Wikipedia:Reference desk/Mathematics: Difference between revisions

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

January 27

Rotating a Right Traingle 180 Degrees

I was trying to fiugre how do I rotate a right triangle with the base facing to the right 180 degrees from the third quadrant to the fourth quadrant. The points of this triangle are A=-2,-4, B=-6,-4, and C=-2,-7. I thought I would make a mistake picturing the rotation in my head. Does anyone know of a simpler method?

Best Reagrds,

--Writer Cartoonist (talk) 02:55, 27 January 2009 (UTC)

Draw it on graph paper, and it will be fairly obvious how to rotate it. StuRat (talk) 03:00, 27 January 2009 (UTC)

How can a rotation of a half turn move anything from the third quadrant into the fourth? --Tardis (talk) 18:38, 27 January 2009 (UTC)

Surely any half turn about a point in the lower half plane moves part of the third quadrant into the fourth. Matthew Auger (talk) 20:00, 28 January 2009 (UTC)

As no center for the rotation was given, I was assuming it to be the origin. Maybe the problem is to identify a center that does move it (entirely) into the fourth quadrant: that would be much more interesting! (The overdone answer $(10^{100},-10^{100})$ is easy, though.) --Tardis (talk) 16:45, 29 January 2009 (UTC)

Rotating 180 degrees about the origin just amounts to multiplying all the coordinates by −1. Michael Hardy (talk) 16:52, 29 January 2009 (UTC)

Weierstrass-Enneper parametrization

Hey,

Could anyone give me a hand rewriting the article about the Weierstrass–Enneper parameterization? I've added some information at this page, but I haven't explained quite a bit, hence why I haven't moved it over yet. It would be especially nice if someone could give explanations about why we can write the composite Gauss map as I did and how minimal surfaces and harmonic functions are related (as I have already described the relation between the Gauss map and holomorphic functions, there's not much left to do. I could do with some reference of the statement that G is conformal if and only if M is minimal.). In fact, something should be written about harmonic functions in the minimal surface article...

I mainly based what I wrote on the article here (PDF), so definitely have a look at that. I'm not sure what the author does with harmonic functions and his map J which rotates the tangent space by 90°, so I ended up leaving that out.

It would also be really nice if someone could provide explicit parametrizations for Costa's surface and Riemann's minimal surface (described in the article I linked). I think the articles about minimal surface are really underdeveloped at the moment, and even worse for surfaces of constant mean curvature. I went to a talk given by Meeks last Friday, where he described much of his work on constant mean curvature and minimal curvature surfaces, and I realized Wikipedia doesn't even have any articles about Delaunay surfaces or constant mean curvature surfaces, which seems quite odd. This seems like a good starting point.

Thanks. --Xedi^Talk 03:47, 27 January 2009 (UTC)

Yes they do need a lot of work. EG-Models and The Scientific Graphics Project might be a good resources for CMC surfaces. --Salix (talk): 08:36, 27 January 2009 (UTC)

College Math Help!

A balloon of total mass 680kg is descending with a constant accelearaiton of 0.4 m/s². Find the upthrust on the ballon. When the ballon is moving at 1.5m/s, enough ballast is released for the balloon to fall with a deceleration of 0.2 m/s². Calculate

(a) how much ballast was releasd

(b) the time for which balloon continues to fall

MY PROBLEM My problem lies in the bold part of the question. When ballast is released, I assume that the mass of the total object in consideration has also changed. But solving with this in mind, it doesn't work. Answers to this question: 6528N (a) 40kg (b) 7.5s People, help a poor guy please! —Preceding unsigned comment added by 202.72.235.201 (talk) 14:20, 27 January 2009 (UTC)

You have to assume that the buoyancy of the ballon does not change when it releases ballast, even though its mass changes. Work out the buoyancy force B from the first part of the question, then work out the new mass m using B - mg = ma, then subtract new mass from original mass to find ballast. I think you also have to take g to be 10 ms^-2 to produce the given answers. Gandalf61 (talk) 14:55, 27 January 2009 (UTC)

Thanks Gandal for your thoughts but I did the math exactly the way you thought , it did not work out. g = 10 m/s2 taken also. —Preceding unsigned comment added by 202.72.235.201 (talk) 15:44, 27 January 2009 (UTC)

Gandalf's method gives exactly the answers provided. Can you explain your working, so we can see what's going wrong? Algebraist 15:48, 27 January 2009 (UTC)

Bouyancy or uptrust equals 6528N as stated in (a) Let the new mass of this balloon be 'm' Weight = mg = 10m Upthrust = 6528-10m N This upthrust causes upward acceleration of 0.2 m/s² thus: upthrust = m x (acceleration)

     6528-10m = m x 0.2
     m = 640 ? 

Woh: Ballast released = 40 kg! Algebraist, I can see my mistake but I did the math for like an hour, I can't see why that time I didn't get my answers! Thanks for your help though! —Preceding unsigned comment added by 202.72.235.201 (talk) 17:04, 27 January 2009 (UTC)

Don't laugh People for my stupid math problem, it was a silly mistake back then, I didn't read the question well, I took a = 0.5 for 0.2. Got to sleep more and clear up my brains!

Math Problem Resolved! How do we get this section removed now! —Preceding unsigned comment added by 202.72.235.200 (talk) 17:16, 27 January 2009 (UTC)

Reading the question carefully is always a good first step - we've all been there! --Tango (talk) 17:24, 27 January 2009 (UTC)

why not fermats last conjecture

if he didnt prove it, why isnt it fermats last conjecture —Preceding unsigned comment added by 82.120.111.130 (talk) 16:11, 27 January 2009 (UTC)

He claimed he had a proof (that didn't fit in the margin) and people gave him the benefit of the doubt. Chances are, he was mistaken (it seems unlikely that nobody would be able to reproduce the proof after hundreds of years if it had existed - the only proof we have involves mathematics Fermat couldn't possibly have known about). --Tango (talk) 16:15, 27 January 2009 (UTC)

After all in maths it is quite common that the "theorem of X" has not been proved by X.--pma (talk) 16:58, 27 January 2009 (UTC)

Seems to be the case more often than not, as far as I can tell. --Tango (talk) 17:23, 27 January 2009 (UTC)

At this point in the thread it's obligatory that someone mention Stigler's law of eponymy (due to Merton). Algebraist 17:25, 27 January 2009 (UTC)

The proof uses algebraic geometry. --PST 12:01, 28 January 2009 (UTC)

Yes, it does, did you have a point? --Tango (talk) 22:31, 28 January 2009 (UTC)

Wake Angle

This page states that the 'angle of the wake of a body moving steadily in deep water is always 2arcsin(1/3)'. How is this result derived? Angus Lepper^{(T, C)} 16:18, 27 January 2009 (UTC)

The angle is called the "Kelvin wake angle" - there is a derivation here. Gandalf61 (talk) 10:16, 28 January 2009 (UTC)

Thanks. Angus Lepper^{(T, C)} 21:23, 28 January 2009 (UTC)

Co-ordinate geometry

Please consider the following....

Supposing y=X^2+3X+1 is an equation...we can draw y against x. Now, if, we draw two tangents at the points where the curve cuts the x-axis and also draw two normals through the same points. Now, lets assume that the two normals meet at A and the two tangents meet at B. My question is.....will the point A be a reflection of point B at x-axis and vise versa or the impression is wrong. I ask this qustion since I rhink that the tangent is a reflection of the normal and vise versa. Please, help me out. —Preceding unsigned comment added by 123.49.43.236 (talk) 17:33, 27 January 2009 (UTC)

For a quadratic, that would only be the case if the slopes of the tangents were 1 and −1 respectively, which they aren't for your function. -- Jao (talk) 17:43, 27 January 2009 (UTC)

You can check this by noting that reflection in the x-axis will negate the slope of a line, while the slope of the normal line is the negative of the reciprocal of the slope. Thus to have the normal equal to the reflection of the tangent, you must have that the slope of the tangent is its own reciprocal. As Jao says, this implies the slope of the tangent must be 1 or −1.

To find all the parabolas for which this occurs, we can translate our parabola to be symmetric about the y-axis, so that its equation is of the form y=Ax²-C. Then the two roots are at ±r, where $r={\sqrt {C/A}}$ (which you can check with the quadratic formula). Then solving the equation y′=2Ax1= at r reduces to 4AC=1 (which happens to be the discriminant of our quadratic polynomial). Thus the general solution for symmetric parabolas is y=Ax²-(1/4A). Tesseran (talk) 17:59, 27 January 2009 (UTC)

saving time

Joe wonders if he can reduce the times it takes for him to test whether a coin is fair by stopping after 500 of his 1000 flips and seeing if a certain ratio isn't met, or doing the same in n segments. —Preceding unsigned comment added by 82.120.227.157 (talk) 19:00, 27 January 2009

It sounds like you want Sequential probability ratio test (SPRT). Sorry didn't read it all as it seemed to ramble and had lots of code in it. there's an acronym for that too tl;dr Dmcq (talk) 09:51, 28 January 2009 (UTC)

I reduced the problem.

I didn't follow Joe's programming either, but consider this. Knowing the probability of tossing head, p, say p=1/2, Joe can forecast the number of heads, i, after tossing n times. The distribution of i is the binomial distribution. The mean value of i/n is p, and the standard deviation is √p(1−p)/n. But this is not Joe's situation. Joe's situation is the reverse one: observing the number of tosses, n, and the number of heads, i, and wanting to estimate the probability p. The distribution of p is the beta distribution. The mean value of p is not i/n, but rather (i+1)/(n+2), and the standard deviation is √P(1−P)/(n+3), where P = (i+1)/(n+2). If the mean value is more than 3 standard deviations away from 1/2, Joe may conclude that the coin is significantly unfair. Joe can never make sure it is completely fair. Bo Jacoby (talk) 12:58, 28 January 2009 (UTC).

hey Joe, I liked the story however ;) pma (talk) 14:02, 28 January 2009 (UTC)

Yes you need to decide what you mean by fair and how sure you want to be that it fits your criteria before you rush off testing it. Sequential analysis is all about cutting the cost of the testing once you've decided on your criteria. It's like those medical trials where they abandon them early if they find too many bad side effects. They could also stop if the effect is obviously good but they normally like to just continue on with the original scheme for that and get better data. Dmcq (talk) 14:16, 28 January 2009 (UTC)

By the way I first came across this sort of thing in a book 'Facts from Figures' by M. J. Moroney which I read whilst I was still at school. I though it was very good, I don't know if there is anything like it around now. Dmcq (talk) 14:26, 28 January 2009 (UTC)

you're right (dmcq) I didn't define what isn't fair. if not fair means it favors heads by at least 3%, can I save any flips but retain my confidence level by not necessarily doing 1000 flips if certain criteria are met? what is the optimal places and criteria to set my early-stops at? —Preceding unsigned comment added by 82.120.227.157 (talk) 14:57, 28 January 2009 (UTC)

I did define "fair" though. "fair", the default option (one I want to VERY rigorously eliminate) means that a 50/50 coin produces the observed experimental results (more than 1 in 100,000 times). So if an experimental result is something that a fair coin (as simulated) produces 3,953 times in one iteration of 100,000 runs, for that test the coin would be considered "fair", since I want to produce experimental results that a fair coin produces less than 1 in 100,000 times -- ie I can bet my life that the coin observed making those experimental results is not behaving as a fair coin does -- even 1 in 100,000 times. —Preceding unsigned comment added by 82.120.227.157 (talk) 15:02, 28 January 2009 (UTC)

If Joe want 1 in 100000 security that the coin is unfair, he must test that the mean value of p is more that 4.4172 standard deviations away from 0.500. (See normal distribution). Bo Jacoby (talk) 16:31, 28 January 2009 (UTC).

can splitting the 1000 flips into two groups and stopping if the first one fails instead of waiting for the total count to reach 1000 help reduce the average number of flips it will take to fail (as it probably will, since he is assuming the coin is probably fair). what abaout n groups? does it help? —Preceding unsigned comment added by 82.120.227.157 (talk) 16:57, 28 January 2009 (UTC)

I guess what I'm really asking is what the least many AVERAGE flips is that it takes to "prove that the mean value of p is more than 4.4172 standard eviations away from 0.500". Is the average better than the wrost-case ? (where you have to do the whole experiment) or is there no way to streamline it that way? —Preceding unsigned comment added by 82.120.227.157 (talk) 17:03, 28 January 2009 (UTC)

Joe may never prove that the coin is unfair. So the worst case is infinite. The average case depend on the unfairness of the coin. The algorithm is

i := 1;

n := 2;

repeat

make a group of flips;

i := i + number of heads in the group;

n := n + number of flips in the group;

P := i/n;

until (P−0.5)² > 4.4172² P (1−P) / (n+1)².

