Wikipedia:Reference desk/Mathematics

Welcome to the mathematics section
of the Wikipedia reference desk.

skip to bottom

Select a section:

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

Want a faster answer?

Main page: Help searching Wikipedia

   

How can I get my question answered?

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

Ready? Ask a new question!

How do I answer a question?

Main page: Wikipedia:Reference desk/Guidelines

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

See also:

• Teahouse
• Help desk
• Village pump
• Help manual

September 3

hilbart space

what is Hilbert Space in metric space? —Preceding unsigned comment added by Debadhi (talk • contribs) 05:55, 3 September 2008 (UTC)

All Hilbert spaces are metrizable (by definition).

Topology Expert (talk) 10:02, 3 September 2008 (UTC)

for a proof

Is any element of a Hilbert Space complet? —Preceding unsigned comment added by Debadhi (talk • contribs) 06:02, 3 September 2008 (UTC)

Could you please make your question clearer? If you are asking whether any element of a Hilbert space is complete then it is meaningless; an element can not be complete but a space can.

Topology Expert (talk) 09:57, 3 September 2008 (UTC)

Similarly, if I have interpreted your question correctly, all Hilbert spaces are complete by definition. Perhaps you should have a look at Hilbert space for discussion on these spaces.

Topology Expert (talk) 10:03, 3 September 2008 (UTC)

Complicated Maths Question

Here's the original question: The Harvard University Examination Board dispatches copies of an examination question paper yearly to schools in 50,20 or 5 copies. The delivery instructions produced by the board's computer are in the form of x/y/z meaning x packets of 50 copies, y packets of 20 copies and z packet of 5 copies. The following rules are observed during the dispatch: one copy of the question paper is supplied to each candidate each school receives at least 3 extra copies. the no. of question papers sent to schools must be as small as possible, satisfying the two above rules. once the number of question papers is determined, the board dispatches the minimum number of packets to the school.

The question: in 1999, the school receives 11 packets of papers for 1 subject. calculate the no. of copies received in that year.

The question seems to have one rule lacking as the model answer is 385 but mine is 55(all the packets may only contain 5 papers). —Preceding unsigned comment added by Invisiblebug590 (talk • contribs) 07:28, 3 September 2008 (UTC)

Your answer violates the rule that the number of packets must be minimized. If they wanted to send 55 papers, they'd send it as 50+5 (2 packets) not 5+5+5+5+5+5+5+5+5+5+5 (11 packets) --tcsetattr (talk / contribs) 07:45, 3 September 2008 (UTC)

385 may be the least number of copies that require 11 packets if packaged to minimise number of packets, but it it obviously not the only number of copies that the school could have received - another possibility is 550 papers in 11 packets of 50. Unless it specifies "least number of copies", I think the question is ambiguous. Gandalf61 (talk) 10:54, 3 September 2008 (UTC)

You are correct. The full list of answers requiring 11 packets at a minimum are: 385, 415, 430, 445, 460, 475, 490, 505, 520, 550; With 385 being the smallest of the answers. Anythingapplied (talk) 15:27, 3 September 2008 (UTC)

A bit of a strange question...

Take this example problem: If a 10 foot tall ladder leans against a wall, and the base of the ladder is 5 feet away from the wall, how far up the wall does the ladder go? The answer using Pythagoras' theorem is:
$5^{2}+b^{2}=10^{2}\,$
$b={\sqrt {100-25}}=5{\sqrt {3}}\,$
Now the square root gives a positive and a negative answer. In this real world example it is quite obvious that the one we want--the correct answer--is the positive one. My question is with things like this where we must pick between 2 or perhaps more mathematically valid answers is there a more "formal" way of deciding of which one to use. For example, in a case where imagining the problem in our heads is difficult common sense may not be of much help to us. --212.120.246.239 (talk) 15:06, 3 September 2008 (UTC)

It really depends on the context. In a physical problem, there are instances where negative solutions are meaningless, or where non-integer solutions are not acceptable.

Consider a family in which the number of children is X. If two less than the number of children, multiplied by five less than three times the number of children, equals 4, then we've got a quadratic equation: $(x-2)(3x-5)=4$, which is equivalent to $3x^{2}-11x+6=0$. Solving this quadratic equation, we obtain two answers: $x=3$ and $x={\frac {2}{3}}$. The latter answer is clearly inapplicable to the problem.

