I'll source all my texts, so in a few days, the articles will be clear and those negative tags will come off the tops of all 6 pages that I plan to work on. Sentriclecub (talk) 00:04, 16 September 2008 (UTC)

Ahhh. I see. I was erroneously taking “is an expression of” to mean “is”. You have my thanks in advance for any clarifications you’re able to make on those pages. GromXXVII (talk) 00:17, 16 September 2008 (UTC)

addendum 1

APR = (Per-period rate) X (Periods per year)

EAR = $e^{APR}-1\,$

APR = $ln(1+EPR)\,$

I'm going to look more into all this for you, and will have a full report back in an hour--it looks as though neither the APR nor EAR are independent of the number of compounding periods. The best way to interpret interest rates is with calculus. It appears that the APR and EAR are dependent on the number of periods per year, which bothers me. I'll re-read everything, and will also consult my HP-10Bii user manual, 2 feet away from me at all times. While you are waiting, check out Khan academy on youtube, this guy is a bona fide expert. The finance videos he has produced are wonderful. Sentriclecub (talk) 13:35, 15 September 2008 (UTC)

To clarify, the EAR is the Effective Annual Rate and in the formula above, the APR is the force of interest or the continuously compounded rate of interest. Zain Ebrahim (talk) 13:43, 15 September 2008 (UTC)

It's both nominal and effective. The opposite of nominal is real. See nominal interest rate, real interest rate and effective interest rate. --Tango (talk) 12:59, 15 September 2008 (UTC)

From reading the above definitions I would say it is always effective but only nominal if the interest period is annual (once a year). -- Q Chris (talk) 13:09, 15 September 2008 (UTC)

For this purpose, you need to look at the second definition at nominal interest rate, in which case a rate can't be nominal and effective. When I studied financial maths, the lecturers intentionally avoided the term APR because it changes from person to person. Zain Ebrahim (talk) 13:12, 15 September 2008 (UTC)

Actually, it depends where you are in the world - see annual percentage rate#Region-specific details. In the US, APR is defined as periodic interest rate times the number of compounding periods in a year, so it is a nominal interest rate. In the UK, however, APR is the effective interest rate, which factors in compounding and is not the same as the nominal rate unless the interest is only compounded once a year (in the US the effective rate is called the Annual Percentage Yield or APY - see this Investopedia article). So a loan with monthly interest payments of 0.5% per month would be described in the US as having an APR of 6%, but in the UK its APR would be 6.17% (this example is used in the nominal interest rate article). As in many instances of financial terminology, US and UK just follow different rules. Gandalf61 (talk) 13:27, 15 September 2008 (UTC)

Here are some sentences straight out of my book. Returns on assets with regular cash flows, such as mortgages and bonds, usually are quoated as an APR. The APR can be translated to an EAR by remembering that APR = (Per-period rate)x(Periods per year). Therefore to obtain the EAR if there are n compounding periods in the year, we solve the equation $1+EAR=(1+{\frac {APR}{n}})^{n}$. Sentriclecub (talk) 14:09, 15 September 2008 (UTC)

So for n large, 1+EAR is approximately e^APR. Algebraist 14:23, 15 September 2008 (UTC)

Here are some sentences straight out of my book. Returns on assets with regular cash flows, such as mortgages and bonds, usually are quoated as an APR. The APR can be translated to an EAR by remembering that APR = (Per-period rate)x(Periods per year). Therefore to obtain the EAR if there are n compounding periods in the year, we solve the equation $1+EAR=(1+{\frac {APR}{n}})^{n}$.

Additionally, I'm glad to see the point made that different sources say different things. I personally think any interest rate which is dependent on the number of periods per year, is a flawed measure. So I think with this information, we can call it a closed case. Finance has a way of trying to simplify formulas which are inherently exponential into algebraic. This is why the APR is probably best described as a nominal rate, not as an effective rate. Since the APR is usually divided into 12 periods per year (afterall, people make mortgage payments, and credit card payments once per month, not once per 30 days) so its good for consumers ultimately. Afterall, what if a homeowner took out a 8% home equity loan to pay a 9% credit card, and one of the APR's was calculuated assuming 12 periods per year, and the other APR was confounded and ended up losing EV on the deal. So with this reasoning, (and in my education, I studied the interest rate schemes which lead to federal standards--probably in the APR article) therefore I would argue that the APR is a nominal interest rate, since the reason the government makes Rent-a-Center publish an APR is so that consumers know they can compare it apples to apples like a mortgage payment, which they are familiar with.

Okay, even further into the philosophy of all this, I have finally concluded that the APR is a number to help consumers from getting duped. Say someone goes to rent-a-center and buys $800 rims, for their car and finances it. That guy can ask virtually anyone if 18% APR is a good deal, or not. He's not actually interested in his monthly payment, or in computing the principle to interest ratio of each minimum monthly payment. He just wants to know if the interest rate, if he's getting ripped off on the financing part of the purchase. In other words, I conjecture that an APR is a way for consumers to know if they are being taken advantage of, under the financing terms. As an anecdote, I heard with my own ears, a Ron Legrand audio lecture, on how to purchase a stream of cash flows from a landlord. Suppose the mark goes up to the landlord (who is a landlord, uses seller-financed a mortgage to the buyer) and if the principle on the mortgage is worth $160,000, make the landowner an offer to buy all the cash flows from now until when the principle on the mortgage is $120,000. You can play around with some of these numbers, but the mark ends up "buying" the $40,000 difference at a negotiated price. These are similar to the Carlton Sheets real estate courses. You'd be amazed at how much profit this comes out to be for the mark. So, I believe that APR is a nominal measure, which most consumers in america know that the default rate on a credit card is extremely undesirable. How do they know that a 30% APR is undesirable? Because APR is a nominal number which is widespread (all mortgages and credit cards I believe are federally required to report APR) so that consumers don't have to take a finance course to understand that banks borrow money at a very low rate, around 4%, and that people with good credit can borrow at around 8%, and that you should only finance at 18% APR, rims which cost $800 if you wish to take on a undesirable interest rate. Therefore, you should only buy them if you make enough utility on the $800 price, to justify the exorbitant interest rate. This would be the case if the rims were of some special case of tremendous value, like maybe a clearance sale. My closing point, which I hope I've made, is that you should ask your self where APR's came from? What is the history of APR's. They are designed to help consumers understand (through the easy to understand simple interest method) a complex formula. Most consumers aren't good at calculating exponents and natural logarithms, but almost anyone that can get a loan, understands how to divide by 12 and multiply by 12. Thus APR's are understood by the majority of consumers. Consumers don't understand EAR in the same proportion as those that would understand simple interest. Therefore the APR which is most helpful, is the one most everyone can understand. APR is nominal. To calculuate the interest for one month, you divide the APR by 12. This is within the abilities of the largest proportion of consumers. Then again, I also question if credit card lobbyists may have been behind this whole enigma. Afterall, it lets them quote an APR of 29.900, but if this value were computed using exponents and non-simple-interest, they would have to report 34.358. But I think its that the greatest proportion consumers understand simple interest. The government (that regulates how consumers are told the interest rate) will guide corporations to quote an interest rate which people will understand. Sentriclecub (talk) 14:42, 15 September 2008 (UTC) Also, I went back and read the APR article and the third sentence is APR is intended to make it easier to compare lenders and loan options. which summarizes my point.

