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January 3

Fractal line

Hi. Does a straight line count as a fractal? I know it would be a really boring example, but it looks the same no matter how much I zoom in. Duomillia (talk) 15:00, 3 January 2009 (UTC)

A fractal has a fractional dimension. A straight line has dimension one. So considering one to be a (special case of) fraction, then yes. Bo Jacoby (talk) 15:40, 3 January 2009 (UTC).

You know that you are likely to start another hot debeate... I personally agree with straight lines, and even points, be fractals.--PMajer (talk) 16:48, 3 January 2009 (UTC)

The Sierpinski pyramid (which is a tetrahedron analog of the Menger sponge or a 3–dimensional analog of the Sierpinski triangle) has the Hausdorff dimension equal 2 — but it certainly IS a fractal.... --CiaPan (talk) 10:34, 5 January 2009 (UTC)

As far as I can see, the pyramid on the picture is not a tetrahedron but a square pyramid, and it has dimension $\log _{2}5$. — Emil J. 13:45, 6 January 2009 (UTC)

Then ignore the picture and base on my words. --CiaPan (talk) 14:00, 6 January 2009 (UTC)

Here is a Sierpinski tetrahedron. Gandalf61 (talk) 14:05, 6 January 2009 (UTC)

OK, so here we have a tetrahedron, and the thing has Hausdorff dimension 2. Now more to the point. The dimensional definition of a fractal does not require the Hausdorff dimension to be fractional, but to be greater than its topological dimension. The Sierpiński tetrahedron pyramid has topological dimension 1,[1] hence it is a fractal according to the formal definition, despite having integral dimension. A straight line is not. — Emil J. 14:37, 6 January 2009 (UTC)

You're right. However Bo Jacoby wrote:
      A fractal has a fractional dimension.
and I called a Sierpinski tetrahedron (sorry for confusing it with a pyramid) as a counterexample to that. --CiaPan (talk) 15:50, 6 January 2009 (UTC)

A fraction is a rational number. Integers are rational. So 2 is a fraction. So the Sierpinsky tetrahedron is a fractal, and the straight line is a fractal too. Confusion is common as to whether trivial special cases should or should not be included in general definitions. Is zero a number? Is the empty set a set? Is a circle an ellipse? Mathematicians answer yes and nonmathematicians answer no. Bo Jacoby (talk) 12:03, 7 January 2009 (UTC).

Oh, really...? So this way the Koch snowflake curve is not a fractal, because its Hausdorff dimension is irrational (log 4/log 3), so it is not a fraction.
OR possibly it is a fraction, namely a quotient of 'log 4' and 'log 3'? But this way any number is a fraction and consequently any figure is a fractal.......
CiaPan (talk) 15:33, 7 January 2009 (UTC)

All depends how you define a fractal. If we use the Falconer definition given in the article then a straight line is not a fractal because it is not "too irregular to be easily described in traditional Euclidean geometric language". It also fails the "fractional dimension" definition - but then so do space filling curves such as the Hilbert curve, and the Heighway dragon curve. As these curves are generally considered to be fractals, I would say that fractional (i.e. non-integer) dimension is a "sufficient but not necessary" condition for a fractal. Gandalf61 (talk) 18:27, 3 January 2009 (UTC)

So, let's say, loosely defined a line is a fractal, but according to the strict/formal definition, it is not. I wonder, is there relevance in the distinction between "self-similar" and "self-identical"? A line is at all magnifications self identical, but not self-similar (if we define "similar" to mean close to but not exactly identical) where as something like the edge of the Mandelbrot set is similar to itself at different scales, but never perfectly identical. Duomillia (talk) 21:42, 3 January 2009 (UTC)

Similarity has a strict mathematical defintion, that if two objects are similar, they are identical apart from scale. It is not the same as similar in the colloquial or lingual sense, as is the case with many scientific or mathematical terms. Strictly the small differences in the mandelbrot set prevent it from being correctly classed as self similar. Strictly the mandelbrot set is quasi-self-similar. Under this definition a line is debatebly self similar, depending on whether you count different segments of the line as scale replicas of each other. —Preceding unsigned comment added by 84.92.32.38 (talk) 21:17, 4 January 2009 (UTC)

