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January 6

Product (not normal) symbol in LaTeX

Hello, I am reading chapter 8 of Royden which is on topological spaces. He uses a big X for product topologies instead of $\prod$. Does any one know the LaTeX symbol for this? StatisticsMan (talk) 01:47, 6 January 2009 (UTC)

The Comprehensive LaTeX Symbol List is the place to look. You seem to have two choices, \bigtimes in the mathabx package or \varprod in txfonts/pxfonts. Base LaTeX has $\bigotimes$. -- BenRG (talk) 01:55, 6 January 2009 (UTC)

See the following macro. 142.103.8.38 (talk) 22:25, 26 November 2009 (UTC)

does 0.999... equal 1 ?

Please bear with me... I'm no mathematician...

If 0.(9) = 1

then 1 = 1.(0)1

and 0.(9) = 1.(0)2

therefore any x = any y

I'm saying that if 0.(9) is accepted to be 1, then any number is equal to any other number. Or from another perspective, the difference between any two numbers is 0.

I guess another tone to put the question would be:

Given any two numbers, one can be equated with the other, by adding or subtracting an (infinite) series of infinitesimal amounts. An infinitesimal amount equates to zero, therefore the difference between the two numbers is an (infinite) sum of zeros.

Yet, in real life, numbers can be conceptualized to be different from each other, and this "discreteness" of them does provide basis for feats of science and engineering.

Meaning, people, where is it ?

Lord KRISHNA (talk) 05:07, 6 January 2009 (UTC)

You went wrong in assuming that any two numbers can be equated with each other by adding or subtracting an infinite number of infinitesimal amounts. That assumption holds in the examples you gave, but how about, say, in the case of 2.1 and 2.2? You can rewrite 2.1 as 2.1(0)9 and 2.2 as 2.1(9), but the difference between those two isn't zero. There is no way to rewrite the two numbers in the way you described. --Bowlhover (talk)

Put in another way, there is not such a real number represented by "0.(9)27" or similar, just because that is not an allowed decimal expansion. The rule is that a decimal representation of the fractional part of a real number has to be a sequence of digits indicized on $\mathbb {N} _{+}$ . If the meaning of "0.(9)27" is that first come countably many 9, then the 2 follows all nines and the last 7 follows 2, we just can't use natural numbers to indexize those digits in that order, so that is not a proper representation of a real number. You are forced to use all natural indices for the 9's, and then you need two further indices like $\omega$, $\omega +1$ for the last two digits. Nevertheless your intuition is meaningful: you can in fact consider systems of numbers represented this way by sequences of digits indexed on ordinal numbers; you will get a non-archimedean field (i.e. containing infinitesimals). Similarly, you can also consider formal power series like $\scriptstyle \sum _{k\leq \alpha }c_{k}x^{k}$, and prove nice algebraic properties of the corresponding constructions, but they are definitely different objects--PMajer (talk) 10:11, 6 January 2009 (UTC)

This:
    'then 1 = 1.(0)1'
makes no sense. The '(0)' part means 'zeros on ALL remaining positions', so there is NO place to put your final '1'. Thus it has no meaning, adds nothing to '1.(0)'. --CiaPan (talk) 10:37, 6 January 2009 (UTC)

We have an article about this: 0.999... -- Aeluwas (talk) 10:45, 6 January 2009 (UTC)

Yea but it's still people who will discuss because they feel it logically incorrect. For example, one thing I've always wondered as it shows in the article why 1/9*9 would be 0.(9) and not 1, in the same way it doesn't go logical that 1/2*2 would equal 0.(9) — chandler — 10:54, 6 January 2009 (UTC)

Indeed. Or, turning the argument around, in ternary, 1/9 (decimal) is 0.01 (ternary) and (1/9)*9 (decimal) is 0.001*100 (ternary). In decimal arithmetic (1/9)*9 evaluates to 0.(9), whereas in ternary arithmetic, 0.001*100 clearly evaluates to 1. So if 0.(9) (decimal) is not equal to 1, then the value of (1/9)*9 depends on which base you are working in. Gandalf61 (talk) 11:20, 6 January 2009 (UTC)

The question "how can one ever get from A to B, if an infinite number of events can be identified that need to precede the arrival at B?" was zenos paradox.Cuddlyable3 (talk) 11:40, 6 January 2009 (UTC)

Anyway I wouldn't claim that "1.(0)1" has no meaning... Sure it has not the current meaning of a real number, but it is at least a representation of an idea of Lord KRISHNA. Nor I would say that there is no place left for the last 1, because in fact he was able to put it :). I would not even say that 1.(0) or 0.(9) is the real number one; it's more properly a representation of it. --PMajer (talk) 12:23, 6 January 2009 (UTC)

Thank you all for replying. So basically I get from your answers that I'd need a (much) better grasp of mathematical language and notation to investigate this matter further and in a rigorous way. But I suppose that 1.(0)1 is at least at conceivable as pi, if not more. I mean, 1.(0)1 is the same as 0.(9) but results from sequences coming from opposite sides of the real axis (real is the most complex my untrained mind can go right now). The notion that all real numbers are joined together "in infinity, by infinitesimals" through additions or subtractions is also not difficult to conceive for me (so much so that this matter has really been bugging me recently).

This is why I gave up on pure logic in high school; clearly to me any logical thought is in itself dry and sterile, and the conclusion is always a reflection of the premise. The only meaningful things are empirical experiences. Said more poetically, the only meaningful thing is communication of reflections. It's like quantum physics or something. Maybe numbers don't actually exist. Maybe I'm just trying to look smart..

Anyway thanks for endulging the cravings of my mind !

---

You are thinking about interesting things. Here's an apparent paradox that seems to be similar to your idea of "infinitely many infinitesimals": Think about the idea of "length". Clearly a single point, by itself, has no length—its length is zero. And if I have five or ten or a million points, the sum of all of their lengths is still zero. Yet somehow, if I have enough points (infinitely many of them—uncountably many, to be technically precise), they can form a line segment, which does have a (nonzero) length. So somehow it must be possible to add together an infinite number of infinitesimals and get something other than zero, which I think is at the heart of your question. This thought experiment leads into a profound and fundamental area of mathematics called measure theory, in which the idea of "length" is formally defined in a logically rigorous and unambiguous way, and leads to some rather surprising results. —Bkell (talk) 00:34, 7 January 2009 (UTC)

For Euclid, points and lines were disparate elements. Imagine his surprise at your thought experiment in alchemistic transmutation. Cuddlyable3 (talk) 10:21, 7 January 2009 (UTC)

Euclid used the word "element" to mean "fundamental"; that is, the theorems he proves in "Elements" are fundamental to an understanding of mathematics at the level that the Greeks had reached in Euclid's time. The title "Elements" in fact comes from the title of a lost textbook by Hippocrates of Chios, "Elements of Geometry". There is no evidence for either Euclid of Hippocrates being atomists; in particular the Greek notion of atomism did not develop until after Hippocrates' time. Thus the title of Euclid's Elements has no connection with the chemical/physical notion of a chemical element.

However, Euclid did think of points and lines are very different "kinds" of things; he would not have regarded putting together an infinite number of points to form a line as a valid geometric operation to be used in proofs. That is not to say that he hadn't thought of it already; the Greeks had come up with the concept of an infinitesmal, but were not able to manipulate it in a rigorous method. Greek geometers, even before Euclid's time, informally calculated areas of two dimensional figures by regarding those figures as being composed of an infinite number of lines; but, after determining the area of a figure they proved their answer to be correct rigorously using Eudoxus' method of exhaustion, without the use of infinitesmals. Eric. 68.18.17.165 (talk) 15:05, 7 January 2009 (UTC)

...and if you want to explore in another direction you may also enjoy this--PMajer (talk) 09:29, 7 January 2009 (UTC)

max size of disconnected graph

Resolved

 – Shahab (talk) 08:49, 6 January 2009 (UTC)

Suppose G is a graph of order n, i.e. has n vertices. What is the maximum number of edges (or order) possible in G so that G is not connected. I have a feeling that the answer is ${\frac {(n-1)(n-2)}{2}}$, (Cnsider the graph obtained by taking a $K_{n-1}$ and adding a vertex but no edges to it.) However I want a mathematical proof that no disconnectd graph with more edges exists. I'll be gateful for any help--Shahab (talk) 06:46, 6 January 2009 (UTC)

First, show that a disconnected n-vertex graph with the maximum possible number of edges must consist of exactly two connected components, each of which is a complete graph (otherwise another edge could be added without connecting the graph). Now, if k is the number of vertices in one of these components, then the number of edges in the graph is $\textstyle {k \choose 2}+{n-k \choose 2}$. Considering n as a fixed constant, this expression is quadratic in k; all that needs to be done is to show that it attains its maximum at $\textstyle k=1$ and $\textstyle k=n-1$ (as k ranges over meaningful values, of course). —Bkell (talk) 07:06, 6 January 2009 (UTC)

Thanks for the quick response and the excellent explanation. (In your last sentence I guess you meant: it attains its maximum at $\textstyle k=1$ or at $\textstyle k=n-1$). The answer is what I expected. Cheers--Shahab (talk) 08:49, 6 January 2009 (UTC)

No, he meant and. It attains its maximum at both ends (since the situation's symmetric). Algebraist 15:29, 6 January 2009 (UTC)

Or, equivalently, consider the complement of your graph of two connected components above. It is a complete bipartite graph between the nodes on one side and the ones on the other. (This is because an edge is not in your graph above iff it crosses between the two components; so the edges in the complement are exactly those that cross between the two components.) The number of edges in this complete bipartite graph is simply (the number of edges on one side)×(the number of edges on the other) = $k(n-k)$. It is easy to see that this quantity (which is a downward pointing parabola centered at n/2) is minimized (and therefore the number of edges in the original graph maximized) at the ends of k = 1 and k = n-1. --Spoon! (talk) 05:15, 7 January 2009 (UTC)

roots of y=0 and definitions

"Find a polynomial with integer coefficients that has ${\sqrt {2}}+{\sqrt[{4}]{2}}$ as a root."