Bo Jacoby (talk) 00:31, 29 January 2009 (UTC).

Evaluate trigonometric definite integral

$\int _{0}^{\pi /4}\sin ^{4}(x)\cos ^{2}(x)dx$

I tried using power reduction identities, but that just left a big mess of sin's and cos's. Pythagorean identities didn't really work either. Just need a hint on what identities to use, no need to do the entire problem.--Yanwen (talk) 20:24, 27 January 2009 (UTC)

${\begin{aligned}&\int _{0}^{\pi /4}\sin ^{4}(x)\cos ^{2}(x)\,dx=\int _{0}^{\pi /4}(\sin ^{2}x)^{2}\cos ^{2}x\,dx=\int _{0}^{\pi /4}(1-\cos ^{2}x)^{2}\cos ^{2}x\,dx\\&=\int _{0}^{\pi /4}(1-2\cos ^{2}x+\cos ^{4}x)\cos ^{2}x\,dx=\int _{0}^{\pi /4}\cos ^{2}x-2\cos ^{4}x+\cos ^{6}x\,dx\end{aligned}}$

etc. Do the power-reduction identities you've looked at fail to work with this? Michael Hardy (talk) 21:11, 27 January 2009 (UTC)

Otherwise, you can integrate by parts repeatedly, starting from $\textstyle \sin ^{4}(x)\cos(x)=\left({\frac {1}{5}}\sin ^{5}(x)\right)^{'}$. Is it clear how to go on?--pma (talk) 21:40, 27 January 2009 (UTC)

For an unpleasant but totally mindless approach, you can always just express sin and cos in terms of e^x, multiply out, and integrate. Algebraist 21:53, 27 January 2009 (UTC)

It's our great utopia, not to have to think! ;) Also, in all cases, starting with a change of variable: $2x=y$, will lower the degree and make it easy. --pma (talk) 23:10, 27 January 2009 (UTC)

Integration by parts works great. Substitution is also good. Thanks!--Yanwen (talk) 03:31, 28 January 2009 (UTC)

Welcome! On these lines, you may like the computation of $\scriptstyle I(n):=\int _{0}^{\pi }\sin ^{n}(x)\,dx$, an integral arising in the computation of the volume of the euclidean n-ball. Integrating by parts you find $I(n)$ as a certain rational multiple of $I(n-2)$. Since $I(1)=2$ and $I(0)=\pi$ , you get a product formula for $I(n)$. There is another interesting fact. You can observe that: 1) $I(n)$ is a decreasing sequence; 2) $I(n+2)/I(n)$ converges to 1, therefore so does $I(n+1)/I(n)$; 3) $I(2n+1)$ is a rational number; 4) $I(2n)$ is a rational multiple of $\pi$. As a consequence, $\scriptstyle \pi I(2n+1)/I(2n)$ is a rational approximation of $\scriptstyle \pi$. It's the Wallis product. --pma (talk) 13:55, 28 January 2009 (UTC)

January 28

Notation used in Wikipedia article "Prime Gap"

Prime Gap Article - Paragraph entitled "Conjectures about gaps between primes". (((Prime gap#Conjectures about gaps between primes)))

The notation used in the first and second formulae is unfamiliar to me. In particular, I could find no explanation of the meaning of the the Capital O function. I have tried to search Wikipedia and google the internet without resolution. Can you explan the notation or direct me to a reference for this usage?208.92.185.250 (talk) 04:33, 28 January 2009 (UTC)

Big O notation. Basically it's just a convenient way of stating that two functions behave similarly when a certain limit (in these cases p_n→infinity) is taken. And if you happen to think it looks like abuse of the equals sign, then you're right. -- Jao (talk) 08:33, 28 January 2009 (UTC)

Resolved

208.92.185.250 (talk) 16:54, 28 January 2009 (UTC)

Thanks Jao - I was searching for Capital O instead of Big O (It sure is easy to run afoul of mathematical notation).208.92.185.250 (talk) 16:39, 28 January 2009 (UTC)

For what it's worth, big O notation is linked from O (disambiguation). Algebraist 17:07, 28 January 2009 (UTC)

January 29

Proven but not constructed

In mathematics, what objects if any have been proven to exist, but have not yet been constructed? NeonMerlin 06:20, 29 January 2009 (UTC)

Well, for example, we know an integer solution to Graham's problem exists, but the bounding limit is stupendously ginormous. For a different type of example, I suppose one could say that we know that uncountable real numbers exist (given the most frequently assumed axioms of mathematics) and yet by definition it would be impossible to explicitly enumerate one. Dragons flight (talk) 06:55, 29 January 2009 (UTC)

We know that uncountably many real numbers exist. Whatever do you mean by enumerating an uncountable real number? --Trovatore (talk) 08:00, 29 January 2009 (UTC)

Specifically, I mean construct an example of a number in the set of real numbers which can not be reached via a countable enumeration of the elements. But since one can always append your new example to any existing covering, that number is self-evidently countable. And yet at the same there are also uncountably many number you missed through any countable covering. We can attempt to enumerate the real numbers in any multitude of countable ways but no matter what your approach, the nature of the problem is such that one can never construct an example of a specific number that a counter would be incapable of saying. Dragons flight (talk) 08:28, 29 January 2009 (UTC)

Perhaps a clearer way to say it is that we can prove that even given infinite time someone picking numbers at random from the unit interval must omit uncountably many of them, and yet it is also impossible to construct any examples of numbers that a random chooser is guaranteed to miss. Dragons flight (talk) 08:42, 29 January 2009 (UTC)

There are no numbers that a random chooser is guaranteed to miss. --Trovatore (talk) 09:26, 29 January 2009 (UTC)

Yes, I know, that's the point. The question asked for things which are known to exist but have not been constructed. It is known that there exist numbers that a random chooser will miss and yet such number can not be constructed. Dragons flight (talk) 10:42, 29 January 2009 (UTC)

There are no numbers that a random chooser is guaranteed to miss, constructible or not. The chooser will miss most numbers, but which ones they are is entirely random; no numbers are off-limits. Algebraist 15:07, 29 January 2009 (UTC)

Hence, why it is impossible to construct one... I think you are both being pedantic to the point of being annoying. The whole point of this diversion was to try and give the poster an example of things which exist that can not be constructed. Dragons flight (talk) 18:10, 29 January 2009 (UTC)

But you didn't give such an example.

We aren't being pedantic. This is important. It's very common to throw around terms like "definable" and "constructible" without really saying what one means, and this leads to all kinds of trouble. There are whole careers dedicated to studying different sorts of definability and constructibility. --Trovatore (talk) 18:27, 29 January 2009 (UTC)

Well either A) you have a notion of "constructible" that allows you to construct specific examples of numbers that fail to be reachable after a countably infinite number of random selections. (Which I'm pretty sure you don't. Since that would appear to be impossible by definition.) Or B) you decide such numbers are "not constructible". Or C) you have a notion of "constructible" that provides that such examples would be neither "constructible" nor "not constructible", in which case I think your notion of "constructible" is sufficiently silly that I no longer care. Dragons flight (talk) 20:02, 29 January 2009 (UTC)

Maybe we're talking past each other; I'm not sure. In context, responding to the question asked, it sounded like you think the elements you have not enumerated have some different character ("cannot be constructed") from the ones that you have.

But they don't; you just stopped too soon. You can't enumerate the reals in countably many steps because there are just simply too many of them; that's really all there is to it. It's like taking a set with six elements, and noting that if you pick any five of them you'll always leave one out (which of course is true), and then phrasing that as "there's an element of the set that can't be enumerated in five steps", which is nonsense. --Trovatore (talk) 09:17, 30 January 2009 (UTC)

Let S be the subset of the unit interval containing those numbers which a random chooser is guaranteed to miss. You appear to be claiming that we can prove that S is nonempty, but we cannot construct any elements of S. However, this is not true, for S is certainly empty. I think you are conflating "the set of numbers which a random chooser is guaranteed to miss" with "the set of numbers which a random chooser happens to miss on one particular trial". Eric. 131.215.158.184 (talk) 23:33, 29 January 2009 (UTC)

One example might be, if we assume the axiom of choice, then there exists a well ordering of the real numbers. But we could never construct such a well ordering. Another example: again assuming the axiom of choice, we can prove the existence of a set with no Lebesgue measure. Eric. 131.215.158.184 (talk) 10:25, 29 January 2009 (UTC)

But does really the OP refer to existence proofs in mathematics that cannot be made constructive, like the various applications of the axiom of choice quoted above? Another popular one is Banach-Tarski paradox, whose utopistic constructive form is sometimes popularized as "how to double a golden ball". It seems to me that the OP refers rather to existence results that only later were given a constructive proof, or that are still demanding for it. In this direction, one example among many, is the envy-free partition in fair division problem: finding an algorithm was considered one of the big problems of the 20th century (well at least by the fair-division people). But the purely existence result is an immediate application of a Lyapunov convexity theorem in measure theory. In fact many other examples are in all math: of course it's commonly easier to prove that an equation has a solution, that exibite it concretely. As we know this fact is an eternal source of jokes on pure mathematicians --only physicists usually find them funny.pma (talk) 11:26, 29 January 2009 (UTC)

Ah, that's a good example, I wasn't aware of the history behind fair division. Here's another example, from a friend of mine, of something that cannot be made constructive: there exists an even number greater than 2, such that if the Goldbach conjecture holds for that number, then it holds for all numbers (the proof is by contradiction). But obviously we aren't going to construct such a number. A similar claim can be made for Fermat's Last Theorem, etc., etc. Eric. 131.215.158.184 (talk) 20:31, 29 January 2009 (UTC)

The busy beaver function Σ(n) has a definite (positive integer) value for all n but no values are known for n>4, it is non-computable, and it is not even possible to compute a bounding function. Gandalf61 (talk) 11:30, 29 January 2009 (UTC)

At probabilistic method there are some examples of graph classes known to exist but not yet constructed. A different type of example: there is a real number which is not the output (as a decimal string) of any Turing machine. We can describe it via a diagonalization argument but we can't actually find it. McKay (talk) 11:33, 29 January 2009 (UTC)

I will assume here that "construct" means "define." In that case, 131's comment needs to be qualified, because if the axiom of constructibility is true then there is a definable well-ordering on R (and indeed on the universe). So 131's statement needs to be clarified. Joeldl (talk) 12:06, 29 January 2009 (UTC)

There is no definition of a well-ordering on R provable in ZFC. — Emil J. 13:19, 29 January 2009 (UTC)

That's true — but that could simply mean that there is a definable formula that wellorders R, but that ZFC is too weak to prove it. I sort of doubt this is actually the case, but I'm not convinced we have enough evidence to firmly rule it out. --Trovatore (talk) 09:59, 3 February 2009 (UTC)

we often use the proof of a number we dont actually construct as part of a proof by contradiction. for example you can use it in the proof that there are uncountably many real numbers. the number so constructed was made as follows. if you could count all the all the real numbers between 0 and 1

1. 0.something
2. 0.something
3. 0.something
4. 0.something

...
then you could:

• express the something parts in binary
• if any of them terminates, pad it with endless zeros afterwards.
• define a new number, 0.somethingnew whose:
  • 1st binary digit is the opposite of the 1st digit of 1
  • 2nd binary digit is the oppose of the second digit in 2

and so on forever.