Well, that is how the term similar is used in geometry. However, the "similar" in "self-similar" has a broader definition, and can generally include any homeomorphism. For example, de Rham curves are self-similar under affine maps, but not generally self-similar under any set of uniform scalings and rotations. An object that is self-similar under a set of geometric similarities is more precisely decribed as scale invariant. Gandalf61 (talk) 09:58, 5 January 2009 (UTC)

Straight lines have been around a long time. The word "Fractal" was derived by Benoît B. Mandelbrot in 1975 from the Latin fractus meaning "broken" or "fractured". A straight line isn't broken so it's not a fractal. You can zoom in for ever on a point and that's not a fractal either. Cuddlyable3 (talk) 12:49, 6 January 2009 (UTC)

Be very careful here as a words etymology and its meaning are in no way recquire to be consistent. —Preceding unsigned comment added by 84.92.32.38 (talk) 13:34, 6 January 2009 (UTC)

My very careful quote mining at [2] shows what Benoît_Mandelbrot means is and is not a fractal:

A fractal is a mathematical set or concrete object that is irregular or fragmented at all scales...

...the infinite sea of complexity includes two islands: one of Euclidean simplicity, and also a second of relative simplicity in which roughness is present, but is the same at all scales.

Smooth shapes are very rare in the wild but extremely important in the ivory tower and the factory, and besides were my love when I was a young man. Cauliflowers exemplify a second area of great simplicity, that of shapes which appear more or less the same as you look at them up close or from far away, as you zoom in and zoom out. Before my work, those shapes had no use, hence no word was needed to denote them. My work created such a need and I coined "fractals."

The last time I checked, straight lines are smooth, of Euclidean simplicity and not irregular at any scale. Cuddlyable3 (talk) 16:49, 6 January 2009 (UTC)

What is the expected value of the lowest of four numbers?

Hi, I'm thinking about how I might write an AI to play a certain board game. After thinking for a while, I realized that I would need a formula that takes a probability density function and gives me the expected value of the lowest of n random numbers. So far what I've come up with is if the number is taken from a uniform distribution from 0 to 1 then the expected value of the lowest of n numbers should be (1/2)^(n-1), but that doesn't seem right at all. Unfortunately, I found this topic somewhat difficult to google for. Thanks for any help. --Tigerthink (talk) 15:43, 3 January 2009 (UTC)

It's $\textstyle {\frac {1}{n+1}}\quad$ (I assume you mean indipendent and uniformly distributed on [0,1]); note that it holds also for n=0, for the infimum is certainly 1. Just integrate. Do you need the details?--PMajer (talk) 16:21, 3 January 2009 (UTC)

Thanks for your help; I have a pretty good idea of how to generalize that result to any probability density function, as was my original question. (You just split it into n+1 portions with equal integrals, right?) I would be interested in reading a proof of your answer if you happen to have one handy. And looking at infimum, it seems you should have used supremum.--Tigerthink (talk) 17:06, 3 January 2009 (UTC)

No, I mean infimum (of the empty subset, being the maximum), so you get 1 with n=0 numbers in [0,1]. In general, just remember that the expectation of a positive real random variable Y writes

$\mathbb {E} (Y)=\int _{0}^{\infty }\mathbb {P} (Y\geq t)dt.\qquad$ If you then have n independent positive random variabls $\textstyle X_{1}\cdots X_{n}$, even not identically distributed, and now $\textstyle Y:=\min(X_{1},\cdots ,X_{n})$, you have of course $Y\geq t$ if and only if $X_{i}\geq t$ for all $\textstyle i=1,\cdots ,n$, therefore by the independence $\textstyle \mathbb {P} (Y\geq t)=\prod _{i=1}^{n}\mathbb {P} (X_{i}\geq t)$. So with n independent and uniformly distributed random variables you get that $\textstyle {\frac {1}{n+1}}\quad$. Notice that the answer will depend on the distribution. Ouh she is going to kill me :( hope is ok --PMajer (talk) 18:56, 3 January 2009 (UTC)

