Welcome to the mathematics section
of the Wikipedia reference desk.

Select a section:

WP:RD/MA

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.

July 16

Solitary numbers

Why is 18 a solitary number even though (18, σ(18)) ≠ 1 ? Same for 45, 48, etc. Thanks, --Think Fast (talk) 01:51, 16 July 2008 (UTC)[reply]

According to Friendly number, a number that is coprime to the sum of its divisors is solitary, but the converse is not true, 18 being a counterexample. --Tango (talk) 01:57, 16 July 2008 (UTC)[reply]

I read the article. Can you explain why 18 is a counterexample? --Think Fast (talk) 02:31, 16 July 2008 (UTC)[reply]

The formulation in Friendly number seemed easy to misunderstand so I tweaked it.[1] PrimeHunter (talk) 02:39, 16 July 2008 (UTC)[reply]

The converse (which is false) would be: If a number is solitary then it's coprime to the sum of its divisors. It is this for which 18 is a counterexample: 18 is solitary (not proved here), but 18 is not coprime to its sum of divisors 39. PrimeHunter (talk) 02:46, 16 July 2008 (UTC)[reply]

This is all good, but why is 18 a solitary number? It is not prime, it is not a prime power, and (18, σ(18)) ≠ 1 (i.e., it is not coprime to the sum of its divisors). But nonetheless, it is a solitary number. Can anyone explain what makes it a solitary number even though it does not satisfy any of these criteria? --Think Fast (talk) 03:04, 16 July 2008 (UTC)[reply]

Presumably it is a solitary number because it is the only positive integer n such that σ(n)/n = 13/6. In other words, your question is "Why are there no positive numbers n (other than 18) such that the sum of their divisors is 13n/6?" or so. JackSchmidt (talk) 03:42, 16 July 2008 (UTC)[reply]

I don't know if it helps, but this was discussed in 2004 at sci.math:

We know that 2*3=6 divides x, so we know that x=2^a * 3^b * n where n is 
relatively prime to 6.  Now it is easy to see that if sigma(x)/x = 13/6 
and x != 18 then a=b=1. (Can you see why?)  Now if x=6*n with n 
relatively prime to 6 then the numerator of sigma(x)/x reduced to lowest 
terms will be even, and the denominator will be relatively prime to 6, 
in particular it will not equal 13/6.

The only other reference appears to be some sort of closed mailing list. JackSchmidt (talk) 04:01, 16 July 2008 (UTC)[reply]

Thank you for finding this. It answers my question. --Think Fast (talk) 05:30, 16 July 2008 (UTC)[reply]

Visitors from the future: If you come across this page and the proof is not clear enough, some additional details are provided here. -- Meni Rosenfeld (talk) 17:39, 20 July 2015 (UTC)[reply]

That's exactly what I wanted (although my math fu is too weak for it to make sense) Ataru (talk) 01:42, 7 January 2019 (UTC)[reply]

Divisor function

This isn't really a reference desk question, but I hoped one of you maths guys could check the table of values I just added to divisor function. Is it ok? (There - I managed to ask a question...) -- SGBailey (talk) 08:24, 16 July 2008 (UTC)[reply]

It doesn't contain any errors, if that's what you're asking. Algebraist 10:09, 16 July 2008 (UTC)[reply]

I added Table of divisors to Divisor function#See also. Your table lists the first 15 of 1000 entries. PrimeHunter (talk) 12:32, 16 July 2008 (UTC)[reply]

For future reference, if you want help with writing/editing/proofreading mathematics articles, the best place to go is Wikipedia talk:WikiProject Mathematics. --Trovatore (talk) 00:58, 17 July 2008 (UTC)[reply]

Numbers

Hello, I do not speak english well and I do not understand the next number: How much is thirty hundred thousand in number? Thank you --Humberto (talk) 14:22, 16 July 2008 (UTC)[reply]

"Thirty hundred thousand" is not how one would phrase that number correctly, so I can't blame you for not understanding it. I think that it is supposed to be 3,000,000; which is three million. Paragon12321 (talk) 14:27, 16 July 2008 (UTC)[reply]

I do not speak English well, either, however... Simple 'googling' returns lots of results:

http://www.google.com/search?q=%22thirty+hundred+thousand%22

which may indicate it is not an ordinary number, but rather a synonym for "lots of", "a large number". --CiaPan (talk) 14:42, 16 July 2008 (UTC)[reply]

It actually sounds like english as spoken by a non-perfect foreign language speaker of english, or possibly a translation from another language written in psuedo-archaic terms.. If that makes sense.

Also is 'thirty-hundred' a normal term in sanskrit or something, as many of the pages turned up by google seem to have a vedic connection?87.102.86.73 (talk) 17:01, 16 July 2008 (UTC)[reply]

Numbers in the teens are often used before "hundred" to indicated thousands, so "thirteen hundred" is the same as "one thousand three hundred". I've never heard numbers greater than 19 used in this sense, though, nor have I heard this technique used on longer numbers. « Aaron Rotenberg « Talk « 16:46, 17 July 2008 (UTC)[reply]

This may be a bad translation of "30 lakh". A lakh equals 100,000 and is a widely used term in India. --169.230.94.28 (talk) 02:56, 22 July 2008 (UTC)[reply]

Dedekind cuts and the proff that 0.999.... = 1

I have a question about these two statements in the proof:

"Every element of 0.999… is less than 1, so it is an element of the real number 1."

I feel that while this states that every cut belonging to 0.999... includes values < 1 -(1/10)^n, it does not necessary exclude the value 1-(1/10)^n on each set. If that is the case, how does the proof jump to the conclusion that "Every element of 0.999… is less than 1, so it is an element of the real number 1" so immediately. Are there any missing steps?