Now you have constructed a new number that differs from EVERY real number between 0 and 1 that you had enumerated. It's not one of them. But it is a real number number between 0 and 1. So, you didn't enumerate every real number in that space after all, however you thought you did. This is a contradiction. As you can see, it involves defining a number that we never construct. (Since we are working in the general case -- we didn't actually ask how you enumerated these reals, we showed that you can't no matter what you do -- we couldn't have constructed it: our proof didn't assume enough information). —Preceding unsigned comment added by 82.120.227.157 (talk) 14:51, 29 January 2009 (UTC)

My favourite example is in computational graph theory. It follows from the Robertson–Seymour theorem that for any class of graphs closed under minors, there is an algorithm (in fact a cubic-time algorithm) for determining whether a given graph lies in the class. But in almost all cases (e.g. the case of embeddability in a surface of genus k>1) we can't actually construct such an algorithm. Algebraist 15:05, 29 January 2009 (UTC)

How about the Riemann mapping theorem? Michael Hardy (talk) 16:57, 29 January 2009 (UTC)

Yes... but of course any construction problem is meaningful if the data are well definite and constructible objects; in this case, at least in case when the boundary is rectifiable, the problem has a variational structure (it minimizes the energy among all H¹ maps $u\scriptstyle :({\bar {D}},\partial D)\to ({\bar {\Omega }},\partial \Omega )$ whose restriction to the boundary is a parametrization), so it can be treated algorithmically via gradient flow -which is maybe not a great construction, ok...--pma (talk) 17:41, 29 January 2009 (UTC)

Some Number Theory

Studying an introductory number theory book, the following problem is given in the very beginning of the book. (Only the GCD, LCM, the division, and the Euclidean algorithm are known.) The problem states that let a and b be positive integers such that $(1+ab)|(a^{2}+b^{2})$. Show that the integer $(a^{2}+b^{2})/(1+ab)$ must be a perfect square. I have tried several techniques from earlier exercises but this is at the very end so I am guessing that this is supposed to be the hardest one and everything seems to fail. Any hints on how to get started? Thanks!-Looking for Wisdom and Insight! (talk) 09:58, 29 January 2009 (UTC)

As a starting point, if nothing leaps to mind, looking for numerical examples can be of help. After hunting around a bit I found that $b=a^{3}$ is always a solution (check it). However, there certainly exist other solutions, such as (8, 30), (27, 240), (30, 112), (64, 1020), (112, 418), (125, 3120), etc., so that's not the end of the story. Unfortunately nothing else came to mind for me; hopefully someone else has pointers. Eric. 131.215.158.184 (talk) 05:13, 30 January 2009 (UTC)

Consider the sequence of polynomials

$p_{0}=0$

$p_{1}=x$

$p_{2}=x^{3}$

$p_{3}=x^{5}-x$

$p_{4}=x^{7}-2x^{3}$

$p_{n}=x^{2}p_{n-1}-p_{n-2}$

Then

$p_{1}^{2}+p_{0}^{2}=x^{2}=x^{2}(p_{1}p_{0}+1)$

$p_{2}^{2}+p_{1}^{2}=x^{6}+x^{2}=x^{2}(p_{2}p_{1}+1)$

$p_{3}^{2}+p_{3}^{2}=x^{10}-x^{6}+x^{2}=x^{2}(p_{3}p_{2}+1)$

and we can prove by induction that

$p_{n}^{2}+p_{n-1}^{2}=x^{2}(p_{n}p_{n-1}+1)$

So if k is any integer greater than 1 then we can take a and b to be any two consecutive terms from the sequence $p_{n}(k)$ to get a solution to

$a^{2}+b^{2}=k^{2}(1+ab)$

For example, with k=2 we have the sequence 2, 8, 30, 112, 418, 1560, 5822 ... and with k=3 we get 3, 27, 240, 2133 ... This gives us a whole bunch of pairs (a,b) such that $(1+ab)|(a^{2}+b^{2})$, and, indeed, $(a^{2}+b^{2})/(1+ab)$ is a perfect square in each case. What I can't (yet) do is prove that these are the only solutions. Gandalf61 (talk) 12:27, 30 January 2009 (UTC)

Spoiler! I found the solution. The hunt for the solution was a bit of an adventure: The way I found this solution was using Gandalf's K=2 sequence and looking it up in the Online Encyclopedia of Integer Sequences (This is a great resource which has often helpped me by allowing me to brute force the first couple solutions of a problem and then find more information about that list of numbers) which gave me sequence A052530 which linked to sequence A061167 which contained the link to the St Andrews Colloquium. Anythingapplied (talk) 18:22, 30 January 2009 (UTC)

(edit coincidence) Very good. And you can start your method from any solution pair, and you can also go down, till you reach the solution (0,k), showing that the solution pair was indeed listed in your sequence (it is the same argument for the recently posed question on markov numbers). I made a search on your polynomials and I also found this: [1], where a discussion of this and other problems is given. --pma (talk) 18:25, 30 January 2009 (UTC)

May I ask how you searched for the polynomials? Google tends to not be great when trying to look up math equations since most symbols are reserved for other purposes. Did you use something else or is there a trick to it? Anythingapplied (talk) 18:47, 30 January 2009 (UTC)

I wish I knew such a trick... No, I started from Chebyshev polynomials and quadratic diophantine, untill I had the same idea as you and reached the link via the Sloane encyclopedia, directly from the k=2 sequence. It's very efficient, yes, you did well to remark it. IN fact it also works for a search of polynomials, using the finite sequence of coefficients. (NOte: I was asked exactly this problem like 10 years ago or so, and I solved on sight. In the last nights, I passed hours trying to remember something, and experienced a horrible sense of darkness!) pma (talk) 21:18, 30 January 2009 (UTC)

Awesome guys, thanks for all this info. No wonder I couldn't crack it. Even in the IMO it was a double starred problem.-Looking for Wisdom and Insight! (talk) 21:10, 31 January 2009 (UTC)

Sum of interior angles of polygons constant - proof?

It is well known that the sum of the interior angles of a triangle equals 180°, and (n-2)·180° for n-sided polygons. I've been unable to find a proof for that. If I had a proof for the sum of interior angles being constant, then the former can easily be deducted. I bet this is an easy one :) - Keta (talk) 19:03, 29 January 2009 (UTC)

The classic proof is in Euclid's Elements (online edition in English here), book I, proposition XXXII. Of course, it doesn't hold in non-Euclidean geometry, where the interior angles can sum up to less or more than 180°. -- Jao (talk) 19:27, 29 January 2009 (UTC)

That's the triangle case. Where does Euclid do the n-gon? I think he does it by cutting the n-gon into n-2 triangles, but I'm not sure. Algebraist 19:40, 29 January 2009 (UTC)

Yes. Corollary 4, just below. -- Jao (talk) 19:49, 29 January 2009 (UTC)

That only covers a special case, though. Does he prove the result for general n-gons? Algebraist 19:54, 29 January 2009 (UTC)

Ah, you mean the construction only works for convex polygons? That does seem to be a problem. -- Jao (talk) 20:04, 29 January 2009 (UTC)

Well, for polygons where one vertex can see all the others. That covers some but not all of the nonconvex case. Algebraist 20:07, 29 January 2009 (UTC)

After you've done the triangle, you can take care of the n-gon by mathematical induction: Every time you add a new side to the polygon, you add another 180 degrees. Or in other words, just triangulate the polygon and add up the angles in all the triangles. Michael Hardy (talk) 23:16, 29 January 2009 (UTC)

Yes, I was just wondering how Euclid proved that polygons can always be triangulated in this way. Algebraist 23:17, 29 January 2009 (UTC)

Thanks for the answers guys — fully satisfied with the info :) - Keta (talk) 21:59, 30 January 2009 (UTC)

Numerical error

Plot showing the numerical error

I am solving a dynamics problem using the Newmark method, and I am getting this strange numerical error (see figure on right) for some reason. For the parameters I am using, I should get a straight horizontal line as the solution, but I am getting these periodic pulses in the solution. The number of pulses does not depend on the total time for which I am doing the integration, but on the number of discretization points (1/Δt). What could be causing this problem? Any ideas would be highly appreciated. Please ask me if further information is needed to answer this question. Thanks! deeptrivia (talk) 22:38, 29 January 2009 (UTC)

Are you doing these calcs on computer ? If so, one thought is that, if Δt is too small, precision errors in the program may start to creep in, especially if the variables used are of low precision. StuRat (talk) 05:17, 30 January 2009 (UTC)

Does look a strange one. The more accurate finite difference methods tend to fall prey to numerical stability more often, but what usually happens is that the error grows exponentially going ping pong up and down. The error in yours does look like it is starting to grow exponentially each time, then it gets to a limit and reverses itself down, then after a while starts up again. Dmcq (talk) 12:08, 30 January 2009 (UTC)

sin(G)

Is there any way to estimate the sine of Graham's number (or has this been done)? It seems to me like it is too big a number to even be able to determine whether its sine is positive or negative, let alone calculate even a roughly accurate value for its sine. —Preceding unsigned comment added by 65.31.80.94 (talk) 23:14, 29 January 2009 (UTC)

I offer the strict (because G is rational) bounds -1 < sin(G) < 1. Anyone who wishes to strengthen this theorem is welcome to calculate the log₁₀(G) digits of π required to determine if it is positive or negative. Fredrik Johansson 10:28, 30 January 2009 (UTC)

January 30

Calc 1 Question

Ok, although this was a homework question its already been handed in and was someone else's and has now gained the intrigue of two math majors. The question was to algebraically solve 0=e^x-3x At the calc one level (i.e. derivatives is as far as they have gone). As far as we can tell, it cannot be solved for algebraically, but it can be solved for graphically. How would you solve it algebraically? --Omnipotence407 (talk) 00:57, 30 January 2009 (UTC)

I'm not quite sure what you mean by solving it algebraically, but I don't think it's possible. Certainly the solution is solutions are not an algebraic number algebraic numbers (by the Lindemann–Weierstrass theorem say). It They can be expressed in terms of the Lambert W function, as −W(−1/3), but I doubt that counts as an algebraic solution. Algebraist 01:06, 30 January 2009 (UTC)

There are two solutions, so the expression in terms of Lambert's W is not the end of the story. Michael Hardy (talk) 01:14, 30 January 2009 (UTC)

Oops, sorry. I wrote W₀ (the single-valued version) when I should have written W for the multi-valued version. The solutions of the given equation are the two real values of −W(−1/3). Algebraist 01:22, 30 January 2009 (UTC)

Sorry, forgot this piece. 0≤x≤1 --omnipotence407 (talk) 01:25, 30 January 2009 (UTC)

Then you do want W₀, for what it's worth. Algebraist 01:29, 30 January 2009 (UTC)

The equation has two real and infinitely many nonreal complex solutions. A J expression for the solutions is this: p.(^-3&*)t.i.9 . The transcendental function ^-3&* is expanded as a taylor series, t. . The first 9 coefficients, i.9 , are used for an approximation, and 8 roots are computed, p. . The real solutions are approximately 0.6190 and 1.512. Bo Jacoby (talk) 12:39, 30 January 2009 (UTC).

Common name for spheres and circles.