Is "$y=0$" a technically correct answer to this problem? Aren't all numbers roots of "$y=0$"? —Preceding unsigned comment added by Metroman (talk • contribs) 07:25, 6 January 2009 (UTC)

Yes, you are technically correct ("the best kind of correct"), though if this question was on an exam or a homework assignment it should be clear that the answer "$y=0$" trivializes the question and thus would not receive full credit (if any). —Bkell (talk) 08:58, 6 January 2009 (UTC)

Okay, by $y=0$ I assume you mean the zero polynomial i.e. the polynomial with no non-zero terms, which evaluates to zero for all values of y (before someone jumps on me - yes, I know that some definitions of "polynomial" specifically exclude this case). Well, obviously the question should read "Find a non-zero polynomial ..." - it is probably assumed that if you are smart enough to think of the zero polynomial then you are smart enough to realise that it is not the intended answer. Gandalf61 (talk) 11:07, 6 January 2009 (UTC)

${\sqrt {2}}$ is an irrational number and it cannot be defined by any function couched in integer numbers.Cuddlyable3 (talk) 11:33, 6 January 2009 (UTC)

Well, that depends on what you mean by "defined" and "function". ${\sqrt {2}}$ is not equal to the ratio of any two integers. But "square root" itself is a function. If you want to stick to rational functions with integer coefficients, then ${\sqrt {2}}$ can be defined algebraically as the positive root of $x^{2}-2$. Or it can be defined analytically as the limit of

$x_{n+1}={\frac {x_{n}+2}{x_{n}+1}}$

for example. Gandalf61 (talk) 11:50, 6 January 2009 (UTC)

All true. The questioner wants a rational function and not a limit. Cuddlyable3 (talk) 14:50, 6 January 2009 (UTC)

The questioner doesn't want a rational function or a limit, they want a non-zero polynomial over the integers with ${\sqrt {2}}+{\sqrt[{4}]{2}}$ as a root. Such a polynomial exists, but we aren't going to give it, because we don't do homework here. Algebraist 15:26, 6 January 2009 (UTC)

Since the Welcome banner says "The reference desk will not do your homework for you." and there is no evidence that we are asked to do that, WP:AGF must apply. Cuddlyable3 (talk) 17:15, 6 January 2009 (UTC)

AGF does not tell us this, since OP did not ask what the answer was. OP asked if y=0 should be considered an answer.Taemyr (talk) 14:53, 7 January 2009 (UTC)

Aside to Gandalf: who defines the zero polynomial to not be a polynomial? Isn't that just really stupid? Algebraist 15:28, 6 January 2009 (UTC)

Didn't say I thought it was sensible, did I ? I vaguely remembered seeing a big debate somewhere about whether the zero polynomial was really a polynomial, and I just thought I would cover my bases in case someone got all pedantic on me. Gandalf61 (talk) 17:21, 6 January 2009 (UTC)

Gods, I hope no-one's teaching such rubbish to students. Algebraist 17:25, 6 January 2009 (UTC)

jeez... questioner's bones everywhere... is there a piece left, for me...--PMajer (talk) 17:35, 6 January 2009 (UTC)

As the cannibals said to the one who came late to dinner, You're too late, everybody's eaten. Cuddlyable3 (talk) 19:24, 6 January 2009 (UTC)

Eliminate y from the equations x = y²+y, y⁴ = 2 to obtain an equation for x. Bo Jacoby (talk) 00:05, 7 January 2009 (UTC).

I asked this question because it was a problem on one of my math tests. I knew how to find the intended solution but I discovered the shortcut answer. I answered with "y=0" but my teacher did not give me any credit. I just wanted to make sure that I was correct.

For those of you that think I'm just trying to trick you into doing my homework, here is how I could have done the problem. $0=x-({\sqrt {2}}+{\sqrt[{4}]{2}})=x-{\sqrt {2}}-{\sqrt[{4}]{2}}$
${\sqrt[{4}]{2}}=x-{\sqrt {2}}$
${\sqrt {2}}=x^{2}-2{\sqrt {2}}x+2$
${\sqrt {2}}+2{\sqrt {2}}x=x^{2}+2$
${\sqrt {2}}(2x+1)=x^{2}+2$
$2(4x^{2}+4x+1)=x^{4}+4x^{2}+4\,$
$8x^{2}+8x+2=x^{4}+4x^{2}+4\,$
$0=y=x^{4}-4x^{2}-8x+2\,$
Metroman (talk) 03:17, 7 January 2009 (UTC)

To be technical, the question did specify "integer coefficients," so you need at least 2. Where's the second one? You could argue that 0 can be considered a coefficient of x^0 (or any other power of x), but that would be a very unreasonable stretch for which you definitely won't receive any marks. --Bowlhover (talk) 04:56, 7 January 2009 (UTC)

Two coefficients? Hope '7×y+3 = 0' would by okay, so I supose '1×y + 0 = 0' is okay, too... --CiaPan (talk) 07:55, 7 January 2009 (UTC)

however FYI even x is a polynomial in x with integer coefficients, and 0 as well--84.221.209.108 (talk) 09:14, 7 January 2009 (UTC)

In what way is $p(x)=x^{4}-4x^{2}-8x+2$ not a polynomial with integer coefficients? Not only has Metroman found a correct answer, he has the "best" answer, in the sense that $p(x)$ is the unique irreducible monic polynomial with $\alpha ={\sqrt {2}}+{\sqrt[{4}]{2}}$ as a root. The set of all polynomials having $\alpha$ as a root is the ideal generated by $p(x)$; that is, every polynomial $f(x)$ with integer coefficients having $\alpha$ as a root can be written in the form $f(x)=p(x)g(x)$ where $g(x)$ is another polynomial with integer coefficients. (In particular, to answer Metroman's original question, yes $f(x)=0$ is such a polynomial that has $\alpha$ as a root.) Eric. 68.18.17.165 (talk) 14:32, 7 January 2009 (UTC)

Yes, however it seems to me that Bowlhover's objection was about the zero polynomial as a solution. The plural thing is most likely a joke, but in this context it can be misleading, so better to repeat once more that a monomial is a polynomial (and a polynomial is a power series, as well as a 1-Lipschitz function is a 2-Lipschitz function, a linear functional is a nonlinear functional, a positive measure is a signed measure, etc). The language of mathematics follows rules of convenience as any other language, and sometimes the evolution brings the use far away from the etymology. More often than not the number of terms of p(x) is unknown and irrelevant, and we don't want to have to say all times: "let p be a polynomial or a monomial or a constant", which would be another PITA, as it is the she/he form for somebody of unknow and irrelevant sex. We simply don't need a term for a "polynomial which is not a monomial", while we do need a term for "real or complex number which is not rational": that's why "irrational" still means "not rational". --PMajer (talk) 19:52, 7 January 2009 (UTC)

Why not have a look at polynomial ring or I would highly recommend having a look at ring (mathematics). At least it is relevant...--Point-set topologist (talk) 20:28, 7 January 2009 (UTC)

Indeed... the very reason of the current meaning of "polynomial" is that K[x] is something, whereas {polynomials-not-monomials} is devil-knows-what. In some sense the shift in meaning of this and other mathematical terms just follows the shift of interest from "individuals" to "classes" in mathematics. --PMajer (talk) 21:06, 7 January 2009 (UTC)

Determining size of label for cylindrical bottle

My high school geometry days are long behind behind me. If I have a cylinder of 'h' height and 'r' radius, how would I determine the linear dimensions (x & y dimensions) of a paper label that would wrap around the object? --70.167.58.6 (talk) 21:34, 6 January 2009 (UTC)