I still don't understand the infimum stuff. How could the empty subset be a maximum of anything? And if there are 0 numbers, isn't the expectation undefined? I think I understand most of the other stuff though. That integral looks like it's impossible to evaluate exactly. How did you get the 1/(n + 1) answer? Will I have to estimate it in most cases? Thanks for your help by the way. --Tigerthink (talk) 21:39, 3 January 2009 (UTC)

The infimum of the empty set is the greatest lower bound of the empty set. Since everything is a lower bound of the empty set (it just has to be less than or equal to every element of the empty set, which is vacuously true), the infimum is just the greatest element of our space, which in this case is 1. Algebraist 23:25, 3 January 2009 (UTC)

Thank you, I think I understand now. --Tigerthink (talk) 04:37, 5 January 2009 (UTC)

• The infimum of the empty set (which is I guess a topic already appeared here): just apply the definition of Greatest Lower Bound for a subset of a given ordered set X: any element of X is a lower bound for the empty set, just because it is in fact less than any element of the empty set; so the Greatest Lower Bound of the empty set in [0,1] is just the max of [0,1]. After all, this is consistent with all properties like "$\scriptstyle A\subset B$ implies $\scriptstyle \inf(A)\geq \inf(B)$". The only point is that $\scriptstyle \inf(A)$, for the special subset $\scriptstyle A=\emptyset$ does depend on the ordered set X, as it does for any other $\scriptstyle A\subset X$. Is it that strange?
• Then, if we have 0 random numbers in [0,1], some objects like their mean is undefined, but their infimum is still a well defined random variable, the constant 1, so the expectation of the infimum is 1. Anyway, that is not a major point.
• To evaluate the integral: you are right, let's say it depends on what are the so called distribution functions or DF, that is the functions (of the variable t), $\scriptstyle \mathbb {P} (X_{i}\geq t)$: for n i.i.d. random variables, with uniform distribution on [0,1], we have $\scriptstyle \mathbb {P} (X_{i}\geq t)=1-t$ for all $\scriptstyle 0\leq t\leq 1$, and 0 for t>1, so $\scriptstyle \mathbb {P} (Y\geq t)=(1-t)_{+}^{n}$ and we get $\scriptstyle \mathbb {E} (Y)=\int _{0}^{1}(1-t)^{n}dt={\frac {1}{n+1}}$ (I understand what you mean with n+1 portions etc, yes, you can do that way also).
• Conclusion: the general rule is that the distribution function of the minimum of n independent random variables is the product of their distribution functions. This holds without assuming positivity of course. Once you know the distribution function, you can compute expectation, moments, absolute moments etc. If we do not assume that Y is nonnegative, we only have to change conveniently the integral formula for the expectation of Y, that is

$\textstyle \mathbb {E} (Y)=\int _{-\infty }^{\infty }\mathbb {P} (|Y|\geq |t|)\operatorname {sgn}(t)dt=\int _{-\infty }^{\infty }\left\{\mathbb {P} (Y\geq t)-\mathbf {1} _{\mathbb {R} _{-}}(t)\right\}dt.\qquad$

I see now the first question has been already answered :)--PMajer (talk) 00:33, 4 January 2009 (UTC)

Alright, I think I understand the infimum stuff. Who knew there was a situation where the greatest lower bound could be bigger than the least upper bound :-) I confess I don't understand a lot of the rest of the stuff you said (which contributes to my late reply, I decided to put off understanding it until later, heh). I think I'll use the n+1 portions technique for my program; in the meantime, can you recommend a decent free online probability textbook? (I am extremely cheap.) --Tigerthink (talk) 04:37, 5 January 2009 (UTC)

Maybe somebody here can give a suggestion about online textbooks better than me. This [3] seems a frendly introduction. A nice introductory course is also Sinai's booklet "Probability Theory". In general, the first difficulty is the language of Probability, which is a bit excentric with respect to the rest of Maths. Anyway the choice of a book also depends on what is your background and what is your scope (mathematical phisics, economics, combinatorics and number theory). As to the stuff I said, it was only two things: first, to write $\scriptstyle \mathbb {P} (Y\geq t)$ as the product, which is just by the indipendence; second, the formula for $\scriptstyle \mathbb {E} (Y)$ in terms of $\scriptstyle \mathbb {P} (Y\geq t)$, that is sometimes taken as a definition and that has a clear graphical interpretation...--PMajer (talk) 10:30, 5 January 2009 (UTC)