combinatorics question

A bridge club has ten members. Four get together to play everyday. Show that in two years one group of four people gets together at least four times.

The only thing that I can think of is that in the first 211 days one group definitely gets together twice. Any other ideas?--Shahab (talk) 19:03, 15 September 2008 (UTC)

Your argument uses the basic version of the pigeonhole principle, but there's also a generalized version, pointing out that since 3×210 = 630 days is the longest possible period of time without any group playing more than three times, in 631 days some group must play four times. -- Jao (talk) 19:36, 15 September 2008 (UTC)

Thank you. Here's another problem which is troubling me. I can't figure out how to apply the pigeonhole principle here: Prove if the numbers 1, 2, 3, . . . 12 are randomly placed around a circle, there must be three consecutive numbers whose sum is at least 19.--Shahab (talk) 07:00, 16 September 2008 (UTC)

Going round the circle, there are 12 sums of three consecutive numbers. You can work out the total of these 12 sums (each number on the circle is counted three times, in three different sums), so you can work out the average value of these 12 sums. The 12 sums cannot all be less than their average, so at least one sum must be greater than or equal to their average. Gandalf61 (talk) 09:23, 16 September 2008 (UTC)

As far as I understand this the total of the 12 sums must be 234, which when divided by 12 would give 19.5 proving more then asked for. Clearly this is an elegant solution. Thanks. But I want to explicitly involve the the pigeonhole principle in the solution coz the problem occurs in that section in my book. (Another way that I thought of was that if each of the sums is less then or equal to 18 then their total can be at most 12 times 18 i.e. 216, contradicting the fact that the total is 234).--Shahab (talk) 09:46, 16 September 2008 (UTC)

Divide the circle into groups of three. If each number occurs once there will be 4 such groups. By the generalized pigeonhole principle if none of these groups sum to more than 18 then the whole circle can not sum to more than 18*4=72, but 1+2+...+12=78.Taemyr (talk) 14:21, 16 September 2008 (UTC)

Another approach, less combinatoric but more direct: try to create an arrangement in which each sum of 3 consecutive numbers is 18 or less. Note that no two out of 9, 10, 11 and 12 can be in the same sum, so no two of these four can be neighbours or have just one number between them. So your arrangement must be a cyclic permutation of abbabbabbabb where the as are 9, 10, 11 and 12. Now show that there is nowhere left to put 8 without creating a sum of 19 or greater. Gandalf61 (talk) 09:33, 16 September 2008 (UTC)

Population and area question

Population and area question

How many times greater is the population of China than Australia? Secondly, how many times greater is the land area of China than Australia? —Preceding unsigned comment added by 81.151.147.129 (talk) 19:17, 15 September 2008 (UTC)

You can certainly find the relevant figures in our articles on Australia and People's Republic of China. -- Coneslayer (talk) 19:29, 15 September 2008 (UTC)

Word Problem- how to setting up the equations

Two pipes feed into a tank. The large pipe can fill the tank in 2 hours, the smaller pipe can fill the tank in 6 hours. If both pipes are used together, how long would it take to fill the tank.

Thank you for your help. —Preceding unsigned comment added by Evansranch (talk • contribs) 20:20, 15 September 2008 (UTC)

A good way to start is by answering the question: "It would take x hours." Then think about it: how many tanks will the large pipe fill in x hours? How many tanks will the small pipe fill? How many will they fill together? -- Jao (talk) 20:29, 15 September 2008 (UTC)

Per hour, the large pipe fills half the tank, the small one one sixth of it. So what fraction together? How many hours to make this fraction into one, i.e. a full tank?—86.148.186.156 (talk) 21:42, 16 September 2008 (UTC)

In one hour, the 'faster' pipe can fill 1/2 of the tank and the 'slower' pipe can fill 1/6 of the tank. (1/2 + 1/6) = 4/6 = 2/3 of the tank. Suppose it takes 'x' hours to fill the tank. Then 2x/3 = 1 which implies that x = 3/2 hours. Therefore the tank is filled in 1 hour and 30 minutes.

Topology Expert (talk) 13:15, 17 September 2008 (UTC)

We usually avoid actually give the answer to questions like this. It's much better to guide the OP towards getting the answer themselves than just giving it. --Tango (talk) 17:29, 17 September 2008 (UTC)

Combinations

A question I am working on reads:

A bag contains 20 chocolates, 15 toffees and 12 peppermints. If three sweets are chosen at random what is the probability that they are:

a) All different,

b) All chocolates,

c) All the same,

d) All not chocolates?

I have calculated the answer to a, to be 0.222 (which is correct), using the following steps:

47C3 = 16215

Chocolates * Toffees * Peppermints = 3600

3600/16215 = 0.222

Firstly, while I know this works, I do not fully understand why, and I am completely stuck on all the other sections. Can anyone point me in the right direction? Thanks. --AFairfax (talk) 20:39, 15 September 2008 (UTC)

What you've done is calculate the number of possible choices of three sweets, that's the 47C3, and the number of choices satisfying the condition that they all be different, the 3600. The reason you can just multiply the number of each type together is because you know there must be one chocolate, one toffee and one peppermint, so the only choice is in which of the chocolates, which of the toffees and which of the peppermints your choose. You then divide one number by the other to get the probability. Does that make any sense? The other questions are answered the same way - you need to calculate how many different ways you can satisfy each condition. So for (b), you need to work out how many ways you can choose 3 chocolates from a bag containing 20 chocolates (you can ignore the other sweets since they aren't allowed) - that's just 20C3. I'll let you try (c) and (d). Good luck! --Tango (talk) 21:11, 15 September 2008 (UTC)