The height of the paper will be the same as the height of the cylinder, h. To figure out the width, imagine looking down the cylinder lengthwise, so that you just see a circle. Then unwrapping the cylinder is like unrolling this circle to form a straight line; thus the width of the paper is the same of the circumference of this circle. Since the circle has radius r, the circumference is $2\pi r$, and the paper label has dimensions $2\pi r$ by h. Eric. 68.18.17.165 (talk) 21:44, 6 January 2009 (UTC)

January 7

Estimation problem

I've been wondering about how to estimate the size of a population if sample members have a unique serial number, these known to be issued consecutively without gaps. Suppose that six enemy aircraft have been shot down in the order 235, 1421, 67, 216, 863 and 429. Assuming that any aircraft is as likely to be shot down as any other, is it possible to say anything about the likeliest number in total? Indeed, what would this estimate be with each successive observation? The problem seems to have something in common with the lifetime estimation of J. Richard Gott, but I can't see any obvious way of tackling it.→81.159.14.226 (talk) 22:21, 7 January 2009 (UTC)

I've heard of this problem before, and can't remember how to construct an unbiased estimator, but have found a real-world application of it in Google Books[1]. In case that doesn't show up for you, Allied statisticians in WWII used a formula that resembled x(1+1/n) to estimate the number of German tanks based on a captured sample, where x was the largest serial number on a captured tank and n was the number of tanks captured. Actually, a google on "estimate largest serial number" gives some more promising results in the first page. Confusing Manifestation(Say hi!) 22:30, 7 January 2009 (UTC)

Even better, this gives the unbiased estimator(s) W(i) = [(n + 1) X(i) / i] - 1, where X(i) is the ith smallest serial number found, with i = n giving the best (lowest variance) estimate. Confusing Manifestation(Say hi!) 22:34, 7 January 2009 (UTC)

This reminds me of a famous tale about Hugo Steinhaus, from Mark Kac's book "Enigmas of Chance" (Harper & Row, 1985):

... My favorite example of Steinhaus's incisive intelligence is the way he estimated the losses of the German army during WWII. Bear in mind that he was hiding under an assumed name and his only contact with the outside world was a rigidly controlled local news sheet that the Germans used mainly for propaganda purposes. The authorities allowed the news sheet to print each week a fixed number of obituaries of German soldiers who had been killed on the Eastern Front. The obituaries were standardized and read something like this: "Klaus, the son of Heinrich and Elvira Schmidt, fell for the Fuhrer and Fatherland." As time went on - late in 1942 and throughout 1943 - some obituaries began to appear which read "Gerhardt, the second of the sons of ...", and this was information enough to get the desired estimate. A friend to whom I told this story had occasion to tell it to a former high official of the CIA at a luncheon they both attended; the official was quite impressed, as well he might have been. --PMajer (talk) 00:41, 8 January 2009 (UTC)

Is there a way to combine the various estimators W(i) into a single estimator of even lower variance than W(n)? It seems reasonable that using more information should give a better estimate (such as using a single sample to estimate the population average versus using an average of a larger sample), but I'm not sure how one would go about this. Maybe a weighted average, with weights chosen carefully to emphasize the lesser variance estimates? I remember something like this being a good idea, since scalars (the weights) get squared (and so smaller), while the variance of a sum is not that much more than the sum of the variances. I'm not at all familiar enough with this stuff to figure out if the W(i) are sufficiently independent for the weights to be chosen well enough, nor to figure out if there is a better way of combining them. JackSchmidt (talk) 19:16, 8 January 2009 (UTC)

See Likelihood_function#Example_2. Bo Jacoby (talk) 21:07, 8 January 2009 (UTC).

It is interesting (to me at least) to compare the unbiased estimator from the ConMan with the maximum likelihood estimator from Bo. The estimates in the first case would have been around 470, 2130, 1900, 1780, 1700, 1660 as the numbers were sampled (rounded to avoid doing homework since this appears to be a standard classroom example). The estimates in the second case are quite simple: 235, 1421, 1421, 1421, 1421, 1421. I'm not sure if the likelihood function article is suggesting to do this, but after 3 samples, one could confuse likelihood with probability and do an expected value kind of thing to get: undefined, ∞, 2840, 2130, 1890, 1780. I think it is particularly interesting that the second (and the third) method "use" all of the information, but don't actually end up depending on anything more than X(n). Since they both seem fishy and bracket the first method, I think I like the first method best. The second method is always a lower bound (on any reasonable answer), but is the third method always larger than the first? Also, I still think it should be possible to use all of the W(i) in some way to get an even lower variance unbiased estimator; can anyone calculate or estimate the covariances well enough? JackSchmidt (talk) 00:51, 9 January 2009 (UTC)

I think that the maximum serial number of the sample is a sufficient statistics for estimating the number of elements of the population. The maximum likelihood value is the lousiest estimate, but in the case N = 1 it is all you have got. If N = 2 you have also median and confidence intervals. If N = 3 the mean value is defined, but the standard deviation is infinite. When N > 3 the standard deviation is also finite, and you may begin to feel confident. Bo Jacoby (talk) 08:37, 9 January 2009 (UTC).

Cool. And "sufficiency" means that even if we use more statistics (like the X(i)), we cannot do better than just using X(n), the maximum serial number, right? Of course what we do with that one X(n) might produce better or worse estimators, but we can ignore the other X(i). This makes sense in a way: the more samples we have the more confident we are that the largest value in the sample is really large. When we got the 1421 on the second try, we weren't sure if the next one would be a million, but after four more tries with 1421 still the largest we begin to have intuitive confidence that it really is the largest.

Can you describe a little how to find the mean, confidence intervals, and standard deviation in this particular case? Feel free to just use n=2 and n=4 and the above numbers; I think I understand the concepts abstractly but have never worked a problem that was not basically setup for it from the start.

For the "median", is this the number S0 such that the likelihood that the population is ≥ S0 (given the observed serials above) is closest to the likelihood that the population is ≤ S0? Since n≥2, both of the likelihoods are finite, so there should be exactly one such integer. I guess I could have a computer try numbers for S0 until it found the smallest (since it should be between 1421 and 10000). Is there a better way?

I'm not sure on the confidence interval. Looking up numbers in tables labelled "confidence interval" is about as far as I've worked such problems. How do you do this?

For the standard deviation, I guess I find the mean (listed above), then do something like sqrt(sum( 1/binomial(i,n)*( i - mean )^2,i=1421..infinity )/sum(1/binomial(i,n),i=1421..infinity)), where n is the sample size and mean is the mean. Is there some sane way to do this, or just let maple do it?

I worry I might have done this wrongly, since for n=4..6, I get standard deviations of 1230, 670, and 460. Since for n=4 the mean was only 2130, that's a heck of a deviation.

For reference, here are my values so far:

• Unbiased: 470, 2130, 1900, 1780, 1700, 1660
• MLE (L-mode?): 235, 1421, 1421, 1421, 1421, 1421
• L-median: ∞, 2840, 2010, 1790, 1690, 1630
• L-mean: undefined, ∞, 2840, 2130, 1890, 1780

How should I measure their accuracy? Confidence intervals? Are they easy to compute from the variances? Interesting stuff. JackSchmidt (talk) 17:29, 9 January 2009 (UTC)

Thank you for asking. I too find this problem fascinating. If the number of items in the sample is N, the (unnormalized) likelihood function is

$L({M=i})={[m\leq i]}{\binom {i}{N}}^{-1}$

where m is the maximum sequence number in the sample, and M is the unknown number of items in the population. The accumulated likelihood is

$L(M\leq k)=\sum _{i=0}^{k}L({M=i})=\sum _{i=m}^{k}{\binom {i}{N}}^{-1}=(1-N^{-1})^{-1}\left({\binom {m-1}{N-1}}^{-1}-{\binom {k}{N-1}}^{-1}\right)$

when k ≥ m ≥ N ≥ 2. See Binomial coefficient#Identities involving binomial coefficients, equation 14. I don't know if the statisticians of WW2 were aware of this simplifying identity. The normalized accumulated likelihood function is

$P(M\leq k)={\frac {L(M\leq k)}{L(M<\infty )}}={\frac {(1-N^{-1})^{-1}\left({\binom {m-1}{N-1}}^{-1}-{\binom {k}{N-1}}^{-1}\right)}{(1-N^{-1})^{-1}{\binom {m-1}{N-1}}^{-1}}}=1-{\binom {m-1}{N-1}}{\binom {k}{N-1}}^{-1}.$

From this you may compute a median M_0.5 satisfying P(M ≤ M_0.5)~ 0.5, and a 90'th percentile M_0.9 satisfying P(M ≤ M_0.9)~ 0.9, and an expected value of the number of items $\scriptstyle \mu =\sum _{i=0}^{\infty }i\cdot P(M=i)$, and a standard deviation $\scriptstyle \sigma ={\sqrt {\sum _{i=0}^{\infty }(i-\mu )^{2}\cdot P(M=i)}}.$

Bo Jacoby (talk) 08:18, 10 January 2009 (UTC).

