# January 1

## Distance

How do I calculate the distance between two latitude/longitude cordinates? Ernest Smith

Haversine formula (Igny 19:41, 1 January 2006 (UTC))
To clarify, do you mean straight line distance (through the Earth's crust) or the great circle distance (along the Earth's surface) ? StuRat 22:42, 3 January 2006 (UTC)

# January 2

## Kurt Gödel

Hi!, does anyone know what Gödel's beliefs where? (ideology, or religion), because I haven't found that in the article. thank you. --Cosmic girl 00:03, 2 January 2006 (UTC)

Gödel was religious and interested in theology, but I haven't read anything saying he actually practiced any form of organized religion. (Or organized anything, really). He did a formalization of the ontological proof of God's existance, but I think that might have been more of a religiously-themed logic puzzle for him than an actual theological argument on his behalf. In any case, he was certainly interested in spiritual matters, which is more than you can say for political ones. He seems to have been a quite apolitical character. He certainly wasn't politically active the way Einstein was. --BluePlatypus 17:05, 2 January 2006 (UTC)
The ontological proof about the existence of El looks hard to swallow and digest, but its talk page contains, if I remember, a controversy worth of Hobbes vs Wallis. --Harvestman 19:23, 9 January 2006 (UTC)
My understanding was that he was an athiest or at least agnostic, but I can't remember offhand where I read that so I may be mistaken. I remember reading that a lot of people thought he was Jewish, but he in fact was not. Meekohi 20:10, 13 January 2006 (UTC)

## How to prove 1+1=2?

can anybody tell me how to prove 1+1=2?

If you're willing to trudge though a few hundred pages of Principia Mathematica, you'll find a proof there. --BluePlatypus 16:52, 2 January 2006 (UTC)
(Please sign your questions.) This can be a fun topic to explore, and very educational. It contains hidden depth. The essential steps are to precisely define "1", "2", "+" and "=" in a formal mathematical sense. A typical approach uses the Peano axioms; but that article is written at a fairly sophisticated level. Perhaps if you could tell us your level of schooling and why you ask the question, we could tailor a more appropriate response. --KSmrqT 18:36, 2 January 2006 (UTC)
Well, how do you define 2? Webster 1913 defines two as "One and one; twice one.", so what you're trying to prove is true by definition. --Keenan Pepper 21:52, 2 January 2006 (UTC)
In Peano arithmetic, 2 is defined as the successor of 1, so proving 1+1 = 2 reduces to proving that x+1 = succ(x). This is easily accomplished from the usual definition of addition: x+1 = x+succ(0) = succ(x+0) = succ(x), OS. --Ilmari Karonen (talk) 23:03, 2 January 2006 (UTC)
Right, the "proof" amounts to just fooling around with definitions because it's so trivial. --Keenan Pepper 02:16, 3 January 2006 (UTC)
Trivial? Humans have been using numbers and arithmetic for millennia, but a satisfactory axiomatic approach is relatively recent, and hardly obvious. Also, Peano axioms are only one alternative. Shall we begin with set theory? First order logic? Recursive function theory/lambda calculus? Topos theory and natural number objects? Physical observation and abstraction? Then what is one apple plus one orange? What is one raindrop plus one raindrop? What about quantum mechanical interference, where it seems that one slit plus one slit physically produces something different from Peano addition? Even in strictly mathematical terms, the theory of cardinal numbers tells us, say, that one infinity (the even naturals) plus one infinity (the odd naturals) equals one infinity (the naturals), with all three of these infinities having the same cardinality. So "trivial" seems like an overly glib dismissal of what may be a thoughtful question. --KSmrqT 10:52, 3 January 2006 (UTC)
Or in other words, truth by definition is highly nontrivial. --- Charles Stewart 17:04, 3 January 2006 (UTC)
Metamath has a proof that 2+2=4 here. —Ruud 19:19, 3 January 2006 (UTC)
It would be very easy to do diagramatically. Proto t c 12:35, 4 January 2006 (UTC)
Really, the proof depends on what axiom system you are working in. Some books for example are Douglas Hofstadter, Gödel, Escher, Bach or Donald Knuth, Surreal Numbers or any introductory book on ZFC set theory (I can't give an English title). – b_jonas 12:51, 13 January 2006 (UTC)

## Statistics

Hi , Can anyone please explain to me when to use permutation and when to use combination? Thanks.219.65.191.160D.R.

I found a good explanation on[1] JeffStickney 17:02, 2 January 2006 (UTC)

Consider a deck of 52 playing cards. Shuffling the deck changes the order of the cards, which is a permutation; it does not change which cards are in the deck, which is a combination. From the shuffled deck deal a poker hand of five cards. Depending on the shuffled state of the deck, many different five-card hands are possible. One such hand is a straight flush in hearts consisting of the four, five, six, seven, and eight of hearts. Although the cards might happen to be dealt in numerical order, from four to eight, it is more likely that they are dealt in some scrambled order. From the point of view of the poker player looking at the hand, the order in which the cards are dealt does not matter; it is the same hand if it contains the same collection of five cards. Thus the hand is a combination. However, to determine the probability of being dealt that hand, it is necessary to count all the permutations of a full deck and also to count the far fewer permutations that result in that hand being dealt (in any order); the ratio of the restricted count to the full count gives the probability. It is hard to give general advice, but the key distinction is that a permutation is ordered and a combination is not. --KSmrqT 18:25, 2 January 2006 (UTC)
Short version: the order in which things are chosen matters in permutations, only which things are chosen matters in combinations. Night Gyr 01:48, 4 January 2006 (UTC)

# January 3

## Lie algebra pronounciation

How do you pronounce "lie" in "lie algebra"? --HappyCamper 03:13, 3 January 2006 (UTC)

[li:]. Rhymes with "knee". -lethe talk

# January 4

## are boundary points critical points?

I had this terminological question from elementary calculus on the talk:derivative page, maybe someone can answer here.

The article states that critical points are where the derivative is undefined or zero, and that local extrema only occur at critical points. But clearly functions on the unit interval can have local extrema at their endpoints, and the derivatives can be nonzero and defined there, using one-sided limits. The statement as it stands is therefore wrong, but I'm unsure as to whether the correct remedy is to allow boundary points of the domain to be critical points, or to change the result to say that extrema occur at critical points or boundary points. -lethe talk 23:54, 3 January 2006 (UTC)

The correct statement is that extrema occur only at critical points or boundary points, unless the function is constant. A linear function on the unit interval has no critical points; its bounds are extrema. A piecewise linear function can have interior extrema at the "joints". A quadratic has either a maximum or a minimum even in the absence of bounds. A bivariate quadratic can have a critical point which is a "saddle", thus neither minimum nor maximum. A cubic can have an inflection point which is not an extremum. --KSmrqT 01:48, 5 January 2006 (UTC)
"...at their endpoints, and the derivatives can be nonzero and defined there, using one-sided limits" - as I understand it, the derivative at a point is not defined unless it's defined from both sides and are equal from both sides. enochlau (talk) 10:38, 5 January 2006 (UTC)
I disagree. I think the derivative is perfectly well-defined at a boundary point. It's true that any limit approaching a boundary point has to be a one-sided limit, and that should be mentioned, but since that's actually entailed in the definition of limit, it doesn't mean you have to change your definition. -lethe talk 10:58, 5 January 2006 (UTC)
Yes, the derivative makes perfect sense at the boundary. Dmharvey 22:18, 5 January 2006 (UTC)
Ah, but I see then that a subtle point was misssed: "The article states that critical points are where the derivative is undefined or zero..." if we insist that the derivative at the boundary point is "undefined" (because its one-sided), then the boundary points are, by definition, critical points, i.e. places where extreme can occur. And so, one may argue that the article is correct. ... Heh. Clearly a bit too subtle. I have no idea how calculus is taught, so its probably best to just be very explicit. linas 00:32, 6 January 2006 (UTC)
I was taught "extrema occur only at critical points or boundary points", as KSmrq said, but I don't quite remember whether points where the derivative does not exist also count as critical point, or are a separate category. -- Jitse Niesen (talk) 17:01, 6 January 2006 (UTC)

## Changing no. of columns in the middle of the page with LaTeX

This isn't exactly math, but I'm sure is the best place to look for a solution. I'm writing an article using latex, and in the middle/bottom of one page, I need to switch to one-column format to write a big matrix, and then switch back to two-column format. If I do that by enclosing the matrix between "\onecolumn" and "\twocolumn", the matrix automatically goes to the next page when I compile it using scientific workplace, and that page has nothing but the matrix. Text following the matrix begins on a fresh page. How can I switch from 2-columns to 1-column and then again to 2-columns while staying on the same page? Thanks! deeptrivia (talk) 22:50, 4 January 2006 (UTC)

I don't know too much about LaTeX, but have you tried using multicols? I think you need to load the \texttt{multicol} package at the beginning as well.
\begin{multicols}{1}{
matrix goes here
}
\end{multicols}


--Canley 23:21, 4 January 2006 (UTC)

Thanks Canley! I get the error message:

! LaTeX Error: Environment multicols undefined.


What should I do? Regards, deeptrivia (talk) 01:01, 5 January 2006 (UTC)

Did you load the multicols package?
\usepackage{multicol}

at the beginning of your document? --Canley 01:23, 5 January 2006 (UTC)

Okay, I got it working with:

\begin{onecolumn}{
matrix goes here
}
\end{onecolumn}


Thanks a lot!! deeptrivia (talk) 01:07, 5 January 2006 (UTC)

However, everything that follows also becomes one-column. I tried overcoming it by enclosing the rest of the document within:
\begin{twocolumn}{
rest of the doc
}
\end{twocolumn}


but then the original problem reappears. deeptrivia (talk) 01:13, 5 January 2006 (UTC)

# January 5

(no questions today)

yet another pitiful day... ☢ Ҡieff

# January 6

## what is supplementary angles? whate acute angle?

An acute angle is an angle with degree less than 90 degrees.

A given angle and its supplement add up to 180 degrees.

An example of supplementary angles: 2 right angles. (Both are 90 degress)--Ali K 12:56, 6 January 2006 (UTC)

## what is a scalen triangle

I assume you mean "scalene triangle", a triangle with three different angles. —Keenan Pepper 03:14, 6 January 2006 (UTC)
All sides are different in length aswell. --Ali K 12:55, 6 January 2006 (UTC)
Which can be shown through the law of sines. Superm401 | Talk 04:22, 14 January 2006 (UTC)

## Finding recursion relations

Say I have a series (which I know only numerically), and I suspect its given by a (probably simple) recursion relation. How can find that relation? Say I have an infinite number of such series, I suspect there all given by closely related recursion relations. How do I find those? linas 15:48, 6 January 2006 (UTC)

To be precise, suppose one was given the numeric values of the bessel function ${\displaystyle J_{n}(z)}$ for some fixed z. These are seen to oscillate a bit and then die off. How can one discover, from this numeric data, that these can be given by the very simple relation ${\displaystyle J_{n+1}=2nJ_{n}/z-J_{n-1}}$ ? linas 16:02, 6 January 2006 (UTC)

One possibility could be to plot the sequence as points in k-dimensional space of ${\displaystyle x_{n},x_{n+1},\ldots ,x_{n+k-1}}$ and discover that the points lie on a plane, or some other simple surface.(Igny 16:12, 6 January 2006 (UTC))
If you have integer sequence values, you can simply plug them into the On-Line Encyclopedia of Integer Sequences. - Taxman Talk 16:30, 6 January 2006 (UTC)

## Compute the uncomputable

After reading article about Chaitin constant, I came to a question on whether it is possible to find sums of very slowly converging series. For example, ${\displaystyle \sum (-1)^{n}(\ln \ln n)^{-1}}$ is a converging alternating series, that is its sum is well defined. But to compute k digits one has to do ${\displaystyle O(e^{e^{k}})}$ operations, right? Is there a way to compute digits of the mentioned series in polynomial time? What is the order and first digit of that number? (Igny 17:57, 6 January 2006 (UTC))