There are some missing steps, but they are arithmetic in nature. If we want to fill in the missing steps we would say something like: What are the elements of 0.999...? Each is some rational smaller than some number of the form 0.999..99. Since the latter number is smaller than one, also the rational is smaller than one. Oded (talk) 19:01, 16 July 2008 (UTC)[reply]

The next question I have regarding this quote: "Conversely, an element of 1 is a rational number \begin{align}\tfrac{a}{b}<1\end{align}, which implies \begin{align}\tfrac{a}{b}<1-(\tfrac{1}{10})^b\end{align}"

How does the first imply the 2nd? If I test out that statement with values for a and b, it could work. But won't I have to test it infinitely to prove to myself that it does work, therefore defeating the purpose of this proof. And then the would "implies" would have a really loose meaning.

You are right: the proof should be improved at this point, ~~and I will improve it~~. (It was not incorrect, but assumed that the reader could fill in some missing steps, which is quite reasonable. But it is better if these steps are more elementary.) Oded (talk) 19:01, 16 July 2008 (UTC)[reply]

Sorry. I tried to improve, but if I try to put in all the details, I would have to prove that

1/b>10^{-b}

. While this can certainly be done, I think it is better to accept on faith that the reader knows these things. Otherwise, this would be just too involved. Oded (talk) 19:19, 16 July 2008 (UTC)[reply]

I am only a student so maybe the proof is rigorous and I just can't understand the jumps. However, these are serious questions and the proof seems to take leaps instead of steps, leaving gaps in the proof. If it is doing so, I hope the author can continue his proof. If it isn't skipping steps, I hope my question would be answered in a way I can understand.

98.210.254.18 (talk) 16:33, 16 July 2008 (UTC)Quang Pham[reply]

Where is this proof you are referring to? I looked at the article Dedekind cut and found no such proof. Oded (talk) 18:08, 16 July 2008 (UTC)[reply]

It's in the 0.999... article - [2]. --LarryMac | Talk 18:11, 16 July 2008 (UTC)[reply]

The infinite set that corresponds to real number

1-(1/10)^{n}

is

U_{n}=\{x\in \mathbb {Q} |x<1-(1/10)^{n}\}

Real number 0.999... is the union of all sets

U_{n}

U_{\infty }=\bigcup _{n=1}^{\infty }U_{n}

and real number 1 is the set

V=\{x\in \mathbb {Q} |x<1\}

What needs to be proven is that the two sets are equal

U_{\infty }=V

.

Clearly every

U_{n}

is a subset of V, thus so is their union

U_{\infty }\subset V

.

For the other inclusion

{\tfrac {a}{b}}<1

implies

{\tfrac {a}{b}}\leq {\tfrac {b-1}{b}}=1-{\tfrac {1}{b}}<1-({\tfrac {1}{10}})^{b}

. Thus a/b is an element of the set

U_{b}

.

{\tfrac {a}{b}}\in U_{b}\subset U_{\infty }

This is true for every element a/b in V, thus V is a subset of

U_{\infty }

. Tlepp (talk) 19:04, 16 July 2008 (UTC)[reply]

Intersecting Spheres

Imagine an (n+1)-simplex in n-dimensional space, with each edge of unit length. I'm pretty sure that the intersection of the unit (n-1)-spheres around all but three of its vertices would form a 2-sphere containing the remaining three. Is this true? And if so, how could I estimate the radius of this sphere? Black Carrot (talk) 20:06, 16 July 2008 (UTC)[reply]

Yes, it is true. The intersection of a sphere with a sphere is a sphere (or a point or empty) you known that each sphere in your intersection reduces the dimension by 1. By symmetry, the center of the 2-sphere will be the average of the vertices which are the centers of the spheres in the intersection. That should facilitate the calculation of the radius. Oded (talk) 20:19, 16 July 2008 (UTC) PS: in calculations involving the m-simplex, it is usually easiest to think of it as the convex hull of the m unit vectors in m-space, rather than embed it in (m-1)-space. Oded (talk) 20:23, 16 July 2008 (UTC)[reply]

July 17

Fractional derivative

Are there any 'real world' applications, meaningful quantities/measurables, etc of fractional derivatives? (that you know of)? thanks.87.102.86.73 (talk) 00:35, 17 July 2008 (UTC)[reply]

[3] might be of interest. --Tango (talk) 00:56, 17 July 2008 (UTC)[reply]

Wow. Where to begin. Let's see. Fractional derivatives are found in the integrodifferential equations corresponding to Levy processes (see Feynman-Kac formula for the duality), which are often used to model a variety of different physical and financial processes (see the article). Existence in a fractional Sobolev space might mean that you can derive higher regularity (and thus better bounds for numerical convergence, etc.) for a function. Fractional powers of the Laplacian can manifest as Dirichlet-to-Neumann operators (see this paper) for other partial differential equations, which means that you can estimate, say, the heat flow through a wall without solving the Laplace's equation inside the wall. There's lots more fun to be had. G'luck! RayAYang (talk) 01:54, 17 July 2008 (UTC)[reply]

Do we have an article on the p-laplacian, Δ^p? We have biharmonic equation, but I didn't see any p other than 1 and 2. JackSchmidt (talk) 20:19, 17 July 2008 (UTC)[reply]

Doesn't look like it. Do you want to create one? RayAYang (talk) 23:27, 18 July 2008 (UTC)[reply]

I'll add it to my to-do list, unless someone beats me to it. siℓℓy rabbit (talk) 01:58, 19 July 2008 (UTC)[reply]

Thanks! It seems like every month some guest speaker is talking about p-laplacian this or that, so I often have some little bit of trivia that might be worth adding, but nowhere near enough perspective to even start a stub. JackSchmidt (talk) 04:05, 19 July 2008 (UTC)[reply]

Is timescales mathematics

Is timescales mathematics? Would it be acceptable as a mathematical document? 122.107.219.245 (talk) 14:10, 17 July 2008 (UTC)[reply]

That article is discussing an area of mathematics, yes. I don't know what you mean by "acceptable as a mathematical document". --Tango (talk) 16:24, 17 July 2008 (UTC)[reply]

If you're asking whether you should use it as a reference or citation in a paper, then the answer is probably no. However, it has links at the bottom that may be acceptable, such as the one to New Scientist. Black Carrot (talk) 23:13, 17 July 2008 (UTC)[reply]

Naive Bayes classifier for non-mutually-exclusive classes?