OK, thanks again for all your help. --Tigerthink (talk) 19:27, 5 January 2009 (UTC)

Details:

1. The probability that a random number, uniformly distributed on [0,1], is less than t, where 0 ≤ t ≤ 1, is = t.
2. The probability that four such numbers are all less than t, is the power t ⁴ .
3. The probability that the highest of the four numbers is less than t, is also t ⁴ .
4. The probability that such a random number is not less than t, is 1 − t.
5. The probability that the lowest of the four numbers is not less than t, is (1 − t ) ⁴ .
6. The probability that the lowest of the four numbers is less than t, is 1 − (1 − t ) ⁴ .
7. The distribution of the lowest of the four numbers is f(t)dt = d(1 − (1 − t ) ⁴ ) = 4 (1 − t ) ³ dt
8. The mean value is ∫₀¹ t · f(t) dt = 4 ∫₀¹ t (1 − t ) ³ dt = 1/5.

Bo Jacoby (talk) 23:19, 5 January 2009 (UTC).

Gawsh, I thunk I gots the answer without thet thar hi falootin' edjumacated talk. Taking uniform distributions I say one number will be the highest and most likely in the middle of the range i.e. 0.5. Then there will be a second highest number most likely in the middle of the remaining range below 0.5, that's 0.25. In the same way the next number down is most likely 0.125. Then the lowest number is most likely 0.0625 = 1/16. Shucks, twernt nuthin hard ter figure out. Cuddlyable3 (talk) 12:31, 6 January 2009 (UTC)

One number is most likely in the middle of the range i.e. 0.5. Well, it is equally likely to be any number in the range. BUT the second number need not be smaller than the first number, and so the highest of two numbers is most likely to be 1, and the smaller number is most likely to be 0. BUT the maximum likelihood estimate is not the same thing as the mean value. The mean value of the smaller of the two numbers is 1/3. See Beta distribution. Bo Jacoby (talk) 20:28, 7 January 2009 (UTC).

Another argument (only for the uniform distribution): n random numbers (u. and in.d.) on [0,1] make n+1 random intervals; the expected length of the intervals is the same --because of the translation invariance of the Lebesgue measure. Since their sum is 1 we get in particular 1/(n+1) for the minimum. (To put in an even more symmetrical way: we may think of n+1 indipendent random points on S¹, where the first is just the choice of the origin. These make n+1 arcs of equal expected length, etc.) --PMajer (talk) 21:41, 7 January 2009 (UTC)

lease money factor

I have always undetstood that to convert a "money factor" to an APR, you first had to divide the money factor by 12 (thus the annual part) and then multiply that answer by 2400 to come up with an equilalent annual percentage rate. In looking it up on your web site, the answers seem to leave out the division part and simply tell you to just multiply the number by 2400. Which is it19:47, 3 January 2009 (UTC) —Preceding unsigned comment added by 63.115.177.81 (talk)

Are you referring to the article titled money factor? If so, it would help to say so and to link to it. Michael Hardy (talk) 20:32, 3 January 2009 (UTC)

Mr. Hardy, thanks for answering. I apologize for not attaching the link(s) I was looking at. 2 things, #1 I can't figure out the math question you posed a+b+c+d+e+f+g+h on your site and I still can't find an answer to my question from the links I have found. Why would you NOT divide a money factor by 12 to create an "annual" percentage rate and subsequently multiply by 2400?

It seems that (as an example) .0035 x 2400 = 8.4

.0035 divided first by 12 =.00029166 and THEN multiply by 2400 = 6.999

It seems to me that the correct APR in this case would be 6.999%(or basically 7%) instead of 8.4%

Your input would be appreciated. Thanks —Preceding unsigned comment added by 63.115.177.81 (talk) 20:59, 3 January 2009 (UTC)

Your last calc's off by a factor of 10, it would be 0.7%, not 7%. StuRat (talk) 21:19, 3 January 2009 (UTC)