I could loosely describe either a rectangle or a cuboid as a 'box'. But I'm looking for an analogous word that could be used to describe both circles and spheres (and maybe also 4D hyperspheres). Our Sphere article offers 'n-Sphere' which is kinda nasty - is there anything better than that? SteveBaker (talk) 14:33, 30 January 2009 (UTC)

How about Ball? Strictly speaking it should be 'n-ball', but plain 'ball' should be clear enough in context, and at least it's easier to pronounce than 'sphere'. I suppose it depends who your audience is... AndrewWTaylor (talk) 16:44, 30 January 2009 (UTC)

If you're talking to mathematicians, though, remember that to us, balls and spheres are completely different things. Algebraist 16:46, 30 January 2009 (UTC)

Yes, as a (former) mathematician I realise that, but it looks as if Steve wants a fairly informal term. AndrewWTaylor (talk) 18:05, 30 January 2009 (UTC)

"Round" can mean circular, spherical, or many other shapes which lack sharp corners, edges, and planar faces (such as ellipses and ovals). StuRat (talk) 18:00, 30 January 2009 (UTC)

Yeah - this is for naming a type in a computer program - so precision of language isn't everything - I just know that if I start talking about 2D spheres it's going to mess with everyone's head. I'm aware of the distinction between a ball and a sphere (the former being the solid inside and the latter being just the surface) - analogous to disk and circle. But my data structure is a center coordinate and a radius and a whole bunch of tests for overlaps and inclusions and such - I'm being deliberately vague about whether it's solid or not because it's going to be used in both situations. So what I REALLY STRICTLY want is a common name for disk,circle,ball,sphere,hyperball,hypersphere and perhaps even the 1D analogs of those things. I guess "Round thingy" doesn't cut it mathematically then? SteveBaker (talk) 18:55, 30 January 2009 (UTC)

I can't help but be reminded of this - "Hey! What's this thing suddenly coming towards me very fast? Very very fast. So big and flat and round, it needs a big wide sounding name like ... ow ... ound ... round ... ground! That's it! That's a good name - ground! I wonder if it will be friends with me?" --LarryMac | Talk 20:55, 30 January 2009 (UTC)

Both "disk" and "ball" are used in any dimension and norm. To be more precise one can say: "euclidean ball" or also " n-dimensional euclidean ball". PS: LarryMac: aren't you talking about The Prisoner, with Patrik McGoohan, are you? --pma (talk) 21:46, 30 January 2009 (UTC)

Not The Prisoner, sorry, I should have put a link in there - List of minor characters from The Hitchhiker's Guide to the Galaxy#Whale --LarryMac | Talk 21:50, 30 January 2009 (UTC)

In computereeze, how about "round object" ? StuRat (talk) 15:52, 31 January 2009 (UTC)

By the way, while today "sphere" sounds quite more technical than "ball", let's recall that the Greek term just coincides with "ball" both in the ordinary sense and in the mathematical one. pma (talk) 17:50, 31 January 2009 (UTC)

The mathematical terms are "ball" and "sphere", you're not going to get any better answer than that on the maths desk. Try the language one! --Tango (talk) 23:41, 31 January 2009 (UTC)

The computing desk might also be appropriate. If you're designing a program to simulate moving objects, represented by circles and spheres, I'd call them "balls". In three dimensions both the surface and the solid can be reasonably called that, and in two dimensions (thanks to cartoons probably) it's natural to extend the terminology downward. If I were programming a sidescroller game, for instance, nobody would complain about a two-dimensional representation of a baseball being a "ball". Natural language hasn't yet produced a word that applies to four dimensions, but I don't think it would hurt to use it there as well. All that matters is that it's readily comprehensible to someone reading your code. Black Carrot (talk) 17:22, 1 February 2009 (UTC)

Rotation along a certain point...

What's the formula used to rotate any given point (2,3) for example, around another point (5,4) any number of degrees in any direction???? Thanks, Saebjorn (talk) 14:47, 30 January 2009 (UTC)

Translate your center of rotation to the origin, rotate around the origin using the formula in rotation (mathematics)#Two dimensions, and translate it back. — Emil J. 15:02, 30 January 2009 (UTC)

You can compose this (in fact any) sequence of translations and rotations into a single matrix using affine transformations. This is very widespread in 3d computer graphics, and also works fine in 2d spaces. —Preceding unsigned comment added by 84.187.97.24 (talk) 01:28, 31 January 2009 (UTC)

Curiosity has revived the calculus - A math homework from yesteryear

Resolved

208.92.185.250 (talk) 21:32, 30 January 2009 (UTC)

This is a nostalgic trip back to a math problem presented in a graduate applied math course in my youth, forty years ago. My professor for that course had a had a lecturing style which can only be be described as succinct. At the front of our lecture hall, there was a stage in front of which were we students, arrayed in orderly rows to the back of the hall. Behind the stage, the wall was a continuous twenty feet of blackboard, and, on either end, there was a door to the exterior hall. The professor would arrive precisely on the hour, walk through the left-hand door, pick up his chalk and begin lecturing (which invariably, was the detailed and annotated solution of a rather mundane math problem but always with an engaging unexpected, even ironic, twist). He would develop his problem and its solution - at all times addressing the blackboard - never acknowledging us, his students. At the conclusion of the lecture, invariably 47 or 48 minutes after his entry, he would lay down the chalk, open the right-hand door and disappear into the hall, not to be seen until the next class.

One particular lecture lives forever in my memory, but I can no longer locate the notebook from that class, and I cannot reproduce the solution - a lapse which continues to haunts me. Perhaps someone here can take the time to outline an approach or provide a solution.

Statement: If a known mass is elevated above a elastic surface and let to drop a known distance, the mass will, of course, acquire momentum during its fall: upon impact with the resilient surface, it will decelerate to zero velocity as a function of the momentum thus absorbed and by the “stiffness” of the surface (expressed as Young's modulus, or, maybe, bulk modulus). The value for the momentum can be calculated from the mass' initial height and from the local gravitational constant.

Further, the interaction with the rebounding surface will cause the mass to decelerate rapidly as the surface distorts to absorb the the transferred momentum until the masses velocity is reduced to zero. The compression caused to the surface during this process is, also, easily calculable.

The question poised, that day forty years ago, was – how much is the apparent weight of the mass at the moment that its velocity is reduced to zero. The answer is to be found in the force that it takes to deflect the elastic material a distance equal to that which it takes the mass to decelerate to zero velocity.

As an extension and through intuition, it is obvious that the apparent weight will be greater if the weight is elevated, say, a meter above the surface compared to the apparent weight when the mass is elevated a centimeter. The “hooker” that day was that, as the professor exited the lecture hall, his last remark was, to the effect -”Gentlemen, please submit as your homework assignment, a solution to this problem when the distance the mass falls is set to zero.”

Thus, this was a pretty standard applied materials problem until the introduction of the “hooker”. In the ensuing discussion, we, the students, reformulated the problem by recasting of the rebounding surface to that of the deck supported by a coiled spring, the deck in turn being “geared” to a pointer to register the degree of spring compression, (analogous to, say, a weighing scale such as typically found in a grocer's shop). Then, if the mass is released above the deck, the momentum it acquires will be “absorbed” in the compression of the spring of the scale mechanism, and when the mass reaches minimum velocity, the scale's pointer will register maximum “weight”. Within the bounds of this model, we then asked the question - if the mass were “raised” off the deck until the “scale” registered zero weight, but remains in contact with the surface, then released, what is the maximum weight the pointer will register.

I dimly recall that the solution was found in taking the limit of the function controlling the deceleration and that the answer was that the apparent “weight” of the mass was precisely twice the rest weight, but I can no longer reproduce the steps to arrive at this conclusion.

Any suggestion or comment would be appreciated. TIA —Preceding unsigned comment added by 208.92.185.250 (talk) 17:58, 30 January 2009 (UTC)

Initially the particle is at rest at height z₀>0. The particle hits the spring at z=0. The particle is again at rest when it has compressed the spring to the maximum at z₁<0. The spring force on the particle, F = −kz₁, is the 'weight'. The conserved energy of the particle is mgz₀ = mgz₁+(1/2)kz₁². The negative solution defines z₁ as a function of m, g, z₀, and k. For z₀=0 you get z₁=−2mg/k and F=2mg. This is independent of k, which is nice, because the value of k was not provided. What was the name of that professor? Bo Jacoby (talk) 21:00, 30 January 2009 (UTC).

Thanks Bo Jacoby. I do not remember the professor's name but have contacted the university with a request for help in his identification. He was a fully tenured professor in the math dept at Rutgers Univ (New Brunswick) in the early 1960's. I believe that he retired in 1967 or 1968. It was widely rumoured that he often collaborated with Einstein, so teaching us first year graduate students must not have been very challenging.208.92.185.250 (talk) 21:32, 30 January 2009 (UTC)

January 31

formula to calculate pi

this formula which has been given by the great indian mathematician srinivasa ramanujam, is the cause of my headache. i am an indian and without understanding this formula, i won't be a true indian. please explain this for a ninth grade student ${\frac {1}{\pi }}={\frac {2{\sqrt {2}}}{9801}}\sum _{k=0}^{\infty }{\frac {(4k)!(1103+26390k)}{(k!)^{4}396^{4k}}}\!$ and this also ${\frac {426880{\sqrt {10005}}}{\pi }}=\sum _{k=0}^{\infty }{\frac {(6k)!(13591409+545140134k)}{(3k)!(k!)^{3}(-640320)^{3k}}}\!$

Do you just want an explanation of the meaning of the terms in the formulae, or do you want an explanation of why the formulae are actually true? The first is substantially easier than the second. Algebraist 01:49, 31 January 2009 (UTC)

Even if I did understand Ramanujan's formulas I would probably not be a true indian. :-) . The formula components are the summation sign Σ , the square root sign √ , the factorial sign ! , and the exponentiation notation, and the Pi symbol π. Click on the colored words to find explanations. You may like to check the formulas numerically, for instance using the J (programming language). The left hand side is

  %o.1
0.31831

and five terms of the right hand side is

  ((2*%:2)%9801)*+/(!4*k)*(1103+26390*k)%(!k)*296^4*k=.i.5
0.31831

the terms of the sum decreases rapidly:

  ((2*%:2)%9801)*(!4*k)*(1103+26390*k)%(!k)*296^4*k=.i.5
0.31831 2.48051e_8 5.31965e_15 4.08817e_21 7.72733e_27

Bo Jacoby (talk) 08:00, 31 January 2009 (UTC).

Some further remarks: The first formula was discovered by Ramanujan in 1910 and published in 1914, with no proof, as his abitude. The first proof appeared in a 1987 book by Jonathan and Peter Borwein, where they reconstructed Ramanujan work on the subject. The other formula is not due to Ramanujan, but to the Chudnovski brothers, following Ramanujan's method. The point of both formulae is that they converge very quickly, as Bo Jacobi shows here above. For instance 4 terms of the second series already give pi with more that 50 correct decimal digits. It was used by its authors in the '80s to compute 4 billions digits of pi. (If you ask why, it seems that computing digits of pi is a sort of competition, I do not know what is the current record; it was over 1 trillion around 2000 by the japanese mathematician Kanada and his team). pma (talk) 17:26, 31 January 2009 (UTC)

Euler's proof that god exists

This is an excerpt from wiki's page on Leonhard Euler:

There is a famous anecdote inspired by Euler's arguments with secular philosophers over religion, which is set during Euler's second stint at the St. Petersburg academy. The French philosopher Denis Diderot was visiting Russia on Catherine the Great's invitation. However, the Empress was alarmed that the philosopher's arguments for atheism were influencing members of her court, and so Euler was asked to confront the Frenchman. Diderot was later informed that a learned mathematician had produced a proof of the existence of God: he agreed to view the proof as it was presented in court. Euler appeared, advanced toward Diderot, and in a tone of perfect conviction announced, "Sir, ${\frac {a+b^{n}}{z}}=x$, hence God exists—reply!". Diderot, to whom (says the story) all mathematics was gibberish, stood dumbstruck as peals of laughter erupted from the court. Embarrassed, he asked to leave Russia, a request that was graciously granted by the Empress. However amusing the anecdote may be, it is apocryphal, given that Diderot was a capable mathematician who had published mathematical treatises.