How does a naive Bayes classifier change if some of the classes are not mutually exclusive? Neon Merlin 17:27, 17 July 2008 (UTC)[reply]

choice of weights for linear regression

A question on choice of weights for weighted linear regression. I need to solve for x a problem of type $Ax=B$ where $A$ is size m by n, $x$ is size n by 1, and $B$ is size m by 1, m >> n > 1, given the weights vector $W$ of length m. In other words, I would like to minimize $\sum _{i=1}^{m}W_{i}\left(B_{i}-\sum _{j=1}^{n}A_{ij}x_{j}\right)^{2}$ . My question is: what is the appropriate choice of weights when different points i, i = 1...m, have different variances of $B$ and different numbers of counts $C$ that went into determination of the variances of $B$ ? Let me explain my question in a bit more detail. I measure $B$ for any given $A$ , many times. $A$ is a discrete variable. $A$ values may be assumed exact for all practical purposes. Measured value of $B$ varies from measurement to measurement even for exact same $A$ . Each possible value $A_{i}$ of $A$ is sampled $C_{i}$ times, $2<C_{i}<10^{5}$ . Note that $C_{i}$ varies by 4 orders of magnitude. What is the best choice of $W$ in that situation? I have tried three choices so far, none of which is really good. Choice 1. If I use $W_{i}=1/var(B_{i})$ I bias my regression towards points that may have low variance beacuse thay have been poorly sampled. Indeed, the lower is $C_{i}$ the higher is the variance of variance of B, obviously; so for some points with low $C_{i}$ the value of $var(B_{i})$ will come out very low by chance. That is bad. Choice 2. If I use $W_{i}=C_{i}$ , I bias my regression towards points with high $C_{i}$ , but these points not necessarily have lower $var(B_{i})$ ; also, any two points with same large $C_{i}$ will have the same weight even though I can reliably claim that $var(B)$ is larger in one than in the other. That is also bad. Choice 3. I can use Bayesian average instead of mean when evaluating $var(B_{i})$ , but, first, I do not know if that is legitimate, and, second, the choice of constant for the Bayesian average is quite arbitrary. So, to repeat my question, what should I use for $W$ ??? Please help! Thank you in advance, --OcheburashkaO (talk) 22:31, 17 July 2008 (UTC)[reply]

Maybe I'll be back later, but weights proportional to reciprocals of variances is the usual thing. 75.72.179.139 (talk) 02:59, 18 July 2008 (UTC)[reply]

You wrote:

Each possible value

A_{i}

of

A

is sampled

C_{i}

times,

2<C_{i}<10^{5}

.

Did you mean each value of B_i? You said observations of A could be taken to be exact. If individual observations of B can be taken to be equally uncertain, then using weights proportional to the number of observations makes sense. But apparently this equality of uncertainty does not hold, according to what you're saying. More later........ Michael Hardy (talk) 03:20, 18 July 2008 (UTC)[reply]

I'm curious; (without giving too much away :-) what is the nature of the measurement that sampling counts vary so much?

And a little confused; in the cases I'm familiar with, A is the variable to be solved for. That is, finding an

A

that minimizes

\sum _{i=1}^{m}W_{i}\left(B_{i}-\sum _{j=1}^{n}A_{ij}x_{j}\right)^{2}

. It's not a constant until you're done. Saintrain (talk) 16:48, 18 July 2008 (UTC)[reply]

I do not think choice 3 is applicable here: as I understand it, you do not have any "prior mean" at hand. I also don't like the arbitrary constant, but I think that if you did have a prior mean, then using the Bayesian mean instead of the true mean when calculating the variance should rather much reduce the chance of getting an unrealistically low estimate for the variance (although not totally eliminate it, of course...), which would help make choice 1 a bit more viable. But until you get this prior mean somehow, the Bayesian mean is not applicable.

Regarding choice 1 vs. choice 2, I prefer choice 2. This is for two reasons. First, the "bad thing" you mention for choice 1 strikes me as worse than the "bad thing" you mention for choice 2. The choice 1 bad thing amounts to a small chance that the results will be way incorrect. The choice 2 bad thing amounts to an admission that you are not fully using all the information available to you. Assuming that this application has a decent tolerance towards not getting the absolutely best possible answer, I would far rather not use all the information I have than to risk getting egregiously bad results some of the time. The second reason is that, even if you don't get unlucky with your estimates for the variances (i.e., even if your estimates of the variances all turn out to be correct), I still dislike the use of the weight

1/var(B_{i})

; if there is a substantial difference (i.e., orders of magnitude) in the true variance from one i to another i, then you're effectively ignoring those is for which the true variance is rather larger, even if your value of

C_{i}

is large enough to get useful information from those samples. But you can attribute this to personal dislike of this weighting; I would like to hear from user 75 if (s)he knows some reasons why this weighting is used.