The annoying thing is I already knew all that really, just didn't put it into practice in the correct way. Thanks for the pointer. --AFairfax (talk) 22:17, 15 September 2008 (UTC)

There are definitely 16215 ways of drawing three sweets and 3600 ways of drawing three sweets of different kinds. But when you divide the one by the other to get a probability you're introducing a new assumption: that all 16215 draws are equally likely. It's a standard assumption in word problems like this, and it's probably meant to be implied by the phrase "chosen at random", but it would be tricky to achieve in reality with a heterogeneous collection like this one. Shaking the bag and then letting one candy fall out would probably favor some types over others by a significant margin. So always think carefully before you treat a ratio of integers as a probability. -- BenRG (talk) 23:55, 15 September 2008 (UTC)

(edit conflict)

$20\cdot 15\cdot 12/{\tbinom {47}{3}}=20\cdot 15\cdot 12\cdot {\frac {3!44!}{47!}}=20\cdot 15\cdot 12\cdot {\frac {3!}{45\cdot 46\cdot 47}}={\frac {20}{47}}\cdot {\frac {15}{46}}\cdot {\frac {12}{45}}\cdot 3!$

Explanation of the derived formula: First we choose among 20 chocolates out of 47 sweets, second we choose among 15 toffees out of 46 remaining sweets, third we choose among 12 peppermints out of 45 remaining sweets. 3! is the number of possible orders of selections (chocolates, toffees, peppermints; or chocolates, peppermints, toffees; or toffees, peppermints, chocolates etc.). According to mathematics of outs in Texas Hold'em, b) is determined by formula ${\frac {20}{47}}\cdot {\frac {19}{46}}\cdot {\frac {18}{45}}=0.070$.

Similarly, c) ${\frac {20}{47}}\cdot {\frac {19}{46}}\cdot {\frac {18}{45}}+{\frac {15}{47}}\cdot {\frac {14}{46}}\cdot {\frac {13}{45}}+{\frac {12}{47}}\cdot {\frac {11}{46}}\cdot {\frac {10}{45}}=0.112$

And d) ${\frac {47-20}{47}}\cdot {\frac {46-20}{46}}\cdot {\frac {45-20}{45}}=0.180$

--Admiral Norton ^(talk) 21:19, 15 September 2008 (UTC)

This is a Hypergeometric distribution.

Topology Expert (talk) 13:10, 17 September 2008 (UTC)

ms^-3

how would you describe the concept of m/s/s/s? 84.13.30.238 (talk) 20:41, 15 September 2008 (UTC)

• Jerk. --PaulTaylor (talk) 21:00, 15 September 2008 (UTC)

  Nitpick: The units of jerk. Zain Ebrahim (talk) 22:02, 15 September 2008 (UTC)

Thanx Paul and Zain! I was just curious... thnx again User:ATMarsden/Templates/Sign/Signature 15:50, 16 September 2008 (UTC)

m/s/s means the change in m/s per second; or the change in velocity per second or acceleration. Therefore, m/s/s/s means the change in m/s/s per second; or the change in acceleration per second. This has no particular name but the third derivative of a function represents this. —Preceding unsigned comment added by Topology Expert (talk • contribs) 13:09, 17 September 2008 (UTC)

As already noted, it has a name: jerk. -- Jao (talk) 19:08, 17 September 2008 (UTC)

Gauss and triangular numbers

from Carl Friedrich Gauss:

"...every positive integer is representable as a sum of at most three triangular numbers..."

A list of triangular numbers taken from triangular number:

1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...

I know I'm missing something here. Integers are in the set {..., -2, -1, 0, 1, 2, ...}, and 1, 3, 6, and 10 are all triangular numbers. 1 + 3 + 6 + 10 = 20 and 20 is an integer. Therefore, the sum of four triangular numbers is equal to an element in the set of integers. What am I missing? Ζρς ι'β' ^¡hábleme! 22:27, 15 September 2008 (UTC)

20 = 10 + 10? Confusing Manifestation(Say hi!) 22:48, 15 September 2008 (UTC)

And just to clarify, what Gauss found was that given any positive integer, you can break it down into a sum of three or fewer triangular numbers. Not that adding more than three triangular numbers together suddenly throws you out of the set of integers entirely. Confusing Manifestation(Say hi!) 22:49, 15 September 2008 (UTC)

Ok, thanks. That's what I was missing. 67.54.224.199 (talk) 23:15, 15 September 2008 (UTC)

Or since 0 is a triangular number then 20=10+10+0 with no change to the statement. Dmcq (talk) 08:19, 16 September 2008 (UTC)

The statement says "at most three", so there is no need to pad it out with zeros. --Tango (talk) 10:44, 16 September 2008 (UTC)

September 16

Constructible numbers

The Wikipedia article on constructible numbers states that :

If z is constructible, then it is algebraic, and its minimal irreducible polynomial has degree and power of 2, or equivalently, the field extension Q(z)/Q has dimension and power of 2. One should note that it is true, (but not obvious to show) that the converse is false — this is not a sufficient condition for constructibility. However, this defect can be remedied by considering the normal closure of Q(z)/Q.

Can anyone tell me how to construct such a number ? Even without an explicit construction, a Galois theoretic explanation of what type of polynomial this number should be a root of.

My understanding is that you need to find a number α such that [Q(α)/Q]=2^k but that, if E is the normal closure of Q(α), then [E/Q]!=2^k for any k. How is that possible ?

I thought that, as [Q(α)/Q]=2^k, there must be an irreducible polynomial P(x) in Q[x] of degree 2^k, of which α is a root. P(x) can't split over Q[x]/(P(x)), otherwise the extension would be normal. But all roots of P(x) must be of degree 2^k so I don't see how we could get a normal closure with degree over Q which isn't a power of two.

So, where does such a non-constructible number α with [Q(α)/Q]=2^k come from ?