"Sir, ${\frac {a+b^{n}}{z}}=x$, hence God exists."

What does that mean? Was it just a joke by Euler to humiliate Diderot, and has no meaning, or does it mean something else? Because no matter how many times I try to interpret it, the sentence makes no sense to me. Could anyone clarify this for me?? Johnnyboi7 (talk) 05:36, 31 January 2009 (UTC)

Isn't this anectode taken from Bell's booklet "Men in mathematics"? I've been also puzzled with its meaning, but first I'd like to check the sources. If it's just the author's version, as I suspect, things could be much different, and easier. I'll have a look to the article you quoted...--pma (talk) 07:34, 31 January 2009 (UTC) PS: Bell quotes the anecdote from De Morgan's book "A budget of paradoxes".PPS: I have read the interesting Gillings' article you mentioned [2], where it is reported the tradition of the anecdote, going back to Thiébault's version (1804) that is probably the origin. (Note that initially the denominator had n instead of z, not that this makes less nonsensical the use of the formula). --pma (talk) 09:03, 31 January 2009 (UTC)

Now that you have convinced me that God exists. I'm going to confront Richard Dawkins and say to his face.

"Sir, ${\frac {a+b^{n}}{z}}=x$, hence God exists."

The look on Richard Dawkins face when I say this will be priceless. 122.107.205.162 (talk) 10:24, 31 January 2009 (UTC)

The difficult thing is to organize everything as a public event, like Catherina the II was able to do, according to the anecdote. Of course, she was particularly interested, as it was God who gave her the kingdom of all the Russias (according to her and to her colligues kings all around Europe). Euler himself was interested, as he got a salary from Catherina (and not by the French Republic, which paid for Diderot, of course). Just to say that nothing has really changed so much. For instance, today a mathematician could gain some founds proving with mathematical authority that Economy is governed by suitable mathematical rules (is it so different from (a+bⁿ)/z=x?). Personally, I suspect that it's always been the same story, in paleolithic times, in Catherina times, and today: just brute force proofs, so to speak. --pma (talk) 16:37, 31 January 2009 (UTC)

To answer the original question, Euler's alleged quote is indeed nonsense. The mathematical sentence $(a+b^{n})/z=x$ is a statement regarding the relationship between a, b, n, z, and x, but in this context these variables have no meaning, so the sentence is meaningless. It would be as if I had said "I am writing a novel where Jane's second cousin is named Richard" -- I've told you absolutely nothing about my novel, because you don't know anything about Jane and Richard, so knowing that they are second cousins is useless. Eric. 131.215.45.82 (talk) 23:32, 31 January 2009 (UTC)

Indeed and, since it is just nonsense, there is no possible logical retort (other than pointing out that it's just nonsense, which is never very impressive). --Tango (talk) 00:42, 1 February 2009 (UTC)

Well we can't reject a proof as nonsense just because you do not understand it. Let say: "the author shoud fill some passage that is not completely clear, or provide a refernce for it. Some variables are not defined. The existence result is in any case quite poor as it is, and demands for further properties of the found solution". pma (talk) 01:44, 1 February 2009 (UTC)

So the ruling monarch comes to you and says - "The guy Diderot is a pain in the ass - please prove to him that God exists so he'll shut up about the atheism thing."...Well, when the ruling monarch tells you to do something - it's generally a good idea to put other matters aside and attend to it right away. Euler knows he can't prove the existence of God - but he knows that Diderot doesn't know squat about math - so he writes down any old mathematical-looking gibberish that's sufficiently complicated that nobody is going to argue about it and challenges Diderot to prove that it's NOT true. With the onus suddenly on him - and with no knowledge whatever about math - poor Diderot can neither challenge nor disprove Euler's assertions. This is nothing to do with math or religion - it's a piece of clever social engineering. SteveBaker (talk) 05:36, 1 February 2009 (UTC)

We have to rely on mathematicians when we do not understand a mathematical proof. So it is sad if Euler really testified that he had a proof of God's existence. I do not understand Andrew Wiles' proof of Fermat's last theorem, and perhaps some day it will be published that an error in Wiles' proof is found and that Fermat's theorem is false after all. SteveBaker's argument applies to Wiles as well: more is earned by providing a proof than by trying in vain. Bo Jacoby (talk) 18:12, 1 February 2009 (UTC).

Localization Morphism

Resolved

 – Shahab (talk) 17:09, 31 January 2009 (UTC)

Hello all. I am currently reading a book on rings. Unfortunately the book is assuming a lot of basic knowledge which I possess only in bits and parts. In an example for showing that a morphism between local rings is not necessarily a local morphism (i.e. doesn't map the maximal ideal into the maximal ideal) the book says:

Let A be a local ring with a prime ideal P, such that $P\subset M\subset A$ where M denotes the unique maximal ideal. If we denote by $\phi$ the localization morphism with respect to the multiplicative system S=A\P, then $\phi :A\to A_{P}$ is not local.

My problem is that I do not understand what is a localization morphism (how it is defined, and what idea it conveys). I believe that $A_{P}$ has S^-1P as its unique maximal ideal but thats about all I understand here. Any help will be appreciated.--Shahab (talk) 14:35, 31 January 2009 (UTC)

A_P is the ring formed from A by adding multiplicative inverses for every element not in P. Thus the elements are of the form a/b for a in A and b in A\P. There is a natural map from A into A_P sending a to a/1. This is the localization map. The canonical example is to take A to be the integers and P={0}. Then A_P is the rationals and the localization map is the natural embedding of Z into Q. Algebraist 15:22, 31 January 2009 (UTC)

Thanks. I don't quite understand your first sentence though. Shouldn't it be: A_P is the ring formed from A by multipliying A by multiplicative inverses for every element not in P. Also if m is in P it is going to be mapped to m/1. In that case it is not a unit and so possibly in the maximal ideal of A_P. How can I conclude that the morphism isn't local?--Shahab (talk) 15:45, 31 January 2009 (UTC)

Perhaps 'adjoining' would have been better than 'adding'. On your last point, yes of course P is mapped into the maximal ideal of A_P. The point is the M is not (we're assuming here that M≠P). Algebraist 15:50, 31 January 2009 (UTC)

An explicit example is A = { a/b : b odd, a,b in Z }, P = 0, M = 2A = { a/b : a even, b odd, a,b, in Z }, A_P = Q = { a/b : b nonzero, a,b in Z }, and φ : A → A_P : x → x. The maximal ideal of Q is 0, and the image of M under φ is much larger than 0. JackSchmidt (talk) 16:14, 31 January 2009 (UTC)

Thanks--Shahab (talk) 17:09, 31 January 2009 (UTC)

elementary graph theory

(This isn't the question... just background...) I was looking at a problem that introduced dropping a node from a connected graph and ensuring that the connected graph is still connected. My first thought was to have the dropped node add connections between all neighboring nodes, but that will be nasty if there are, say 100 neighbors. So, I thought about the minimum number of edges required to ensure connectivity. For 1 neighbor, no edges. For 2 neighbors, 1 edge. For 3 neighbors, 2 edges. For four neighbors, 3 edges (wrong - I noticed it is 4 later). So, it is n-1. Well, I then thought about five neighbors. It takes 7 edges to ensure connectivity. Then, I realized that I don't need this at all and went on to the next step.

(This is the question...) Is there a common proof for the minimum number of edges required to connect n nodes? I don't need it, but now the idea is stuck in my head and I have a lot more pressing things to work on. -- kainaw™ 22:23, 31 January 2009 (UTC)

There are many ways of showing that n-1 edges are required to connect n vertices. I don't know which, if any, is most common. The only textbook I have to hand is Bollobás's Modern Graph Theory, which does it by observing that the two algorithms for finding spanning trees he's given obviously make a graph with n-1 edges. Algebraist 22:33, 31 January 2009 (UTC)

I know that n-1 edges can create connectivity, but that doesn't ensure connectivity. With 4 nodes, ABCD, I can have vectors AB, BC, AC. That is n-1, but it is not connected. I must have 4 vectors to ensure connectivity. With 5 nodes, I must have 7 vectors. I was wondering about ensuring connectivity. -- kainaw™ 00:52, 1 February 2009 (UTC)

Oh, I see. We had that question here a little while ago. The answer is that with the disconnected n-vertex graph with most edges is a complete graph on n-1 of the vertices with an extra isolated vertex. Thus (n-1)(n-2)/2+1 edges are required to ensure connectivity: not much of an improvement on having all edges. Algebraist 00:57, 1 February 2009 (UTC)

Previous discussion. Algebraist 00:58, 1 February 2009 (UTC)

Thanks. Luckily, all the nodes in the program I was writing have unique IDs. So, when dropping a node with n neighbors, I only need n-1 vectors. If I consider the neighbors a line of unconnected nodes and put a vector between each pair along the line, I've ensured connectivity. The unique IDs makes it very easy to do that. -- kainaw™ 01:10, 1 February 2009 (UTC)

For completeness, it should be noted that you are assuming the graph is simple - that is, contains no self-loops or duplicated edges. --Tango (talk) 20:56, 1 February 2009 (UTC)

Modular arithmetic/number theory

Hi. For some work I'm doing, I need to work with PSL(3,2ⁿ). Trying to find the centre of SL(3,2ⁿ) involves solving the equation

$x^{3}\equiv 1{\pmod {2^{n}}}$

which some straight calculations for low n show to be only true for x=1. Is this true for any n? I was never any good at number theory...SetaLyas (talk) 23:28, 31 January 2009 (UTC)

Yes. The group of units of Z mod 2ⁿ has 2^n-1 elements, so every element has order a power of 2. If x³=1, then x must have order dividing 3, so the order must be 1, so x is 1. Algebraist 23:30, 31 January 2009 (UTC)

Wow, thanks for the speedy answer! SetaLyas (talk) 00:21, 1 February 2009 (UTC)

Vectors & Potentials

A particle at position vector r experiences a force (a(R⁻³) + b(R⁻⁴))r, where R=|r|. How would one find the function (say V(r)) of the potential in this case? I know with 1-D work done it's just the integral of F with respect to x, but how does one approach it in 3 dimensions? Can you simply integrate the function before the r with respect to R and ignore the r to get -(a/2(R⁻²) + b/3(R⁻³))r? It seems horribly wrong to me to just ignore the position vector in the integral but I'm not honestly sure where to go otherwise - what's the correct method?

On a similar vector-y calculus-y note, how would one differentiate 1/R with respect to time, when again R=r for the position vector?

Thanks for the help,

131.111.8.104 (talk) 23:40, 31 January 2009 (UTC)Zant

While the force is a vector, the potential is a scalar. As the force is in the radial direction and numerically depend on the distance only and not on the direction, it is the negative gradient of a potential which also depend on the distance only. The potential −a/(2R²)−b/(3R³) will do. Bo Jacoby (talk) 08:47, 2 February 2009 (UTC).

Sorry about the terse response. Unless I am confused (a strong possibility), the first question has no solution for nonzero b. See conservative vector field and scalar potential for information. See scalar potential#Integrability conditions for how to calculate the potential -- it requires a line integral.