Regarding options outside of the three choices given, I'm going to suggest a quick-and-easy linear combination of 1 and 2:

W_{i}=k\cdot \lambda {\frac {1}{var(B_{i})}}+(1-\lambda )C_{i}

where

\lambda

is an appropriately chosen function of

var(B_{i})

and

C_{i}

that reflects how certain you are of your value for the variance (closer to 1 means more certain), and where k is just a constant (think of it as necessary to make the units come out right). Of course, I've just changed the problem to choosing a function

\lambda

(and an arbitrary but important constant k!), but I don't think the results are too overly sensitive to your choice of

\lambda

... for example:

\lambda ={\begin{cases}0&{\text{if }}C_{i}<30\\1&{\text{otherwise}}\end{cases}}

is easy and not totally unreasonable.

However, I'm sure there are better definitions of W lurking out there waiting to be found. Eric. 83.173.235.68 (talk) 20:30, 20 July 2008 (UTC)[reply]

Rational Angle

Is the angle between two faces of a regular tetrahedron, arccos(1/3), a rational multiple of pi? If not, how would you prove it? Black Carrot (talk) 23:21, 17 July 2008 (UTC)[reply]

Typing "arccos(1/3)" into my calculator, dividing by pi and pressing the fraction button fails to give a fraction, which would suggest it's not rational (it could be rational with a denominator too large for my calculator to handle, I suppose). I wouldn't know where to start with proving it analytically. --Tango (talk) 00:01, 18 July 2008 (UTC)[reply]

One thing to be careful about with many calculators is that they have much more precision when dealing with fractions. That is for instance, the CAS in the TI line will do fractions with hundreds of digits, but if you start out with a decimal they only store around 9-15 digits, depending on the model – so getting a fraction from a decimal can be tricky.

I don’t remember what model, but I used a calculator once that I actually got to round “0.3333333333333” or so to 1/3, so all of them might not be entirely accurate either. GromXXVII (talk) 11:14, 18 July 2008 (UTC)[reply]

I don't think that arccos(1/3) is a rational multiple of pi. And one way to prove it would be by contradiction. Assume that this arccos(1/3)/pi is rational and get a contradiction like pi being a rational number also.--A Real Kaiser...NOT! (talk) 04:26, 18 July 2008 (UTC)[reply]

It isn't. Let θ=arccos(1/3). Then the complex number

\alpha =e^{i\theta }={\frac {1}{3}}+i{\frac {2{\sqrt {2}}}{3}}

has degree 2 over the rationals, and so cannot be a root of any cyclotomic polynomial since the minimal polynomial of α is not cyclotomic. siℓℓy rabbit (talk) 06:00, 18 July 2008 (UTC)[reply]

How does one tell that the polynomial is not cyclotomic? 24.227.163.238 (talk) 17:03, 18 July 2008 (UTC)[reply]

It's pretty obvious. There are only three cyclotomic polynomials of degree two, since there are only three integers whose totient function is two: 3, 4, and 6. The corresponding cyclotomic polynomials are

\Phi _{3}(z)=z^{2}+z+1,\quad \Phi _{4}(z)=z^{2}+1,\quad \Phi _{6}(z)=z^{2}-z+1.

whereas minimal polynomial of α is 3z²−2z+3. siℓℓy rabbit (talk) 18:53, 18 July 2008 (UTC)[reply]

Is arccos(1/3) an irrational algebraic [i.e. not transcendental] multiple of pi (or pi^2 sqrt(pi) 1/pi etc)? I.e. could you describe it using a finite number of operations on integers and pi? --Random832 (contribs) 18:40, 18 July 2008 (UTC)[reply]

All I can say is that your number "arccos( ${\tfrac {1}{3}}$ )" is not a rational multiple of $\pi$ , nor is any other rational number. $\pi$ is an irrational number, so any multiple of $\pi$ will be irrational as well. Earthan Philosopher (Talk) 19:30, 19 July 2008 (UTC)[reply]

arccos(1/3) is not rational (if it were, e would be algebraic). Algebraist 19:34, 19 July 2008 (UTC)[reply]

It still wouldn't be a rational multiple of $\pi$ then because it wouldn't be rational at all. Earthan Philosopher (Talk) 19:42, 19 July 2008 (UTC)[reply]

The sentence 'a is a rational multiple of b' means 'a/b is rational'. Algebraist 19:59, 19 July 2008 (UTC)[reply]

Right, I intended "a rational multiple of pi" to mean "pi times a rational number." Thanks for the help. Black Carrot (talk) 03:31, 20 July 2008 (UTC)[reply]

July 18

Exponential Exponents

Exponents or powers are the superscripts that in math mean to multiply the base the exponent's number of times. There are also numbers with exponents that have exponents. If you have an infinite number of exponents, does that make the number infinity or another infinite number?

i.e.

$2^{2^{2^{2^{2^{2^{2^{2}...}}}}}}=\infty$

or

$2^{2^{2^{2^{2^{2^{2^{2}...}}}}}}\neq \infty$

Earthan Philosopher (talk) 18:57, 18 July 2008 (UTC)[reply]

I don't think I've ever encountered an author who gave any meaning at all to such expressions. Algebraist 19:03, 18 July 2008 (UTC)[reply]

I think it would be like to asking if 1+1+1+1+1+1...=inf. Exponents are just a faster way to get there. I think.--Xtothe3rd (talk) 19:15, 18 July 2008 (UTC)[reply]

Is there any other way this be interpreted other than the limiting sense (you start with a finite tower, and exponentiate once for each additional term in the sequence)? RayAYang (talk) 19:49, 18 July 2008 (UTC)[reply]

Not that I can think of. Of course, you have to decide whether those twos are reals, complexes, ordinals, cardinals, 2-adics, or whatever, the answers being