Thanks. --Xedi^Talk 05:05, 16 September 2008 (UTC)

I believe the article is just saying that having the highest power of your polynomial a power of two isn't sufficient, for instance x in x⁴+x+1=0. However if you extend the rationals with the roots of this equation you'll get cube roots as well coming up. Dmcq (talk) 08:43, 16 September 2008 (UTC)

By the way if you like this then you might like Mathematics of paper folding where trisection of angles can be done. Dmcq (talk) 08:46, 16 September 2008 (UTC)

Well, the thing is, x⁴+x+1 is an irreducible polynomial of degree 4, so all its roots must be of degree 4 over Q, no ? And then, taking normal closure is not going to affect this degree, so I don't see how "the defect can be remedied" that way. I did have the impression something could go wrong if cube roots were involved somewhere, but I don't see how taking normal closure remedies this. --Xedi^Talk 16:10, 16 September 2008 (UTC)

In general, if p is an irreducible degree n polynomial over Q, then any root of p generates a n-dimensional extension. The normal closure of any of these extensions is the full splitting field of p. The degree of this splitting field is divisible by n and divides n!. In this case, the degree of the splitting field is 24. Algebraist 16:27, 16 September 2008 (UTC)

So, at which point does the factor of 3 in 24 come in then ? And how do you know that in this case the degree of the splitting field is 24 (Other than just giving the order of the Galois group of the extension) ? --Xedi^Talk 16:50, 16 September 2008 (UTC)

I calculated it using standard tricks for computing Galois groups using reduction mod m. The factor of 3 comes in because when we adjoin a single root of the polynomial, this splits the poly into a product of a linear poly and an irreducible cubic, and we have to adjoin a root of that cubic to make the poly split further. For a simpler example, consider x³-2. This is irreducible, so we adjoin a root a. The polynomial then splits as (x-a)(x²+ax+a²). The second factor is irreducible; to split it, we need to adjoin a square root of -3. Thus the splitting field has degree 6. Algebraist 17:15, 16 September 2008 (UTC)

Ah yes ok. Thanks a lot ! --Xedi^Talk 17:21, 16 September 2008 (UTC)

Baffling Proportions Problem

The other day I was helping my younger brother with a maths problem. His teacher had given him the problem in a quiz. "Working 8 hours a day, 24 men do a certain task in 15 days. Find how many days it would take 16 men working 9 hours a day to do the same task"

It is a inverse proportion problem but there are 5 pieces of information here. Usually these kind of problems involve 3 pieces of information. This problem has got me very confused and I am feeling pretty embarrassed. Would be glad if I could get some help here! Thank you! —Preceding unsigned comment added by 203.81.220.213 (talk) 16:33, 16 September 2008 (UTC)

Try breaking it all down into man-hours (the amount of work done by one man in one hour). Use the first scenario to work out how many man-hours the task requires, and then use that to work out the time required for the second scenario. --Tango (talk) 17:23, 16 September 2008 (UTC)

The 5 pieces of information fall into one base piece (the given time to do the task, i.e. 15 days), and two pairs of adjustment pieces (the number of men, and how many hours they work each day), where each pair forms a fraction to multiply the base piece. So the fraction for men is either 24/16 or 16/24, and for hours either 8/9 or 9/8. Which fraction to use in each case will come by considering the direction of change to the base figure resulting from (a) fewer men (b) more hours per day. When that has been decided, the answer will come by multiplying the original time by the two fractions in turn. This approach lends itself to there being more than two factors in the problem, say a rate of efficiency as well. If the 24 men worked at 80% and the 16 men at 90%, there would just be a third fraction (80/90 or 90/80, to be decided) in the multiplicative chain.—86.148.186.156 (talk) 22:15, 16 September 2008 (UTC)

Random trivia: hapless schoolboys once solved these problems using the "double rule of three," immortalized in Lewis Carroll's Mad Gardener's Song:

He thought he saw a Garden-Door that opened with a key:

He looked again, and found it was a Double Rule of Three:

‘And all its mystery,’ he said, ‘Is clear as day to me!’

The "mystery" was a reference to the fact that this rule was famously difficult to teach. I don't recommend teaching it to your brother—far better to teach general problem-solving techniques and real understanding of the problem. But here's an explanation of the rule from an 1830 textbook. It's interesting to note that the word problem they chose as their example can't correctly be solved this way—if $100 gains $6 in one year, it's not likely to gain $6n in n years. I wonder if any of the students ever noticed that. For some real soul-killing awfulness, check out this page. "Three sailors having been aboard 9 months 1/4, received 40ℓ 3/15; I demand how much 100 sailors must receive for 28 months 3/7 service? Answer 4118ℓ 6s 0d 1qrs 1305/11655." -- BenRG (talk) 23:13, 16 September 2008 (UTC)

Is there really any 'problem' in this?

Input data: "Working 8 hours a day, 24 men do a certain task in 15 days."

Twenty-four men do the work, so each of them does a 24-th part of the work in 15 days. He works 8 hours per day, so he completes his 1/24 of work in 15×8=120 hours — that means he does a 1/120 part of his work per hour, so one worker does 1/120×1/24 = 1/2880 of the task per hour.

Question: "Find how many days it would take 16 men working 9 hours a day to do the same task"

Each man, working 9 hr/day, does 9×1/2880 = 1/320 of the task per day. Sixteen workes do 16×1/320 = 1/20 of the taks a day. So they need 20 days to complete the work. Is anything obscure here? --CiaPan (talk) 07:19, 17 September 2008 (UTC)

This is a problem which requires more logical thinking than calculations. The answer from my calculations is: 23 days, 7 hours and 30 minutes. Try to see whether you get the same answer. Topology Expert (talk) 12:56, 17 September 2008 (UTC)

Nope. But this way is non-baffling (as outlined by Tango above):

• 8 hours/day $\times$ 15 days $\times$ 24 men = 2880 man-hours of work to be done (calculated by doing it the original way).
  
• 9 hours/day $\times$ 16 men = 144 man-hours/day available (by doing it the new way)
  
• 2880 man-hours (of work to be done) $\div$ 144 man-hours/day (by doing it the new way) = 20 days. -hydnjo talk 13:00, 17 September 2008 (UTC)

Huh? 'but this is non-baffling'...??? I apologize then, must have missed something important. I understand 'problems' are sometimes given in a complex, sophisticated form, but thought the solution should always be made as simple as possible, not baffling. At least that's what I remember from schools. --CiaPan (talk) 14:03, 17 September 2008 (UTC)

Yes, and Hydnjo's solution (which is the one I recommended at the top of this section, although I didn't actually give the answer) is simpler than yours. At least, it is to me. --Tango (talk) 17:27, 17 September 2008 (UTC)

Yeah, I spelt it out because no one seemed to be picking up on your simple explanation. -hydnjo talk 23:17, 17 September 2008 (UTC)

September 17

Functions

I'm given a graph, however I am confused about how to solve something like $2f(x+3)=52$ algebraically for x. Could someone please point me in the right direction. (By the way, this is a homework question.) I have to show my work, or I would just use the graph. TIA, 67.54.224.199 (talk) 01:45, 17 September 2008 (UTC)