On the other hand, if we're just solving the equation $(x-2)(3x-5)=4$, with no thought of application, then we'll keep both answers. The key is in your question: "is there a ... way of deciding which one to use?" If there's an application, then it may provide constraints on acceptable solutions. Determining which solutions apply and which do not apply to a particular application is part of the challenge of applied mathematics. I don't know of a general method, simply because applications can vary widely. Sometimes, a solution that seems meaningless in a physical problem will lead to new physical ideas. If memory serves, the existence of Black holes was predicted when some equations "blew up", with an infinite term appearing, which was at first assumed to be extraneous. I know there are other examples of this sort of thing, but that's the one that comes to mind just now.

In other cases, simply checking whether our solution satisfies the original problem will lead to some solution being thrown out. Consider: ${\frac {x+1}{x-2}}=x+9+{\frac {3}{x-2}}$. If we multiply both sides by $x-2$, we obtain another quadratic equation: $x+1=x^{2}+7x-15$. We can solve this to obtain $x=-8$ or $x=2$, but the latter solution clearly does not work in the original problem. (We could have caught this earlier by being more careful, and noting that $x\neq 2$ is implied by the original equation.)

I realize this isn't quite a succinct answer, but I hope it helps illuminate your question somewhat. -GTBacchus^(talk) 16:02, 3 September 2008 (UTC)

Whenever you use mathematics for a real world problem, you have to come up with a mathematical model that is basically a way of converting your real world concepts into mathematical concepts. In your example you are modelling lengths as non-negative real numbers. The fact that they are non-negative is part of the model, so acts as a constraint to the mathematics. There is no absolute way of working out what constraints you will need when modelling any given situation, coming up with a model is the first, any often hardest, part of solving the problem (once you have a good model all that's left is often basic arithmetic). --Tango (talk) 18:00, 3 September 2008 (UTC)

• (ec) You ask 'how FAR UP the wall does the ladder go?', so you define a vertical axis, along which you measure the 'height', and the axis is oriented upwards. The negative result then means simply a point below the floor. An algebraic equation, which you used to express the problem, does not contain a condition of 'only positive ordinate allowed', so the Pythagorean formula describes two triangles: the one between the ladder, the floor and the wall, and the other one, which is the reflection of it in respect to the floor. Eventually you get two solutions, defining the upper and the lower vertex of the upper and the lower triangle, respectively. --CiaPan (talk) 18:11, 3 September 2008 (UTC)

• In 'formal' mathematics the values which satisfy the equation are the ones that are in the problem's domain. Going from a real world problem to the mathematics involves deciding the domain of the variables as well as the equations that apply to them. For number of children the domain would be the non-negative integers (except in statistics of course!). If you want only heights and don't allow negative heights then you can either have a condition that the height is greater than or equal to zero and use the domain of real numbers (but of course negative ones won't satisfy the condition), or else you could use a domain consisting only of the non-negative reals. Dmcq (talk) 18:46, 3 September 2008 (UTC)

• The quadratic equation is a tool you can use to find out the answer you want. But the tool does not know the context of the question, so it just spits out all the possible answers that could, in some context, satisfy it. An analogy is this question: How do I get from A to B? Well, the options include walking, crawling, driving a car, swimming, flying by plane, taking a bus, train, taxi etc etc etc. All those answers could apply to specific As and Bs. But if your A was your kitchen and your B was your bedroom, then only one of these options (walking) would actually apply, and you'd know which one it was. (Ok, maybe crawling could be an option, too, but not the others). -- JackofOz (talk) 01:26, 5 September 2008 (UTC)

Peanut shaped

Is there a name for that geometric shape? If so, what is it?--Honeymane_{Heghlu meH QaQ jajvam} 22:07, 3 September 2008 (UTC)

As far as I know, there isn't a specific name for a "peanut shaped" object. If you draw a Cassini oval with b/a slightly greater than 1 - say between 1.1 and 1.2 - then you get a two-lobed curve that resembles the cross-section of a peanut. Gandalf61 (talk) 11:02, 4 September 2008 (UTC)

Really? I was under the impression that there was names for everything. Oh well.--Honeymane_{Heghlu meH QaQ jajvam} 20:48, 4 September 2008 (UTC)

This Prolate spheroid certainly looks pretty much like a peanut to me. I got linked to the article from Rugby ball because, well, I find it comical that some people refer to rugby-players as peanut-huggers. 194.221.133.226 (talk) 11:18, 5 September 2008 (UTC)

I was thinking more about the outer shell of the peanut, the part that looks sort of like dumbbells.--Honeymane_{Heghlu meH QaQ jajvam} 16:02, 5 September 2008 (UTC)

I think a mathematician would most likely call it "peanut shaped". Playing around with Greco-Roman roots I came up with "arachiform", for which a Google search actually turns up a few uses alongside such terms as "obpyriform". But I can't deduce from those few references whether "arachiform" refers to the shape of a peanut shell or of the nut inside or neither. -- BenRG (talk) 18:44, 5 September 2008 (UTC)

September 4

Plot request

Can someone plot an excel line graph for me and send it to me by e-mail or attach a rapidshare link ? I am an absolute wally when it comes to computers so please help me. Here is the table of data. It is a distance-time graph. Here it is:
distance (km) time (min)

You'll need your own pencil, though.