+\infty

, unsigned

\infty

,

\omega

,

\aleph _{0}

and zero respectively. The only context I've seen infinite towers of exponentials is ordinal arithmetic, which I doubt is what the OP cares about. Algebraist 20:09, 18 July 2008 (UTC)[reply]

The power tower

a^{a^{a^{\vdots }}}

converges iff

e^{-e}\leq a\leq e^{1/e}

and is equal to

{\frac {W(-\ln(a))}{-\ln(a)}}

, where W is the Lambert W function. There are further explanations on the Mathworld article on "Power Tower". --Xedi^Talk 20:12, 18 July 2008 (UTC)[reply]

Look at the Wikipedia article on tetration, which is the equivalent of the Mathworld article on "Power Tower", I would assume. J kasd 20:19, 18 July 2008 (UTC)[reply]

Quite right, indeed, I should've thought of going on that article ! --Xedi^Talk 21:28, 18 July 2008 (UTC)[reply]

The formal expression $x=2^{2^{2^{2^{2^{2^{2^{2}...}}}}}}$ satisfies the equation $x=2^{x}.$ So x has better be a solution to that equation. One of the solutions is the complex number x≈0.824679+1.56743i. Bo Jacoby (talk) 05:52, 19 July 2008 (UTC).[reply]

The W function is analytic, right? Can that expression be extended to give an answer for 2? Black Carrot (talk) 02:35, 19 July 2008 (UTC)[reply]

Yes, it does give an answer, albeit not a very sensible one (approximately 0.8246785461-1.567432124i), sort of in the same way you could say

\sum _{n=1}^{\infty }n=\sum _{n=1}^{\infty }{\frac {1}{n^{-1}}}=\zeta (-1)=-{\frac {1}{12}}

--Xedi^Talk 03:00, 19 July 2008 (UTC)[reply]

\sum _{n=1}^{\infty }n=-{\frac {1}{12}}

? Does that have any deeper meaning? In similar cases (1+2+4+8+...=-1, for example), the strange answer is actually because you're working in a strange metric without knowing it (2-adic in the case of that example), but I can't think of a metric in which 1+2+3+4+... would converge. --Tango (talk) 04:01, 19 July 2008 (UTC)[reply]

I'm not really aware of how to put this into context, but, in this particular example, -1/12 is the Ramanujan sum of

\sum _{n=1}^{\infty }n

, which might give a few clues. I would, too, like to know what mathematical theory is best suited to these kind of tricks, which seem to mainly involve a sum convergent on only part of its domain, extending it analytically and getting a "sum" for where the sum doesn't initially converge. --Xedi^Talk 04:39, 19 July 2008 (UTC)[reply]

See the article on Analytic continuation. The sum 1+2+4+8+... is the value f(2) = −1 of the function f(x) = 1+x+x²+... = (1−x)⁻¹. The sum 1+2+3+4+... is the value f '(1) of the function f '(x) = 1+2x+3x²+... = (1−x)⁻², so f '(1) is not the Ramanujan sum (which I do not understand). Bo Jacoby (talk) 06:00, 20 July 2008 (UTC).[reply]

\left({\frac {1}{(1-x)^{2}}}\right)\!\!{\Bigg |}_{x=1}

is not going to help as a value for

\sum _{n=1}^{\infty }n

seeing as it isn't defined... --Xedi^Talk 15:58, 20 July 2008 (UTC)[reply]

Right! Is Ramanujan's −1/12 consistent with analytic continuation? Bo Jacoby (talk) 08:17, 21 July 2008 (UTC).[reply]

Well, using another method, you define the Riemann Zeta function for Re(s) > 1 as

\zeta (s)=\sum _{n=1}^{\infty }{\frac {1}{n^{s}}}

. This doesn't converge for Re(s) < 1 but you can use that to give values for the sum, hence giving

\sum _{n=1}^{\infty }{\frac {1}{n^{-1}}}=\sum _{n=1}^{\infty }n=\zeta (-1)={\frac {-1}{12}}

which is what the Ramanujan summation gives, not totally sure why though, there must be some argument to explain that if we define that sum to have a value, -1/12 is the only sensible one to give. --Xedi^Talk 01:32, 22 July 2008 (UTC)[reply]

OK. The sum 1+2+3+4+5+... appears in two ways: either as the infinite value for x=1 of the differential quotient f '(x) = 1+2x+3x²+... = (1−x)⁻² of the geometric series f(x) = 1+x+x²... = (1−x)⁻¹, or as the value −1/12 of the zeta function ζ(s) for s = −1. Does a sum like 1−2+3−4+5−6+... = f '(−1) = (1−(−1))⁻² = 1/4 have a different Ramanujan sum? Bo Jacoby (talk) 11:28, 22 July 2008 (UTC).[reply]

Well now that it is explained to me I suppose that it doesn't have to be just $2^{2^{2^{\vdots }}}$ or any exponent, it could be like $1+1+1+1...$ , and thus I have my answer, because $2^{2^{2^{2^{2^{2^{2^{2}...}}}}}}=1+1+1+1+1+1...=1\infty =\infty$ and so there for $2^{2^{2^{2^{2^{2^{2^{2}...}}}}}}$ must equal $\infty$

Earthan Philosopher (Talk) 03:20, 19 July 2008 (UTC)[reply]

Reversi

This question was originall posted at wp:rd/m --Shaggorama (talk) 20:39, 18 July 2008 (UTC)[reply]

Hello. What is the largest number of possible moves on anyone's turn at anytime for Reversi? For example, dark has four legal moves on its first turn. Thanks in advance. --Mayfare (talk) 18:19, 14 July 2008 (UTC)[reply]

One challenge is working this out is that you would need to know if a given situation is actually possible during a game. Situations could be constructed to yield a high number of possible moves (16 is easy, more than that shouldn't been too hard), but verifying that they are achievable under the rules of the game would be much harder. --Tango (talk) 21:14, 18 July 2008 (UTC)[reply]