Wow, is money factor a poorly written article. Nowhere does it tell you what it is. It says it's given as a decimal, "for example .0035", but doesn't say what ".0035" tells you. So at the end of the day all I get is that when I see some money thing that's given as a decimal, it's called a "money factor"? Real useful. --76.167.241.238 (talk) 00:15, 4 January 2009 (UTC)

Money factor is a VERY badly written article. It doesn't define its terms at all. I can't tell what it says. "The finance charge you end up paying"? The only way to understand that would be to have information about rental arrangements that is not in the article. Michael Hardy (talk) 00:25, 4 January 2009 (UTC)

I have somewhat re-written the article (which is currently nominated for deletion) and added more sources. Basically, "money factor" is half of the monthly interest rate or 1/24 of the equivalent APR (but usually quoted as a decimal, not a percentage, so numerically it is APR percentage/2400). Why half of the monthly interest rate ? Well, the amount outstanding on the car loan starts out at a value of C, the initial cost of the car, and ends at a value of F, the value of the car at the end of the lease, so the average monthly interest payment is

${\frac {(C+F)r}{2}}$

where r is the monthly interest rate. The factor r/2 is called the "money factor" - I guess "money factor 0.0030" sounds more attractive than "APR 7.2%". Gandalf61 (talk) 09:11, 4 January 2009 (UTC)

January 4

Sir Isaac Newton

Just so that this date doesn't go unnoticed, Sir Isaac Newton, FRS, was born on this date in 1643 (NS), (OS: 25 December 1642) :)) hydnjo talk 20:56, 4 January 2009 (UTC)

In some calendars. Algebraist 21:03, 4 January 2009 (UTC)

Just wanting to be sure that this wasn't a null date here at the good ol' /Math. So, more completely:

Gregorian (New Style) 1643-01-04 (Sunday, January 4, 1643)

Ordinal Calendar 1643-004

Week Date (ISO commercial calendar) 1643-W01-7

Julian (Old Style) 1642-12-25 (Sunday, December 25, 1642)

Islamic (Moslem) 1052-10-13 (Shawwal 13, 1052)

Hebrew (Jewish) 5403-10-13 (Teveth 13, 5403)

Mayan Long Count 12.1.4.11.14

Mayan Haab --10-7 (7 Yax)

Mayan Tzolkin --14-13 (13 Ix)

French Revolutionary *

Old Hindu Solar 4743-09-26 (Dhanu 26, 4743)

Old Hindu Lunar 4743-10-14 (Pausha 14, 4743)

Achelis' World Calendar 1643-01-04 (Wednesday, January 4, 1643)

Coptic 1359-04-29 (Kiyahk 29, 1359)

Ethiopian 1635-04-29 (Takhs'as' 29, 1635)

Jalaali *

Japanese Traditional "Kyureki" with CE 1642-11-14 (Shakku, Shimotsuki 14, 1642)

chronological Julian day number (JD) 2321157

chronological modified Julian day number (MJD) -78844

Lilian day number (LDN) 21997

and others I'm sure :)) hydnjo talk 21:52, 4 January 2009 (UTC)

Positivist calendar: day of Ulysses, month of Moses, 146 before the great crisis

Illuminati calendar: 4th January 5643—Preceding unsigned comment added by Algebraist (talk • contribs) 22:03, 4 January 2009

Oops! How the f*ck did we manage to miss that non-obscure calender reference riveted in its glorious historical perspective? :) hydnjo talk 00:56, 5 January 2009 (UTC)

Discordian: Prickle-Prickle, Chaos 4, Year of Our Lady of Discord 2809 :-) the wub "?!" 15:17, 5 January 2009 (UTC)

---

Without this thread, this would've been a null day :( Thanks Algebraist for your contributions, and Sir Isaac Newton's birthday is a null day not :) hydnjo talk 02:25, 5 January 2009 (UTC)

---

I was just coming over here to procrastinate, and now you've given me a ton of strange calendars to look up! Excellent. No homework will be done until tomorrow. Eric. 68.18.17.165 (talk) 04:49, 5 January 2009 (UTC)

Welcome to procrastinator central! Shoot, we can always find something worthwhile and humanity saving to do here instead of <whatever>. Bless you man! hydnjo talk 04:58, 5 January 2009 (UTC)