0.2 0.5
0.2 1
1.2 3
1.2 3.5
2.5 5
2.5 5.5
4.4 7
7.3 11
7.3 11.5
8.3 13
9.2 13.5
9.5 16
Please send it to me at [email address redacted to prevent spam] or upload it and give me a link. Thank you in advance.

Saintrain (talk) 20:52, 4 September 2008 (UTC)

You would be far better off learning to use Excel (or some other spreadsheet program). There are plenty of courses and online tutorials out there. We're not going to do it for you, that's not the purpose of this desk. We're here to help you learn and doing your work for you won't help that at all. --Tango (talk) 22:05, 4 September 2008 (UTC)

Is this some kind of zig-zag thingy with all the vertical lines? – b_jonas 20:29, 6 September 2008 (UTC)

Calculator for student in college

Hello, I need to buy a graphing calculator for a student in college. I understand he is using a TI-89 (not titanium) now. He is a Math major and will most likely go to do graduate study in Math. What graphing calculator should I get for him? The TI-83 plus is significantly cheaper than a TI-89 Titanium (and there is a newer model already). However, TI-89 has CAS which earlier models don't. Is the TI-89 accepted in major tests like the GRE (general, Physics, Math, Chemistry) and so on?

Do you think the additional cost of the TI-89 is justified? How about a TI-89 Titanium? I also wonder if he is going to end up doing most of his work on a notebook computer or something. Thanks a lot. You guys are awesome! Kushal (talk) 23:18, 4 September 2008 (UTC)

I'm studying maths at Uni and have never needed a graphing calculator. They are banned from all exams offered by my department (few exams allow a calculator at all and those that do only allow a basic scientific calculator) and outside exams a computer is usually much better. Are you sure he really needs it? If it's for a particular course, whoever is taking the course will be able to recommend one, or at least give you a list of features he'll need. --Tango (talk) 00:20, 5 September 2008 (UTC)

He is taking a bunch of math classes with different professors. I don't want to overkill (or get a calculator so sophisticated that he cannot use in tests) but I want to buy him a decent graphing calculator so that it does not become worthless in his advanced graduate work. I am in the US, if it helps any. Kushal (talk) 00:41, 5 September 2008 (UTC)

He'll need to find out what the rules are if he wants to use it in exams. What area of Maths does he expect to do graduate study in? Unless its stats, I can't see him needing a graphics calculator for that any more than he needs one for undergrad work, and even then it's unlikely. --Tango (talk) 00:50, 5 September 2008 (UTC)

I don't know if this will help, but I'm studying Maths + Physics at a Uni in the UK. I bought a graphical calculator when I started at College (which is 16-18 here) and found it very useful, I was also allowed to use it in all of my exams. At university, I have very rarely had any use for it, and I am not allowed to use it in any exams (the university has a ban on some of its features, such as being able to store text). Another thing to consider is that his university might well have some CAS software like Matlab or Maple on its computers, which can do anything a graphing calculator can, plus much more (and a lot faster).84.12.252.210 (talk) 10:31, 5 September 2008 (UTC)

Yes, 84, we do have Maple in all departmentally owned computers. He could get Maple for his Macbook too. However, I am not sure any teacher in the department would allow using a MacBook in finals. This brings me to another question. Is the USB thingy for the titanium even compatible with macs? We have Intel Macbook with Tiger on it. Thanks. Kushal (talk) 12:32, 5 September 2008 (UTC)

I am a graduate student in math at Nebraska, and I use my TI-89 all the time for various things. On the other hand, in "advanced graduate work" any calculator is fairly useless, since most advanced math classes focus much more on logical reasoning and proof-writing rather than numerical calculation. At this point in my studies an advanced graphing calculator with a CAS like the TI-89 is mostly just a convenient luxury, but since I use it primarily for its CAS capabilities it's much more useful to me than a TI-83 or TI-84. —Bkell (talk) 10:47, 5 September 2008 (UTC)

Bkell, do you ever use custom apps or use your TI-89 with a computer? How useful has the USB thingy been to you? Kushal (talk) 12:32, 5 September 2008 (UTC)

This might be too late to be helpful, but personally I don't ever use that stuff, mostly because I haven't really ever tried it and haven't felt that I'm lacking anything. When I used to write programs on my TI-86 I would use the computer link to save them so they wouldn't be lost if my calculator died or something, and I transferred games like Tetris the other way, but that was high school and Tetris is the primary reason to have a graphing calculator in high school ;-) —Bkell (talk) 13:37, 6 September 2008 (UTC)