As the OP, let me thank everyone for their thoughts. No, it wasn't homework, I just made up the serial numbers to give an example, being long past the age of taking courses and being given set work.→81.151.247.41 (talk) 19:20, 10 January 2009 (UTC)

January 8

Group Axioms

My lecturer stated (without proof, naturally) that if we take the following defintion of a group:

A set S with a binary operation "$\cdot$" which maps $S\times S\rightarrow S$ and obeying the following,

- Associativity: $a\cdot (b\cdot c)=(a\cdot b)\cdot c$

- Identity: $a\cdot e=a$

- Inverse: $a\cdot a^{-1}=e$

(where all the elements exist and belong to S etc.)

Then $e\cdot a=a$ follows as a theorem.

Now, in the books that I've seen and the wikipedia article this property is included in the definition of the identity. So, the question for any helpful RefDeskers is: can this be shown from the above or must it be assumed? 163.1.176.253 (talk) 00:29, 8 January 2009 (UTC)

I guess they mean $e\cdot a=(a\cdot a^{-1})\cdot a=a\cdot (a^{-1}\cdot a)=a\cdot e=a$. First one has to prove $a^{-1}\cdot a=e$ however, which is another little trick. I have to confess that I do not see the point of this making the axioms as ecomonical as possible (in this case). C'm on, we do not have to pay for them. I prefer the more simmetrical ones :) --PMajer (talk) 00:55, 8 January 2009 (UTC)

The article, elementary group theory, might be helpful. --Point-set topologist (talk) 09:59, 8 January 2009 (UTC)

Yes, thank you. There's a very relevant section on alternative axioms providing the proof. 163.1.176.253 (talk) 10:37, 8 January 2009 (UTC)

(ec) Let's start from $a\cdot e=a$. For $a=e\,\!$ it gives us $e\cdot e=e$. We replace the first $e\,\!$ with $a\cdot a^{-1}$ and get $(a\cdot a^{-1})\cdot e=e$ and by associativity $a\cdot (a^{-1}\cdot e)=e$. I'm not sure, however, if this is enough to conclude that $(a^{-1}\cdot e)=e$.... --CiaPan (talk) 10:46, 8 January 2009 (UTC)

It certainly isn't - if that were true then you would have a.e=e, which only holds in the trivial group. --Tango (talk) 15:52, 8 January 2009 (UTC)

I think that he was being sarcastic. --Point-set topologist (talk) 17:02, 8 January 2009 (UTC)

That would seem odd. I think a simple mistake is more likely. (That last e should probably be an a^-1, in which case it isn't quite enough - it uses the fact that if both left and right inverses exist then they are equal and unique, but that needs proving.) --Tango (talk) 19:35, 8 January 2009 (UTC)

Original research and WP:OR

This is not really a mathematics question but I thought that I would post this here because it is relevant. Most of the time, the questions here are either trivial or of the same difficulty as a textbook exercise. Sometimes the questions are simply posted out of interest and the answer depends on opinion. But is there a possibility that someone asks a problem here that is publishable and includes some of his/her findings with the question? I remember seeing such problems here before and it is possible that such a thing can be done by an inexperienced mathematician (someone who is learning mathematics and is not really familiar with reading journals; for example, a university student). I think it would be appropriate to include, in the guidelines at the top of the page, that people should consider this when asking a question. Any opinions (and how one can add to the guidelines at the top if people agree)?

Thanks!

--Point-set topologist (talk) 13:37, 8 January 2009 (UTC)

This should be on the talk page. Algebraist 13:39, 8 January 2009 (UTC)

Thanks for the quick reply but which talk page? --Point-set topologist (talk) 13:42, 8 January 2009 (UTC)

WT:Reference desk. Algebraist 13:44, 8 January 2009 (UTC)

Thanks. I posted the message there. Any opinions would be greatly appreciated. --Point-set topologist (talk) 14:45, 8 January 2009 (UTC)

Seeking a definition of "Positive Terms"

The linguist Ferdinand de Saussure in his Course in General Linguistics states that "...language is a system of differences..." He goes on to state:

"Even more important:a difference generally implies positive terms between which the difference is set up; but in language there are only differences without positive terms."

Somehow, the clause "a difference generally implies positive terms between which the difference is set up" sounds to me like it refers to some sort mathematical idea.

I would appreciate any comments which might shed light on the meaning or references of this clause. Thwap (talk) 15:18, 8 January 2009 (UTC)

As far as I can see, it has nothing to do with mathematics. I'd understand "positive" here as "factually existing", in a similar sense as in positive statement, positive science or positivism. Anyway, you may have better luck at the language reference desk when it comes to interpreting de Saussure. — Emil J. 16:09, 8 January 2009 (UTC)

I've posted this also in the Language Ref Desk but get the same type of responses.

When I consider the original context of the statement and the phrase "a difference generally implies positive terms", the operative word being "difference", there is nothing specific to language being stated here. I get the sense that the author is referring to some kind of mathematical operation used to derive differences whereby is matters whether the terms are positive or negative.68.157.93.254 (talk) 20:40, 8 January 2009 (UTC)

More context is needed in order to decipher de Saussure. Bo Jacoby (talk) 21:18, 8 January 2009 (UTC).

The statement occurs in "A Course in General Linguistics" on page 118, in the section entitled "The Sign as a Whole", near the beginning of the first paragraph. The document can be found here:[2]Thwap (talk) 11:44, 9 January 2009 (UTC)

He is not trying to refer to any sort of higher mathematics. The wikipedia article on positive statement is about the term in economics, but rather one should use 7th definition on wikt:positive. There is something in logic and philosophy (related to positivism) that tries or tried to distinguish between positive and negative statements. "This is a dog." is positive, and "This is not a dog." is negative. The reference to differences may be to the idea that a difference, as a "lack", is a negative statement associated to two positive statements. "I had 20 dollars yesterday. I have 10 dollars today. I lack 10 dollars today." The first might be called positive, the second is positive, and the third might be called negative.

I learned about this stuff in a modal logic course, and one of the tasks was to decide if a given collection of symbols was a positive statement or a negative statement (I think the answer was, there is no such decision procedure since the law of the excluded middle does occasionally hold).

At any rate, the point of p116-118 is that the symbols used in language are arbitrary because their primary purpose is not to carry meaning, but rather to be distinguished from each other. The mathematical version of this is called coding theory. It does not matter what bit-strings are used to encode a message, only that it can be reliably decoded. Some Chinese characters consist of two symbols, one that refers to pronunciation and one to meaning, but in some sense the pronunciation part is to distinguish from other words that relate to the same meaning, and the meaning part is to distinguish it from other words that have the same pronunciation. The fact that both are sometimes related to pictures of real things has by this time become only a mnemonic device.

Some people disagree with this sort of thing, and think that the letters, numbers, sounds, etc. we use to communicate have intrinsic meaning. The only easy examples I know of these are somewhat silly or old-fashioned: divine language as in, before the tower of babel meaning and symbol were the same, only after were they separate and confusing; Pseudoscientific metrology#Charles Piazzi Smyth and the Egyptian inch. Probably the textbook here is trying to clearly state that in its analysis of language, there is no "mystical" meaning in language, and one does not need to examine the symbols themselves, only how they differ from other symbols (or earlier versions of the same symbol, etc.).

At any rate, I think the language ref desk answer is sufficient to understand the passage. When a linguist refers to mathematics by analogy, the linguists should explain, not the mathematicians. To a mathematician, it just seems like an expression of the basic explanation of the applications of coding theory to communication. JackSchmidt (talk) 19:12, 9 January 2009 (UTC)

Thanks for your help.Thwap (talk) 11:22, 10 January 2009 (UTC)

Question relating to the discriminant

Im given a quadratic equation $x^{2}+kx+k=0$ and Im asked to write down the discriminant in terms of k. I did this and got $k^{2}-4k$.

Then Im asked to find the set of values k can take so I did the following:

$k^{2}-4k<0$ $k(k-4)<0$ Im pretty sure this is all right so far but I dont know what to do next. Is it that the 2 possible solution are k<0 and k<4 so overall k<4 or what? --212.120.247.244 (talk) 21:03, 8 January 2009 (UTC) perhaps you are searching for those value of k ,for which the given equation has real roots,for this take k(k-4) > 0 instead of k(k-4) < 0. and check the values of k(k-4) " any value in this range " , in intervals k<0 , 0 < k < 4 and k > 4 then you will get the required values . —Preceding unsigned comment added by Khubab (talk • contribs) 21:28, 8 January 2009 (UTC)

I'm afraid that this question is really meaningless. For a start, k can be anything; I don't see the problem with complex numbers. If you want real roots, then k is less than or equal to 0 or k is greater than or equal to 4. This is easy to see because k * (k-4) is greater than or equal to 0, when either both factors are greater than 0, both less than 0 (the product of two negative numbers is positive), or one of the factors is 0.