Thanks for reminding me. I'd meant to make a new years resolution about not procrastinating. Dmcq (talk) 18:45, 5 January 2009 (UTC)

I hope you weren't waiting to the fifth to make your resoutions, then? Eric. 68.18.17.165 (talk) 00:23, 6 January 2009 (UTC)

Sir Isaac Newton was not born on any of those dates. Perhaps you are thinking of Isaac Newton? He had to wait until Queen Anne knighted him in 1705. Cuddlyable3 (talk) 12:12, 6 January 2009 (UTC)

Are you claiming Sir Isaac Newton is a different person from Isaac Newton? If so, that's an interesting philosophical position. It does have the problem, though, that Sir Isaac was probably not named at all until somewhat after his birth. I myself rejoiced in the name of <surname> Twin Two for a brief period. Does this mean <my real name> was never born? Algebraist 15:32, 6 January 2009 (UTC)

People would snigger if the Queen of England dubbed someone who was already a knight. Sniggering at the Queen is unthinkably dastardly so that cannot be. Q.E.D. My compliments to your older twin who can rejoice in being your senior. I guess you arrived before a name was prepared for you, but you seem to have turned out all right. Cuddlyable3 (talk) 17:06, 6 January 2009 (UTC)

In that date, neither Isaac Newton was born indeed... --PMajer (talk) 18:55, 6 January 2009 (UTC)

On this date (as defined by some calendars), a male was born to Hannah Ayscough. He was named after his deceased father Isaac Newton. -hydnjo talk 02:49, 7 January 2009 (UTC)

January 5

Discussion for On-Line Algebra

How do you feel about the format of mathematics test; should they be multiple choice, open response, or a combination of both methods? Provide evidence to support your claims. —Preceding unsigned comment added by Johnny51R (talk • contribs) 11:47, 5 January 2009 (UTC)

Do you think this homework question is multiple choice,open response or a combination of both? Oh and answer in approved english (or what pases as that as education people keep inventing weird terms) as that seems to be more important than getting the maths right. And by the way I'd like to see more collaboration in answering which isn't tested in exams. Dmcq (talk) 13:28, 5 January 2009 (UTC)

Since this happens to be a topic which interests me, I'll include some arguments each way:

1) Pro-multiple choice:

a) Easy and quick to grade, especially if you use ScanTron forms (what, no article ?). Quick feedback to students is important in the learning process.

b) Objective. That means the teacher can't give better grades to students they like, or be so accused.

c) Allows students to solve the problem in any way they see fit, not just the way they were taught in class.

d) Reduces the likelihood of having a correct answer marked wrong. I once had an entire division test marked wrong because I'd answered correctly in decimals, rather than fractions. I suppose a particularly sadistic teacher could put both 10 1/2 and 10.5 in the answer list and only accept one, but that would at least make the students aware that they were being graded on the format of the answer, not just it's correctness.

2) Pro-open ended questions:

a) Teachers seem more likely to review this type of test afterwards, and show how the correct answer was determined. With some ScanTron tests I've just been told my grade, with no opportunity to challenge mistakes the teacher may have made or to learn from my own mistakes.

b) This allows the teacher to review the student's work to determine where they went wrong and let them know. This may be helpful to find each student's weaknesses and work on them. It might also help to identify learning disabilities, such as if the student is constantly transposing digits.

c) Partial credit can be given for answers that are almost right, like if they just got the sign of the answer wrong.

d) It's harder to cheat if you must show your work.

e) If a student gets a question right, and their work supports this, you can be sure they learned the material. On a multiple choice Q they may just have guessed right.

f) This doesn't allow students to work backwards. That is, they can't just plug in each of the possible answers to see which works. While this is a valuable skill for checking their work, students must also learn to solve problems on their own.

g) This is the only way to do some type of math tests, like proofs. You could list 4 different proofs and ask the students to select the correct one, but that's not the same thing as being able to write your own proof.

h) Tests often contain too much material for students to do a good job and double-check their work. This encourages them to learn to do things "quick and dirty", as opposed to "slow and carefully". The latter is what we should be teaching, at least in math, if we want to avoid spaceships crashing because of basic math errors. If the teacher must spend hours grading open-ended tests, they are less likely to make them too long for students to do properly in the time alloted. StuRat (talk) 17:05, 5 January 2009 (UTC)