Calculators never helped me in math classes (and we were rarely able to use them anyway). Where I did find the 89 very helpful though was in science classes. For instance I had physics tests which involved solving systems of sometimes three or four linear equations. They weren’t testing that skill, but you had to know how to do it. Thus I gained a significant time advantage to anyone with a lesser calculator by merely plugging it into the calculator and moving on with the physics. GromXXVII (talk) 22:13, 5 September 2008 (UTC)

When I was studying for my Math and Computer Science degree, I had a TI-86. It wasn't much use in math, but I found it quite handy in physics (among other things, I learned the hard way about numerical instability while writing a simulation program), and absolutely essential during humanities classes (writing a Minesweeper clone saved my sanity during some of the more boring lectures). --Carnildo (talk) 22:29, 5 September 2008 (UTC)

Thanks. I am getting him a TI-89 Titanium after all. Kushal (talk) 03:31, 6 September 2008 (UTC)

September 5

How many different song variations are there?

It's probably an astronomical number, but the question came to me as I was reading about how one star supposedly was "inspired by" another star to write a similar song and successfully was sued. (It was the one with My Sweet Lord by George Harrison, and He's So Fine)

My question - which is why it's in the math section - is how many possible variations are there on a song? If 10,000 performers/bands each wrote, say, 100 songs a year, thus making a million songs a year, how long before they ran out of variations that were suitably different from each other, so no one performer could say, "Well, that sounds almost exactly like my song." Note that I'm not saying "the exacty same" - obviously, there are subtle differences in songs, but yet "My Sweet Lord" and "He's So Fine" were seen by a court to be similar enough.

I'm just amazed at how people on Name That Tune can guess a song after one note, or even seven; I'm part deaf, but even so, I'm amazed anormal hearing person can tell the difference between songs that seem to start so similarly.Somebody or his brother (talk) 00:22, 5 September 2008 (UTC)

There are enormous numbers of possibilities, the vast majority of which, however, will sound absolutely terrible. Calculating the total number of songs that anyone is likely to enjoy listening too is probably extremely difficult due to problems defining what people like. --Tango (talk) 00:59, 5 September 2008 (UTC)

Quite true. If the OP wants an idea of the magnitude, let's assume that there are 88 notes (the number of keys on a piano), that only a single note is played at any given time, that every note is held for the same amount of time, that there are no rests, and that the song is exactly 100 notes long. Given these restrictions, more "songs" can be composed than there are atoms in the known universe. Wikiant (talk) 01:36, 5 September 2008 (UTC)

There are more "songs" than the number of atoms in the observable universe squared. --Tango (talk) 01:47, 5 September 2008 (UTC)

Awww, but what fraction of those songs seems similar? Let's be more concrete. Lets songs be defined by sequences of notes (a_i) and (b_i), and define two songs as "similar" if there exists p,q,r such that (a_p,...,a_p+9) = (b_q + r,...,b_q+9 + r). In other words, they are similar if the same pattern of 10 notes occur in both songs modulo a change in pitch. Now for each real song there would be quite large space of disallowed similar songs. I'll leave estimating that number to someone with more free time ;-). Dragons flight (talk) 01:59, 5 September 2008 (UTC)

My extremely rough estimation (so rough I daren't share the details!) puts it somewhere in the trillions, at least. --Tango (talk) 02:17, 5 September 2008 (UTC)

Assume that a song can be recorded on a CD containing N bits. The total number of possible songs cannot exceed 2^N. A lower bound is obtained by using the observation that skilled people can identify a song after hearing M≈7 notes. Let A≈24 be the number of different pitches and B≈5 the number of different durations of a single note. Then the number of different songs are ≤(AB)^M≈120⁷=358318080000000. Bo Jacoby (talk) 07:23, 5 September 2008 (UTC).

I believe that part of what makes a song easily recognizable after just a few notes is its general "sound" (instrumentation, accompanying chords, tempo, echoes or lack of them and so on). So even songs that are "formally" very similar (say, sharing a subsequence of notes in their main melody) can be easily told apart. (On the other hand, they are not different in the sense of the OP.) If we could strip a song of all these factors, probably more notes would be necessary to recognize them. Goochelaar (talk) 11:28, 5 September 2008 (UTC)

You might want to reduce the number of songs by taking conventions of western music into account. Much of western music is in four-four time and the range of notes maybe span two/three octaves, further if we restrict to songs in a major key, the notes will tend to comprise the root note, the 3rd and the 5th, which sound more harmonious. (This is due to fact that 2^4/12≈5/4, 2^7/12≈3/2 see Just intonation). So if we ignore all the fancy jazz stuff and stick to major keys in four four time we will get a much reduced but still large number. I've read somewhere that there is actually quite a small number of common baselines or grooves which tend to be more repetative.