Sure. Here's a hand-waving way to do this... you'll have to flesh out the details and navigate around a few poles and cuts.
${\displaystyle \int _{0}^{\infty }\sum _{n}{\frac {(-1)^{n}}{(\ln n)^{s}}}ds=}$
${\displaystyle \sum _{n}(-1)^{n}\int _{0}^{\infty }e^{-s\ln \ln n}ds=}$
${\displaystyle \left.\sum _{n}(-1)^{n}{\frac {e^{-s\ln \ln n}}{\ln \ln n}}\right\vert _{0}^{\infty }=\sum (-1)^{n}(\ln \ln n)^{-1}}$
The problem is now reduced to finding the somewhat simpler sum, and integrating it. To get the simpler sum, we do a relaed trick:
${\displaystyle {\frac {d^{k}}{dt^{k}}}\sum _{n}{\frac {(-1)^{n}}{n^{t}}}=\sum _{n}(\ln n)^{k}e^{-t\ln n}}$
And so we have a sum that's defined on positive integers k. However, we want to extend this so that its valid for fractional k (and negative k in particular), and take the limit ${\displaystyle t\to 0}$. There are several ways of taking something like this, defined on integers, and extending it to any k valid on the complex plane. One is Newton's divided differences, another (and usually better) is by working in fourier space. The Fourier works because d^k/dx^k in position space becomes p^k in momentum space, and so in momentum space, its "trivial" to make k be any complex number.
There are various subtle points above, about convergence, poles, etc. etc. But I think it should work; the basic techniques are not uncommon. linas 01:21, 7 January 2006 (UTC)
Oh, and the last sum,${\displaystyle \sum _{n}(-1)^{n}n^{-s}}$ can be mangled into the Dirichlet beta function, its actually the first non-trivial Dirichlet L-function. These articles could be expanded, there are identities not listed here. See also Hurwitz zeta function for related tricks and hints. linas 01:32, 7 January 2006 (UTC)
Chaitin's constant is much much harder. I beleive one can map Turing machines to grammars, and then explore the sentances that the grammar can generate. Grammars in turn can be understood in terms of tilings and Lindenmayer systems; (non-terminating programs generate infinite tilings; those that terminate are finite). One can take stochastic averages over all possible tilings, see Potts model and subshift of finite type for examples of taking stochastic averages over some very very simple grammars. Now if this was easy, it would have been done; in fact its probably really really hard. linas 01:52, 7 January 2006 (UTC)
Linas, thanks for the elaborate response.. (Igny 01:49, 9 January 2006 (UTC))

# January 7

(no questions today)

aww, no... ☢ Ҡieff 23:14, 9 January 2006 (UTC)

# January 8

## long math problems

1 × 2 × 3 × … × 98 × 99 × 100 ≈ 9.33262154 × 10^157

Is there any easier way to solve this problem than to write the entire thing into a calculator (any formulas?)?

Is there a name for this kind of math problem? --172.130.196.207 03:39, 8 January 2006 (UTC)

Yes, that's 100 factorial, or 100! in math notation. A scientific calc will allow you to enter 100 then hit the "!" key to get the answer. StuRat 03:41, 8 January 2006 (UTC)
A very good approximation can be made very fast using Stirling's approximation:
${\displaystyle 100!\sim {\sqrt {2\pi 100}}\left({\frac {100}{e}}\right)^{100}}$ deeptrivia (talk) 04:19, 8 January 2006 (UTC)
Also, you wrote the answer wrong. It's not 9.33 times 10157, it's 9.33 times 10 to the power of 157, which is much bigger. You can write it like this: 9.33 × 10^157 or like this: ${\displaystyle 9.33\times 10^{157}}$Keenan Pepper 05:51, 8 January 2006 (UTC)

3is the answer provided by bignums. Don't believe there is any trick as there is for the equivalent summation (sum of all numbers from 1 to 100). Sdedeo (tips) 23:40, 12 January 2006 (UTC)

There are many tricks to calculate factorials faster with a computer, based on splitting the subproducts into smaller pieces, but they don't help if you're entering commands into a calculator by hand. There is indeed no closed formula like the one for sums. Fredrik Johansson - talk - contribs 13:45, 13 January 2006 (UTC)

## Largest number

Infinity is the largest number possible, but it is not a number. So would infinity minus 1 to the power of negative infinity be the largest number that is a number?

No. There is no largest (natural/rational/real/complex) number. Proving that rigorously, of course, requires first rigorously defining what a number is, which can get difficult. For natural numbers, a common definition uses the Peano axioms, which explictly state as given that each number has a successor, which is greater than it.
Questions like this don't make sense unless you explicitly specify the set of objects you're working with. The terms "integer", "rational number", "complex number" and so on all have rigorous mathematical definitions, but the word "number" by itself does not. Are quaternions numbers? You can call them numbers if you want, it doesn't really matter. Same with infinity. You can call it a number if you want, but it is not an integer, a complex number, or even a quaternion, so if you want to talk about things like "infinity minus 1 to the power of negative infinity", first you have to define a number system that includes infinity and define a way to raise something to a power in it. —Keenan Pepper 05:45, 8 January 2006 (UTC)
For examples of number systems that do include infinity, see extended real number line, point at infinity, cardinal number, and ordinal number. —Keenan Pepper 05:56, 8 January 2006 (UTC)
However, note that neither the cardinals nor the ordinals have a largest number. The extended real numbers do have a largest number — positive infinity — but not a second-largest one. And the projective line is not totally ordered. —Ilmari Karonen (talk) 06:44, 8 January 2006 (UTC)
If you have some spare time, I can recommend you this page. All you ever wanted to know about big numbers. Some wikipedia articles: Ackermann function, Tetration --Joris Gillis 13:19, 8 January 2006 (UTC)
An article on Busy beaver claims "Both of these functions are noncomputable, because they grow faster than any computable function." However, I do not see the proof of this statement. Yes, the sequences are noncomputable by Turing Machines, but it does not say anything about the growth rate, does it? After all, the problem may be noncomputable, yet the answer could be 42.(Igny 01:49, 9 January 2006 (UTC))
Nevermind, if such Turing-computable function f(n)>S(n) existed, it would solve the halting problem. (Igny 03:21, 9 January 2006 (UTC))
Infinity is a number of sorts (depending on your definition of a number), however it isn't representable by digits, because of its immense (dare I say, infinate) length. The same can be said for the smallest (in magnitude) known real number, which, as far as I know (feel free to say if I'm speculating here), is representable only by ${\displaystyle 1-0.{\dot {9}}}$ (I call it epsilon; came about because the smallest floating-point number representable on a computer is called machine epsilon): because of its length, it is unrepresentable in a purely digital form (using solely digits--and the decimal point, of course--and no mathematical symbols).
The smallest real number in magnitude is zero. Paul August 15:46, 11 January 2006 (UTC)
The largest singularly-named (that is, uses only one word for its name) number (in terms of magnitude) is the googolplex, which is represented as ${\displaystyle 10^{10^{100}}}$. --JB Adder | Talk 22:26, 10 January 2006 (UTC)
Thankyou, Paul, for picking out my error there; I completely forgot to put non-zero in. --JB Adder | Talk 05:50, 16 January 2006 (UTC)
Another very interesting article about large numbers and infinities can be found here. It's a long article, so be sure to check out all the pages. --DannyZ 04:39, 11 January 2006 (UTC)

# January 9

## Standard Deviation

If I recall correctly, in a normal distribution, each point of inflection is located one standard deviation away from the mean. Why isn't this in the normal distribution and standard deviation articles? --JianLi 01:30, 9 January 2006 (UTC)

It's mentioned here, at the bottom of the section: Normal distribution#Probability density function. —Keenan Pepper 02:34, 9 January 2006 (UTC)
I didn't see that, thanks :) I also added the info to Standard deviation.

It doesn't seem correct. --JianLi 05:45, 9 January 2006 (UTC)

It looks fine to me at first glance. What specifically is wrong with it? —Keenan Pepper 17:22, 9 January 2006 (UTC)
There is a total that does not amount to 1OO% ? --Harvestman 19:17, 9 January 2006 (UTC)
no, the points of inflection seem to be off. though it is hard to judge, because it is pretty flat near those points
They add up to 100.4%, which isn't bad. Black Carrot 18:44, 11 January 2006 (UTC)
Are you sure? They seem to add up to 99.8, which is more accurate as - theoretically - we can never account for 100%. Typically, this is further illustrated by never completely closing the tails.
Yeah, I think it's 99.8%. But this is not because we can never account for 100%, as the "0.1%" takes into account everything past 3 st dev. Rather, the reason it is not perfectly 100% is just roundoff error -JianLi 23:48, 15 January 2006 (UTC)

## pi

Can someone estimate how many years until those super-computers calculate the full extent of pi?

${\displaystyle \pi /0}$ years. StuRat 04:07, 11 January 2006 (UTC)
Pi has infinite digits , they will never calculate it. helohe (talk) 19:55, 9 January 2006 (UTC)
What is the point, anyway? 100 digits are enough to circumscribe a circle around the visible universe accurate to the Planck length. —Keenan Pepper 22:29, 9 January 2006 (UTC)
The points of calculating many digits are as far as I can tell 1) prestige, 2) investigation of the digits' statistical properties, 3) testing/benchmarking computer hardware and software. Fredrik | tc 22:36, 9 January 2006 (UTC)
For a while in the 80s and early 90s it yeilded some interesting results in parallelization, but not as much as things like factoring or protein folding. —James S. 09:13, 11 January 2006 (UTC)
Contact by Carl Sagan told me that there are secret messages in pi. Proto t c 09:41, 12 January 2006 (UTC)
But since pi is conjectured to be a normal number, any message imaginable is contained on its digits. (hence this joke). ☢ Ҡieff 00:26, 14 January 2006 (UTC)
So you're saying there are secret messages, right? :) Superm401 | Talk 04:35, 14 January 2006 (UTC)
There was a show on the history channel explaining the bible code, the idea that there are secret messages in the bible that can be decoded by, for instance, taking every tenth letter. Of course, things that look like words can be found in any series of letters if it's long enough. Someone apparently ran the bible code system on Moby Dick and found just as many messages from god. According to the announcer, there are now people who believe God wrote Moby Dick. Black Carrot 02:29, 21 January 2006 (UTC)

## Pearson r

Can the Pearson r be used to analyze a univariate distribution? I say no, but my friend says yes. --Neutralitytalk 22:56, 9 January 2006 (UTC)

Pearson's ρ is a measure of correlation which is used to describe the linear dependence of two distributions. I suppose you could correlate a univariate distribution against an ideal uniform, normal, lognormal, or other distribution, if you really wanted a univariate measure, but I've never seen that done. Although the more I think about it, the more I believe that you might want to compare a distribution to the uniform distribution with Pearson's ρ. So, I say you are both right, but you are more right than your friend. —James S. 09:34, 11 January 2006 (UTC)

# January 10

## how much?

How much wood can a woodchuck chuck if a woodchuck could chuck wood?

A woodchuck would chuck as much wood as a woodchuck could chuck, if a woodchuck could chuck wood. (In other words, he would do his very best.) StuRat
ZOT!Ilmari Karonen (talk) 17:49, 10 January 2006 (UTC)

As much wood that a woodchuck can chuck wood!