For the second question, use the chain rule. You may find $R=|r|={\sqrt {\mathbf {r} \cdot \mathbf {r} }}$ helpful. (See dot product#Derivative for taking the derivative of a dot product.) Eric. 131.215.158.184 (talk) 09:38, 2 February 2009 (UTC)

February 1

Power Series Expansion

I'm trying to find an approximation for the following formula: $kT=(1+\epsilon )(1-e^{-kT})$

and I want to "obtain by expanding kT as a power series in $\epsilon$", the approximation $kT\approx 2\epsilon (1-{\frac {\epsilon }{3}})$

I've expanded the exponent which I presume is the correct approach but there are so many ways it seems I could go from there - dividing by kT, subtracting kT from each side and so on. I've got as close as $kT\approx 2\epsilon (1-{\frac {\epsilon }{2}})$ but no better - can anyone see what I should be doing? On an unrelated formatting note, how comes my first LaTeX formula is smaller than the other two?

Thanks,

Spamalert101 (talk) 00:02, 1 February 2009 (UTC)Spamalert

The first formula is smaller because it only contains simple symbols so can be displayed in HTML. The others contain more complicated symbols (the fractions, probably), so have to be done as an image, which for some reason is always bigger. I'm a little confused by your main question, though, have you copied out the first formula incorrectly? That formula expanded as a power series in epilson is simply $kT=(1-e^{kT})+(1-e^{kT})\epsilon$, no approximation. I can't see any way you can get any higher powers from that formula - it's just a polynomial in epsilon (with complicated, but constant, coefficients). --Tango (talk) 00:40, 1 February 2009 (UTC)

Tango, I think you got confused. You need to solve for kT. That means you shouldn't have kT anywhere on the right side; it should appear only on the left side. You can't do that in closed form, but you can give as many terms of the power series as you want. See below.... Michael Hardy (talk) 02:41, 1 February 2009 (UTC)

OK, start with

$u=(1+\varepsilon )\left(1-e^{-u}\right).\,$

Separate the two variables:

${1 \over 1+\varepsilon }={1-e^{-u} \over u}.$

Expand both sides as power series:

$1-\varepsilon +\varepsilon ^{2}-\varepsilon ^{3}+\cdots =1-{u \over 2}+{u^{2} \over 6}-{u^{3} \over 24}+\cdots .$

Differentiate with respect to ε:

$-1+2\varepsilon -3\varepsilon ^{2}+\cdots =\left({-1 \over 2}+{u \over 3}-{u^{2} \over 8}+\cdots \right){du \over d\varepsilon }$

Since u = 0 when ε = 0, setting ε to 0 gives us

$\left.{du \over d\varepsilon }\right|_{\varepsilon =0}=2.$

Differentiating again (applying the product rule to the right side), we get

$2-6\varepsilon +\cdots =\left({1 \over 3}-{u \over 4}+\cdots \right)\left({du \over d\varepsilon }\right)^{2}+\left({-1 \over 2}+{u \over 3}-{u^{2} \over 8}+\cdots \right){d^{2}u \over d\varepsilon ^{2}}$

When ε = 0 then u = 0 and du/dε = 2, so we have

$2={1 \over 3}\cdot (2)^{2}+\left({-1 \over 2}\right)\left.{d^{2}u \over d\varepsilon ^{2}}\right|_{\varepsilon =0}.$

Therefore

$\left.{d^{2}u \over d\varepsilon ^{2}}\right|_{\varepsilon =0}={-4 \over 3}$

The power series we seek is

$u=a+b\varepsilon +{c \over 2}\varepsilon ^{2}+{d \over 6}\varepsilon ^{3}+{e \over 24}\varepsilon ^{4}+\cdots$

where a, b, c, d, e, ... are the values of the 0th, 1st, 2nd, 3rd, 4th, ... derivatives of u with respect to ε at ε = 0. So a = 0, b = 2, c = −4/3, and that gives us

$u=2\varepsilon -{2 \over 3}\varepsilon ^{2}+{\text{higher-degree terms}}=2\varepsilon \left(1-{\varepsilon \over 3}\right)+{\text{higher-degree terms}}.$

Michael Hardy (talk) 01:51, 1 February 2009 (UTC)

The general theory behind Michael's approach can be found at Lagrange inversion theorem. Another approach is to find a contraction mapping. Start from Michael's step

$1-\varepsilon +\varepsilon ^{2}-\varepsilon ^{3}+\cdots =1-{u \over 2}+{u^{2} \over 6}-{u^{3} \over 24}+\cdots .$

Rearrange it like this:

$u=2\varepsilon -2\varepsilon ^{2}+2\varepsilon ^{3}-\cdots +{u^{2} \over 3}-{u^{3} \over 12}+\cdots .$

Call the right side F(u). Now define a sequence

$u_{0}=0;\quad u_{1}=F(u_{0});\quad u_{2}=F(u_{1});\quad \cdots$

You will find the sequence converges at least one term per iteration. McKay (talk) 09:32, 1 February 2009 (UTC)

On the LaTeX issue: try \textstyle and \scriptstyle. --pma (talk) 15:01, 2 February 2009 (UTC)

Number sequence help

What is the next number in this sequence (thankfully this isn't homework)

1 11 21 1211 111221 312211 ?

thanks —Preceding unsigned comment added by 70.171.234.117 (talk) 07:59, 1 February 2009 (UTC)

Haha. This is an old riddle. Each sequence is describing the previous sequence. How would you say 111221? It has three ones, two twos, and one one. Which gives you 312211. If it is any consolation I had to have this one explained to me too when I first saw it. Anythingapplied (talk) 08:38, 1 February 2009 (UTC)

wow my friend really did me over then... this isn't even a mathematical sequence at all -__-

Not sure why it's disqualified from being mathematical, but the sequence does appear in the Encyclopedia of Integer Sequences. --Ben Kovitz (talk) 22:18, 1 February 2009 (UTC)

just for fun, what would be a quadratic function that would actually produce:

f(1) = 1 f(2) = 11 f(3) = 21 f(4) = 1211 f(5) = 111221 f(6) = 312211

? —Preceding unsigned comment added by 70.171.234.117 (talk) 08:52, 1 February 2009 (UTC)

I very much doubt there is anything as simple as a "quadratic function" for this sequence. For more information see look-and-say sequence. Gandalf61 (talk) 09:22, 1 February 2009 (UTC)

If you want a polynomial that will go through a given 6 points then, in general, you need at least a quintic (degree 5). --Tango (talk) 13:39, 1 February 2009 (UTC)

Polynomial interpolation would give you a polynomial of degree at most 5.--Shahab (talk) 13:48, 1 February 2009 (UTC)

According to [3], the unique polynomial of least degree that fits those six points is -(11597/6)x^5+(100285/3)x^4-(416905/2)x^3+(1766885/3)x^2-(2247664/3)x+337211. I haven't checked whether that's right. Black Carrot (talk) 16:57, 1 February 2009 (UTC)

Indeed, when I say that in general you need at least degree 5 I mean that there is a way of getting a degree 5 or lower solution for all such problems. You can, however, do it with higher degree if you like (although the solution ceases to be unique), that's why it's "at least" not "precisely". --Tango (talk) 20:52, 1 February 2009 (UTC)

LOL sorry I meant polynomial function not "quadratic". Thanks anyways —Preceding unsigned comment added by 70.171.234.117 (talk) 18:17, 1 February 2009 (UTC)

Probability function

Hi there - was hoping to get a hand with this question, I'm useless with probability and it's really doing my head in!

N doctors go into a meeting, leaving their labcoats at the door (they all have coats). On leaving the meeting they each choose a coat at random - what is the probability k doctors leave with the correct coat?

Would I be right in thinking that you have $n \choose k$ selections of the k doctors with the correct coats multiplied by the number of arrangements of remaining doctors to wrong coats, over 'n!' ? If so, how do you find the latter?

Thanks a lot, 131.111.8.98 (talk) 09:59, 1 February 2009 (UTC)Mathmos6

:My knowledge about probability is limited but I'd say you have a binomial distribution here. So the answer should be ${\binom {n}{k}}{\frac {1}{n}}^{k}{\frac {n-1}{n}}^{n-k}$--Shahab (talk) 11:01, 1 February 2009 (UTC)

Oh I think its not the binomial distribution after all. Since the doctors will most certainly pick up coats one by one so the probability of success is changing for a particular trial.--Shahab (talk) 11:19, 1 February 2009 (UTC)

It's more a problem of enumeration, what you want are the rencontres numbers (for the probability, of course, divide by n!). You also have to decide if you mean "exactly k" or "at least k", the two answers being immediately related with each other. --pma (talk) 12:58, 1 February 2009 (UTC)

We have a Wikipedia article about this problem: rencontres numbers. Michael Hardy (talk) 17:42, 1 February 2009 (UTC)

There is also an article on rencontres numbers, that may be interesting too.--pma (talk) 10:57, 2 February 2009 (UTC)

Zero divisors in polynomial rings

Resolved

 – --Shahab (talk) 11:31, 1 February 2009 (UTC)

Hello all. I am trying to prove the following theorem: Let f(x) be a polynomial in R[x] where R is a commutative ring with identity and suppose f(x) is a zero divisor. Show that there is a nonzero element a in R such that af(x)=0.

Now I start by letting f=(a₀,a₁,...a_n) and g=(b₀,b₁,...b_m) where g is of least positive degree such that fg=0. I can see that a_nb_m=0 from here and so I can conclude that a_ng must be zero (for else a_ng would contradict g's minimality). The hint in the book I am reading asks to show that a_n-rg=0 where 0≤r≤n. Equating the next coefficient in fg=0 gives me a_nb_m-1+a_n-1b_m=0 but I can't figure out what to do next.

Can anyone help please?--Shahab (talk) 10:12, 1 February 2009 (UTC)

If a_ng is the zero polynomial, what does this tell you about each a_nb_k (0 ≤ k ≤ m) ? So what does this tell you about a_n-1b_m ? And what can you conclude about a_n-1g ? And if you repeat this argument, what can you conclude eventually about a_n-rg ? Gandalf61 (talk) 11:19, 1 February 2009 (UTC)

Thanks. I proved it.--Shahab (talk) 11:31, 1 February 2009 (UTC)

what is the most important mathematical operation

if you could only have 1 mathematical operation, which one would you have? —Preceding unsigned comment added by 82.120.227.157 (talk) 14:36, 1 February 2009 (UTC)

I'd go with addition. As long as you restrict yourself to the integers, multiplication is just repeated addition and exponentiation is just repeated multiplication, so if you have addition you can do all three, it just takes longer. Even if you work in larger number systems, or even things without numbers, most of the standard operations are ultimately based on addition (eg. in the rational numbers, multiplication of fractions is defined in terms of multiplication of integers, which is defined in terms of addition). --Tango (talk) 14:42, 1 February 2009 (UTC)

I would have to agree, addition is most important. However, it should be stated that integer exponentiation is repeated addition, non-integer exponentiation gets rather more complicated. -mattbuck (Talk) 14:59, 1 February 2009 (UTC)

I did state that... --Tango (talk) 20:50, 1 February 2009 (UTC)

You can only get multiplication, exponentiation, etc. from addition if you allow yourself recursion, which for this reason I would say is a more important operation. Algebraist 17:15, 1 February 2009 (UTC)

If you can do something once, you can do it lots of times - I think you get recursion for free. (As with any question like this, there are slightly different interpretations which get different answers.) --Tango (talk) 20:50, 1 February 2009 (UTC)

In that case, why not go the whole way and start with the successor function? Algebraist 21:20, 1 February 2009 (UTC)

A few thoughts:

1. The question is not entirely well-formed, because an operation requires some objects to apply it to. But that's easy to fix: we just include your choice of objects as part of your choice of operation.

2. If all you could do was recursion, then you couldn't do anything, because recursion is just the ability to do some other thing any number of times.