Maybe the correct mathematical way to workout the chance of two songs sounding the same would be more a bassian approach. Sample songs in a given genera, rescale so they are all in the same key, find frequency of each note and do the sums. --Salix alba (talk) 11:30, 5 September 2008 (UTC)

What does "bassian" mean here? Cuddlyable3 (talk) 10:34, 10 September 2008 (UTC)

All you need now is a way to extrapolate the tune from the first seven notes taking into account these various rules and put the lot on the web so when a person asks for tune number 52933031492 your web server returns the tune with an appropriate copyright. Having filtered out all the current tunes of course. You're gong to make a fortune using the RIAA. Dmcq (talk) 20:03, 5 September 2008 (UTC)

For an estimate of how many different themes there are in real music, see Dictionary of musical themes. The book has 655 pages, and lists themes from classical music, transposed to the key of C. I don't know how many themes there are per page, but can hardly imagine there being room for more than 15 (it's written in music notation). This would imply that the book contains about 10,000 entries. The book has a companion volume with themes from opera and songs (lieder?), which has about 8000 entries. So I would guess that the number of reasonably well known themes in classical music is probably smaller than 50,000. --NorwegianBlue ^talk 20:08, 5 September 2008 (UTC)

Consider coding melodies this way: 1) Decide how many notes distinguish melodies, say $N$. 2)Give each note one of these 3 values: +, -, or =. The value says whether the note is above, below or equal to the preceding note. There will be 3^$N$-1 possible codes since the first note is not coded. Example: Happy Birthday song is =+-+-. This code, which could also be expressed as a number by giving the notes arbitrary numerical weights, is not unique to this song but can serve as a hash for compactly cataloging the vast number of possible melodies.

Further, no human can sing all the notes of a piano and we can restrict our estimae to melodies that stay on key i.e. can be played on just the white piano keys, key of C, and have maximum one octave difference between successive notes. Our 3^$N$-1 hash codes then suggest (3 x 8)^$N$-1 different songs. Error sources in that estimate can be that note changes coded "=" are just that (rather than 8 possible sizes), and that we ignore that melodies can have off-key notes, exceed the octave difference or vary note lengths and intermissions. But since the first error is in opposite sense to the others we can expect some cancellation of errors.

If I suppose $N$ = 7 then my estimate is (3 x 8)^$7$-1 = 2E8 different songs. The 10,000 bands would have work for 20 years.

However the contestants identifying songs work with a much smaller number of songs encompassed by culture, known recordings and suitability for the TV show, its singers, orchestra, etc. Cuddlyable3 (talk) 11:27, 10 September 2008 (UTC)

Differential equation

Hello. I've done another exam question and, as usual, I would like you kind people to check it for me. Same old story, please tell me if I'm right but if I'm wrong, let me know but don't tell me how, when, where or why.

Find functions f, g and h such that the equation

${\frac {d^{2}y}{dx^{2}}}+f(x){\frac {dy}{dx}}+g(x)y=h(x)$

is satisfied by $y=x$, $y=1$, $y=x^{-1}$ for 0<x<1.

If f, g and h are the functions from above, what condition must the real numbers a, b and c satisfy in order that

$y=ax+b+{\frac {c}{x}}$

should be a solution of the above differential equation?

I have

$f(x)=-{\frac {2}{x(x-1)}}$

$g(x)=h(x)={\frac {2}{x(x-1)^{2}}}$

and the final condition is a+b+c=1. Thanks 92.3.59.243 (talk) 19:09, 5 September 2008 (UTC)

Correct. Algebraist 19:19, 5 September 2008 (UTC)

Cheers. Really appreciate it. 92.3.59.243 (talk) 19:21, 5 September 2008 (UTC)

September 6

heights and distance

I would like to have a brief information on this topic'heights and distance' as i have to use the information as my class 10th board examination 2008 project —Preceding unsigned comment added by 117.197.48.170 (talk) 15:15, 6 September 2008 (UTC)

Can you be more specific? It is impossible to be brief in response to such a general question. Wikiant (talk) 16:56, 6 September 2008 (UTC)

Is trajectory of interest?…86.132.239.22 (talk) 17:45, 6 September 2008 (UTC)

Which board? Also, try height and distance. – b_jonas 20:26, 6 September 2008 (UTC)