Suppose a woodchuck can chuck, say x, amount of wood. then the woodchuck can after chucking that chuck a small amount, say p, of wood. so that he chunks x+p amount of wood. this is a contradiction to the satement that he can chuck only x amount. Thus Proved ;-) —Rohit_math 1:45 11 January 2006 (IST)

Not really. Suppose, as a counterexample, that the woodchuck drops dead of a heart attack after chucking x amount of wood. He can't very well chuck any amount of wood, however small, after that, now can he?
Besides, even if your proof was correct, you would only have proven the nonexistence of a maximum amount of wood. A supremum, however, could still exist. —Ilmari Karonen (talk) 20:42, 10 January 2006 (UTC)

Two cords. Night Gyr 20:40, 10 January 2006 (UTC)

Definitely less than two chords 'cause a woodchuck aint that good at playin' music. And besides which, this mere nursery rhyme would seem to agree 'bout that ('cause they lack opposin' thumbs) which adds to their being musically challenged. So, whereas a 'chuck might get away with an Em(open) there's no way he could do a Cm7. hydnjo talk 01:22, 11 January 2006 (UTC)
Aach... I underestimated our pet 'chuck who at this very moment is riffin' the opening chords of Stairway To Heaven (geesh, who knew?) ;-) I guess a woodchuck can... hydnjo talk 03:01, 11 January 2006 (UTC)
If he had a saw in his little paw, a ton of wood he could! Black Carrot 18:47, 11 January 2006 (UTC)
And what if another woodchuck came chucking wood ?
Two woodchucks would chuck twice as much wood as a woodchuck could chuck, if woodchucks could chuck wood. --Harvestman 21:57, 11 January 2006 (UTC)
How much wood could two woodchucks chuck if two woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...
How much wood could three woodchucks chuck if three woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...
How much wood could four woodchucks chuck if four woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...
How much wood could five woodchucks chuck if five woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...
How much wood could six woodchucks chuck if six woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...
How much wood could seven woodchucks chuck if seven woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...
How much wood could eight woodchucks chuck if eight woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...
How much wood could nine woodchucks chuck if nine woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...
How much wood could ten woodchucks chuck if ten woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...
How much wood could eleven woodchucks chuck if eleven woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...
How much wood could twelve woodchucks chuck if twelve woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...
How much wood could thirteen woodchucks chuck if thirteen woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...
How much wood could fourteen woodchucks chuck if fourteen woodchucks could chuck wood? hydnjo talk 07:58, 12 January 2006 (UTC)...
How many roads must a man walk down?b_jonas 18:18, 12 January 2006 (UTC)
What is six times nine? Night Gyr 09:56, 14 January 2006 (UTC)

## Alligation

I've placed requests for an article on Alligation (alligation medial and alligation alternate) [2]. It is all explained in that source, but with the archaic language, I'm likely to mess it up. Someone with more experience in mathematics or the history of mathematics would be better for the job. If you use that source in creating the article, please add {{1728}} to the page. Thanks! — 0918BRIAN • 2006-01-10 21:23

I'm working on an article. I never knew what this was called. It's hard thinking of examples... I keep thinking of ones that involve drugs for some reason. =P —Keenan Pepper 03:41, 11 January 2006 (UTC)
Well, thank you KP for making that red link blue. We never heard of it either, that's one of the wonders of the WP:RD. hydnjo talk 04:10, 11 January 2006 (UTC)
Thanks for all your help! — 0918BRIAN • 2006-01-11 04:23
What's fun is that with that latin origin word alligation one could generate an english word like adlinking, which is of widespread use over the net. --Harvestman 21:54, 11 January 2006 (UTC)
Huh? hydnjo talk 07:30, 12 January 2006 (UTC)

# January 11

## derivation-circumcenter

can anyone give me the derivation for finding the coordinates of the circumcenter of a triangle?

(Please sign your questions with ~~~~.) Did you read the explanation at circumcircles of triangles? If there is something additional you need, please let us know. --KSmrqT 15:56, 11 January 2006 (UTC)
Let's see. Knowing about barycentric coordinates makes this proof really easy. If you have a triangle ${\displaystyle ABC}$ and a point ${\displaystyle P}$ anywhere on the plane, and the area of the triangles ${\displaystyle PBC,PCA,PAB}$ are ${\displaystyle \lambda _{a},\lambda _{b},\lambda _{c}}$ respectively than you can get the vector ${\displaystyle \mathbf {p} }$ pointing to ${\displaystyle P}$ can be calculated from the vertices by the formula ${\displaystyle \mathbf {p} =(\lambda _{a}\mathbf {a} +\lambda _{b}\mathbf {b} +\lambda _{c}\mathbf {c} )/(\lambda _{a}+\lambda _{b}+\lambda _{c})}$. Here, ${\displaystyle \lambda _{a}}$ is considered to be negative if the line ${\displaystyle BC}$ separates the points ${\displaystyle A}$ and ${\displaystyle P}$, and the sign of ${\displaystyle \lambda _{b}}$ and ${\displaystyle \lambda _{c}}$ is assigned similarly.
Knowing this, it's easy to get the coordinates of the circumcenter. Let ${\displaystyle P}$ be the circumcenter, the radius of the outscribed circle be ${\displaystyle r}$, and the angles of the triangle ${\displaystyle \alpha ,\beta ,\gamma }$. Now the angle ${\displaystyle BPC}$ is ${\displaystyle 2\alpha }$, so its area is ${\displaystyle \lambda _{a}=r\cos \alpha }$. Similarly, ${\displaystyle \lambda _{b}=r\cos \beta }$ and ${\displaystyle \lambda _{c}=r\cos \gamma }$. Thus, ${\displaystyle \mathbf {p} =(\cos \alpha \mathbf {a} +\cos \beta \mathbf {b} +\cos \gamma \mathbf {c} )/(\cos \alpha +\cos \beta +\cos \gamma )}$. – b_jonas 18:02, 12 January 2006 (UTC)

## what is the formula for the diameter of a circle f

What information is given? —Keenan Pepper 15:34, 11 January 2006 (UTC)
Is the radius is given? The area? The circumference? Three points on the circle? Each produces a different formula. --KSmrqT 15:51, 11 January 2006 (UTC)
In any case, see Circle. —Ilmari Karonen (talk) 22:46, 11 January 2006 (UTC)
Let diameter be d, radius r, circumfrence c, and area a.
d=2r
d=c/pi
d=2*sqrt(a/pi)
--Superm401 | Talk 04:39, 14 January 2006 (UTC)

## 17/23 Correlation

Has anyone ever heard of the above, supposedly related to the Illuminati? What is it? --Black Carrot 23:23, 11 January 2006 (UTC)

Hmm.. well apparently, 23 is a significant number among crackpots numerologists, I mean.. and that page says 17 is popular too, so it's not surprizing if some conspiracy nut Illuminati researcher, thinks there's some connection. --BluePlatypus 07:50, 12 January 2006 (UTC)

Anybody know something more specific than that? Black Carrot 01:06, 17 January 2006 (UTC)

My 2 cents: An example is the "3172" that can be seen in the bushes at the left side of the Lincoln Memorial on the reverse side of the five-dollar bill. It's a numerical anagram of the 17/23 correlation. [1]

# January 12

## lines that converge

The preceding unsigned header was added by 68.82.121.86 (talk • contribs) .

If I recall correctly, Myrna Minkoff was opposed. --George 01:57, 13 January 2006 (UTC)

## What do you call lines that are not parallel?

The preceding unsigned header was added by 68.82.121.86 (talk • contribs) .

I'm confused. Did this person answer their question before they asked it? Black Carrot 00:49, 12 January 2006 (UTC)
(Hmm, I see what you mean, Nah, just jumpin' in before anybody had a chance (I think)) hydnjo talk 04:39, 12 January 2006 (UTC)
In any event, lines that are not parallel can either intersect (share a point in common) or be skew lines. —Keenan Pepper 01:12, 12 January 2006 (UTC)
To be more specific, if the lines are in the same plane, then they must either intersect or be parallel. If the lines are not coplanar, then they must be skew. StuRat 01:55, 12 January 2006 (UTC)

## Parabolas

• How can you determine the equation of a parabola when given only a point on the parabola and the x-intercepts?
• When given a point on the parabola and the vertex?

Thanks, anon.

In general, it is not possible to determine a parabola given only three points, or given a single point and the vertex. You probably want to know how to find the parabola whose axis is perpendicular to the x axis.
• If the two x intercepts are x1 and x2, the equation of the parabola must have the form y = k(x-x1)(x-x2) for some k. Find k using the third point on the parabola.
• If the vertex is at (x1,y1), the equation must have the form y = k(x-x1)2 + y1. Again, find k using the other point. —Keenan Pepper 01:22, 12 January 2006 (UTC)
I could be mistaken, but your preamble assertion is not correct. The Lagrange interpolating polynomial always gives you a parabola when given three points (not all collinear). If you know the vertex of the parabola, the other point gives you some idea of its concavity. By an argument appealing to the parabola's symmetry, you have three points and thus you can determine the parabola exactly. Dysprosia 09:50, 12 January 2006 (UTC)
... but by using Lagrange polynomials you are implicitly assuming that the axis of the parabola is perpendicular to the x axis, so its equation takes the form y = <second degree polynomial in x>. This is probably what the original question meant, but it was not explicitly stated. Without this assumption, three points or one point + vertex is not sufficient to define the parabola. For example, the parabolas y = x2 and x = y2 both pass through (1,1) and have their vertex at (0,0). Two general parabolas can intersect in up to 4 different points (see Bézout's theorem) so you need 5 points to define a unique parabola. Assuming a given direction for the axis of the parabola implicitly provides two of these points, because the direction of the axis determines where the parabola is tangent to the line at infinity. Gandalf61 15:35, 12 January 2006 (UTC)
If one knows the orientation of the parabola in advance, one could simply change variables when necessary to obtain a description in Lagrange polynomials. However, you're right, you need to know this orientation ahead of time. Dysprosia 22:52, 12 January 2006 (UTC)
You might also want to see Wikipedia:Reference_desk_archive/Mathematics/December_2005#Two_parabola_questions which is a generalization of this question. – b_jonas 16:26, 14 January 2006 (UTC)

# January 13

I know that one. A decade is ten years long. Dmharvey 02:33, 13 January 2006 (UTC)
(Or a factor of ten, as in "100 is two decades away from 1," but probably not in the context above. Sdedeo (tips) 05:04, 13 January 2006 (UTC))
Some decades seem to last longer than others. The 1970's, for example, seemed to refuse to end. :-) StuRat 11:13, 13 January 2006 (UTC)
That's a good point. Technically a decade is ten years, but when referring to generational/cultural decades, they can be anywhere from, say, 6 to 15 years. Black Carrot 07:33, 16 January 2006 (UTC)

## number theoretic transform

I have a prime p, say around 107 or so. I have a vector of values ${\displaystyle x_{k}\in \mathbb {Z} /p\mathbb {Z} }$ for ${\displaystyle k=0,\ldots ,p-2}$. (i.e. the vector has length ${\displaystyle p-1}$). I would like to efficiently compute the number theoretic transform,

${\displaystyle y_{j}=\sum _{k=0}^{p-2}\omega ^{jk}x_{k},}$

where ${\displaystyle \omega }$ is some primitive ${\displaystyle (p-1)}$th root of unity in ${\displaystyle (\mathbb {Z} /p\mathbb {Z} )^{\times }}$. I've been reading about the fast fourier transform, number theoretic transform etc on and off wikipedia but not having much luck finding a clean, fast way to do it. It would be really nice to be able to do this without leaving the world of arithmetic mod p. Any suggestions? Dmharvey 12:27, 13 January 2006 (UTC)

If you can get on IRC, I would recommend asking that one on efnet's #math channel. You could send them the URL to this section and see who answers. --James S. 21:49, 13 January 2006 (UTC)

# January 14

## characterictic function

I have such a problem. Decide whether the following teorem is true and explain why. Let |φ(t)| ≤ φ(0) = 1 and |Ψ(t)| ≤ Ψ(0) =1 for each t, t is an element of R. If (φ+Ψ)/2 is a characteristic function of a random variable, then φ and Ψ are also characteristic functions of some random variables. Please help me. Thank you in advance Julia

Our article on Characteristic function is fairly incomplete, and sadly I forgot the details of probability theory as well. I guess it exists by the inversion theorem, which is mentioned in the article, but in general case it's somewhat hairier. The right-side integral is additive in regards to φ, so I guess you'll just get FZ=(FX+FY)/2, which should be a cumulative function of some random variable Z.  Grue  11:26, 16 January 2006 (UTC)