3. Opposing #2, take a look at the lambda calculus. The lambda calculus contains only one kind of object: functions that take a single argument. There is only one operation: function-application. All you can give to a function is a unary function, and all a function can return is a unary function. It is easy in the lambda calculus to define integers, addition, multiplication, recursion, Boolean operations, etc. Once you have the integers, you can define real numbers, irrational exponents, and anything else you like. So, I guess I'll take "function application" as my sole operation, with "unary functions" as my objects. That buys me everything.

Are there other known ways in math to get everything with just one operation?

--Ben Kovitz (talk) 21:06, 1 February 2009 (UTC)

I suppose the project to embed all of mathematics in set theory can be seen as reducing everything to the single operation the takes a property φ and outputs the class of all objects which φ. (This idea works better in some set theories than others) Algebraist 21:18, 1 February 2009 (UTC)

But you still need a way of combining properties (AND, OR, etc [or unions and intersections, depending on point of view]). Is there a version of set theory in which those aren't taken as undefined? --Tango (talk) 21:23, 1 February 2009 (UTC)

If you can have ω-recursion for free, I can have first-order formulae for free. Algebraist 22:02, 1 February 2009 (UTC)

It was my understanding that much of modern set theory intended to build up the rest of mathematics from this sort of "single operation" approach. The section Set_theory#Axiomatic_set_theory explains some of the efforts, which seem to have been largely successful. "Nearly all mathematical concepts are now defined formally in terms of sets and set theoretic concepts. For example, mathematical structures as diverse as graphs, manifolds, rings, and vector spaces are all defined as sets having various (axiomatic) properties. Equivalence and order relations are ubiquitous in mathematics, and the theory of relations is entirely grounded in set theory." Nimur (talk) 21:24, 1 February 2009 (UTC)

That's also my understanding of the main goal of set theory. What would you say is the single operation of set theory? Function-application actually depends on sets for its definition, since a function is a kind of set (of ordered pairs, themselves sets). "One operation" and "defined in terms of" are different, at least as I understand them. --Ben Kovitz (talk) 21:54, 1 February 2009 (UTC)

As I said above, I take the basic operation of set theory to be 'take a property, form the set/class of all sets/objects with that property'. Of course, some work needs to be done to avoid paradox. Algebraist 22:02, 1 February 2009 (UTC)

Thanks for repeating your point, Algebraist. I hadn't given it proper consideration the first time, because I was thinking that "property" is too vague for a mathematical operation. For example, the property of "good government" or "wise choice". What do they call that operation (mapping a property to the set/class of all the things that have it)? (Most folks I've talked with about this say that a property is that set/class, so "having a property" simply means "being a member of that set/class", but those folks were philosophers, and I think that's a dumb theory, anyway.) I had been thinking of operation as meaning a function mapping a tuple of elements from a set A to the set A, or something close to that; so, for example, integer addition is an operation that maps to integers to an integer, etc. Am I being too narrow? --Ben Kovitz (talk) 23:02, 1 February 2009 (UTC)

Are there other known ways in math to get everything with just one operation?: for instance (talking a little bit more about constructions rather than operations) in category theory every universal construction turns out to be a particular case of an initial object, the simplest concept in the theory. The trick of course is that the category changes -and becomes possibly more complicated. --pma (talk) 00:38, 2 February 2009 (UTC)

Ratio of binomial coefficients

Hiya,

Having shown that ${\binom {n}{0}}+{\binom {n}{1}}+{\binom {n}{2}}+...{\binom {n}{m-1}}\leq {\binom {n}{m}}$ for all $m\leq {\frac {n}{3}}$, and supposing $n=3m$, how would one show that the limit of the ratio of the two sides of the above inequality as $n->\infty$\infty }"/> equals 1?

Many thanks for the help!

Spamalert101 (talk) 16:02, 1 February 2009 (UTC)BS

I'm not sure, but here's the first thought that comes to mind. It might or might not be fruitful. "Ratio = 1" means "they're equal", so just prove that the difference between them gets smaller than any epsilon. You might be able to do that with an inductive proof. --Ben Kovitz (talk) 22:03, 1 February 2009 (UTC)

Hiyatoo! Look at the LHS: it is, starting the sum from the last term (and the largest):

${\binom {3m}{m}}\left[{\frac {m}{2m+1}}+{\frac {m(m-1)}{(2m+1)(2m+2)}}+{\frac {m(m-1)(m-2)}{(2m+1)(2m+2)(2m+3)}}+...\right]$.

Notice that it is a finite sum, although with an increasing number of terms. The kth term into the sum converges to ${\frac {1}{2^{k}}}$ as $m\to \infty$. In general, this should not be enough to conclude that the inner sum converges to

$1={\frac {1}{2}}+{\frac {1}{4}}+{\frac {1}{8}}+...$,

as you want, BUT it's also true that each term is less than the corresponding term ${\frac {1}{2^{k}}}$. Then you conclude applying the dominated convergence theorem for series (a toy version of the usual dominated convergence theorem of integrals; its a particular case, as $\ell ^{1}$ is a particular case of $L^{1}$). Is it ok? in case ask for further details. Note that in the same way you can prove (try it) an analogous asymptotics for your sum in the more general case of an integer multiple of m, that is $n=pm$, instead of $3m$. If you got the geometric series & dominated convergence thing, you can write down immediately the limit of the ratio in terms of p. pma (talk) 23:58, 1 February 2009 (UTC)

This is Question of sequence & series

This is Q of sequence & series hlp me 1+2+4+8+16+........find nth term of the series & find sum of first n ter? —Preceding unsigned comment added by 117.196.34.27 (talk) 16:11, 1 February 2009 (UTC)

n'th term is 2ⁿ⁻¹. Sum of first n terms is 1+2+4+8+...+2ⁿ⁻¹=2ⁿ−1. Bo Jacoby (talk) 17:10, 1 February 2009 (UTC).

Hi. Have you actually tried doing this yourself? -mattbuck (Talk) 18:03, 1 February 2009 (UTC)

Seemingly straightforward problem

Hi, I'm trying to solve the following, seemingly simple, problem but I'm stumped by something. I want to find the maxima and minima of the following function:

$f(x)=8{\sqrt {x}}-4x{\sqrt {x}}+{\frac {1}{2}}x$

I find the following derivative:

$f'(x)={\frac {4}{\sqrt {x}}}-6{\sqrt {x}}+{\frac {1}{2}}$

Which, when set equal to zero, I rewrite to this polynomial:

$144x^{2}-193x+64=0$

Using the quadratic equation gives me the solutions x = 0.738 and x = 0.6020. Plotting these functions shows that the first is actually the maximum of the function, but the second makes no sense at all. I've gone over it a million times, and I can't find any errors. I was thinking that there might be some complex business that I'm not aware of (like when I assume that ${\frac {\sqrt {x}}{\sqrt {x}}}=1$). Can anybody elucidate? risk (talk) 20:19, 1 February 2009 (UTC)

The second term in your derivative is (fairly obviously) wrong: I haven't worked it through, but fixing that should help. AndrewWTaylor (talk) 20:30, 1 February 2009 (UTC)

Sorry, I copied it out wrong. I've fixed it in my original post (to avoid confusion). risk (talk) 20:33, 1 February 2009 (UTC)

Where did you get that polynomial from? I get $12x-{\sqrt {x}}-8=0$, which has only one solution. Algebraist 20:45, 1 February 2009 (UTC)

I used the following steps

${\begin{aligned}{\frac {4}{\sqrt {x}}}-6{\sqrt {x}}+{\frac {1}{2}}=&\;0\\4-6x+{\frac {1}{2}}{\sqrt {x}}=&\;0\\6x-4=&\;{\frac {1}{2}}{\sqrt {x}}\\12x-8=&\;{\sqrt {x}}\\144x^{2}-192x+64=&\;x\\144x^{2}-193x+64=&\;0\end{aligned}}$

In the first step I multiply by ${\sqrt {x}}$. In the fourth, I square both sides. Any illegal moves? risk (talk) 20:55, 1 February 2009 (UTC)

You will see the error if you try to plug the solution $x=0.6020$ into the equation $12x-8={\sqrt {x}}$; the left side becomes negative .776, and the right side becomes positive .776.

You didn't make any algebraic errors, but rather a subtle logic error. What your algebraic manipulations show is the following: if x is a solution to $4/{\sqrt {x}}-6{\sqrt {x}}+1/2=0$, then x is a solution to $144x^{2}-193x+64=0$. However, because you squared both sides in one step (and squaring is not an injective function), your proof does not go in reverse; it is not necessarily true that all solutions to the latter equation must also be solutions to the former equation. In fact, you have even constructed an example of a solution to the latter equation which is not a solution to the former equation. Eric. 131.215.158.184 (talk) 21:26, 1 February 2009 (UTC)

See extraneous solution. --Tango (talk) 21:27, 1 February 2009 (UTC)

Of course. Thank you both. I should take some time to read the Extraneous Solution article. Could you tell me how you would solve it from $12x-{\sqrt {x}}-8=0$, or how you could tell that it had only one solution? risk (talk) 21:34, 1 February 2009 (UTC)

It's just a quadratic in ${\sqrt {x}}$. Use your favourite way of solving quadratics, and remember that ${\sqrt {x}}$ is by definition non-negative. Algebraist 21:39, 1 February 2009 (UTC)

I solved it by mapping $sqrt(x)$ to a dummy variable, s, and solving the cubic equation in s. $f(s)=8*s-4*s^{3}+0.5*s^{2}$; taking the derivative, $-12*s^{2}+s+8=0$, solve s by quadratic formula. Then note that one of the zeros is negative and so mapping back into x yields the square root of a negative number. That is the extraneous root. Nimur (talk) 21:37, 1 February 2009 (UTC)

That approach relies implicitly on the chain rule and the fact that sqrt(x) has no stationary points. Algebraist 21:47, 1 February 2009 (UTC)

Although your approach can produce extraneous solutions, it definitely won't omit any correct solutions. So you can just take both solutions that you found and plug them into the original equation to verify their correctness, and throw out any extraneous solutions. But Algebraists' approach is better. Eric. 131.215.158.184 (talk) 22:33, 1 February 2009 (UTC)

Summarizing, it seems worth repeating Nimur's remark. Since you are looking for maxima and minima, it is convenient to make the substitution $\scriptstyle {\sqrt {x}}=s$ from the beginning and look for max & min of $\scriptstyle g(s)=8s-4s^{3}+{\frac {1}{2}}s^{2}$ over all $s\geq 0$ --pma (talk) 00:24, 2 February 2009 (UTC)

February 2

"Complement" of an automorphism group

I am confused as to how the notion of a "complement" (in the group-theoretic sense) can apply to non-group-elements. In the example I have, we have a quotient $L=Q/[Q,Q]$, and it is written "Let L^- be the complement in L under the action of $\rho$" where $\rho$ is some automorphism of Q. How can L^- be defined as the complement of a group of automorphisms - what is it the complement in, and if it is $L$ how can that even be defined?? I'm lost! SetaLyas (talk) 00:34, 2 February 2009 (UTC)

Try giving a little more context; perhaps there is some minor typo. It is not uncommon for Q to be a q-group, ρ to be an automorphism of coprime order, and to ask about a complement of the centralizer of ρ in Q/[Q,Q], which is likely equal to image of [ Q, ρ ] in L, also known as L^1−ρ, another important subgroup. JackSchmidt (talk) 03:44, 2 February 2009 (UTC)

Thanks ^_^ I'm reading from a paper filled with typos, so that's not unlikely! You're correct in some of your guesses... Q is a 2-group, $\rho$ is an automorphism of order $2^{n}-1$ (so of coprime order). It is to do with considering the associated Lie ring of the group, and then just says "Let $L^{-}$ be the complement in L under the action of $\rho$". So you are saying by "the complement in L under the action of ρ" means the complement of the centralizer of ρ (or <ρ>?) in Q/[Q,Q], which could equal [Q,ρ]/[Q,Q]? Are there any texts where this terminology is used that you know of? SetaLyas (talk) 12:27, 2 February 2009 (UTC)

Especially if the Lie ring methods are being used to talk about fixed points or fixed point free automorphisms, I think the text means the complement of the fixed points/centralizer. I don't think leaving out the words "of the fixed points" is standard terminology, but if it is a preprint, it could be a very plausible typo. The index of the centralizer in Q of ρ should be equal to the (group theoretic, not vector space) index of L⁻ in L (because ρ has coprime order; Khukhro's p-group book, page 81) if this is the case, so that might be something to check.