Also note that the discriminant can be precisely 0 for real roots. Hope this helps. --Point-set topologist (talk) 22:00, 8 January 2009 (UTC)

to find the set of values k can take in order to what? You forgot to turn the page, maybe? ;) --84.220.230.137 (talk) 22:14, 8 January 2009 (UTC)

I imagine there is an unstated assumption that k is real, so the question is intended to be "what is the range of the function f:R->R, f(k) = k² − 4k". Gandalf61 (talk)

January 9

a question regarding liters and milliliters

can anyone answer me how many milliliters are in a 1/16 liter? i dont think it could be answered; my friend thinks it is approx 62.5(?) would appreciate a quick response. thank you04:28, 9 January 2009 (UTC)Anilas (talk)

1 litre is 1000 millilitres, by definition. Thus 1/16 of a litre is 1/16 of 1000 millilitres. For more, see long division. Algebraist 04:38, 9 January 2009 (UTC)

Or calculator. ;) --Tango (talk) 12:49, 9 January 2009 (UTC)

It is exactly 62.5 mL; your friend is a genius (I am dumbfounded as to why he is still solving these trivial problems).

P.S What I am trying to say is that you can just evaluate 1000/16 on google (search 1000/16 on google). Please don't ask trivial problems at the reference desk.

PST —Preceding unsigned comment added by Point-set topologist (talk • contribs) 12:59, 9 January 2009 (UTC)

Google can do more than that. [3]. Taemyr (talk) 13:38, 9 January 2009 (UTC)

Is this equation solveable without a computer/brute force?

let a,b & c exist in set of natural numbers { 0,1,2 ... }
a < b < c
a^2 + b^2 = c^2
a + b + c = 1000

Find a,b and c.212.23.11.198 (talk) 11:49, 9 January 2009 (UTC)

Find a right triangle composed of three line segments; the sum of the lengths of which equals 1000 and such that lengths of the line segments form a set of three distinct natural numbers. --Point-set topologist (talk) 11:51, 9 January 2009 (UTC)

Yes, it is. There is a unique solution which can be found in under ten minutes of thought. Algebraist 12:19, 9 January 2009 (UTC)

(After edit conflict - maybe it took me 11 minutes !) Yes, there is a straightforward algebraic approach. The numbers a, b and c form a Pythagorean triple, so their sum has a particular factorisation, which allows you to find candidate triples quite easily if you are given their sum. For a sum of 1000 there is, I think, only one solution. Gandalf61 (talk) 12:26, 9 January 2009 (UTC)

Hint: use Euclid's formula for generating pythagorean triples (see the first section of Pythagorean triple) and figure out what k, m, and n have to be for the sum a+b+c to equal 1000. You should get two answers for triples k, m, and n, but these two will turn out to give the same triple a, b, c. kfgauss (talk) 17:58, 9 January 2009 (UTC)

Algebra II word problem

I've hit a brick wall on this word problem:

"In his job at the post office, Eddie Thibodeaux works a 6.5-hr day. He sorts mail, sells stamps, and does supervisory work. One day he sold stamps twice as long as he sorted mail, and he supervised 0.5 hr longer than he sorted mail. How many hours did he spend at each task?"

Word problems are my Kyptonite. --24.33.75.35 (talk) 23:23, 9 January 2009 (UTC)

Please do your own homework.

Welcome to the Wikipedia Reference Desk. Your question appears to be a homework question. I apologize if this is a misinterpretation, but it is our aim here not to do people's homework for them, but to merely aid them in doing it themselves. Letting someone else do your homework does not help you learn nearly as much as doing it yourself. Please attempt to solve the problem or answer the question yourself first. If you need help with a specific part of your homework, feel free to tell us where you are stuck and ask for help. If you need help grasping the concept of a problem, by all means let us know. Algebraist 23:25, 9 January 2009 (UTC)

I only need help with writing the exact equation. After that, I can solve it myself. I know there are 3 jobs or 3 "x's". All of which = 6.5. Should it be x + x(2) + x(0.05) = 6.5? --24.33.75.35 (talk) 23:33, 9 January 2009 (UTC)

There are three quantities involved, as you correctly observe, so calling them all by the same letter is a bit silly. Why not call them x, y and z, or perhaps m, s, and w (for Mail, Stamps, and supervisory Work)? Algebraist 23:34, 9 January 2009 (UTC)

Ok. m + s(2) + w(0.05) = 6.5. Even with the newly labeled quantities, that equation does not look right to me. --24.33.75.35 (talk) 23:47, 9 January 2009 (UTC)

It's dangerous to write down symbols without knowing what they are supposed to mean. What, for example, does w(0.05) mean in your formula? Algebraist 23:51, 9 January 2009 (UTC)

It's where he worked 0.5 hr longer supervising than he did sorting. Just multiply the two and add the difference to one of the quantities. However, I'm starting to think this word problem has more than one equation. --24.33.75.35 (talk) 23:56, 9 January 2009 (UTC)

Very good. Why don't you forget about equations for a second and think about what relations between your quantites are given by the question? Algebraist 23:59, 9 January 2009 (UTC)

I've got to go to work. I'll toy with it on my breaks and tell you what I come up with. Thanks for the help so far. --24.33.75.35 (talk) 00:07, 10 January 2009 (UTC)

I'll give you a hint - in order to solve a problem with 3 variables you need 3 equations. Read through the information given and try and find 3 relationships. --Tango (talk) 00:49, 10 January 2009 (UTC)

Using x to represent "sort", 24.33...'s original approach seems fine but his/her 3rd term is wrong. It shouldn't be (x(0.05)) but should be (x+0.5), thus using one variable and one equasion:

(x)+(2*x)+(x+0.5)=6.5

4*x+0.5=6.5

4*x=6

x=1.5

hydnjo talk 03:06, 10 January 2009 (UTC)

True, although I think it is best to do it step by step when learning rather than substituting the 2nd two equations into the first without writing them down. It makes it less likely to make mistakes like the OP did with the final term. --Tango (talk) 03:13, 10 January 2009 (UTC)

Also true except that the OP grasped the concept but made a bit of a beginner's mis-step with that third term which could happen regardless of method. I'd encourage the original method as he/she did understand the simplest notation (forgiving that 3rd term). hydnjo talk 03:21, 10 January 2009 (UTC)

As I said, doing it step by step and writing down all the steps makes it less likely that you will make such a mistake. --Tango (talk) 23:34, 10 January 2009 (UTC)

Now this thread is a bit silly. The OP was just driven away and is not going to learn anything (despite your efforts to make him learn something). I am not saying that you didn't follow the right path in teaching the OP, but I think that you could have done it a lot quicker and more easily. PST —Preceding unsigned comment added by Point-set topologist (talk • contribs) 12:56, 11 January 2009 (UTC)

January 10

Does this continued fraction converge?

I was playing around with [1;2,1,3,1,4,1,5....] at school today, and I'm fairly sure it converges to a value around 1.44. Is there a test that I can run to see if it does converge? Does anyone know if this particular series turns out to equal something interesting? Thanks. 24.18.51.208 (talk) 01:44, 10 January 2009 (UTC)

For a start you may find continued fraction interesting. Every continued fraction with positive integer coefficients will converge; yours converges to about 1.35804743869438. It is irrational (because the continued fraction is infinite), and furthermore is not the root of any quadratic equation (because the continued fraction is not periodic). The continued fraction looks similar to the continued fraction for ${\frac {\tan(1)-1}{2-\tan(1)}}=1.259415845$, which has continued fraction [1;3,1,5,1,7,1,9,...]. Perhaps if you can figure out how to calculate the continued fraction for tan (1) (which has continued fraction [1;1,1,3,1,5,1,7,1,9,...]) then that might give some ideas of how to find the value of [1;2,1,3,1,4,1,5,...]. Eric. 68.18.17.165 (talk) 04:23, 10 January 2009 (UTC)

You may also find this calculator interesting if you just want an evaluation. It gives 641/472 = 1.3580508474576272 = 1, 2, 1, 3, 1, 4, 1, 5 hydnjo talk 04:59, 10 January 2009 (UTC)

Pentagons

With only a sheet of paper, a ruler and a pencil, how might I go about drawing a regular pentagon of which each side is 150mm long, without having any construction lines around it afterward? The easiest way seemingly might be to start with one line 150mm long and work out how far away horizontally and vertically from that the end of each other line would be, but how should I do that? 148.197.114.165 (talk) 11:52, 10 January 2009 (UTC)

Can you use a compass? If so, see Pentagon#Construction. You'll need the formula in the previous section to work out what radius to use. Zain Ebrahim (talk) 12:00, 10 January 2009 (UTC)

Actually, this is better because it gives links to details for each step. Zain Ebrahim (talk) 12:13, 10 January 2009 (UTC)

I think I've worked it out. Would this give the right shape?: To draw regular pentagon ABCDE, where s is the length of any side of the pentagon, create a rectangle of size (1+(5^1/2)/2)*s by (((1+(5^1/2)/2)*s)^2 - s/2^2)^1/2. Draw then the line AB, of length s along one of the longer sides of the rectangle, so both lines have their centres at the same point. Draw the line AE from A to the nearest of the shorter sides of the rectange such that its length equals s, then do the same for BC from B to the other side. Find the centre of the long side opposite line AB, mark this as point D and draw lines from this to both C and E.