Out of interest, what's your 16? My trivialist idea yields 28. Algebraist 21:19, 18 July 2008 (UTC)[reply]

Columns of one colour in the sequence .BWB.BWB or similar - white can go in any of the 16 blank spaces. I don't doubt that there are far better arrangements, that's just the first one I thought of. --Tango (talk) 23:03, 18 July 2008 (UTC)[reply]

July 19

Mod Inverse Formula

What is the mod inverse formula is in simple terms?--Melab-1 (talk) 18:03, 19 July 2008 (UTC)[reply]

I don't know any formula by that name, and google has nothing. Can you try to explain what formula you're talking about? Algebraist 18:20, 19 July 2008 (UTC)[reply]

Are you talking about:

$57\equiv 7\quad mod~50$

$x~57\equiv 1\quad mod~50$

where x is the inverse of 57 (mod 50)

122.107.219.245 (talk) 00:13, 20 July 2008 (UTC)[reply]

You must mean the inverse of Modulo operation?

When mod (a,b) = c,d where c=integer (a/b) d = remainder = a-(cb)

Then a = bc+a that's the nearest I can get to an inverse

If you only have c and d, then you get a range of values depending on what the divisor is.87.102.86.73 (talk) 23:33, 19 July 2008 (UTC)[reply]

Maybe Modular multiplicative inverse has what you want. PrimeHunter (talk) 00:07, 20 July 2008 (UTC)[reply]

Pretty much anything with "mod" in the name involves a loss of information, so it won't have a single-valued inverse. The best you can get is a family of inverses, like the one 87.102.86.73 mentions. We would need to know exactly what you mean by "mod" to help further, the term is used for various things. --Tango (talk) 00:21, 20 July 2008 (UTC)[reply]

July 20

Vector Analysis

Hi I saw a question in a model paper which is mentioned below. I am bit confused in the way that answer got in the paper. Please get mt the answer with a explaination

If the position at time t of a particle is given by, r=(2t²,t²-4t,3t-5) Find the particle's velocity and an accelaration components to the direction i-3j+2k? —Preceding unsigned comment added by Mufleeh (talk • contribs) 13:16, 20 July 2008 (UTC)[reply]

The position is (x,y,z) so r_x=2t² , r_y=t²-4t , r_z=3t-5 (those are cartesian coordinates)

the direction i-3j+2k means the line travelling in the direction given by the line from (0,0,0) to (1,-3,2) (i,j,k are used to represent the x,y, and z axis respectively)

The velocity is given by differentiating the position with respect to time

eg v_x=d/dt(r_x)=d/dt(2t²)=4t etc

The acceleration is given by differentiating the position with respect to time twice ie the rate of change of velocity. eg a_x=d/dt(v_x)=d/dt(4t)=4 etc

To find the components of these in the direction i-3j+2k you need to find the angle between the direction of velcoity (or acceleration) and that vector.

You can use the dot product to do this.

If the angle between the two vectors is A then the component in that direction is V_total cosA along the vector 1-3k+2k. (ie consider the triangle formed by the velocity and the line i-3j+2k)

and you should already know that: V_total² = V_x²+V_y²+V_z² (pythagorus)

Is that enough to help you?87.102.86.73 (talk) 15:16, 20 July 2008 (UTC)[reply]

Random Formula

Moved to Earthan Philosopher and Earthan Philosopher talk by Philosophia X Known(Philosophia X Known) 03:06, 21 July 2008 (UTC)--Earthan Philosopher[reply]

Algebraic numbers

Is it true or false that any arithmetic expression involving roots is an algebraic number?

For instance, this expresion;

${\frac {i+{\sqrt {3}}}{{\sqrt[{7}]{2}}+{\sqrt[{5}]{7}}}}=x$

I cannot solve to an algebraic expression. Of course that doesn't prove much - most likely it is my limited mathematical abilities. I can eliminate the 7th root by making it the subject and then raising to the 7th power. The 5th root is then a problem as the equation is now quartic in the 5th root of 7. Of course, one could solve the quartic equation but I don't see how that is going to help to get an algebraic expression. SpinningSpark 18:56, 20 July 2008 (UTC)[reply]

Yes, x is a root of a polynomial, of degree dividing 140. It is in general true that adding, subtracting multiplying and dividing algebraic numbers gives algebraic numbers. The usual proofs are fairly indirect, but I'm sure you could find an explicit polynomial if you really wanted to (do you?). Algebraist 19:01, 20 July 2008 (UTC)[reply]

No, I am not trying to solve a real problem, I was just interested in knowing whether I failed to find the polynomial because it was impossible or I did not have the skill. SpinningSpark 19:07, 20 July 2008 (UTC)[reply]

If you have access to mathematica, the command RootReduce will produce a polynomial with the given root. I believe maple has similar functionality, and there are other freely available computer algebra systems which should be able to do the same thing. It possible, but slightly tricky, to code this on your own however. If you want, I have some references on how to do this. siℓℓy rabbit (talk) 19:23, 20 July 2008 (UTC)[reply]

Assuming that you know a little bit of linear algebra, then the following will make sense to you. An algebraic number x is characterized by the fact that the numbers