## convergence in distribution

Help me please to solve my homework please. Let X1,X2,...Xn be independent random variables such that P(Xk = k) = P(Xk = -k) = 1/(2k^2) and P(Xk = 1) = P(Xk = -1) = 1/2(1-1/k^2). Let Sn=∑Xk (k goes from 1 to n). Prove, that (1/√n)Sn converges in distribution to N(0,1) and var((1/√n)Sn) converges to 2. (n goes to infinity) Thanks;o)

You'll notice at the top of this page, it says "please do not post entire homework questions and expect us to give you the answers". If you want to ask homework questions, you have to be more devious. But you might want to look at convergence in distribution. Not sure exactly what "convergation" is, but it probably has something to do with "convergence". Dmharvey 23:34, 14 January 2006 (UTC)

## Most-used math functions/symbols?

Will anyone please go to the list of math codes and give me a list of any they would like to see added to the Special characters box on edit pages? I'm looking for the top 25-50 characters/functions/formula/etc. Thanks! — 0918BRIAN • 2006-01-14 13:47

• ≠ ≤ ≥ < > ≡ ≈ ≅ ∝   − × ÷ ± ⊥ ⊕ ⊗ ∗   …   ¼ ½ ¾ ¹ ² ³ °   ∂ ∫ ∑ ∞ ∏ √ ∇   ← → ↔ ⇐ ⇒ ⇔   ⌈ ⌉ ⌊ ⌋ 〈 〉   ¬ ∧ ∨ ∃ ∀   ∈ ∉ ∋ ∅ ⊆ ⊇ ⊃ ⊂ ⊄ ∪ ∩ ℵ (mostly copied from Wikipedia:Mathematical symbols) -- Jitse Niesen (talk) 16:04, 14 January 2006 (UTC)
• I'm specifically looking for TeX/LaTeX expressions, but this is a start. — 0918BRIAN • 2006-01-14 16:11
• I see, I should have answered the question you actually asked instead of the question I thought you asked. Well, let's see. \nabla \partial \forall \exists \emptyset \in \not \subset \cap \cup \setminus \sqrt \sim \approx \cong \equiv \le \ge \ll \gg \pm \to \mapsto \iff \ldots \times \cdot \circ \sum \prod \int \frac \choose \begin{matrix}\end{matrix} \begin{bmatrix}\end{bmatrix} \begin{cases}\end{cases} \mbox \pi \mathbb \mathbf \left \right \langle \rangle \| \quad \qquad \infty \hbar. But, at least for me, I'd prefer to have the first list, as I know the TeX/LaTeX commands rather well. -- Jitse Niesen (talk) 17:51, 14 January 2006 (UTC)
• Alright, if you try editing a page, and go to "Math/TeX" under "Special characters", you should see what I've put in so far. You might have to press CTRL+F5 on the edit page to see the latest version. — 0918BRIAN • 2006-01-14 18:25
I have a large table of Unicode mathematics characters at User:KSmrq/Chars. I'm fond of infty ("∞"), mapsto ("↦"), frasl ("⁄", as in 57), lang ("〈"), and rang ("〉"), among others. --KSmrqT 22:02, 14 January 2006 (UTC)
Special characters is a bit of a travesty. In non-Unicode compatible clients, it mucks the edit process up rather seriously. Dysprosia 11:07, 16 January 2006 (UTC)

# January 15

(no questions today)

# January 16

## Partial harmonic sum

Greetings:

Does anyone know how to compute the following expression exactly without using a computer? :

1 + 1/2 + 1/3 + 1/4 + ... + 1/2004 + 1/2005

note that there are only 2005 terms in the above question. It's a question taken out of a certain 2005 math competition.

Regards,

129.97.252.63 04:31, 16 January 2006 (UTC)

Without using a computer? Wikipedia is not paper
http://www.macalester.edu/aratra/chapt2/chapt2_4_1.html
--James S. 06:14, 16 January 2006 (UTC)

Okay, let's see...each rational in the sequence is ${\displaystyle {\frac {1}{x}}}$, where ${\displaystyle 1\leq x(\in \mathbb {N} )\leq 2005}$. Now, we must get a common denominator—we'll use ${\displaystyle 2005!}$ for simplicity—and each fraction become ${\displaystyle {\frac {\frac {2005!}{x}}{2005!}}}$. This thus makes the equation ${\displaystyle \sum _{x=1}^{2005}{\frac {\frac {2005!}{x}}{2005!}}}$...which in reality is not much improvement from where we started.
In reality I don't think there is a non-computer way other than by brute force, which will take a while. However, the computer (yes, I cheated) brought it to 8.18086436047126000000 (to 20dp). --JB Adder | Talk 06:33, 16 January 2006 (UTC)
But if it's a maths competition, then presumably you can do it by hand. enochlau (talk) 08:31, 16 January 2006 (UTC)
Well, it is not much of a simplification, but
1 + 1/2 + 1/3 + 1/4 + ... + 1/2004 + 1/2005 = H(2005) = ψ(2006) + γ
Was the task really to compute the expression exactly? Do you have the exact formulation of the question?
Rasmus (talk) 10:55, 16 January 2006 (UTC)
I strongly suspect the solution here is not by a general method, but rather some clever tricks of algebraic manipulation. --137.205.236.45 11:32, 16 January 2006 (UTC)
I plugged it into Mathematica and it was not a pretty result, so I don't think there's any trick that'll help us. Perhaps they wanted an approximate answer only - in that case, upper and lower Riemann sums would do the trick? enochlau (talk) 11:40, 16 January 2006 (UTC)
Thanks, I also suspected that this question might not be attainable by traditional means after checking with Maple. 129.97.252.63 13:56, 16 January 2006 (UTC)

Last night's Maple (c) results for the above problem:

825419062412938618065172093089947889397149613536805433233293531037345374983607845
607168571303751022983376044774946356625609868037459266524315881348303251988897297
264702267356771192306405854938327434861393197790709922461870578674170520718259858
709208867298356740872930086641532477167774390192178199597628909379208113940191580
795063697454416576692622505998280631510167317360912683033791803968444942384758990
858234347547510716553741330866608178304475580453652608327809900363947607113897164
452225510715716886988749112270863058179358139930700335029925139304914516700451064
337264806479832238026346350115626220638452291756962627332973452014220212662317566
133773747505598340212718763148981187341324675165293976681254362615682179088701625
329089664066770381767365995112667658279228474516706783621733174277607689316898498
294088334946622546427438165210457801275834481235079191000093
/
100896314379840321186998408533386120937997942211239805942055164219806651357887170
338069896071979300241604627960528226608198826833104802676509391191233622864697953
746131105044797407517764321203308711546496890490543206638668744705597392728340013
118518193575666739999190087664589823668755903919618121564163432457507043319012050
368704876735996000856162502069906637807469380429233657356927717928175747560331309
853759376077421934183746322086962398625735054481422211469583634354868579997531326
324730964497861549892549888791063696999263322606486002338059407329131457462819758
458471125333723038281994099496140309993163410926769370241232610047663184932185829
917494388279088327072506676178218231921632353872338260712580851072644683225000157
132703283102083111875042687135934903664056637859033901806260787677436158789532946
406020520294300450258958671116562100185384730286114583680000

now, if that was the final fraction that could not be reduced further, then it could rightly be argued that no clever algebraic trick exists for miraculously solving the above problem. However, if it could be shown that the above fraction does reduce to a far simpler fraction then the disturbing prospect remains that a clever algebraic method exists that would circuitously produce the same result.

So I guess the grand question is this:

Does Maple automatically reduce all of its fractional outputs? Does it even have the capability of reducing humongous fractions like this??????

Thanks.

129.97.252.63 17:50, 16 January 2006 (UTC)

Yes, the above fraction is reduced. Reducing a fraction simply amounts to cancelling the numerator and denominator by their greatest common divisor, and the GCD is inexpensive to calculate even for humongous numbers. Fredrik Johansson - talk - contribs 17:59, 16 January 2006 (UTC)
AHA! Via the Euclidean Algorithm!!! In fact I hit upon that insight while going to lunch. So the fraction Maple displays MUST have been reduced to the simplest form. Thank you so much Fredrik Johansson! and everyone else for your responses. 129.97.252.63 18:37, 16 January 2006 (UTC)

I think that the original person must bave misread the question, it actually asks for one to show that the sum is greater than a certain number. But inexperienced ppl might think that in order to prove the sum to be greater than some number the sum itself must be sought of in the first place, which is not necessarily the case. 129.97.252.63 18:37, 16 January 2006 (UTC)

Hello and thank you for lending your time to help improve Wikipedia. If you are interested in editing more often, we suggest you create an account to gain additional privileges and continue to participate in our WikiCommunity. 129.97.252.63 21:38, 16 January 2006 (UTC)

## Is it just me?

Or are these questions getting much harder compared to last week? I guess school is back in session. --James S. 08:38, 16 January 2006 (UTC)

## mathematics

what is the value of root4 is it only +2 or +2 and -2

Both +2 and -2 are square roots of 4. However, in certain circumstances, only the positive root may apply, say if the height of an object launched vertically from the ground varies with time acording to H = T^2. Here the time could only be positive. The positive root is sometimes called the principal root. StuRat 15:17, 16 January 2006 (UTC)
If you want to define a square root function, it can only have one value for each argument (that's the definition of a function). The convention is to define it to be nonnegative. —Keenan Pepper 17:50, 16 January 2006 (UTC)

# January 17

## Popes

What is the chance of four popes having died on the same day ?

There has been 265 popes. Each having died a different year.

Since a new pope is elected only after the current one dies, I don't see how that's even theoretically possible. (Therefore the probability would be exactly zero.) —Keenan Pepper 01:17, 17 January 2006 (UTC)
My guess is, the questioner wants to know the probability that four have died on the same day of the year, a routine but tedious calculation I don't personally feel like making at the moment. (Of course it's "routine" only under the assumption that all dates except February 29 have the same probability and that the probabilities are independent; I suspect neither assumption is exactly right but that they're close enough for the purposes in question.) --Trovatore 01:36, 17 January 2006 (UTC)
Actually, the "routine" answer completely ignores February 29th, assuming all years have 365.0 days. Bah! On that basis alone, I refuse to offer anything more than the link given. --James S. 01:45, 17 January 2006 (UTC)
Questions at this desk enjoy an extremely w-i-d-e latitude but a bit of context would help me to feel a bit less like I was being quized (or trolled). hydnjo talk 02:40, 17 January 2006 (UTC)
Sorry, it's very similar to the birthday problem to which I linked, but as a matter of policy I won't do anything more than link to the page because whenever I've tried to get people to consider February 29th in the past, they tell me the best answer is just to ignore it. Maybe that's true. If someone else wants to do the whole problem for the questioner, fine, and I don't care whether they accommodate Feb 29 or not. --James S. 02:56, 17 January 2006 (UTC)
Sorry James S., no offense intended. With or without the inclusion of February 29, I thought that the question lacked context. I realize that the numerical odds would be affected by the inclusion (or exclusion) of that particular date but not so much as to alter my comment (a generalization) as to the nature of the question. hydnjo talk 03:20, 17 January 2006 (UTC)
Oh, you were asking for the questioner's context instead of my comment's context. I'm guessing it's either a homework problem or someone trying to decide whether a coinsidece is uncanny. Indentation modified. --James S. 03:35, 17 January 2006 (UTC)
Exactly. :-) hydnjo talk 03:45, 17 January 2006 (UTC)

## Coterminal angles

What are coterminal angles? Maybe someone should make a page on this?--Urthogie 08:51, 17 January 2006 (UTC)

Coterminal angles are angles that coincide (when placed in standard position). For example 20°, 380° and -340° are coterminal angles. I am not sure we could make much more than a dictionary definition out of this. Rasmus (talk) 09:14, 17 January 2006 (UTC)
OK thanks that answers my question...whats a reference triangle, though?--Urthogie 09:16, 17 January 2006 (UTC)
Perhaps you mean what we call the "standard triangles" -- the 30-60-90-degree triangle (where sin of the one angle is 0.5, and so is cos of the other) and the 45-45-90-degree triangle (where the tan of both angles is 1). --대조 | Talk 16:02, 22 January 2006 (UTC)