If you are given an actual formula for f, then you can solve algebraically, starting with appropriate substitution. For example, if you were given f(x) = x^2, then f(x+3) = (x+3)^2 and from that you then solve 2 (x+3)^2 = 52. If you are only given a graph, then you would need to first use the graph to find an approximate solution to 2 f(x') = 52, and then using that approximate solution to solve x' = x + 3 for x. Confusing Manifestation(Say hi!) 03:50, 17 September 2008 (UTC)

In more details, locate the intersections of your graph and the horizontal line y=26 (52/2) to find x' = x+3. Then x = x'-3 are your solutions if there are more than one point in the intersection. twma 10:08, 17 September 2008 (UTC)

Need x so that: f(x+3) = 26. Therefore, you find x' such that f(x') = 26 and subtract 3 from x' to get the desired x (since f(x+3) = f(x'-3+3) = f(x') = 26 as desired).

The most important thing to remember is that f(x-a) translates the graph a units to the right (to get the same value of f(x), you need to add 'a' to the x-value). Similarly, f(x+a) translates the graph 'a' units to the left (to get the same value of f(x), you need to subtract a from the x-value so the graph must be translated 'a' units left). Similarly, -f(x) reflects f about x-axis and f(-x) reflects f about y-axis. In general a*f(x) dilates graph parallel to y-axis; scale factor 1/a. Similarly f(ax) dilates graph parallel to x-axis scale factor 1/a (since you need 1 a^(th) the same value of x to get the original f(x) so the graph must shrink horizontally) —Preceding unsigned comment added by Topology Expert (talk • contribs) 13:03, 17 September 2008 (UTC)

differential equation

Hi, is there a solution to xy" + y' - y = 0 that uses a finite set of well known functions?(Not the power series solution--I got the d.e. as something a certain power series satisfied and I want to express the function repped by the p.s. in terms of well known functions if possible. The function is y = the sum of (x^n)/[(n!)^2] from n=0 to infinity, which I got from the linear operator that sends the p.s. for 1/(1-x) to the one for e^x. When you do the same on the p.s. for e^x, you get y.)Thanks in advance.Rich (talk) 02:54, 17 September 2008 (UTC)

See Bessel-Clifford function with n=0. Dmcq (talk) 09:49, 17 September 2008 (UTC)

good referral, thank you.130.86.14.25 (talk) 05:10, 19 September 2008 (UTC)

combinations and permutaions

what is value of nC3 —Preceding unsigned comment added by 59.184.240.35 (talk) 13:13, 17 September 2008 (UTC)

See Combination. --Tango (talk) 13:19, 17 September 2008 (UTC)

Use the definition of binomial coefficient, substitute 3, and simplify. (Alternately, use the theorem that nCk is a polynomial of n of degree k for any fixed nonnegative integer k, compute it for a few small values, then interpolate. Alternately, compute for the first few values of n and look up in OEIS for a formula. Alternately rewrite the binomial to a falling power and use the equation about falling powers and Stirling numbers to get the result in expanded forms rightaway.)

If you don't know what binomial coefficients are in general then a good introduction might be the first few chapters of Graham–Knuth–Patashnik, Concrete mathematics. – b_jonas 15:15, 19 September 2008 (UTC)

September 18

Formula for planet temperature

I'm having trouble converting this formula from "FUNDAMENTALS OF ASTRONOMY" page 318 (QB43.3.B37 2006) by Dr.Cesare Barbieri, professor of Astronomy at the university of Pauda, Italy. It is not simplified for the Sun/Earth, it is the full formula with data for Gliese 581/581 c. Can someone help?

=(((((0.0000000567051)*(3840^4))/(4*PI()*((0.0613*149597876600)^2))) * ((4*PI()*((0.29*695500000)^2))/(4*PI()*((11162)^2)))*((PI()*((11162)^2)*(1-0.64))))/0.0000000567051)^0.25

$T_{p}={\frac {(5.67051E-8)*(3840^{4})}{(4*pi*(0.0613*1AU)^{2})}}*\ {{4*pi*(0.29*{R_{\odot }})^{2}}{(4*pi*(R)^{2}}}{pi*(R)^{2}*(1-0.296){}0.0000000567051)^{0}.25}$

--GabrielVelasquez (talk) 04:09, 18 September 2008 (UTC)

I think you will have to provide a few more details here. You have written two different expressions - which one (if either) is correct ? And in what way are you trying to "convert" the formula - what are you converying from and to ? Gandalf61 (talk) 08:59, 18 September 2008 (UTC)

I found the reference on Google books. You appear to be trying to use the second equation on the page, which looks like this (rewriting quantities that refer to the Sun to refer to a general star e.g R_⊙ becomes R_∗ and marking those which refer to the planet explicitly, e.g. T becomes T_p):

$\sigma \cdot T_{\mathrm {p} }^{4}={\frac {\sigma \cdot T_{\ast }^{4}}{4\pi \cdot d^{2}}}\cdot {\frac {4\pi \cdot R_{\ast }^{2}}{4\pi \cdot R_{\mathrm {p} }^{2}}}\cdot \pi \cdot R_{\mathrm {p} }^{2}(1-A_{\mathrm {p} })$

However I don't see why you are trying to use this equation, as many of the terms in it cancel each other out, making it unnecessarily cumbersome to work with. In fact it is rearranged on the next line as follows:

$T_{\mathrm {p} }=T_{\ast }\left[\left({\frac {R_{\ast }}{d}}\right)^{2}{\frac {1-A_{\mathrm {p} }}{4}}\right]^{\frac {1}{4}}$

This expression is far easier to work with (note for example that the radius of the planet cancels out) and gives exactly the same results! If you want to try it for yourself, I recommend Google calculator, which helpfully includes various units conversions and the constant r_sun, so for a planet orbiting, say, 0.1 AU from a 3500 K star with a radius 0.3 times that of the Sun and a bolometric albedo of 0.5, you could do a Google search for: 3500 K*((0.3*r_sun/0.1 AU)^2*(1-0.5)/4)^(1/4). Icalanise (talk) 12:53, 18 September 2008 (UTC)

Finding an average flood height overtime

dates of floods                   height of each flood
1965..........................    5.22
1972..........................    5.14
1993.........................     6.31
1997...........................   6.13