1,x,x^{2},\dots ,x^{n}\,

are linearly dependent over

\mathbb {Q}

for some n. In other words, the field

\mathbb {Q} (x)

generated by x is finite dimensional over the rationals. If x and y are algebraic,

e_{1},\dots ,e_{k}

is a basis for

\mathbb {Q} (x)

over

\mathbb {Q}

and

f_{1},\dots ,f_{m}

is a basis for

\mathbb {Q} (y)

over

\mathbb {Q}

, then the collection of products

e_{j}\,f_{\ell }

, where

j\in \{1,\dots ,,\}

and

\ell \in \{1,\dots ,m\}

contains the sum

x+y

as well as the product

x\,y

and is closed under addition and multiplication. From this you can deduce that x+y and

x\,y

are algebraic. To see that also

1/x

is algebraic you can just use the linear relation between

1,x,x^{2},\dots ,x^{n}

, multiply it by

1/x

(a few times if necessary), and get that

1/x

is a linear combination of

1,x,\dots ,x^{n}

. I hope this clarifies the picture. Oded (talk) 19:49, 20 July 2008 (UTC)[reply]

Whoops. I misread the above post as saying "I am ~~not~~ trying to solve a real problem". Indeed, it is fairly easy to see abstractly (as Oded argues) that algebraic numbers are closed under arithmetic operations. With some effort, one can write down an algorithm for determining a polynomial equation that x+y, xy, and 1/x must satisfy, given the minimal polynomials of x and y. siℓℓy rabbit (talk) 22:19, 20 July 2008 (UTC)[reply]

Who is the world's most famous mathemetician?

Past or present?........I'll give you a little longer.........See, even people into math get stumped. The souls who discover and eloquently express reality don't get the props they deserve. Will there ever be a day when mathematicians (and scientists in general) are as household nameable as Ashton Kucher, Mariah Carey, and Santa Claus? I'll bet if they ever have to calculate with a sliderule the missile trajectory to blow an Earth shattering rock into smithereens! --Hey, I'm Just Curious (talk) 19:30, 20 July 2008 (UTC)[reply]

I nominate Count von Count.87.102.86.73 (talk) 20:15, 20 July 2008 (UTC)[reply]

I don't agree with the premise that people "into math" couldn't name famous mathematicians. There are tons of possible answers, but a couple of choices are Gauss, Euler and Euclid, for historical figures, and Andrew Wiles and Terry Tao for active mathematicians. 69.106.57.217 (talk) 22:21, 20 July 2008 (UTC)[reply]

Many math historians rank Archimedes among the top mathematicians, not necessarily the most famous though. In my opinion Erdős would be among the most famous recent mathematicians – and after all, he has the most publications among them. GromXXVII (talk) 23:04, 20 July 2008 (UTC)[reply]

IMO, the most well known by the public is Pythagorus, The most famous to those who know a bit about it is probably Euclid and within the maths community, it is probably Paul Erdos -- SGBailey (talk) 22:19, 20 July 2008 (UTC)[reply]

I'd agree with Pythagoras, with a nod to Isaac Newton, although he's known more amongst the public for his work in the physical sciences than for his contributions to mathematics. Confusing Manifestation(Say hi!) 22:36, 20 July 2008 (UTC)[reply]

One of the first that sprung to my mind was Nikolai Ivanovich Lobachevsky, in part because of the Tom Lehrer song by the same name. Also because he helped discover hyperbolic geometry. Black Carrot (talk) 23:51, 20 July 2008 (UTC)[reply]

Other somewhat famous ones are Kurt Gödel, Alan Turing and John Nash. 84.239.160.166 (talk) 10:23, 21 July 2008 (UTC)[reply]

All the ones I initially thought of have already been listed, but I think Gottfried Leibniz and Blaise Pascal also deserve mention. Oliphaunt (talk) 10:52, 21 July 2008 (UTC)[reply]

Are there no patriots here - if you go to Poland it will be quite clear who the greatest mathematician is Nicolaus Copernicus.. (also the greatest scientist, philosopher, physicist, astronomer. etc)87.102.86.73 (talk) 11:15, 21 July 2008 (UTC)[reply]

He's not even the greatest Polish mathematician. That accolade probably goes to the great Stefan Banach. Algebraist 11:27, 21 July 2008 (UTC)[reply]

According to this poll, 89% of people could identify Einstein's face (although only 4% classified him as a mathematician). Although it also reported that "was pleased to see that even 2/3 of the kids under 15 could spot Einstein" so his recognizability it falling somewhat. Newton is a good pick but most wouldn't recognize him for his contributions to mathematics (sort of like Kant's contributions to gravity/science aren't widely known, but definitely not to the same degree). Descartes also is a recognizable mathematician but probably not for his mathematics.--droptone (talk) 12:21, 21 July 2008 (UTC)[reply]

Surely the combination of "89% of people can identify Einstein's face" and "2/3 of the kids under 15 could spot Einstein" only shows that the recognition increases with age, as one might expect. I think the latter figure is quite impressive, given some of the things they (and their elders) don't know about. All is not yet lost... AndrewWTaylor (talk) 15:06, 21 July 2008 (UTC)[reply]

Srinivasa Ramanujan. --LarryMac | Talk 15:24, 21 July 2008 (UTC)[reply]

(The question was "most famous", not greatest..)87.102.86.73 (talk) 21:01, 21 July 2008 (UTC)[reply]

Functions

I've just done an exam question that I can't check because I don't have the answers; if I give the question and my answer, could someone here please tell me if I'm right?

If $f(x)=3x^{2}+4x$ and $g(x)=x^{2}+p$ and it is given that there is only one value of t, a scalar parameter, for which $f(x)+tg(x)$ can be written in the form $a(x+k)^{2}$ for some constants a and k, find p.