## Euler characteristic

Wikilutations!
I wanted to know how one could find out the Euler characteristic of a given surface. Say atleast for simple surfaces like Torus, Disc, etc.
Rohit_math 18:24, 17 January 2006 (UTC)

Divide the whole surface to triangles, and use F − E + V formula to calculate the characteristic. Also, if the surface in question is orientable and has g holes, Eurer's characteristic is 2-2g.  Grue  18:43, 17 January 2006 (UTC)

Since it was requested, here's detailed description of triangulation, maybe it'll be more helpful:

Take a sphere for example. Draw an equator and four meridians. The sphere is now divided into 8 octants, which are, for topological purposes, triangles (they have three sides and three vertices). So we have the whole sphere divided into triangles. This is called a triangulation. Triangulation is proper when each side of triangle is connected only to one triangle (i.e. there are no vertices on the side, except for its ends). When we have a proper triangulation, we can finally calculate Euler's characteristic. The formula is F-E+V, where F is number of triangles, E is number of edges and V is number of vertices. In the case of the proper triangulation E=3*F/2 (each triangle has three edges, but each edge is used by two triangles), so the final formula is V-(F/2). Now, let's look at the sphere: there are eight triangles and six vertices. So the answer is 6-8/2=2.  Grue  19:35, 18 January 2006 (UTC)

## Smarties lids

File:Smarties (Candy).jpg
Smarties. Good to eat, but how many tubes before I get the entire alphabet? Only Smarties know the answer

Before the new, horrendous "hexatube" packaging, tubes of Nestlé Smarties used to contain a letter of the alphabet under the cap. Assuming the cap letters are chosen uniformly at random, how many Smarties tubes would I expect to have to buy before I had collected the entire alphabet? — Matt Crypto 18:33, 17 January 2006 (UTC)

I believe we had an equivalent question some time ago. Ah, here it is. The probability of finding all the n = 26 letters of the alphabet in m tubes is
${\displaystyle P(m,n)=\sum _{k=0}^{n}(-1)^{k}{n \choose k}\left({\frac {n-k}{n}}\right)^{m}}$.
This is the number of surjections from M = {1 … m} to N = {1 … n} divided by the total number of functions from M to N (= nm). Using the Perl code on the page I linked to above, I find that 94 tubes is enough to give you just barely more than even odds of getting all the letters. I even made a graph of the probabilities for 0 ≤ m ≤ 200. —Ilmari Karonen (talk) 19:33, 17 January 2006 (UTC)
I am not sure about the wording of the problem. I think that the question is not how many tubes would guarantee (with some probability) that I have all 26 letters in them. You start buying the tubes, and in X-th tube you get the last needed letter (you have 25 letters in X-1 tubes and 26 in X tubes). What is expected value of such X?
Is this the right question?(Igny 22:11, 17 January 2006 (UTC))
(edit conflicted, but this relates nicely to Igny's posting)
So
${\displaystyle {\begin{matrix}E[{\mbox{Smarties tubes}}]&=&\sum _{m=0}^{\infty }mP({\mbox{tubes}}=m)\\\ &=&\sum _{m=0}^{\infty }m(P({\mbox{tubes}}\leq m)-P({\mbox{tubes}}
(Estimation made with the nice Perl code) Rasmus (talk) 22:23, 17 January 2006 (UTC)
Thanks for the answers! I meant to ask the latter question as clarified by Igny, but both answered questions are interesting. I'd give you all a smartie if I could. — Matt Crypto 23:39, 17 January 2006 (UTC)
While I am not 100% convinced that formula for P(m,n) is correct, there is a recursive way to compute P(m,n). Introduce
${\displaystyle f_{m}(x)=\sum _{k=0}^{n}{n \choose k}\left({\frac {n-k}{n}}\right)^{m}(-1)^{k}x^{n-k}}$.
Notice that
${\displaystyle P(m,n)=f_{m}(1)}$
${\displaystyle f_{0}(x)=(x-1)^{n}}$
${\displaystyle f_{m+1}(x)={\frac {x}{n}}{\frac {d}{dx}}f_{m}(x)}$
(Igny 13:55, 18 January 2006 (UTC))
Smarties? Isn't that just Communist for M&Ms?'--George 19:41, 17 January 2006 (UTC)

You get your first letter after an average of 26/26 packets, your second letter after an average of 26/25 packets, and your last letter after an average of 26/1 packets. Since all these are independent, the expected total number of packets is

${\displaystyle \sum _{i=0}^{26}{26 \over 26-i}}$

Which is about 100.2, just as Rasmus calculated. Gdr 18:37, 18 January 2006 (UTC)

In other words, ${\displaystyle 26H_{26}\,\!}$. (Note, by the way, that your top index is off by one.) Fredrik Johansson - talk - contribs 21:09, 18 January 2006 (UTC)
My congratulations, it is a very elegant solution. Indeed, all previous solutions were based on trying to find P(X=x) and EX without realizing that ${\displaystyle X=X_{1}+X_{2}+...+X_{26}}$, such that ${\displaystyle P(X_{i}=x_{i})}$ and ${\displaystyle EX_{i}}$ are easy to compute. And you don't even need independence, since ${\displaystyle EX=EX_{1}+...+EX_{26}}$ even for not independent ${\displaystyle X_{i}}$ (Igny 13:17, 23 January 2006 (UTC))

# January 18

## Implication by association

What's a technical term for "implication by association" in logic? --HappyCamper 00:19, 18 January 2006 (UTC)

Dunno, but I suspect you'll find it in Category:Logical fallacies -- maybe Association fallacy -- but sadly, unlike the others, this one doesn't seem to have a weighty Latin associated it. Ask the lawyers, maybe? linas 01:16, 18 January 2006 (UTC)
One of Aristotle's 13 fallacies is generally known by the latin term Secundum quid et simpliciter which covers fallacious inference from properties of particulars to properties about more general groups: it can cover certain kinds of implication by association. I don't know if we have any good coverage of this here. Ex Concessis means guilt by assocaition; I think it is, as linas thought, a legal term and is covered (not very well) in the association fallacy article he gave, and maybe there's something in one of the law articles. If you mean something else, I'm not sure. --- Charles Stewart(talk) 15:58, 18 January 2006 (UTC)
See also Dicto simpliciter. —Ilmari Karonen (talk) 19:39, 19 January 2006 (UTC)
If the association is temporal (one thing happened near another in time), then it could be Post hoc ergo propter hoc or Cum hoc ergo propter hoc Insulanus 01:10, 5 July 2006 (UTC)

## Logarithm Question

If log 5 = m and log 7 = n

Determine:

log (35/10) in terms of m and n

(log is in the base of 10)

Hmm, don't try to fool your teacher. Read the Logarithm article (and your textbook on this subject) and if you get stumped then please come back with a question about what it is that you don't understand. From the top of this page, don't just dump your homework here (or something like that). ;-) hydnjo talk 03:38, 18 January 2006 (UTC)
Anyway, it's useful to know that ${\displaystyle \log(xy)=\log x+\log y}$, and that ${\displaystyle \log(x/y)=\log x-\log y}$. – b_jonas 12:18, 18 January 2006 (UTC)

## Representation of orders of magnitude

I am looking at a good graphical representation for presenting these orders of magnitudes of units of time on a scale. Unfortunately with the tools I have (Excel, Matlab), I'm not able to come up with a good way. Any ideas/support will be appreciated.

Unit of Time		Duration (s)

truti			3.08642E-07
renu			1.85185E-05
lava			0.001111111
leekshaka		0.066666667
paramanu		4
vighati		24
ghati			1440
muhurta		2880
nakshatra ahoratram	86400
paksha			1296000
masa			2548800
ruthu			5097600
aayan			15292800
year			30585600
mahayuga		1.3213E+14
manvantara		9.38122E+15
day of brahma		1.3213E+17
cycle of brahma	9.51212E+21


deeptrivia (talk) 05:39, 18 January 2006 (UTC)

Do you want something like this ?

10^22...................................+.
10^21...................................|.
10^20...................................|.
10^19...................................|.
10^18...................................|.
10^17.................................+.|.
10^16...............................+.|.|.
10^15...............................|.|.|.
10^14.............................+.|.|.|.
10^13.............................|.|.|.|.
10^12.............................|.|.|.|.
10^11.............................|.|.|.|.
10^10.............................|.|.|.|.
10^09.............................|.|.|.|.
10^08.............................|.|.|.|.
10^07.......................+.+.+.|.|.|.|.
10^06...................+.+.|.|.|.|.|.|.|.
10^05.................+.|.|.|.|.|.|.|.|.|.
10^04.................|.|.|.|.|.|.|.|.|.|.
10^03.............+.+.|.|.|.|.|.|.|.|.|.|.
10^02.............|.|.|.|.|.|.|.|.|.|.|.|.
10^01...........+.|.|.|.|.|.|.|.|.|.|.|.|.
10^00.........+.|.|.|.|.|.|.|.|.|.|.|.|.|.
10^-1.......+.|.|.|.|.|.|.|.|.|.|.|.|.|.|.
10^-2.......|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.
10^-3.....+.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.
10^-4.....|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.
10^-5...+.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.
10^-6...|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.
10^-7.+.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.
10^-8.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.
10^-9.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.|.
------------------------------------------
^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^ ^
| | | | | | | | | | | | | | | | | |
truti | | | | | | | | | | | | | | | | |
| | | | | | | | | | | | | | | | |
renu | | | | | | | | | | | | | | | |
| | | | | | | | | | | | | | | |
lava | | | | | | | | | | | | | | |
| | | | | | | | | | | | | | |
leekshaka | | | | | | | | | | | | | |
| | | | | | | | | | | | | |
paramanu | | | | | | | | | | | | |
| | | | | | | | | | | | |
vighati | | | | | | | | | | | |
| | | | | | | | | | | |
ghati | | | | | | | | | | |
| | | | | | | | | | |
muhurta | | | | | | | | | |
| | | | | | | | | |
nakshatra ahoratram | | | | | | | | |
| | | | | | | | |
paksha | | | | | | | |
| | | | | | | |
masa | | | | | | |
| | | | | | |
ruthu | | | | | |
| | | | | |
aayan | | | | |
| | | | |
year | | | |
| | | |
mahayuga | | |
| | |
manvantara | |
| |
day of brahma |
|
cycle of brahma


StuRat 08:39, 18 January 2006 (UTC)

Yeah, thanks a ton! I was looking for something that's more compact, and goes only in one dimension (only vertically, so all names are stacked one on the top of the other at an appropriate height.) This is meant as an illustration on Hindu units of measurement. I'll highly appreciate any help! deeptrivia (talk) 14:42, 18 January 2006 (UTC)

Is this the kind of thing you were looking for? Just knocked it up in Excel in a couple of minutes. --MathGraphGuy 22:14, 18 January 2006 (UTC)

Thanks, but I was look for something that is one-dimensional (two columns, one for powers of 10 and the other for names). Roughly like Image:Edge of Space.png with powers of 10 on the left and the time units one on the top of the other at different heights (rather than spread on the x-axis.) Thanks for your effort though. deeptrivia (talk) 23:04, 18 January 2006 (UTC)

This looks like a job for... EasyTimeline! I'm not too familiar with it myself, but you could probably knock something up based on the "Soviet leaders" example on that page. —Ilmari Karonen (talk) 08:45, 19 January 2006 (UTC)
Alternatively, try something like Image:HinduMeasurements.svg. —Ilmari Karonen (talk) 09:45, 19 January 2006 (UTC)
Yeah! Exactly what I was looking for! Thanks!!! How did you make it Ilmari? deeptrivia (talk) 12:21, 19 January 2006 (UTC)
Drew it in Inkscape. The scale has 1 cm spacing, so you can easily add new units by cloning (and unlinking) the "←Second" entry, translating it vertically by log10duration centimeters and changing the text appropriately. (Or just tell me the units and I'll edit it myself.) Actually making the scale itself took some playing around with the tiling tool, but it's rather straightforward once you know how to do it. —Ilmari Karonen (talk) 13:40, 19 January 2006 (UTC)