From this information iam required to find the average height of these floods for a 1 in 10 year event and a 1 in 100 year event

please help thankyou

if you need to email me it <email removed by Ζρς ι'β' ^¡hábleme!> the 0 is a zero —Preceding unsigned comment added by 124.170.38.135 (talk) 04:51, 18 September 2008 (UTC)

To find the average of values, just add them up, and divide by the number there are. For example, the average of 1, 2, 3, 4, and 5 is 3 because 1+2+3+4+5=15 and 15/5=3. Ζρς ι'β' ^¡hábleme! 21:53, 18 September 2008 (UTC)

Probably the best thing to do for real world problems is ask someone who deals with this in practice. Both the cases of flooding by the sea and of inland areas are quite complicated. You really don't have enough figures to come to any sort of even inaccurate conclusion without some model to plug them into. If you just want to use flood heights with nothing else you'd need many more figures from lower level floods so you can fit a probability distribution to them. Your figures above would be some way along one side and highly skewed and there's too few to see how quickly they go down. And anyway Global warming will make a mess of any figure just derived from historical data. Personally I'm rather surprised at the difference between 1972 and 1993 - I'd guess someone changed the environment of the area between the two. Dmcq (talk) 18:28, 20 September 2008 (UTC)

Even if the trend is towards higher floods, there is still going to be random fluctuations from year to year, it's not at all surprising that a later flood was a little lower than an earlier one. --Tango (talk) 19:51, 20 September 2008 (UTC)

Word for the "terms" in a union

Is there a word along the lines of "summand", "integrand", "radicand" for the individual "terms" in a union? For example, in $A\cup B\cup C$, what word describes the role of $A$, $B$, and $C$? Is "term" acceptable, or is there a niftier word like "unand" or "unitand"? While we're at it, what about intersections? —Bkell (talk) 05:01, 18 September 2008 (UTC)

I think terms in acceptable. As far as I know, there is no particular word associated to this. Thenub314 (talk) 07:31, 18 September 2008 (UTC)

I also have never heard such a specific term. If you wanted to invent one, then some people call union set summation, so you could call them summands. Or you could use the analogy with the logical OR operator and call them disjuncts. Algebraist 11:00, 18 September 2008 (UTC)

0 dimensional object in 4 dimensional world

In a tesseract, if the cube (3d) serves the function analogous to a 2d surface for a cube, and the plane serves the function analogous to the edge of a cube in 3d, what would be the status of a point(0 dimensional) in regard to four dimensions? It couldn't be the -1st dimension, could it?Leif edling (talk) 07:03, 18 September 2008 (UTC)

The analogy breaks down, or was never really more then an analogy. In the 4 dimensional cube you gain one more thing then you had before, you don't just translate all the things from a 3 dimensional cube. So in a 3 dimensional cube you have squares (2 dimensional cubes), line segments (1 dimensional cubes, this is a bit degenerate), and points (0 dimensional cubes, very degenerate). In a 4 dimensional cube there are, 3 dimensional cubes, squares, line segments and points. Thenub314 (talk) 07:40, 18 September 2008 (UTC)

Yeah, because if you stepped into 4d land, you'd have almost no mass and no weight, because you are infinitesimal thin. See Paper mario that would be you. Sentriclecub (talk) 11:48, 18 September 2008 (UTC)

My second answer, is just like a point in 3d. A point is so thin, so small, that even though I can look on a map and see a point 37 degrees noth, 48 degrees east. If I go to that actual location in the world, I still would not be able to even begin to see the "point" which exists, its smaller even the planck lengths. So my answer is that its just the same as in a 3d world. Remember, a point has no volume, so as soon as you give it a relative size, you instantly defined a cube (which is a 3d structure, which has volume). Sentriclecub (talk) 11:53, 18 September 2008 (UTC) My third answer is that it is a line, because given my second answer, that would say that a cube in 3d is not analogous to a tesseract (as you actually asked--I failed your original question) because a cube can exist in 4d just as a sheet of paper (to me at least) represents a 2D plane in my 3d world.

I read this "analogy" perspective:

Cells, Ridges, Edges: The upshot of all this is that in 4D, objects have a much richer structure than in 3D. In 3D, a polyhedron like the cube has vertices, edges, and faces, and fill a 3D volume. The cube is bounded by faces, which are 2D. Every pair of faces meet at an edge, which is 1D, and edges meet at vertices, which are 0D.

In 4D, objects like the hypercube not only has vertices, edges, and faces, but also cells. A 2D boundary is insufficient to bound a 4D object. Instead, 4D objects are bounded by 3D cells. Each pair of cells meet not at edges, but at 2D faces, also called ridges. The ridges themselves meet at edges, and edges meet at vertices.

The point here is that in 4D, 3D volumes play the role analogous to surfaces in 3D, and 2D ridges play the role analogous to edges. Because of this, it is important to visualize 4D objects by thinking in terms of bounding volumes, and not 2D surfaces. A 2D surface only covers the equivalent area of a thin string in 4D! When you see a 2D surface in the projection of a 4D image, you should understand that it is only a ridge, and not a bounding surface.

I got it here:[3]

So, do the above answers signify that a tesseract does not have 2d planes analogous to the edges of 3d cube? (As is said in the last paragraph above)Leif edling (talk) 15:49, 18 September 2008 (UTC)

Elliptic Curves Isomorphism

What is an isomorphism when it comes to elliptic curves? I have an idea of what it is based on reading about it in two different books, but neither seems very clear to me.

Knapp's book "Elliptic Curves", which I don't have with me so I can't tell you exactly what it says, basically says two elliptic curves are isomorphic if they are related by an admissible change of variables. That's the definition of isomorphic. But, it never says anything about what that means. I assume it means the group structures are the same. But, is that all it means?

The other book I looked at is Dale Husemoller's "Elliptic Curves". Again, I don't have it with me, but I believe one theorem says two elliptic curves are isomorphic if and only if they have the same j-invariant. This is why I am not entirely sure on the isomorphism meaning only group structure. If this is what it means, then since there are an infinite number of j-invariants, there must be an infinite number of elliptic curve groups possible. The problem is Knapp's book doesn't seem very clear on this. It says only that elliptic curves for ranks up to 12 are known but a fact like there are an infinite number of such groups means clearly elliptic curves of much higher rank exist (higher than any given number).