I worked out p as $-{\tfrac {16}{9}}$ . If I'm wrong, please tell me but don't give me the answer; I want to get that by myself. Thanks. 92.2.122.213 (talk) 20:51, 20 July 2008 (UTC)[reply]

Yes, that's correct. Algebraist 21:04, 20 July 2008 (UTC)[reply]

Geesh, a mere 13 seconds later - I'm still reading the question - give us a break here ;-) -hydnjo talk 02:07, 22 July 2008 (UTC)[reply]

Those are minutes, Hydnjo. Algebraist 11:34, 22 July 2008 (UTC)[reply]

July 21

Equally puzzling as prime numbers

Are there other phenomena on mathematics that are equally puzzling as the distribution of prime numbers? Mr.K. (talk) 10:10, 21 July 2008 (UTC)[reply]

I think a few phenomena in chaos theory can be very puzzling, for example that such complicated patterns as the Julia set and the Mandelbrot set arise with simple recursions. I'm sure there are many other examples. --Xedi^Talk 13:19, 21 July 2008 (UTC)[reply]

You might want to look at Unsolved problems in mathematics. When I read your question, the first one I thought of was the Riemann hypothesis which is one of the most famous unsolved problems ones, and is also related to the distribution of prime numbers. J kasd 14:51, 21 July 2008 (UTC)[reply]

Look at our page on the Collatz conjecture. The quote from Paul Erdős is rather entertaining, if also a bit scary. « Aaron Rotenberg « Talk « 10:26, 22 July 2008 (UTC)[reply]

Vectors

Please help me with the below question

A,B,C and D are four points in the space. They have position vectorsa,b,c and d respectively.

a=(2,4,-1) b=(4,1,0) c=(1,2,2) d=(3,3,3)

Find the value of K where, K=(AB*BC).(CA) using the value of K what can we say about points A,B and C? —Preceding unsigned comment added by 202.124.160.212 (talk) 14:20, 21 July 2008 (UTC)[reply]

The dot product and cross product articles have useful information on how these operations work (I assume the "*" in your question is the cross product), and should help you figure out the meaning of K. We can't do your homework for you, though! :-) --_{tiny plastic} Grey Knight ⊖ 14:48, 21 July 2008 (UTC)[reply]

Thanks for your advice, the value of K is zero. But I have no idea about the points by this value of K? —Preceding unsigned comment added by 202.124.160.212 (talk) 05:20, 22 July 2008 (UTC)[reply]

See also Triple product. Hope that helps. --CiaPan (talk) 06:23, 22 July 2008 (UTC)[reply]

Interpretation

I'm doing a question and I need a bit of help interpreting the meaning of it.

This is for a specific $f(x)$ and is not just a statements about all functions. One part of the questions says 'Either prove or disprove by means of a counterexample that $f(pq)=f(p)*f(q)$ if p and q are distinct prime numbers.' The next part of the question says 'Either prove or disprove by means of a counterexample that $f(pq)=f(p)*f(q)$ only if p and q are distinct prime numbers.'

What is the difference between these statements? 92.3.187.235 (talk) 21:27, 21 July 2008 (UTC)[reply]

I believe the second statement means you must prove (or disprove) the case where p and q are NOT distinct prime numbers (meaning they are either the same prime number or one or both are not prime). StuRat (talk) 21:40, 21 July 2008 (UTC)[reply]

Yes. The two claims to be proven or disproven are:

"p and q are distinct prime numbers" ⇒ "f(pq)=f(p)*f(q)"
"f(pq)=f(p)*f(q)" ⇒ "p and q are distinct prime numbers"

—Ilmari Karonen (talk) 21:54, 21 July 2008 (UTC)[reply]

One says "if"; the other says "only if". "A if B" means if B is true then so is A. "A only if B" means if A is true then B is true. Michael Hardy (talk) 04:31, 22 July 2008 (UTC)[reply]

July 22

Hexagonal lattices and coordinate systems

Please be gentle. It is forty years since I last used Maths in anger, so if this a well known topic, a pointer to a book or web page would be appreciated.

I have developed a coordinate system for working with hexagonal/triangular lattices that allows simple formulae for rotation about a centre, and for mirror image flipping. I use this instead of trig formulae because (a) I cannot afford the inevitable inaccuracies that would creep in in a computer system (b) I need a way that, given a specific point on the lattice, of determining its neighbors.

It works because the two primary axes are at 60 degrees to each other, rather than 90 degrees.

For example, a point at a,b can be rotated 60 deg anticlockwise by

a' = a+b

b' = -a

I cannot find any references to a better system to use. Is there a better one ?

Is there a known way of extending this to three dimensions, for example a lattice built from a face-centered-cubic packing of spheres ?

Dave Joubert 82.71.76.203 (talk) 15:17, 22 July 2008 (UTC)[reply]

I'm a bit confused and not sure if you made a mistake above? is the rotation about (0,0)?

You could use a=i, b=icos60+jcos60

ie a is a vector from (0,0) to (x,0) where x is the length of the side of the hexangon

b is a vector from (0,0) to (xcos60,xsin60)

So a > b > b-a > -a > -b > a-b > a where '>' indicates 'maps onto' by 60 anticlock rotation

Is this the same as what you are doing?

As for FCC - yes I think so - you need the vector corresponding to the upward diagonal edge of a tetrahedron 'call it c'. Where a and b are the vectors corresponding to the base edges of the tetrahedron

ie if Tet. = ABCD a=vector AB, b=vector AC, c=vector AD.87.102.86.73 (talk) 15:36, 22 July 2008 (UTC)[reply]

I'll let you read that and if it makes sense (it should) then please ask any further questions .. as you've already said using a,b instead of the (1,0) and (1/2,sqrt(3)/2) vectors prevents a computer messing up with rounding errors etc.87.102.86.73 (talk) 15:48, 22 July 2008 (UTC)[reply]

I believe such things may be described as "lattice vectors" eg http://www.matter.org.uk/diffraction/geometry/lattice_vectors.htm (as an example) I can't find a comprehensive link.87.102.86.73 (talk) 15:58, 22 July 2008 (UTC)[reply]