For a pentagon with sides length 150mm, this rectangle would be 242.7mm by 230.82mm, with A and B 46.35mm from the corners.

Something like that? 148.197.114.165 (talk) 17:48, 10 January 2009 (UTC)

Try it and see. It's easy enough to tell if you have the right shape at the end. --Tango (talk) 23:32, 10 January 2009 (UTC)

max size of a k-component graph

Resolved

 – --Shahab (talk) 14:24, 11 January 2009 (UTC)

This is related to a question I asked a few days back. What is the maximum number of edges possible in a n vertex graph having k connected components where each component has $n_{i}$ vertices. Obviously $\sum n_{i}=n$ and I need the maximum value of ${\frac {1}{2}}\sum n_{i}(n_{i}-1)$. How should I proceed further? I want the final answer to be in terms of n and k. Thanks--Shahab (talk) 13:06, 10 January 2009 (UTC)

As with the k=2 case, the maximal case is when all but one component has 1 vertex. Algebraist 17:00, 10 January 2009 (UTC)

I came to the same conclusion by the following procedure: As $\sum n_{i}^{2}=n^{2}-(k-1)(2n-k)-2\sum _{i<j}(n_{i}-1)(n_{j}-1)$ so if all but component has 1 vertex then the number of edges is ${\frac {1}{2}}(n-k)(n-k+1)$. Since this acts as an upper-bound on the number of edges too (because of the negative sign in $-2\sum _{i<j}(n_{i}-1)(n_{j}-1)$) hence this is the maximal achievable value for the number of edges. Is this proof correct? Secondly, is there a way to maximize the function ${\frac {1}{2}}\sum n_{i}(n_{i}-1)$ subject to the constraints $\sum n_{i}=n,n_{i}\in \mathbb {N}$. Cheers--Shahab (talk) 17:31, 10 January 2009 (UTC)

Shahab asks "is there a way to maximize the function $\textstyle {\frac {1}{2}}\sum n_{i}(n_{i}-1)$ subject to the constraints $\textstyle \sum n_{i}=n,n_{i}\in \mathbb {N}$?" The answer is to take $n_{1}=n$. But perhaps Shahab meant to ask the related question: how to minimize the function $\textstyle \sum n_{i}$ subject to $\textstyle {\frac {1}{2}}\sum n_{i}(n_{i}-1)=N$. Bob Robinson (UGA) and I solved this problem more than 20 years ago but never published it. Consider the greedy approach: choose $n_{1}$ as large as possible so that $\textstyle {\frac {1}{2}}n_{1}(n_{1}-1)\leq N$, then continue in the same fashion with what is left. Our theorem was that this works (i.e. the greedy choice of $n_{1}$ is correct) for all but a finite set of values of N, which we determined. If I recall correctly, there were something like 20 exceptions and the largest was about 86,000. In the exceptional cases, $n_{1}$should be chosen 1 or 2 less than the maximum possible. McKay (talk) 01:59, 11 January 2009 (UTC)

Thanks for the extra info. But my initial question was about maximizing ${\frac {1}{2}}\sum n_{i}(n_{i}-1)$ subject to $\sum n_{i}=n,n_{i}\in \mathbb {N}$ where I meant $\mathbb {N} :=\{1,2\cdots \}$. I cannot take $n_{1}=n$ for then the graph is just $K_{n}$ and I need k connected components in the graph. Cheers--Shahab (talk) 05:15, 11 January 2009 (UTC)

But that question is already answered above: take k-1 components equal to 1 and one component equal to n-k+1. The fact that it is best for k=2 proves that it is best for arbitrary k also. McKay (talk) 11:11, 11 January 2009 (UTC)

I am probably being dense here (& for this I ask you to indulge me) but I cannot understand your last statement. Isn't it possible for the max value to be different then ${\frac {1}{2}}(n-k)(n-k+1)$ and yet be equal to ${\frac {(n-2)(n-1)}{2}}$ for k=2. Can you please clarify?--Shahab (talk) 12:46, 11 January 2009 (UTC)

McKay probably has in mind the following simple proof (also the proof I had in mind above, for what that's worth): suppose we have values n_i subject to $\sum n_{i}=n,n_{i}\in \mathbb {N}$, and suppose there are two values of i for which n_i is not 1. Then by the k=2 case, we can increase ${\frac {1}{2}}\sum n_{i}(n_{i}-1)$ by changing one of these two values to 1 (and the other to their sum minus 1). Hence in the maximal case, all but one n_i must be 1. Algebraist 13:32, 11 January 2009 (UTC)

OK. Thanks--Shahab (talk) 14:24, 11 January 2009 (UTC)

Multiplying integers by digit-swapping.

I've been playing around with integers which are multiplied by a factor (≤9) when the RH digit is moved to the LH end. For example, 102564 is quadrupled when the 4 is moved to the start. There is obviously an infinite number of these, seen by considering 102564102564, so I'm interested in the smallest integer for each multiplier. For multiplier k, my analysis gives that the integer comes from an expression with 10k-1 in the denominator, so 102564, for example, must have something to do with thirty-ninths. And 1/39 = 0.0256410256410..., so as a hypothesis the smallest integer for a given multiplier is found by looking at the decimal expansion of 1/(10k-1) and selecting the cycle starting 10. Which brings me to the question - acting on the basis outlined, I've found that the 58-digit integer 1,016,949,152,542,372,881,355,932,203,389,830,508,474,576,271,186,440,677,966 is multiplied by 6 when the final digit is moved to the front, but is there a smaller one?→81.151.247.41 (talk) 19:51, 10 January 2009 (UTC)

There is certainly a name for an integer whose digits get swapped (as you have described) when multiplying with a positive integer less than or equal to 9. Although I can't seem to remember... Can any number theorist help out here? —Preceding unsigned comment added by Point-set topologist (talk • contribs) 20:38, 10 January 2009 (UTC)

According to Sloane's, these are called k-parasitic numbers and the OP's number is indeed the least 6-parasitic number. Algebraist 20:42, 10 January 2009 (UTC)

3x3 Antisymmetric Matrices

Let A be a real 3×3 non-zero antisymmetric matrix. How does one show that there exist real vectors u and v and a real number k such that Au = kv and Av = −ku? What relation do these u,v and k bear to the exponential of the matrix A?

I know the exponential of the matrix A is a rotation matrix, but having not seen an appropriate approach to the first part of the question, I'm unsure as to how to progress with this - thanks.

86.9.125.104 (talk) 23:40, 10 January 2009 (UTC)Godless

If Au = kv and Av = -ku, then A²u = kAv = -k²u, and similarly A²v = -kAu = -k²v. So one approach would be to see what you can show about the eigenvalues and eigenvectors of A², given that A is 3x3 antisymmetric. Gandalf61 (talk) 09:31, 11 January 2009 (UTC)

Three '2's Problem

You have 3 twos, and you must use all of them in any expression of a number. Operations allowed are:

addition (e.g 2+2+2=6)

subtraction (e.g 2-(2-2)=2)

multiplication (e.g 2*(2+2)=8)

division (e.g 2*(2/2)=2)

raising to a power (e.g 2^(2^2)=16)

rooting - i.e. raising to the power of 1/n, where n is an integer (e.g 2^(1/(2+2))=2^(1/4))

logarithm to base 2 or base e - if to base 2, you have to indicate, and that would mean using one of your three twos: otherwise it would be to the base e. (e.g log2(2*2)=2, or log(2*2*2)=3ln(2), although the latter is not an integer)

concatenation - i.e combining say 3 and 4 to get 34 (Using "~" for this: e.g 2~(2/2)=21)

How many distinct positive integers can you form in this way? (note that for example 2+(2+2)=2+(2*2) so this only counts as 1 distinct positive integer).

Thanks,

Mathmos6 (talk) 23:58, 10 January 2009 (UTC)Mathmos6

This problem is fairly trivial: I suggest that you have a go yourself. We won't give you answers here; we can only give you hints at the most. Please first show us your progress (tell us what is the smallest and highest possible positive integer that you can obtain). —Preceding unsigned comment added by Point-set topologist (talk • contribs) 13:01, 11 January 2009 (UTC)

Without concatenation, it's pretty trivial. With concatenation, however, you can get significantly larger integers. --Tango (talk) 16:47, 11 January 2009 (UTC)

See also four fours. -- SGBailey (talk) 19:33, 11 January 2009 (UTC)

January 11

Group of points on the unit sphere

Let G be a finite non-trivial subgroup of SO(3). Let X be the set of points on the unit sphere in R^3 which are fixed by some non-trivial rotation in G. G acts on X - show that the number of orbits is either 2 or 3. What is G if there are only two orbits?