Can any one help me understand this better? Thanks. StatisticsMan (talk) 13:46, 18 September 2008 (UTC)

You seem to be a bit confused about what base field (or whatever) you are working over. If one has a change of variables making two curves isomorphic, the group structure will be the same, sure. But if I understand your question correctly, when you say "the group structures are the same" this means that points over some field form groups which are isomorphic. But this is not enough to say that the curves are necessarily isomorphic. If one looks at things over the complex numbers, all elliptic curve groups are isomorphic as abstract groups, or even as real Lie groups, being tori, the product of two circle groups. (But they aren't isomorphic as complex Lie groups / Riemann surfaces unless they have the same j-invariant.)

When Knapp? says that elliptic curves with ranks up to 12 are known, he is talking about the Q-rational points on the curve as a finitely generated abelian group (cf Mordell–Weil theorem). Again, this fga group does not characterize the curve. The infinitude of different groups doesn't follow from the infinitude of j-invariants, as many curves could have the same group of Q-rational points. A theorem of Barry Mazur shows that there are only finitely many possible torsion subgroups of the Q-points, so the possible infinitude of different groups must come from there being (possibly) infinitely many possible ranks, but as Knapp says, only ranks up to 12 are known.John Z (talk) 01:34, 19 September 2008 (UTC)

Well, you are correct that I meant over the rationals, so I apologize for not saying that. But, I still do not think I understand what an isomorphism for elliptic curves over Q really means. You told me that it's not just the group. Can you please tell me what it is since the books I have read don't bother to explain this? My question was based on Mazur's theorem which says there are only 15 possible torsion subgroups of Q-points. So, since I do not understand what an isomorphism means, I made a guess that it meant only the group. If it means only the group, it follows there must be an infinite number of different ranks. But, Knapp says only up to 12 are known. So, it's either not just the group or it is and his statement does not say all it should. StatisticsMan (talk) 15:24, 19 September 2008 (UTC)

I'm not familiar with this stuff at all, but it probably means they're isomorphic as algebraic varieties. Algebraist 22:03, 19 September 2008 (UTC)

Yes, that is all that it is, an invertible morphism of varieties. In StatisticsMan's words "related by an admissible change of variables" - we just need to change variables in an invertible way. An elliptic curve may be given by an equation like y^2=x^3+ax^2+bx+c. If we did a simple invertible change of variable, say replacing x by x+1, we get an equation of the same type that thus gives an isomorphic curve embedded trivially differently in the plane. The j-invariant can be defined algebraically from the coefficients (a,b,c) of such an equation and is invariant under any such coordinate change. An isomorphism of the groups of rational points of two different curves with different j-invariants is not necessarily induced by an algebraic change of coordinates relating the two curves. With elliptic curves it is practical to see modern algebraic geometry at work in a very concrete fashion. For this one might want to look at the lucid paper of John Tate, The Arithmetic of Elliptic Curves, Inventiones Math.(23) 1974, 179-206, which later expository works on elliptic curves owe much to. Things like Mazur's theorem are much harder.John Z (talk) 06:10, 20 September 2008 (UTC)

So, just to be clear, we are saying that curves with isomorphic groups may have different j-invariants, so they are not isomorphic as curves, even thought their groups are isomorphic. So, although it is conjectured that there is no upper limit on the possible rank of a curve group, the fact that we can construct curves with an infinite number of different j-invariants does not prove this conjecture.

Incidentally, the search for curves whose groups have high rank has moved on a bit - this page says that the highest rank of an elliptic curve whose group's rank is known exactly is 18, and the highest known lower limit on a curve's group's rank is 28 (i.e. there is a curve whose group's rank is known to be at least 28, but it may be higher). Both examples were found by Noam Elkies in 2006. Gandalf61 (talk) 09:04, 20 September 2008 (UTC)

Alright, thanks a lot everyone. This has been very helpful. I appreciate that you took the time to answer my question.StatisticsMan (talk) 15:15, 20 September 2008 (UTC)

Is there any limitation to defining all x in Q as...

$x={\frac {m}{n}}$, where $n,m\in \mathbb {Z}$ and $n>0\,$? I do not mean we need to show that Q is a field, but I have seen other "loose" definitions such as $n,m\in \mathbb {Z} ,n\not =0$ and I just thought that having n and m as negative was redundant, since we only need the numerator, m to be in Z.

Yes, a common way of doing this is defining $\mathbb {Q}$ as the set of equivalence classes of these formal fractions ${\frac {m}{n}}$, defining two fractions as equal if $m_{1}n_{2}=m_{2}n_{1}$. When doing this, it does not matter at all if you require $n$ to be positive or just non-zero. The reason for choosing just non-zero is probably that then the same process will work for non-ordered domains. In other words: for constructing rationals from integers, your approach is just as good; but it's less adaptable. -- Jao (talk) 16:30, 18 September 2008 (UTC)

One use for the signs of the integers in computational work is to specify that $(-a)/b$ and $a/b$ (ie not $a/(-b)$ and $(-a)/(-b)$) mean that the fraction is not known to be in its lowest terms. It's a nice convenient flag that means "you do not need to try and reduce this fraction" Robinh (talk) 07:03, 19 September 2008 (UTC)

Algebra

What are some applications of groups, either to other mathematical structures or to the real world? Other than a little bit of Galois theory and a mention of homology, my textbooks haven't given much indication of what it's for. Black Carrot (talk) 22:49, 18 September 2008 (UTC)

Quantum chromodynamics, The Standard Model, and related areas of particle physics are littered with group theory. Dragons flight (talk) 22:52, 18 September 2008 (UTC)

Group theory is used extensively in chemistry, too. I don't know very much chemistry, but I believe groups are used in crystallography at least. Perhaps a chemist will come along and can provide more information. Another critically important application of group theory is the analysis of the Rubik's Cube. ;-) —Bkell (talk) 04:29, 19 September 2008 (UTC)

Correct about chemistry/crystallography. Spectroscopy as well. The types of events that give rise to spectroscopic signals are determined by group theoretical properties of the molecule's symmetry. See the book mentioned here for several such applications. Baccyak4H (Yak!) 16:20, 19 September 2008 (UTC)

Groups have lots of applications in functional analysis, harmonic analysis and algebraic topology (although vector spaces are also used).