In general if ρ acts coprimely on Q, then Q = [Q,ρ]·C_Q(ρ), and if Q is abelian, then this is a direct product (so true for instance in the quotient Q/[Q,Q]). This can be found in Aschbacher's or Kurzweil and Stellmacher's or Gorenstein's group theory textbooks (I think always under "coprime action"). Again, you could check this in each factor to see if this is what the author means; it should basically just be finding the Fitting decomposition of 1−ρ.

For Lie ring methods: Chapter VIII of Huppert-Blackburn 2 is probably reasonable place to compare to, as it covers the regular automorphism case, which is similar (especially section 9). Khukhro's p-automorphisms of finite p-groups has a reasonable description of Lie ring methods, but focused on the not-coprime case. Vaughan-Lee's book on the Restricted Burnside Problem had some good material if I recall correctly, but I don't think it focused on automorphisms at all. Leedham-Green and McKay's structure of p-groups book I think uses similar techniques to the other books, but is fairly specialized and fast paced. If the paper is online, I can probably check if this is reasonable. JackSchmidt (talk) 18:52, 2 February 2009 (UTC)

mathematics

Who is the father of geometr —Preceding unsigned comment added by 74.125.74.37 (talk) 02:55, 2 February 2009 (UTC)

In the classical western tradition, this is usually ascribed to Euclid. You may want to evaluate History of geometry to define your question more precisely, as well as consider a more world-wide perspective. Nimur (talk) 03:23, 2 February 2009 (UTC)

y? —Preceding unsigned comment added by 82.120.227.157 (talk) 15:13, 2 February 2009 (UTC)

Euclid is listed in Fathers of scientific fields#Mathematics which also lists fathers of some subfields of geometry. PrimeHunter (talk) 18:01, 2 February 2009 (UTC)

Name of a type of puzzle

When I was in elementary school, teachers often gave us math puzzles during lunch for whatever reason. They were the types of puzzles where you had to form a number after being presented with a set of numbers. For example, if I am given the numbers 1,2,4,5 then I would have to manipulate them in a way where the result would be 24. The answer, of course would be 4(5+2-1). Is there a name for this? Thanks, Vic93 (t/c) 04:28, 2 February 2009 (UTC)

Your variant sounds like 24 Game. The best known variant may be four fours. I don't know whether there is a general name for this type of puzzle, but see Krypto (game). PrimeHunter (talk) 04:37, 2 February 2009 (UTC)

Hi, thanks. I certainly didn't know about that game before. However, looking back now, I may have misworded my original query. While 24 does sound a lot like what I described, I neglected to mention that the end result could be any integer, not just 24. For example, I would be given the numbers 3,7,5,2 with the goal being say, 8. The answer would be ((7*3)-5)/2. Perhaps this is just another variant of the game, but I'd like to know if it has another name (if it is not indeed 24). Vic93 (t/c) 22:15, 2 February 2009 (UTC)

Four fours

Is it possible to generate any integer, as constrained in the Four fours article (using addition, multiplication, concatenation, factorial, exponentiation)? No logs, as it states that it is trivial to do with them. Nadando (talk) 05:37, 2 February 2009 (UTC)

• You may have to allow division, subtraction, square roots and floor function as well, to get many numbers. I am guessing it is not possible to generate any integer, but stated without proof! How would you make 0 1 2 3 and other small numbers with out some operation that would produce smaller numbers. Graeme Bartlett (talk) 06:22, 2 February 2009 (UTC)

The book Mathematical Recreations and Essays (mentioned in the article) is available on google books. Please check out page 14. Depending on what opperations are allowed you can go up to different numbers. Anythingapplied (talk) 17:13, 2 February 2009 (UTC)

What is this type of problem called?

For example, you have a container of capacity 3 and another of capacity 5, and have to use fill and transfer operations to measure out, for example, 2 units. One solution is to fill the 5 container, then transfer 3 units to fill the other container, leaving 2 units behind. My question:

(a) Does this type of problem have a name? (b) For general containers of capacity m and n units, which values of quantity can be produced, and what sequence of steps are required in each case?→86.132.164.81 (talk) 13:11, 2 February 2009 (UTC)

(b) Values: all multiples of the greatest common divisor d of m and n, which is also characterized as the smallest positive integer of the form d=xn+ym over integers x,y; the corresponding algorithm is old Euclid's one. Not by chance: if you replace "containers & capacity" with "segments & length" the geometric origin of the problem appears. --131.114.72.215 (talk) 13:23, 2 February 2009 (UTC)

I don't see how to implement Euclid's algorithm with two containers. If you have a units in one container, and b in the other, how do you replace a with a mod b or a − b while leaving b intact? I don't think you can produce gcd(m, n) in general without using emptying a container as another operation (and even in that case the algorithm is not Euclid's, but a sort of "counting x times n modulo m" for a suitable x). Furthermore, you obviously cannot fit more than m + n units in containers of size m and n, so arbitrary multiples of the gcd are out of question. — Emil J. 14:14, 2 February 2009 (UTC)

Example: let m = 5 and n = 7. If we denote by (a, b) the state where a units are in the smaller container and b units in the larger, then it is easy to see that the following set of states in closed under the operations of filling and transfering: (0, 0), (5, 0), (0, 7), (5, 7), (0, 5), (5, 2). Thus you cannot produce gcd(5, 7) = 1. — Emil J. 14:31, 2 February 2009 (UTC)

Yes, I was just wondering whether to add a remark... If you are allowed to use only the two containers, filling and emptying them (e.g. you are at the sea), then of course you get exactly all multiples of the gcd up to n+m (and potentially any multiple if e.g. you drink it). If you impose the constraint that you can't waste water, then it is your situation (and the answer is different as you say). The OP refers to "fill" and "transfer" operations indeed (86: is unfill=-fill allowed??). But your ecological version is somehow more attractive. --pma (talk) 14:44, 2 February 2009 (UTC)

I believe this is called the Die Hard with a Vengeance problem. -mattbuck (Talk) 15:35, 2 February 2009 (UTC)

Fill 7; Pour 7 into 5 leaving 2; Throw 5 away; Pour 2 in 7 into 5; Fill 7; Pour 3 of 7 into 5 leaving 4 in 7; Throw 5; Pour 4 in 7 into 5; Fill 7; Pour 1 of 7 into 5 leaving 6 in 7; Throw 5; Pour 5 of 6 in 7 into 5 leaving 1 in 7; Throw 5. You have 1 in 7. -- SGBailey (talk) 20:47, 2 February 2009 (UTC)

Expanding an integral

I am trying to expand the following expression for small ξ:

$\int _{0}^{\infty }t^{2}\ln \left(1-e^{-{\sqrt {t^{2}+\xi ^{2}}}}\right)dt\;.$

Just expanding the integrand and integrating term by term does not work, since it runs into ever more divergent integrals. I guess the expansion will involve log terms and the like... Does anybody have an idea of how to do it? Thanks, MuDavid 15:00, 2 February 2009 (UTC)

Capitalization conundrum

Hello all. I'm writing because my office mate and I were having a discussion concerning the capitalization of terms in mathematics that use a person's name. In particular, we're concerned with those names that get turned into adjectives. We're thinking "Boolean", "Abelian", "Cauchy", "Lipschitz", things like that. (For instance, a function can be Lipschitz, but no one would ever say "the graph is Petersen").

We noticed that almost everyone gets their name capitalized except for Abel. The word "abelian" appears in lowercase all over the place. What gives? Can anyone explain this to us? Also, does anyone have other examples of lowercase typeset names?

Thanks! –King Bee ^{(τ • γ)} 17:42, 2 February 2009 (UTC)

I always think of it as it being an extra honour, not getting a capital letter (a bit like members of the Royal College of Surgeons going by Mr. not Dr.). Abel is such a great mathematician that something named after him has being a word it its own right and is no longer thought of as eponymous. (Take a group of students that have just finished a first year algebra course and see how many of them even know where the term "abelian" comes from!) --Tango (talk) 18:09, 2 February 2009 (UTC)

The late professor Børge Jessen said that the highest honor for a mathematician is to become an adjective spelled without capitalization. He mentioned, apart from abelian, also galois groups, and hermitean and pythagorean and euclidean. I think he also wrote 'hilbert space'. Bo Jacoby (talk) 18:16, 2 February 2009 (UTC).

I definitely capitalise Galois group. I may be inconsistent with the others. Abelian is the only one that I would think it odd to see capitalised. --Tango (talk) 18:19, 2 February 2009 (UTC)

I agree with Tango (although the honor is slightly diluted by having an exact synonym -- commutative -- in common use), but I also never capitalize boolean (largely because of programming). I don't have any explanation for why these conventions are used, though... maybe "Abel" and "Boole" just don't sound like typical western last names? By the way, the term for being names after something is eponym. Eric. 131.215.158.184 (talk) 18:50, 2 February 2009 (UTC)

It's also worth noting your sample set there was quite strange. Lipschitz and Cauchy will always be capitalised, as they are the surnames themselves, not adjectives derived from them. Adjectives derived from names are uncapitalised through increased usage, there is no "correct" way... SetaLyas (talk) 21:22, 2 February 2009 (UTC)

Golden ratio tesselation?

<img src="http://www.cantonese.ca/quilt-model.png"> I want to adjust the ratios and angles in this tesselation to make the square bigger, possibly into a golden rectangle, and adjust the rhombi accordingly. Can anyone work out what dimensions and angles to use for it to "work" mathematically? --Sonjaaa (talk) 20:14, 2 February 2009 (UTC)

What ratio? All the edges are the same length. All you can do is alter the angle between the squares. At one extreme (0 degrees) the rhombii will cease to exist and you'll have tessellating squares at the other extreme (90 degrees) the rhombii will become more squares so you'll have tessellating squares again. An obvious compromise is to use 45 degrees. -- SGBailey (talk) 20:52, 2 February 2009 (UTC)

Is it possible to turn the square into a rectangle, a golden rectangle, and adjust the rhombi accordingly? Or would that be geometrically impossible?--Sonjaaa (talk) 21:42, 2 February 2009 (UTC)

Function proof, hard one

g(g(k)+g(m))=k+m.

k,m are greater or equal to zero, g(x) has x greater or equal to 0, and g(x) is greater or equal to 0.

(all defined to be greater or equal to zero)

Prove or Disprove:

g(x)=c (with restrictions for greater or equal to 0, constant c) and g(x)=x (with restrictions) are the ONLY TWO FUNCTIONS that satisfy this.

Its easy to see that they satisfy.

if g(x)=x, then g(g(k)+g(m))=k+m. because g(g(k)+g(m))=g(k+m)=k+m if g(x) =x, the left side is obviously k+m so they satisfy...

its also easy to tell for constant c.

However, HOW DO YOU FIND OTHER FUNCTIONS!?/ (like exponential, log ones)

Thanks —Preceding unsigned comment added by 208.119.135.108 (talk) 21:00, 2 February 2009 (UTC)

I don't understand. If g(x)=c, then g(g(k)+g(m))=c, which is not necessarily equal to k+m, surely? Algebraist 21:03, 2 February 2009 (UTC)

about question

^but K+M=constant (since they are both integers) so it is a constant. —Preceding unsigned comment added by 208.119.135.108 (talk) 22:14, 2 February 2009 (UTC)