## linear programming

can any body give me a brief definition for feasible conditions in a linear programming problem??? thanks

A feasible solution to a linear programming problem (or any other optimisation problem) is any solution which meets the constraints of the problem - it does not necessarily optimise the objective function. The set of feasible solutions is sometimes called the feasible region. I would imagine that feasible conditions refers to the constraints themselves, but I don't think this is common terminology. Gandalf61 15:36, 18 January 2006 (UTC)
I believe the constraints are sometimes called feasibility conditions. -- Jitse Niesen (talk) 16:04, 18 January 2006 (UTC)
A linear programming problem is to minimize (maximize) a linear objective function in one or more variables, with the variables subject to linear equality or inequality constraints. For example, minimize x+yz with 0≤x≤5, 0≤y≤17, and x+2y+z=4. For a solution to be feasible, it must satisfy the constraints. The point (x,y,z)=(1,1,1) is thus feasible. It is clearly not optimal, because there the value of the objective function is 1; but at (0,0,4), also feasible, the objective function is −4, which is less. The point (−4,−3,5) gives an even smaller objective value, −12; but it satisfies none of the constraints, so is not feasible. As Jitse suggests, the constraints could be called feasible conditions for a solution. Positivity and other simple bounds on individual variables are quite common constraints in optimization problems that model physical systems. It is possible that no feasible solution exists, or that some conditions are redundant, or that the feasible region is not sufficiently bounded to guarantee a minimum. In algorithms for solving linear programming problems, the feasibility constraints typically play a greater role than the objective function, and contribute more to the complexity. --KSmrqT 21:01, 18 January 2006 (UTC)

# January 19

## Summation

I'm trying to solve for N: ${\displaystyle 4n^{2}-4n+1=2\sum _{i=1}^{n}(4n-4)+1}$ but I don't know what to do with the summation. I've been trying to figure this out for a week now and I've been checking all my textbooks but I still can't figure out how to isolate the n. Please help! Thanks!

Should the n inside the sum be an i? Otherwise the sum is trivial. -- SCZenz 00:39, 19 January 2006 (UTC)
Hint: sum of Arithmetic progression. enochlau (talk) 00:50, 19 January 2006 (UTC)
In light of SCZenz's observation, either the summation should use 4n−4 or 4i−4, with the latter being more likely. Either way, the summation has a closed form. First, notice that the −4 added to each term can be pulled outside the sum, contributing −4n. The factor of 4 in 4n also multiplies each term, and the distributive law allows that to be pulled out as well, contributing a factor of 4 to the sum from 1 to n of i. Suppose n is 3, so that the terms are 1, 2, 3. Reversing a copy of these and adding them to the original gives 1+3, 2+2, 3+1; and since the stepping up and stepping down are equal, this is 4, 4, 4. This generalizes, so that ∑i=1,n i is (1+n)n/2, where the division by two compensates for the fact that we added the series to itself. Thus the original sum simplifies to 4(1+n)n/2−4n, or 2n2−2n. When this is used in your original formula, the equality is trivially true for every value of n; in which case solving for n makes no sense, and it's likely you've misunderstood the problem. --KSmrqT 01:08, 19 January 2006 (UTC)
Alternative derivation ...
${\displaystyle 2\sum _{i=1}^{n}(4i-4)+1}$
${\displaystyle =8\sum _{i=1}^{n}(i-1)+1}$
${\displaystyle =8\sum _{i=0}^{n-1}(i)+1}$
${\displaystyle =8{\frac {n(n-1)}{2}}+1}$
${\displaystyle =4n(n-1)+1}$
${\displaystyle =4n^{2}-4n+1}$
- or you could even prove it by induction - but either way, it is still an identity. Evaluate both sides for a few different values of n to see it in action. Gandalf61 13:41, 19 January 2006 (UTC)

You're right, it was supposed to be an "i". Thanks a lot!

## The multiplicative inverse of an imaginary number

How, in standard form ${\displaystyle a+bi}$, would you write the multiplicative inverse of ${\displaystyle 5+6i}$?

Solve the two simultaneous equations given by ${\displaystyle (a+bi)(5+6i)=1}$, for a and b. Paul August 02:29, 19 January 2006 (UTC)
It might be quicker to just find ${\displaystyle {\frac {1}{5+6i}}}$ by rationalising the denominator. enochlau (talk) 03:38, 19 January 2006 (UTC)
Please be more precise about what you seek. One possible answer is (5+6i)−1, if all you wish is a notation. If, instead, you want to know the actual complex value of the inverse, say so. Lastly, if you want to know how to find that value, then you should make that intent clear. To address this final case, note that the product of any complex number with its complex conjugate is a non-negative real number, the square of its modulus or magnitude: zz = (a+bi)(abi) = a2+b2 = |z|2. So ask yourself, "By what must I multiply z to get 1 instead of |z|2?" I think that's enough hints for what sounds suspiciously like a homework question. --KSmrqT 04:02, 19 January 2006 (UTC)
Or let Google do it. hydnjo talk 03:49, 19 January 2006 (UTC)
The poster did say "in standard form ${\displaystyle a+bi}$", which presumably means that the poster wants an answer in that form. Dysprosia 04:05, 19 January 2006 (UTC)

Do it this way: ${\displaystyle {\frac {1}{5+6i}}={\frac {1}{5+6i}}*{\frac {5-6i}{5-6i}}={\frac {5}{61}}-{\frac {6}{61}}i}$ deeptrivia (talk) 18:39, 19 January 2006 (UTC)

Many routes lead to the same solution. The one I suggest has the advantage that it works equally well with any algebra in which a quantity and its "conjugate" multiply to give a real "norm". For example, it works with quaternions. --KSmrqT 00:34, 21 January 2006 (UTC)

## log for frequency

Does anyone know where to find the logarithmic function used to determine frequency in music?

And if it is not too much trouble, where would I find such information?

Thank you so much, any help is greatly appreciated.

-Rachel

See if any of these sites help. hydnjo talk 03:59, 19 January 2006 (UTC)
Up to a multiplicative factor there is only one logarithm function for real or complex numbers, and it alone is not sufficient to determine pitch. A likely candidate for what you really want is the cepstrum of a signal. Wikipedia currently appears weak in relevant information. You may find helpful information among the pages of Stanford professor Julius Smith, or the music research center, CCRMA, where he works. For example, this paper may be helpful. Finally, search the web for cepstrum and 'pitch detection'. --KSmrqT 04:27, 19 January 2006 (UTC)
No, no, no. I'm pretty sure Rachel is looking for cents. —Keenan Pepper 05:09, 19 January 2006 (UTC)
Ah, that does sound plausible. My hunch was guided by the phrase "used to determine frequency"; I don't think of cents that way. On the contrary, cents are intervals, like perfect fifths or half steps, only smaller and independent of tuning. Of course logarithms are also used in a peculiar way to describe volume, in decibels. --KSmrqT 06:44, 19 January 2006 (UTC)

No, "the formula that was created to find frequency" adds no new information about what you mean by "the logarithmic function used to determine frequency in music", so you have cleared up nothing. Please read the suggested answers and try to revise the details of your question to make it clear if you want cepstrum or cents or something completely different. We have some clever people here but we can only read what's written, we can't read minds. --KSmrqT 05:17, 21 January 2006 (UTC)

Oh.. I'm sorry. Cents makes the most sense out of everything above. Thank you all so much for your help. -Rachel

## Zero

To whom it may concern,

1. Is the number "Zero (0)", infinite?
2. Is the number "Zero (0)", the beginning, the point of origin of infinity?
3. Can the number “Zero (0)”, be the beginning to where a number goes to negative or positive to infinity and back to “Zero (0)”?

Sincerely,

To answer your first question, no. I am unable to make enough sense of the other two questions to answer them. —Ilmari Karonen (talk) 14:51, 19 January 2006 (UTC)
For the first, ranges (not individual numbers) are finite or infinite. So the answer is "no" but neither is 0 "finite" in any common sense.
Zero is not a particular endpoint of infinity. Any infinite range can terminate (one-ended) at a number. The integers counting from 0 up are infinte in number, but the integers counting from three billion and two are equally infinite in number.
For the third, no. As above, that really doesn't make any sense. Positive and negative infinity do not wrap around, though their inverses asymptotically approach zero. — Lomn Talk 14:54, 19 January 2006 (UTC)
Sure positive and negative infinity can wrap around. See projective line. —Keenan Pepper 15:18, 19 January 2006 (UTC)
(I've taken the liberty of numbering the questions.) These are peculiar questions to someone who understands mathematics, and I'm curious about their motivation. They are awkward to answer, because they are not entirely meaningful.
1. No, zero is not infinite. Nor is one or two or three or any natural number we can count to.
2. It is hard to give any definite meaning to the phrase "point of origin of infinity". Mathematics deals with a diversity of infinities, describing size, order, geometry, and so on. Essentially, this question asks us to make sense of nonsense.
3. For this question there is hope. When we study the geometry of the real number line, it is commonly helpful to augment the standard real numbers with a "point at infinity". Likewise, when we study the geometry of the Euclidean plane we commonly add a "line at infinity". In the augmented projective plane, two distinct lines always intersect in a single point; for parallel lines, that is a point at infinity. When we augment the real line alone, for geometry purposes, the projective line does "wrap around" at infinity, rather like going around a circle with zero diametrically opposite infinity. However, for other purposes we may choose to augment the real line with two distinct infinities, one positive and one negative. So, yes, we may go from zero through positive/negative infinity and back again in some mathematical contexts, but in other contexts that may not be possible. --KSmrqT 19:36, 19 January 2006 (UTC)

## knots revision

Hello RD. I'm not quite sure on what a skein relation is (even after reading the page), and know what a Jones polynomial is. There's a question that always pops up in exams. "Describe the skein relation for the Jones polynomial". I mean, how should one word the answer to that question? Is it just describing how to get the Jones polynomial of the link diagram? Thanks, --Knotted 17:59, 19 January 2006 (UTC)

Skein relations are used to recursively define certain knot invariants like knot polynomials. The basic idea is to start with some given knot diagram and pick a particular crossing. If one were free to cut and reglue the strand at that point there are three different knots (actually links) that arise. These are labelled ${\displaystyle L_{+},L_{0},L_{-}}$. One of these is the knot you started with. The other two will, in general, be simpler than the one you started with (it's a complicated argument as to why). This is why the recursion relationship works. This base case is the unknot or unlink.
As to the answer to the question, the Jones polynomial satisfies the following skein relation:
${\displaystyle t^{-1}V(L_{+})-tV(L_{-})+(t^{-1/2}-t^{1/2})V(L_{0})=0}$
where t is the polynomial indeterminant. I forget what the base case is, but I think its V(unknot) = 1. It's very eductaional to try and work it out for the Hopf link or the trefoil knot. Everything should become clear after that. -- Fropuff 00:16, 20 January 2006 (UTC)

# January 20

## New article

For anyone who was interested in the Alligation article I requested before, I've started another article at Apotome which could use some expanding. I'm not sure if the content is right, because it's from a book from 1728 (and one from 1828). Thanks. — 0918BRIAN • 2006-01-20 22:56

You just have books from 1728 lying about ? StuRat 05:01, 21 January 2006 (UTC)
Damn, I just saw it and thought the same thing. ☢ Ҡieff 09:14, 22 January 2006 (UTC)
I corrected the musical definition. Unless I'm mistaken, 128/125 is always called a diesis, never an apotome. I'll ask the guys on the tuning list if any other interval is called an apotome, but the only one I know of is 2187/2048, the Pythagorean apotome. —Keenan Pepper 04:59, 21 January 2006 (UTC)

# January 21

## Zero

To whom it may concern,

Question: Is Zero (0) equal to infinity?

Example: 0 = Infinity

Thank you for help.