117.197.48.170: This is an experiment. If you see this then just let us know somehow. Thanks, -hydnjo talk 23:27, 6 September 2008 (UTC)

Surds

If a and b are positive integers and not both square, is it possible for sqrt(a)+sqrt(b) to be rational? Black Carrot (talk) 22:02, 6 September 2008 (UTC)

No. Algebraist 22:08, 6 September 2008 (UTC)

Let d be the HCF of a and b. Then we have sqrt(d)(sqrt(a')+sqrt(b')) rational, with a' and b' coprime. Squaring we find that sqrt(a'b') is rational, so a' and b' are both squares. But then sqrt(d) is rational, so d is a square, so a and b are both squares after all. Algebraist 22:11, 6 September 2008 (UTC)

September 7

solenoidal and irrotational field

Is it possible for a vector field to be simultaneously solenoidal and irrotational, that is, divergence and curl free. I don't think it us, but I can't think of a good reason why. This seems to be the essence of the Helmholtz Theorem, but I don't quite see how I could prove that. Thanks very much for any help you can provide! Nathan12343 (talk) 00:39, 7 September 2008 (UTC)

Well, there's the zero field, and for that matter all constant vector fields. Algebraist 00:49, 7 September 2008 (UTC)

See Cauchy-Riemann equations. siℓℓy rabbit (talk) 00:59, 7 September 2008 (UTC)

Calculus of variations spring problem

Hi, I am trying to learn to use calculus of variations to solve a simple hypothetical problem I made up for fun; but I am getting stuck and I don't understand why.

So I have a spring. One end of the spring is moved around so that it is at position $g(t)$ at time $t$. The other end of the spring I can control its position $f(t)$ at time $t$. The goal is to find a $f(t)$ so that I get the maximum work I can get out of my end. I hope that this problem makes sense. It seems that at worst, $f(t)=0$ should give a solution of zero work; and it seems to me that you should be able to do better than that. Perhaps my intuition is wrong here.

So I start with the definition of work:

${\frac {dW}{dt}}=F(t)\cdot f'(t)$

And then apply Hooke's law, where $k$ is the spring constant:

${\frac {dW}{dt}}=-k(f(t)-g(t))\cdot f'(t)$

${\frac {dW}{dt}}=k(g(t)-f(t))f'(t)$

The total work is then

$W=\int _{t_{1}}^{t_{2}}{\frac {dW}{dt}}\,dt$

This is in the same form as needed for the Euler–Lagrange equation:

$W[f]=\int _{t_{1}}^{t_{2}}L[t,f,f']\,dt$

where

$L[t,f,f']=k(g(t)-f)f'$

So now I plug this into the Euler–Lagrange equation:

$-{\frac {d}{dt}}{\frac {\partial L}{\partial f'}}+{\frac {\partial L}{\partial f}}=0$

$-{\frac {d}{dt}}\left[k(g(t)-f)\right]-kf'=0$

$-k(g'(t)-f')-kf'=0$

The $f'$ now cancel and I am left with:

$-kg'(t)=0$

Now this equation is simply a condition on my given $g(t)$ function! And the function I am trying to optimize ($f(t)$) completely vanished! How is this possible? Did I do something wrong? Is there some assumption I am violating? I hope this problem makes sense. If not this way then is there another way I can use to solve this problem? Thanks, --71.141.132.142 (talk) 01:27, 7 September 2008 (UTC)

Here are two points. First, regarding your physical model, the distance in Hooke's is distance from equilibrium position, not from the other end: in 1D this is simply a difference of a constant, and so the mathematics works out the same, but if you are working in more than 1D then there is more than one equilibrium, and you have a bit of mess. Second, regarding your mathematics, offhand I can't find any mistakes. Assuming that you have correctly applied the Euler-Lagrange equation, the conclusion you reach is not necessarily impossible: it would simply indicate that if both $k\neq 0$ and there exists t for which $g'(t)\neq 0$ (i.e., g is non-constant), then there does not exist any such extremal f. Unfortuantely I do not understand the mathematics well enough to be more helpful. Eric. 213.158.252.98 (talk) 13:18, 7 September 2008 (UTC)

Indeed, I can create an example where arbitrarily high work can be extracted. Let t1 = 0, t2 = 1, g(t) = t, and k = 1. Here g represents the equilibrium position. We subject f to the constraint that the spring starts and stops at equilibrium, i.e., f(0) = 0 and f(1) = 1.

Consider the following f. First, just after t = 0, very quickly bring f to the point -A, where A is a constant. Leave your end of the spring at that position until just before t=1, when you very quickly restore f to equilibrium at 1. The work gained is approximately equal to Ak times one unit of distance. Since A is a arbitrary, we can get both arbitrarily high and arbitrarily low amounts of work: thus in this situation we have no global extrema.