It's Scantron, not ScanTron... but I made a redirect. -- Coneslayer (talk) 17:34, 5 January 2009 (UTC)

Thanks. Without the caps that could be taken as "Scant Ron", a web site concerned with the lack of Reaganomics in the current US Republican Party ? StuRat (talk) 17:49, 5 January 2009 (UTC)

3) Con-multiple choice

a) A monkey i.e. a total idiot can get on average a score of the reciprocal of the average number of choices in multiple questions just by picking at random. That makes low exam scores increasingly meaningless.

b) Some multiple choice questions invite multiple answers (e.g. "Tick every number that is a prime.") That raises a quandary about how to score when an answer contains both right and wrong "ticks".

I am not convinced by 2 f) above. Nothing stops a student investigating a problem "from both ends". Cuddlyable3 (talk) 11:57, 6 January 2009 (UTC)

I believe your 3a is similar to my 2e. For 3b, you have the same problem for such a question on an open ended exam, although if using an automated method for grading the multiple choice exams, they may not have any way to handle partial credit. In such a case that portion would need to be graded manually. As for working problems backwards, this is appropriate for some problems, like proofs, but not for others, like finding the solutions to a polynomial equation. If 4 answers are given to choose from, such a problem becomes trivial to work backwards, while solving it without multiple choice answers is not trivial. StuRat (talk) 20:46, 6 January 2009 (UTC)

I agree. My 3a develops the consequences of your 2e. You have developed 2f convincingly.

c) A problem that arose with automatic scoring of multiple choice math answers at a junior school where I taught was that students could use 3.16 or 22/7 as approximations to pi. They had not yet been taught about correct rounding of calculations involving pi. The problem was to accomodate the many different but acceptable answers which should include rounded and non-rounded versions using these or any better approximations (and also accept exact answers where pi is kept unevaluated as a symbol). Other irrational numbers could cause similar difficulty. (strikeouts by Cuddlyable3)

I've had a similar problem with multiple choice tests, where the number of decimal places wasn't listed in the question, and at least one answer was correct, but to a low number of decimal places, and there was also a "none of the above" option. For example: "What's the value of pi ? A) 22/7 B) 3.14 C) 3.1416 D) 3.1415926 E)None of the above". Of course, this isn't an inherent problem with all multiple choice exams, just with poorly written ones. Similarly, open-ended questions can have a problem where the answer is irrational and yet the number of decimal places is not specified. In such cases, I'd give as many decimal places as I, or my calculator, could manage. StuRat (talk) 19:16, 7 January 2009 (UTC)

I struck out some words in my post above because it described a problem with automatic scoring in general, not only multiple choices. One might even see an advantage in having multiple choices that "nudge" the examinee towards a particular answer form or resolution. StuRat, I would tick your "E" unless there are ellipses that I can't see after the digits in B, C or D. (What fun to lure students into the trap of believing 3.1415926 is closer to pi than 3.14... .) As for getting many decimal places from a calculator that's fine except for the last place. Calculators can't be relied upon to round that digit. Just push $10\div 3X3$ in that order on a cheap calculator and see whether it gets back to 10 or 9.9999999. Cuddlyable3 (talk) 19:59, 7 January 2009 (UTC)

c) Another problem with multiple choice questions is that they don't allow for multiple interpretations of a question. One I once had was "What's the perimeter of a 100 meter square yard ?". Not knowing if they meant 100 meters on a side, and thus a perimeter of 400 meters, or 100 square meters in area, hence 10 meters on a side and 40 meters in perimeter, I gave both answers. Had this been multiple choice, I might have gotten lucky if only one of the possible answers was listed and there wasn't a "none of the above", but I might not. StuRat (talk) 19:31, 7 January 2009 (UTC)

The heading says On-Line Algebra, so multiple choice seems the only choice, unless there is a live human teacher at the other end of the line. Cuddlyable3 (talk) 20:12, 7 January 2009 (UTC)

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)

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[4]. 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).

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:[5]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. [6]. 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)

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 247.7mm 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)

max size of a k-component graph

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)