Sincerely,

No,
but confusion, may arise as in different contexts, zero, equal and especially infinity have very different meanings and definitions in maths, so even more in general understanding or religion. In geometry, zero is an rather arbitrarily choosen point.
You have to understand that maths is only concerned with sets of rules and the inherent consequences thereof. The connection with the "real" world is that we choose to see some of these sets of rules close enough to realworld-problems (such as arise in mechanic, physic, economy etc.) that we take maths as a model for these problems.
The Infidel 08:49, 21 January 2006 (UTC)
You have posted this question before, and it was answered. Please do not post a question repeatedly. --KSmrqT 16:08, 21 January 2006 (UTC)
There was a repost to the question, but I don't think the question was (or is now) answered. If you asked someone "Could you please tell me what time it is?" and the answerd was "Yes, I could" that would be a completely true and logical answer. I think we don't understand what Guadalupe Guerra is really asking, so we can only guess what helps to put a more precise question. So I don't consider it undue to ask again. The Infidel 17:14, 21 January 2006 (UTC)
Nevertheless, it is inappropriate. If the previous answers were unclear or otherwise unsatisfactory, if the question seemed to have been misunderstood, then Guadalupe should have posted additional information in the thread of the original discussion. Instead, the exact same question was posted a second time in the exact same forum. That is a rude waste of everyone's time. --KSmrqT 05:06, 22 January 2006 (UTC)

## Infinity

I have a curiosity... isn't infinity just nothing? and if somehow I can't concieve, it isn't and it is in fact infinity.. I guess the only thing that can contain it is nothing ... am I right? because infinity can't exist anywhere that isn't nothingnes It sounds really crazy, but I need to know the maths of this... what I mean is that infinity depends on nothingness to exist and vice versa, in fact I think that nothingness is even more complete and abolute that infiniteness since nothingness is the same thing all over and infiniteness is full or interrelations inside it... never ending interrelations indeed. thanx. --Cosmic girl 19:52, 21 January 2006 (UTC)

And God or the universe share the same problems in our limited minds. First there was nothing. With an infinity of potentialities. Then mental evolved towards something finite - the universe we live in and look at - energy and matter. Also, don't feel the only one to concieve that, it was a matter of knowledge long before science showed the end of its nose. --DLL 20:09, 21 January 2006 (UTC)
Would you be a bit more specific? An infinity of what? An infinity of states can be contained within certain particles (according to quantum mechanics, unless I'm wrong). An infinity of time could have elapsed before our universe sprang into being and/or god got bored. An infinity of space may or may not exist beyond our atmosphere, and it may or may not contain infinite stars and planets. Infinitly many angels can dance on the point of a needle. What specifically are you looking for? Black Carrot 20:14, 21 January 2006 (UTC)
I tried to answer to a point of view about things related to mathematical concepts but far enough from modern definitions. The infinity question sounds more like a philosophical one. Maybe in "Humanities" you would find more appropriate answers. --DLL 20:28, 21 January 2006 (UTC)
Can I buy some pot from you? =P —Keenan Pepper 20:29, 21 January 2006 (UTC)

hahaha! no dude I have NOTHING left... but that means I have an infinite quantity of potential pot, so let's hope some of it becomes 'actual pot'. ( I don't really smoke pot)

to answer black carrot ... what I mean by infinity is the opposite of nothingness... I mean, not infinite matter nor particles, maybe not even time, what I mean is the infiniteness of like... going smaller and smaller and never ending ( I'm not sure if that can be done since I'm aware of the plank scale's limit, but propably we can get smaller than that) and infiniteness of thougth also, and the infiniteness of relations you can find in things and specially events. --Cosmic girl 21:57, 21 January 2006 (UTC)

You have not asked a mathematical question. In fact, you have not formulated a coherent question of any sort, even philosophical. Mathematics deals with precise formal definitions and rules of inference and the theorems that can be so derived. Often the early stages of inquiry are vague, intuitive, and difficult to verbalize. Unfortunately, it is rather difficult for us to be of much assistance to you during this stage through the reference desk. You might have better luck with a chat room or newsgroup for now. Also you might want to explore Wikipedia and the web to find ways in which "infinity" and "nothing" are used and defined, and to find concepts similar to yours. --KSmrqT 05:25, 22 January 2006 (UTC)

I think it might be more of a philosphical question...are opposites actually the same ? For example, the tangent curve goes to infinity and negative infinity at the same pt, depending on the direction of approach. So, one could argue philosophically (but not mathematically) that +inf and -inf are one and the same. Perhaps if this is true of additive opposites it would also be true of multiplicative inverses, like 0 and either +inf or -inf. In the political arena, the extreme right (like NAZIs) and the extreme left (like Soviet communists) seem almost identical regarding their cults of personality, supreme power of the state, and genocidal policies. In the realm of perception, violet and red seem almost the same, despite being at opposite ends of the visual frequency spectrum. StuRat 08:08, 22 January 2006 (UTC)

Actually you don't know how right you are... ${\displaystyle +\infty }$ and ${\displaystyle -\infty }$ can be considered one and the same point, an unsigned infinity or the point at infinity, denoted just ${\displaystyle \infty }$, which is exactly the topic of an article I'm planning on writing. But this is only as long as we agree to treat them this way; If we don't, they are two separate entities. And there are other examples of opposites being identical, such as being on any point on Earth and going all the way east, you get to a point which is to your west. But this really is more of a philosophic argument; In math, things depend on how you define them and nothing else - What you define is what you get. So you can talk about the extended real number line, in which ${\displaystyle +\infty \neq -\infty }$, or about the real projective line, in which ${\displaystyle +\infty =-\infty }$. -- Meni Rosenfeld (talk) 18:37, 22 January 2006 (UTC)

Isn't - infinity just absolute nothingness and + infinity absolute somethingness? or am I confussing stuff? or maybe both - infinity and + infinity are absolute somethingness and zero just is the middle of both? so...it's so confusing!has math actually prooved the concept of infinity? or is it just something our brain requires to have as an axiom which doesn't have any reality in the actual universe?--Cosmic girl 20:46, 22 January 2006 (UTC)

For a simple example of how they differ in common use, let's use money. If you owe me 0, then neither of us owes the other anything. If you owe me +inf, then all your money goes to me and you still owe more. If you owe me -inf then all my money goes to you and I still owe you more. StuRat 21:36, 22 January 2006 (UTC)

thank you StuRat!!:D that example really helped! :) --Cosmic girl 02:21, 23 January 2006 (UTC)

Don't confuse math with physics. Physics is about the actual universe, and you should ask physicists about whether there is infinity in the universe (though I am inclined to think that there is). Math has nothing to do with the actual universe (though it is useful in describing it), it is about making definitions and axioms and seeing what logically results from them. So there's no such thing as "proving that infinity exists", there is "defining infinity in some way, and see what results". You can for example define ${\displaystyle \infty =banana}$, but it is unlikely that this definition will get you anywhere. There are many definitions for infinity, each one leads to different results. -- Meni Rosenfeld (talk) 07:35, 23 January 2006 (UTC)

I disagree because if (like you supose) the universe turns out to be infinite, then there IS such thing as proving that infinity exists. and we may someday be able to do that.--Cosmic girl 20:06, 23 January 2006 (UTC)

Once again, you are confusing math with physics. The nature of the physical universe has no influence whatsoever on mathematics. "Proof" is a concept that really exists only in math. Physics is an experimental science and you can't really prove anything in it, you can at best give convincing evidence to a fact. Newton thought he "proved" that gravity is a force that is inversely proportional to the square of the distance, but Einstein showed that this isn't accurate. And now Einstein's theory is also being questioned. As experimental evidence is accumulated, a theory can be either strengthened or weakened, but it can never be "proved". -- Meni Rosenfeld (talk) 07:50, 24 January 2006 (UTC)

I see your point, but do you think it will be ever possible to 'PROVE' any physical theory? I mean, when we have more evolved theories? or do you think there will always be some doubt left? --Cosmic girl 19:58, 24 January 2006 (UTC)

As far as I know, it is extremely unlikely that any physical theory can ever be proved. It simply defies the concept of natural science. The point is that experiments are made, and people try to formulate a theory that explains the results. If experiments show different results than expected by a theory, that theory is usually thrown to the scientifical recycle bin. But just because a theory fits the experiments doesn't prove that it's true. Since we can only make a finite number of experiments, no matter how many we do and no matter what theory we formulate to explain them, there are always many other theories, indeed, uncountably infinitely many theories, that could also explain them. For example, about infinity, current knowledge suggests that it is much more plausible for the universe to be infinite (in some way), and that any attempt to explain the universe as finite will be hopelessly complicated. But the fact stands that, since we only have a finite amount of experimental data, there can always be a theory that assumes that the universe is finite and yet is able to accurately predict each and every result of our experiments (even if we, as humans, may have difficulty formulating such a theory). So no, I do not believe a physical theory can be proved. In mathematics, in contrast, we can create whichever "universe" we'd like, and prove statements about it. -- Meni Rosenfeld (talk) 08:31, 25 January 2006 (UTC)

Thank you :D your explanation really helped me. :) --Cosmic girl 18:20, 25 January 2006 (UTC)

Perhaps Martin Heidegger's Being and time might be interesting too.? linas 19:12, 25 January 2006 (UTC)

# January 22

## Cantor's diagonal argument

Could someone explain Cantor's diagonal argument in terms that a relative layman could understand? I tried reading the article, but repeated kept running into terms like "countably infinite" - which links to "countable set," which would make more sense to me if it was "countably finite" instead of "infinite" - and, at any rate, countable set, soon gets into bijections, etc., and before I know it, I'm lost. What does it mean that "real numbers are not countably infinite"? And in what manner is that counterintuitive? Zafiroblue05 03:55, 22 January 2006 (UTC)

I guess people see it as counterintuitive because when you first start thinking about the concept of infinity, it's not at all obvious that there can be more than one size of "infinity". To say that the "real numbers are not countably infinite" basically means that you can't write down the real numbers in a list. No matter how carefully you arrange your list, you'll always miss out some real numbers (in fact you'll miss most of them). Cantor's argument explains why this happens. This situation is in contrast to the natural numbers (0, 1, 2, 3, etc); you can write them down in a list. (I just did.) The term countably infinite pins down more precisely what we mean by "able to write down in a list". Dmharvey 05:15, 22 January 2006 (UTC)
Hmmm... I think I'm starting to get it - what threw me off was the term "countably infinite." But could you explain one more thing - in the actual article - in point number 9 in the proof, it says "x differs in the nth decimal place from rn." How do we know that? Zafiroblue05 01:20, 23 January 2006 (UTC)
Because that's how we defined it. We defined x by saying that it differs on the nth place from rn, by choosing each digit to be different from the corresponding digit on that particular rn. An intuitive way of considering it is that Cantor basically says that we can't make a rule to order the real numbers in such a way that when we are given a real number, we can say that it's the fifth, sixth, or 49th real number, or something like that. (Note this is different from saying that we can make a rule to order the real numbers in a way that when given a number, we can find the 'next' number. That actually is possible)--Fangz 02:58, 25 January 2006 (UTC)
Cantor's diagonal argument compares two sets, and finds that one is strictly larger than the other. Necessarily, it depends on machinery for comparing the sizes of sets. Consider the set S={a,b} and the set T={x,y,z}. It is "obvious" that these two sets are of different sizes, and that the second is larger than the first. But how can we formalize our intuition? A standard approach is to attempt a pairing. Thus we might pair 〈a,x〉 and 〈b,y〉, so that distinct members of S are paired with distinct members of T. This defines a function, more precisely a monomorphism, from S to T. We can conclude that T is larger than S because our monomorphism is not also an epimorphism; that is, at least one member of T is left unpaired.
Such an argument is straightforward and uneventful for finite sets. For infinite sets, things can get much more interesting. We discover that the set of even postive integers, {2,4,6,8,…}, is exactly the same size as the set of odd postive integers, {1,3,5,7,…}. We also find that the set of even postive integers is exactly the same size as the set of all postive integers! Nor are we restricted to integers. The set of rational numbers, all ratios of integers to positive integers, is again the same size as the set of positive integers alone. In fact, we can define a set as infinite if it contains a subset of the same size.
Having discovered this much, it is tempting to assume that all infinite sets are the same size. Cantor's argument shows that this assumption is false, because the set of real numbers is strictly larger than the set of positive integers. The method of proof is powerful, but peculiar; it is proof by contradiction. For simplicity, restrict attention to positive real numbers less than 1. Cantor says, suppose we're wrong, suppose we have such a real number for every positive integer (a monomorphism) with none left over (an epimorphism). Then we can use the integer paired with each real number to sort them, to list all the real numbers in order. Now write each real number as a (possibly infinite) decimal expansion, such as 0.8537 or 0.333… or 0.7170… or the like. Here the first number has first digit 8; choose a different digit (other than 9, for technical reasons), say 7. The second number has second digit 3; again choose a different second digit, say 2. And so on, through all the real numbers on our (supposedly complete) list. We are constructing a number as we go, something like 0.726…, which is necessarily different from the first number, and from the second, and in fact from any number on the list. And this number is a perfectly valid positive real number less than 1 (the latter because we exclude 9s). Oh, dear. (Actually, oh joy!) We have a contradiction: Our "complete" list necessarily omits our constructed number. Thus our monomorphism (the list) is not an epimorphism (because of the omission), and the set of real numbers is therefore larger than the set of positive integers. --KSmrqT 06:15, 22 January 2006 (UTC)