A similar construction can be used to make arbitrarily high or low amounts of work in any situation provided that k is nonzero and g is nonconstant. Eric. 213.158.252.98 (talk) 13:38, 7 September 2008 (UTC)

Oh I see. So I can force the other end to put in arbitrarily large amounts of work if they move at all; because if the other end went up, I could just hold the spring arbitrarily low to get him to put in work. It is only when $g'(t)=0$ that the other end doesn't put in work at all; and then it doesn't matter what I do. --71.141.132.142 (talk) 21:06, 7 September 2008 (UTC)

Precisely. You could put in realistic constraints on f (disallowing going too far from equilibrium) to force a maximum to exist, but in that case I doubt that the Euler-Lagrange equation will help you find this maximum. Alternatively, you could use a force equation other than Hooke's Law (keep in mind that Hooke's Law is an approximation best for near equilibrium) to close this loophole, and then proceed again with the Euler-Lagrange equation. I bet you can find something interesting with the latter (athough I don't know if it'd be physically realistic -- go find a physicist and tell us what (s)he says!). Eric. 84.215.155.88 (talk) 20:31, 8 September 2008 (UTC)

Help

How do you get from the left hand side to the right: ${\frac {x^{3}}{1+x^{2}}}=x-{\frac {x}{1+x^{2}}}$ --RMFan1 (talk) 20:51, 7 September 2008 (UTC)

Use polynomial division (or whatever) to get $x^{3}=x(1+x^{2})-x$. Then just cancel a $1+x^{2}$ and you're there. Algebraist 20:56, 7 September 2008 (UTC)

Multiply both sides by ${(1+x^{2})}$ .Cuddlyable3 (talk) 23:41, 8 September 2008 (UTC)

That proves equality, but is no use if you have the left hand side and are trying to get to something like the right. Algebraist 23:43, 8 September 2008 (UTC)

September 8

The Hawthorne Effect and "improvements due to monitoring" of performance

Hello,

I am an engineer - working in the area of energy efficiency improvements in industrial processes.

We are attempting to build evidence or provide past performance reports where we can show people the value of measruing watts or power consumption - due to something parallel to the Hawthorne Effect. Essentially, we know that management principals state that "what gets monitored gets improved".

Thus, we would appreciate any reports or resources which document the fact that installing watt meters and monitoring the level of energy consumption - continuously - will change the amount of watts/energy being consumed simply due to the effects on human nature to improve upon what is being monitored.

The Hawthorne effect appears to indicate that a temporary improvement will occur when people are "watched". We are looking for long term improvements due to continuous feedback and an awareness that the watts being consumed will be compared to others - thus, taking advantage of the natural competitive nature of people to be the best/most efficient.

Thank you for your input/help.

Sincerely,

Don Voigt, P.E. —Preceding unsigned comment added by 75.86.190.159 (talk) 12:37, 8 September 2008 (UTC)

There is a list of books, papers, scholarly works, etc, located in the Further Reading section of the article. Have you investigated these? Nimur (talk) 14:00, 8 September 2008 (UTC)

Thank you, Nimur, for your suggested readings. I was hopeful that someone had already made a review of the literature and may have had a similar experience - looking for statistical correlation of "what gets monitored and recorded, gets improved upon"....without going through a survey of all papers, I'd like to know where to focus my attention to find the answer.

Don —Preceding unsigned comment added by 75.86.190.159 (talk) 14:21, 8 September 2008 (UTC)

Electric power is comparatively cheap in Norway which has a small population served by hydroelectric generators. At one time domestic electricity was charged at a fixed cost per household installation (main fuse rating) ignoring actual consumption. Then household meters were introduced with charging by watthour; the electricity was still cheap. Only recently has there been much incentive to reduce household electric consumption, partly due to the national interest in exporting more electric power to foreign markets rather than "wasting" it at home. Household electric consumption typically has strong summer/winter variation due to more power being used for heating in winter. Electricity charging in Oslo is now by quarterly invoices that present each household with a graph of the previous 12 months consumption, thereby focusing attention on how one's latest quarter consumption is relative to "par" (as in golf). My understanding of Hawthorne effect and the foregoing is that to reduce domestic consumption consumers need regular fresh reminders that their micro-choices e.g. whether to bother turning off unnecessary lights, etc. matter, or the old relaxed mindset returns. Cuddlyable3 (talk) 23:35, 8 September 2008 (UTC)

Differentiation

Can someone show me, in really small steps, how to get ${\frac {dP}{dx}}$ given that $P=2x+2{\sqrt {d^{2}-x^{2}}}$. --RMFan1 (talk) 13:38, 8 September 2008 (UTC)