Could really just use a hand getting started on this one if nothing else. If the group is finite, I think all rotations have to be of the form 2pi/k for some integer k, and all axes of rotation have to be at an angle of 2pi/k to all other axes for a (different) integer k, right? That's about as much as I've accomplished so far. Cheers guys,

Spamalert101 (talk) 00:22, 11 January 2009 (UTC)Spamalert101

If the elements of G are not all rotations about the same axis then G must be the rotational symmetry group of one of the Platonic solids. There are three different "types" of points that are fixed by some non-trivial element of G and permuted by the other elements of G - these form the three orbits of G acting on X. One orbit is the vertices of the Platonic solid - I'll let you work out what the other two orbits are (hint: think about dual solids which have the same symmetry group).

On the other hand, if the elements of G are all rotations about the same axis then all non-trivial elements of G have the same two fixed points (what are they ?) and hence each of these two points is an orbit, and we have just two orbits. Gandalf61 (talk) 13:44, 11 January 2009 (UTC)

Is your first sentence obvious? Algebraist 14:13, 11 January 2009 (UTC)

Trivial to be precise (I was joking). :) PST —Preceding unsigned comment added by Point-set topologist (talk • contribs) 16:11, 11 January 2009 (UTC)

Having brushed up my geometry with our article Point groups in three dimensions, I can now say that it is not obvious, or even true. Gandalf has forgotten the rotation groups of the regular prisms. In this case, there are 3 orbits. Algebraist 17:22, 11 January 2009 (UTC)

To be more precise, if the elements of G are not all rotations about the same axis then G must be a subgroup of the rotational symmetry group of one of the Platonic solids. Moreover, considering only the icosahedron and octahedron suffice, as the dodecahedron has the same symmetry as the icosahedron, the cube the same symmetry as the octahedron, and tetrahedral symmetry is a subgroup of octahedral symmetry. Eric. 68.18.17.165 (talk) 16:27, 11 January 2009 (UTC)

(Reply to Algebraist) Not entirely obvious, I suppose - there is a proof in Chapter 15 of Neumann, Stoy and Thompson's Groups and Geometry. But maybe there is a simpler (although less geometric) solution to the original problem using Burnside's lemma. Each element of G fixes just 2 points of X, apart from the identity element which fixes all |X| points, so the sum over G of the points of X fixed by each element of G is |X| + 2|G| - 2. But by Burnside's lemma this is a multiple of |G| - in fact it is |G| times the number of orbits. I'll let the OP take it from there. Gandalf61 (talk) 17:25, 11 January 2009 (UTC)

Thanks for the reference. Google books reveals that the correct result is that the finite subgroups of SO(3) are exactly those you mentioned above plus the rotation groups of prisms (including the rectangle as a degenerate prism). The subgroups of these groups Eric alludes to in fact turn out to be on the list already. Algebraist 17:38, 11 January 2009 (UTC)

I agree with all of that - so which part of my original reply do you still think is incorrect ? Gandalf61 (talk) 17:54, 11 January 2009 (UTC)

The first sentence, since it excludes the prism/dihedron case, with dihedral symmetry groups. Algebraist 17:58, 11 January 2009 (UTC)

All of the non-trivial elements of the rotational symmetry group of a prism/dihedron have the same axis. So my first sentence "If the elements of G are not all rotations about the same axis then G must be the rotational symmetry group of one of the Platonic solids" is correct. The prism/dihedron case is covered in my second paragraph which starts "On the other hand, if the elements of G are all rotations about the same axis ...". In short, 3 orbits<=>Platonic sold, 2 orbits<=>prism/dihedron. Gandalf61 (talk) 18:07, 11 January 2009 (UTC)

No, they do not. For the prism associated with the regular n-gon, there are n-1 non-trivial rotations about an axis through the centres of the ends, plus 1 non-trivial order 2 rotation about the centre of each other face and about the midpoint of each edge running between the ends. Thus there are 3 orbits. Algebraist 18:13, 11 January 2009 (UTC)

If one took the prism with vertices (±1,±2,±3), then it has a symmetry group whose intersection with SO(3) has 8 elements, right? It is generated by rotations of 180° about the x, y, and z axes, right? It is not a platonic solid, right? I don't really get the geometric ideas though. I like your counting proof, since it only uses the fact that the elements fix at most 2 points. JackSchmidt (talk) 18:18, 11 January 2009 (UTC)

No, that group has 4 elements. It's the same as the rotation group of the rectangle-in-space, which I included in my list as a degenerate regular prism (the prism on a 'regular 2-gon'). It may gratify you to know that Gandalf's counting argument is used as the basis of the classification of finite rotation groups in the reference he supplied. Algebraist 18:24, 11 January 2009 (UTC)

I agree, 4 elements: the identity and the three 180° degree rotations about the x,y,z axes. I did get that it was the same as the flat rectangle, but I just misremembered you got the dihedral group of order 8. Having no mind for geometry, this seemed reasonable to me, though I was a little concerned that the group appeared to be abelian. I took out my very expensive model of a rectangular prism and applied the rotations carefully to check. Who says a geometry textbook is useless? Glad to know the classification uses ideas I could grasp. Even with 4 elements, this is one of your (A's) counterexamples to G's first sentence, right? Not that it matters that much, since the structure of G's proof is right, and checking the details should always be a part of reading someone's answer. JackSchmidt (talk) 18:37, 11 January 2009 (UTC)

It's not my counterexample, I'm hopeless at geometry. I lifted it from our article. Algebraist 19:58, 11 January 2009 (UTC)

(Reply to Algebraist) Yup, dumb mistake on my part. But it's always good to know that you are waiting to catch any little slips like that from us amateurs. Gandalf61 (talk) 18:44, 11 January 2009 (UTC)

Unidentified trig functions/identities?

Where $\Delta \theta =\theta _{f}-\theta _{s};\quad \Sigma \theta =\theta _{f}+\theta _{s}\,\!$, are

${\frac {\cos(\Sigma \theta )+\cos(\Delta \theta )}{\sin(\Delta \theta )}}={\frac {2\cos(\theta _{s})\cos(\theta _{f})}{\sin(\Delta \theta )}}={\frac {2}{\tan(\theta _{f})-\tan(\theta _{s})}}=?;\,\!$

and

${\frac {\cos(\Sigma \theta )-\cos(\Delta \theta )}{\sin(\Delta \theta )}}={\frac {2\sin(\theta _{s})\sin(\theta _{f})}{\sin(\Delta \theta )}}={\frac {-2}{\cot(\theta _{f})-\cot(\theta _{s})}}=?;\,\!$

(or their reciprocals) recognized as special functions, particularly in regards to spherical trigonometry? ~Kaimbridge~ (talk) 01:13, 11 January 2009 (UTC)

These are elementary consequences of the formulas for sin(A+B) and cos(A+B). What is your question exactly? McKay (talk) 01:26, 11 January 2009 (UTC)

Just if there is some elementary relationship/identity, like the "sine for sides", or in the same way that

${\frac {\lambda _{f}-\lambda _{s}}{{\mbox{arccosh}}(\sec(\phi _{f}))-{\mbox{arccosh}}(\sec(\phi _{s}))}}={\frac {\Delta \lambda }{\int _{\phi _{s}}^{\phi _{f}}\sec(\phi )d\phi }}=\tan(\alpha ){}_{\color {white}.}\,\!$

for loxodromic azimuth, which involves the inverse Gudermannian function. ~Kaimbridge~ (talk) 15:25, 11 January 2009 (UTC)

Conic section slope of directrix

Given a conic section $Ax^{2}+By^{2}+Cxy+Dx+Ey+F$ what is the slope of its directrix? Borisblue (talk) 01:21, 11 January 2009 (UTC)

Sadly, we don't do homework for people. Furthermore, this problem is trivial. Please think about it a bit more and tell us what you have done on the problem so far. PST —Preceding unsigned comment added by Point-set topologist (talk • contribs) 16:10, 11 January 2009 (UTC)

Cubic functions

I'm trying to use cubic (and similar) functions for a little model, but I really don't have much maths. Could a mathematician please explain whether one changes a, b or c to make the curves flatter/steeper, change the y intercept, etc.? The article on quadratic equations does this graphically through a clever image, but a few lines of text would be great. I'd like to add this info to the relevant article as well. --Matt's talk 14:09, 11 January 2009 (UTC) Edited to clarify which article is relevant --Matt's talk 14:17, 11 January 2009 (UTC)

I presume you mean that the cubic is of the form:

$f(x)=ax^{3}+bx^{2}+cx+d\,$

In that case, "flatness" of the curve is measured by its derivative which is (at x):

$f'(x)=3ax^{2}+2bx+c\,$

So only the coefficients a, b and c have an impact on the steepness of the curve (the greater these values are, the greater the steepness; the smaller these values are, the greater the flatness). The y-intercept is given by the image of 0 under f so the value of d equals the y-intercept. If d is 0, the curve passes through the origin. PST

The article elliptic curve might also be of interest to you. PST

And by the way, mathematicians usually use one branch of mathematics in another branch of mathematics. There are numerous examples of this (I might as well let someone else list these examples; there are so many that I can't be bothered!). One interesting example is applying graph theory and the theory of covering maps to prove the well known Nielson-Schreier theorem; i.e every subgroup of a free group is free.