Topology Expert (talk) 05:51, 20 September 2008 (UTC)

Elliptic curve cryptography. Gandalf61 (talk) 08:51, 19 September 2008 (UTC)

September 19

probability

what is the probability of 1.0 means? and between probability of 1.0 and o.4 which the best to present the survival without parasitemia —Preceding unsigned comment added by 41.221.34.123 (talk) 07:53, 19 September 2008 (UTC)

A probability of 1 means that it is certain (100% likely) to occur. A probability of 0 means it is certain to never occur, 0.5 means the event will happen half (50%) of the time. A probability cannot be less than 0 nor can it be greater than 1. -- SGBailey (talk) 07:58, 19 September 2008 (UTC)

This is wrong. If an event has probability 1, it does not mean that it is certain. For example, consider the event of randomly picking a number in [0,1]. The probability that you pick a number other than 0 is 1 but it is not certain that you will not pick a 0.

See Almost surely.

Topology Expert (talk) 05:44, 20 September 2020 (UTC)

Please clarify your second question. Are you asking whether 1 or 0.4 is a more reasonable measure of a survival rate without parasitemia? Note that survival rates are usually associated with a certain length of time. Zain Ebrahim (talk) 14:15, 19 September 2008 (UTC)

Though maybe it should be noted, from a mathematical and not at all practical stand point, that probability 1 is not quite certain and probability 0 is not quite "certainly not". Thenub314 (talk) 14:57, 19 September 2008 (UTC)

Eh? From a mathematical point of view, 1 is certain and 0 is certainly not. From a colloquial point of view then anything goes. Use "Acme's gadget for 110% success" is gibberish mathematically. -- SGBailey (talk) 15:16, 19 September 2008 (UTC)

1 is not certain; see my previous comment.

Topology Expert (talk) 05:47, 20 September 2008 (UTC)

What Thenub314 means is that any set with measure zero has probability zero, not just the empty set. Pick a random real number between 0 and 1. The probability of your picking that number was 0, yet you picked it. -- BenRG (talk) 16:28, 19 September 2008 (UTC)

We have an article: almost surely. Algebraist 16:31, 19 September 2008 (UTC)

September 20

Planes and Spheres

I need a bit of help understanding a problem I'm doing and Wikipedia seemed like the best place.

A plane $P_{1}$ has equation

$3x-y-1=0$

and a sphere $S_{1}$ has equation

$x^{2}+y^{2}+z^{2}=7$.

I have to determine the equation of the circle, $C_{1}$, defined by the intersection of $P_{1}$ and $S_{1}$.

I have tried this in a number of ways and nothing seems to help. From the equation for $P_{1}$, I can see that $y=3x-1$. Substituting this into the equation for $S_{1}$ gives $x^{2}+(3x-1)^{2}+z^{2}=7$, which is simplified to

$10(x-{\frac {3}{10}})^{2}+z^{2}={\frac {69}{10}}$

This clearly isn't a circle and plotting it in my graphing application confirms it.

My next method was to rearrange the equation for $P_{1}$ to arrive at $21x-7y=7$. Substituting this into the equation for $S_{1}$ gives $x^{2}+y^{2}+z^{2}=21x-7y$, which eventually simplifies to

$(x-{\frac {21}{2}})^{2}+(y+{\frac {7}{2}})^{2}+z^{2}={\frac {490}{4}}\,\!$.

This is even more confusing because somehow the intersection of plane and a sphere has become a sphere, defying all logic. Could someone please tell why these two methods have failed and point me in the right direction, ie don't explicitly tell me what to do, just give a hint? I want to try and do as much of this by myself as possible. Thanks 92.2.212.113 (talk) 15:22, 20 September 2008 (UTC)

A circle, a 1-dimensional thing, cannot be represented by 1 equation in 3-D, because each equation only constrains one dimension; so you will need 2 equations to represent your circle. --71.147.13.131 (talk) 17:17, 20 September 2008 (UTC)

Indeed. Baring degenerate cases, each equation (or "constraint") reduces the dimension by one. You've started in 3D space, so one constraint gives you a 2D subset, another gives you 1D and a third would give you a single point (0D). You can't really get much better than just giving the equations of the sphere and the plane - the question seems to be flawed. Perhaps it wants the equation of the projection of the circle on to one of the co-ordinate planes? --Tango (talk) 19:49, 20 September 2008 (UTC)

Do the two constraints come from the fact that the equation for the plane involves both x and y? 92.2.212.113 (talk) 19:54, 20 September 2008 (UTC)

Perhaps clarity will be aided with the complete question. $P_{1},S_{1},C_{1}$ are the same as above. $P_{2}$ is th eplane with equation $x-y+1=0$, $S_{2}$ is the sphere with equation $x^{2}+y^{2}+z^{2}-6y-4z+10=0$, $C_{2}$ is the circle of their intersection and L is the intersection of $P_{1}$ and $P_{2}$. "Find the points where L $C_{2}$ are linked." Does that help? 92.2.212.113 (talk) 20:02, 20 September 2008 (UTC)

Since a circle has an area, I see no way that it can be a 1-dimensional object. Also, if a plane intersects a sphere, not if it just touches the sphere, the points of intersection must produce a circle, see here Plane-sphere intersection. 92.2.212.113 (talk) 19:50, 20 September 2008 (UTC)

Hmm I think my confusion stems from thinking that the intersection produce a disc as well as a circumference. It's now clear that it only produces a circumference. This question comes from a well established exam though, so there must be something I've missed out that would make it work. 92.2.212.113 (talk) 19:53, 20 September 2008 (UTC)

Radius vs. radian

The mean circumference of Earth is about 40041.5 km, which means 1° of arc ≈ 111.2263 km, or 1 rad of arc ≈ 6372.8 km, as 1 rad = 57.295779513...°. A radian ("rad") is the arc equivalent of radius. So why is the inverse of curvature called "the radius of curvature", and not "a radian of curvature"? Since 1° of arc, or "arcdegree", is 1/360° of the circumference, isn't 1/(2π), here 6372.8, the "arcradian", rather than "arcradius".  ~Kaimbridge~ (talk) 17:06, 20 September 2008 (UTC)

The radius of curvature is so called because it is the radius of a circle with the same curvature as the given curve. It would be silly to say that a curve has x radians of curvature rather than radius of curvature x for at least two reasons. Firstly, radius of curvature is a dimensional quantity, in fact a length. Do you want to say that the surface of the earth has 6372.8 km radians of curvature? Secondly, the 'of curvature' suggests that a higher number corresponds to higher curvature, which is of course not the case. Algebraist 17:11, 20 September 2008 (UTC)

Is it accurate to use "1° of arc" like that? I though the "of arc" was just to distinguish between other things measured in degrees (temperature, for example). "1° of arc" is still a measure of angle, not length, by my understanding. --Tango (talk) 19:44, 20 September 2008 (UTC)