It's unlikely that you mean $d^{2}$ since you're going to wind up with confusion over what 'd' is. You should look to differentiate term by term. Mrh30 (talk) 15:01, 8 September 2008 (UTC)

And you'll want the chain rule, the power rule and the linearity of differentiation. Algebraist 15:06, 8 September 2008 (UTC)

Well yop can use another letter if you like but I don't think using a d would be confusing. Anyway, there was a mistake: it should have been d^2 - x^2 not +. Ive gone ahead and changed it. The answer should come to
${\frac {dP}{dx}}=2-2{\frac {x}{\sqrt {d^{2}-x^{2}}}}$. I just don't know why? I doubt there's anything wrong with the question though--it's an example from a the Cambridge STEPs. You can see it yourself here [1], Question 9 --RMFan1 (talk) 15:54, 8 September 2008 (UTC)

Well, what do you know? Do you know what differentiation is? Do you know the rules of differentiation I linked to above? What have you done in attempting to answer the question? (the given answer is correct, btw) Algebraist 15:59, 8 September 2008 (UTC)

And if you don't know the chain rule etc. you could do this from first principles - write down an expression for P(x+ δ), use the binomial theorem to expand the second term in powers of δ, subtract P(x), divide by δ, then use

${\frac {dP}{dx}}=\lim _{\delta \rightarrow 0}{\frac {P(x+\delta )-P(x)}{\delta }}$

(but knowing and using the rules makes life easier). Gandalf61 (talk) 16:18, 8 September 2008 (UTC)

mathematical representation of geography

hi,,can you help me to understand what a mathematical model mean? i want to know the mathematical representation of : the rotation of earth,its longitudes and latitudes59.95.205.94 (talk) 14:58, 8 September 2008 (UTC)chhaupa@yahoo.com

You could start by reading our mathematical model article. The "latitude and longitude" model represents the Earth by a sphere, and uses two spherical co-ordinates to model the location of any point on the surface of the Earth. We know it isnt 100% accurate - the Earth isn't a perfect sphere, and the model doesn't represent altitude, for example - but it is good enough for many purposes. Gandalf61 (talk) 22:07, 8 September 2008 (UTC)

questions about polynomials

I realise that this falls somewhat under the category of 'begging for homework help', but I have been trying for quite a while with two problems from my homework (IB Higher Maths, if you're interested), and I'd appreciate any help you feel you can ethically give me. I apologise also for my mangling of the correct ways to put maths notation into Wikipedia.

14. A cubic polynomial gives remainders (5x+4) and (12x-1) when divided by x^2-x+2 and x^2+x-1 respectively. Find the polynomial.

24. If the polynomial P(x)=x^2+ax+1 is a factor of T(x)=2x^3-16x+b, find the values of a and b.

For the first, I attempted to put them in the form p(x)=d(x)q(x)+r(x) (with the quotients replaced by (ax+b) and (cx+d)) and to set them equal to one another, in the hope of equating the coefficients, but this didn't work. I have a niggling feeling that the solution has to do with some use of the remainder or factor theorem that I haven't fully understood, but I'm not sure how to proceed. For the second I again tried to put them into the form p(x)=d(x)q(x), infering that q(x)=(2x+b), but this led to answers that were wrong.

So if anyone has any helpful suggestions, I would be very grateful.

Thank you,

Daniel (‽) 20:24, 8 September 2008 (UTC)

Your method for the first question seems right - it worked for me. What went wrong when you tried it? --Tango (talk) 21:18, 8 September 2008 (UTC)

I don't have it to hand, but as far as I recall I simply got results to do with relating (ax+b) and (cx+d) to each other (ie c=d and so on). I couldn't see a way from there to working out the starting polynomial. Daniel (‽) 06:08, 9 September 2008 (UTC)

As Tango said, your method is correct. You might just be making silly errors because you're expanding a fair number of terms by hand (I'll admit I had to do the system of equations twice to get the right answer). But in case you're setting up the initial equation incorrectly, here is what you should be starting with:

p(x) = (ax + b)(x² – x + 2) + (5x + 4)

. . . = (cx + d)(x² + x – 1) + (12x – 1)

Then proceed to expand, simplify, and move all the terms without coefficient variables to one side of the equation. You should get 7x – 5 on that side. Just keep checking your algebra until you get a solution where a, b, c, and d are all integers. You will get some relations like a = c, but if you keep working through all the equations you will find a single unique solution set.

Regarding the second problem, note that there are two solution sets (it is a plus/minus pair of solutions). 75.63.57.93 (talk) 06:41, 9 September 2008 (UTC)