On the same note, there are mathematicians who would prefer not applying mathematics to another field (theoretical mathematicians) and those who would prefer applying mathematics to another field (applied mathematician). From experience, applied mathematicians are generally not so interested in the theoretical parts of mathematics and thus do not choose to learn much theoretical mathematics. But there are special cases. PST —Preceding unsigned comment added by Point-set topologist (talk • contribs) 16:05, 11 January 2009 (UTC)

Elliptic curves aren't really relevant to what the OP is doing, and what does that last paragraph have to do with anything? --Tango (talk) 17:00, 11 January 2009 (UTC)

Have a look at his links to see the relevance of that part. PST

The steepness of the graph for large (either positive or negative) values of x is determined primarily by a. For smaller values of x, the graph will change direction a lot so it's rather more complicated. You may find it helpful to write the cubic as y=a(x-u)(x-v)(x-w), then the steepness for large values of x is, again, given by a, and u, v and w are the x-intercepts. The y-intercept would be -auvw. --Tango (talk) 17:00, 11 January 2009 (UTC)

Re-write equation as:

$y=a\left(x+{\frac {b}{3a}}\right)^{3}+\left(c-{\frac {b^{2}}{3a}}\right)\left(x+{\frac {b}{3a}}\right)+e$

where e is a function of a,b,c and d that I can't be bothered to write out. Then change co-ordinates:

$v=y-e\,;\,u=\left(x+{\frac {b}{3a}}\right)$

$v=au^{3}+\left(c-{\frac {b^{2}}{3a}}\right)u$

so we have placed the cubic's centre of symmetry at the origin. Now we can see that a determines the slope of the cubic far from its centre and $\left(c-{\frac {b^{2}}{3a}}\right)$ determines the slope at its centre, and the number of turning points. Gandalf61 (talk) 17:44, 11 January 2009 (UTC)

Sounds to me like you might be interested Bézier curve and splines in general.Dmcq (talk)

Manifolds

I have a question for those people at the reference desk. This is not homework so you don't need to worry about that. I would like to know why manifolds (and more generally differentiable manifolds) are so important objects of study. I mean, they are in a sense so restrictive in nature (why exclude so many important topological spaces from study?). To me, it seems odd that the figure 8 is excluded from study (why can't you do calculus at the center of the figure 8?). I know that there are more general types of manifolds such as orbifolds and Banach manifolds but is every topological space some type of manifold? If not, what are the minimal conditions required for a space to be some type of manifold? I am pretty sure, for instance, that one can do calculus on the indiscrete space (any function between indiscrete spaces should have derivative 0). But on the other hand, I am also sure that the indiscrete space is not a type of manifold. I certainly understand that manifolds are important but unlike continuity and integration, I don't understand why differentiation cannot be generalized to arbitrary topological spaces (integration to measure spaces). Or it maybe just my lack of knowledge of the subject that I don't know that one can do calculus on arbitrary topological spaces. Thankyou for you help! —Preceding unsigned comment added by 129.143.15.142 (talk) 20:53, 11 January 2009 (UTC)

Note that the indiscrete space is indeed a (0-dimensional) manifold if the space in question has precisely one-point or vacuously a manifold if the space in question is empty. But those are just fine details for you. PST

It's not an area I've ever studied properly, but I can't see how you could define derivatives on a space unless the space was locally metrizable. That rules out some spaces (but not things like a figure of 8). Also relevant is that, depending on your definitions, the long line is not a manifold, but can have a differential structure put on it. --Tango (talk) 22:00, 11 January 2009 (UTC)

Thanks. The definition I follow is the "locally Euclidean" one so I allow the long line to be a manifold. I think one other important thing to consider is that in manifold theory there are important facts about the derivative of a smooth function between smooth manifolds (namely it is a linear isomorphism between tangent spaces). But on one of your notes, the indiscrete space is not locally metrizable (when it has more than one point as someone above mentioned) but I would expect that the derivative of a function between such indiscrete spaes to be 0. You could argue similarly that continuity needs a metric although that is not the case. Isn't there an 'open set formulation' of differentiability? —Preceding unsigned comment added by 129.143.15.142 (talk) 22:16, 11 January 2009 (UTC)

Just to clarify something I said, I think that people don't study calculus on locally metrizable spaces because often in manifold theory, one considers vector spaces also. These are basically my questions (by the way, a real-valued function defined on the figure 8 is differentiable if and only if it extends to a differentiable function on a open set containing the figure 8 but this cannot be generalized to topological spaces that cannot be embedded in Euclidean space; this brings me to another question):

1) Since all Hausdorff, second countable manifolds can be embedded in Euclidean space, why not define differentiability on them as I have just done (i.e differentiable iff extends to a differentiable function on an open superset?)? I know this would exclude all those interesting theories of differentiable structures (like one interesting theorem that there are only 28 non-equivalent differentiable structures on S^7), but for the most part, this definition would suffice.

2) Is the figure 8 some type of manifold?

3) What are minimal conditions for a space to be some sort of manifold?

4) Is the indiscrete space (with more than one point) some sort of manifold (I would expect this naturally although I don't think that it is)?

I have not been getting sleep over this since I first learnt about manifolds (why are they so restrictive in nature?) so I would appreciate any help from the knowledgeable people at WP! —Preceding unsigned comment added by 129.143.15.142 (talk) 22:26, 11 January 2009 (UTC)

1) While such manifolds can be embedded in Euclidean space, they can be embedded in lots of different ways which wouldn't yield the same definition of differentiability (for example, a 2-sphere can be embedded in R³ as a cube, in which case the image of a great circle wouldn't be differentiable since it would have corners in it, but with the standard embedding it clearly is). While you may be able to improve things by requiring the embedding to be smooth, you would end up with a circular definition. 2) A figure of 8 isn't a manifold simply because there is no open neighbourhood of the central point that is homeomorphic to the real line. Mathematicians define things the way they do because those definitions are useful. For example, if a space is locally Euclidean at a point you can define its tangent space at that point (which is, itself, a useful thing to do for all kinds of purposes), you can't define the tangent space to a figure of 8 at that central point (it has two tangents there, so you would end up with the union of two lines, which isn't a vector space). 3) A manifold is a space that is locally Euclidean everywhere (possibly with some extra conditions, depending on who you ask), that is the minimal condition (I'm not really sure what you mean by "some sort" of manifold, there are generalisations of manifolds, but they are really manifolds any more even if the word may appear in their name, you could argue that "topological space" is a generalisation of "manifold", but that doesn't mean much). 4) No, the only neighbourhood of any point in an indiscrete space is the whole space, which can't be homeomorphic to any Euclidean space because no Euclidean space (beyond R⁰, I guess) is indiscrete. --Tango (talk) 00:04, 12 January 2009 (UTC)

January 12

A Pascal's triangle question I've never seen before

Dear Wikipedians:

I'm used to the following type of questions which applies Pascal's triangle, it asks how many ways to form the word "November":

    N
   O O
  V V V
   E E
    M
   B B
  E E E
 R R R R

as you can see each line is always 1 character more or less than the line above it, however, recently my teacher gave the following variation, which totally stumped me:

    N
   O O
  V V V
  E E E
   M M
 B B B B
  E E E
   R R

as you can see there are now lines with same number of characters as previous line, and also line with 2 characters more than the previous line, how do I handle this?

Thanks,

76.65.14.151 (talk) 00:09, 12 January 2009 (UTC)

I assume the goal is to move from top to bottom without "jumping too far". If no rules are specified then I would guess (and state that as an assumption if possible) that if an M is directly above a B then you can only move directly down to that B, so two of the B's cannot be reached. If the question form allows it then you could also make two answers where the other assumes you can move left-down, straigth down, or right-down, so from each M you can move to 3 B's. PrimeHunter (talk) 00:27, 12 January 2009 (UTC)

I'm not sure how to interpret the question in the second case - how do you form the word? Are you allowed to move diagonally when letters are directly on top of each other? If not, then the question is easily reduced to one like the first case. If you are, then it depends on precisely what is and isn't allowed. Incidentally, I don't see how Pascal's triangle comes into it - for the first case the answer is just 2ⁿ where n is the number of lines which are longer than the line before (since, when going to those lines, you have a choice of 2 letters to go two, for all the other lines you just have 1 choice). --Tango (talk) 00:32, 12 January 2009 (UTC)

The question is understated, can we have some more information please; such as an explicit copy of the said question. In your post you only reference the question you are trying to answer without actually stating what it is that you are trying to do. I too fail to see any relation to pascals triangle. Or the fact that the layers form a word. Or what you mean by 'handle this'. —Preceding unsigned comment added by 92.16.196.156 (talk) 01:41, 12 January 2009 (UTC)