A hog weighs 250 pounds. A high-yield diet allows the animal to gain 6 pounds a day at a cost of $0.56 a day. The market price for hogs is currently$0.75 per pounds, but it is falling at a rate of $0.01 per day, and that price decline is expected to remain steady for the foreseeable future. When should the hog be sold in order to provide the farmer with the highest financial gain? --66.81.193.109 14:19, 22 January 2006 (UTC) Using the variable t, measured in days, let t=0 be 6 a.m. today. After t days, the hog will weigh 250+6t pounds. The price of hogs after t days will be (0.75-0.01*t) dollars per pound. The cost of the feed for t days of feeding will be 0.56*t. So, the profit after t days will be ${\displaystyle P(t)=(250+6t)\times (0.75-0.01\times t)-0.56\times t}$ ${\displaystyle P'(t)=(250+6t)\times (-0.01)+(0.75-0.01t)\times 6-0.56=0}$ Thus, t=12 days. Here the profit function has the shape of the downward parabola because of the (-t^2) term, and as a result, solving the marginal profit function for t gives the optimal input for the profit function. [I received help from Dr. Math.] (hmm... answered your own question did you? hydnjo talk 18:28, 22 January 2006 (UTC)) hmm...that was an offensive question. deeptrivia (talk) 15:25, 22 January 2006 (UTC) ## modus tollens vs. proof by contradiction Is modus tollens the same thing as proof by contradiction? They seem to mean the same thing intuitively, but they have separate articles, and they're phrased differently. -- Creidieki 19:02, 22 January 2006 (UTC) I think the difference is mainly of scope. Modus tollens usually means the specific axiom that if p→q then ~q→~p, and proof by contradiction usually means proving an entire argument by negating its conclusion. The basic idea is the same though. -- Meni Rosenfeld (talk) 19:32, 22 January 2006 (UTC) Yes. Modus tollens is used in the key step of proof by contradiction. —Keenan Pepper 19:36, 22 January 2006 (UTC) Just a quibble here: Modus tollens usually denotes, not the axiom (p→q)→(¬q→¬p), but rather the rule of inference ${\displaystyle {\frac {A\rightarrow B,\lnot B}{\lnot A}}}$ It's a fine distinction, of course, but there are contexts where it matters at least a little. --Trovatore 19:42, 22 January 2006 (UTC) • Okay, that distinction makes some sense. So then modus tollens wouldn't be included as an inference rule in constructive logics? I'm a bit confused, though; why would you need this as both an axiom and an inference rule? -- Creidieki 19:56, 22 January 2006 (UTC) • I don't think intuitionists have any objection to modus tollens. They object when you try to prove positive statements by contradiction; proving negative statements by contradiction is fine with them. • As for what you "need", I suppose that depends on what you need it for. Keep in mind that we don't actually do mathematical reasoning in formal deductive systems; rather, we study them as comprehensible analogs of (much messier) actual mathematical reasoning. If you start with a deductive system that includes modus ponens and the axiom schema (AB)→(¬B→¬A), you won't get any new theorems by adding modus tollens, but some proofs will get slightly shorter. On the other hand, if you add the rule, you can dispense with the axiom, at least as long as you have the rule (I forget what it's called) that allows you to infer (AB) from a proof of B using A as an assumption. --Trovatore 20:10, 22 January 2006 (UTC) Modus tollens is an admissible rule of intuitionistic logic: it's the converse, if ~q→~p then p→q which the intuitionists reject (and adding this rule to intuitiontistic logic gives classical logic). Trovatore forgot what the implication introduction rule of natural deduction was called. --- Charles Stewart(talk) 17:58, 25 January 2006 (UTC) ## to decrypt a message I have a message to decrypt, is there someone can help, thank you. a message: tcbat jcbet cjwjr lybrk pwjad fdzzp orfba by using an affine cipher to encrypt it, and there is no key. the hint is the first and last characters of the plaintext are w and e, respectively. also, a function f(x)=ax+b (mod 26) has been used in it, and I want to find the value a and b.  thanks again.  If you know it's a linear function, and you know two values (f(w) = t and f(e) = a), then you can determine the function. —Keenan Pepper 20:17, 22 January 2006 (UTC) Actually that's not true for modular arithmetic. Forget I said it. —Keenan Pepper 20:37, 22 January 2006 (UTC) Well, there may be more than one possible solution, but it sure narrows it down. —Keenan Pepper 20:47, 22 January 2006 (UTC) is "tcbat" a code if a word? I could not find english words of the form "w..ew". Besides, the equations f("w")="t", or 22a+b=19 mod 26, and f("e")="a" or 4a+b=0 mod 26 can not be solved for a and b, can they? (Igny 21:39, 22 January 2006 (UTC)) Maybe there are extra letters on the end, to make a full group of 5? —Keenan Pepper 22:01, 22 January 2006 (UTC) I'm with Keenan on the possibility of padding, or alternatively all spaces and punctuation have been removed and the letters grouped into blocks of 5. However (assuming a simple letter<=>mapping a<=>0, b<=>1, ..., z<=>25, or anything else with the letters in alphabetic order), you're right that there cannot be values of a and b which satisfy both equations, since the first implies b is odd while the second requires b to be even. Maybe a typo in the question? -- AJR | Talk 23:15, 22 January 2006 (UTC) Let's see if we can do this in a mathematical fashion...Letting ${\displaystyle a=1}$ and so on (ie excahnging each letter with the number of its position in the alphabet), then the functions become ${\displaystyle f(5)=5a+b(\mathrm {mod} 26)=1}$ and ${\displaystyle f(23)=23a+b({\textrm {mod}}26)=20}$. Now, we'll let ${\displaystyle 5a+b=27}$ for the moment. Now, using Cramer's Rule, we have...${\displaystyle a=-{\frac {1}{3}},b=27{\frac {2}{3}}}$. Okay, so a little bit ofbrute force may be required here, but you have a starting point. --JB Adder | Talk 23:23, 22 January 2006 (UTC) I think ${\displaystyle a}$ and ${\displaystyle b}$ in ${\displaystyle f(x)=ax+b{\pmod {26}}}$ must be integers, for the following reason: Since we are working mod 26, ${\displaystyle x\in \mathbb {Z} _{26}}$ i.e. ${\displaystyle 0\leq x\leq 25,x\in \mathbb {Z} }$, so when ${\displaystyle x=0}$, ${\displaystyle f(0)=b{\pmod {26}}}$, and so ${\displaystyle b}$ must be an integer. Now consider ${\displaystyle x=1}$, giving ${\displaystyle f(1)=a+b{\pmod {26}}}$ so ${\displaystyle a+b}$ is an integer, and since ${\displaystyle b}$ is an integer, so is ${\displaystyle a}$. So, ${\displaystyle a}$ and ${\displaystyle b}$ must be integers. But, mapping the message letters to numbers in alphabetical order, A->0, B->1, ..., Z->25 (which letter we call 0 doesn't matter, it just changes the value of ${\displaystyle b}$) we get from the hint in the original question f(W) = T i.e. ${\displaystyle f(22)=22a+b=19{\pmod {26}}}$(1) and f(E) = A i.e. ${\displaystyle f(4)=4a+b=0{\pmod {26}}}$(2). Equation (2) tells us that ${\displaystyle 4a+b=26x}$ for some unknown integer ${\displaystyle x}$, and since ${\displaystyle 4a}$ and ${\displaystyle 26x}$ are both even (because they are integer multiples of even numbers) ${\displaystyle b}$ must also be even. From equation (1), we get ${\displaystyle 22a+b=19+26y}$ for some unknown integer ${\displaystyle y}$ i.e. ${\displaystyle 22a+b}$ is odd, which implies ${\displaystyle b}$ is odd. But ${\displaystyle b}$ clearly cannot be both even and odd, so there must be an error in one of: the supplied ciphertext, the supplied hint, or my reasoning. And I'm fairly confident I haven't made a mistake - if I am wrong, I'll give a small wikiprize to whoever shows how. -- AJR | Talk 01:51, 23 January 2006 (UTC) There is a possibility that the position of the letter is also important, ${\displaystyle y_{n}=f(x_{n},n)=ax_{n}+bn+c\mod 26}$ (Igny 13:09, 23 January 2006 (UTC)) The qeustion specifies that it is an affine cipher, which tells us that the general form of the encryption function is ${\displaystyle e(x)=ax+b{\pmod {m}}}$, where ${\displaystyle m}$ is the size of the alphabet. Given that it is an affine cipher, it should be breakable from the information we have - "The cipher's primary weakness comes from the fact that if the cryptanalyst can discover the plaintext of two ciphertext characters then the key can be obtained by solving a simultaneous equation." -- AJR | Talk 17:59, 23 January 2006 (UTC) Keenan Pepper: you're right, there's indeed an extra letter at the end which decrypts to "e". The last given letter "a" decrypted is "t". It's not difficult to find the solution knowing only that the first letter of the message is "w", as that leaves only ${\displaystyle \varphi (26)}$ possibilities. As an addittional clue, the most frequent letters after a "w" beginning a word are (in order starting with most frequent) "hiaeorw". Most of those are impossible for the same parity reason as mentioned above. That leaves two: "wrgvwqrgnwrqdqamzgabedqvcyckketaygv", and the real solution which I won't spoil here. – b_jonas 21:16, 23 January 2006 (UTC) Oh, and there seems to be one more typo in the middle of the cyphertext. The correct one should be "tcbatjcbejcjwjrlybrkpwjadfdzzporfbaj" I guess. – b_jonas 21:24, 23 January 2006 (UTC) # January 23 ## Name of a kind of 3D space I was wondering if there's a name for this 3D space I thought of when playing with some functions. In a way, the space would be ${\displaystyle \mathbb {R} \times \mathbb {R} \times (\mathbb {C} -\mathbb {R} )}$ (probably wrong notation), that is, a complex plane with an extra real axis perpendicular to it. This space would allow functions returning imaginary numbers such as ${\displaystyle f(x)={\sqrt {r^{2}-x^{2}}}}$ to be seen in three dimensions. ☢ Ҡieff 01:24, 23 January 2006 (UTC) "A complex plane with an extra real axis perpendicular to it" should be just ${\displaystyle \mathbb {C} \times \mathbb {R} }$, no? —Keenan Pepper 01:34, 23 January 2006 (UTC) I wasn't so sure about this point (hence the "wrong notation" note), because one of the axis would be exclusively for the imaginary part, where the other two would be real. If ${\displaystyle \mathbb {C} }$ automatically defines a 2D space (which seems to be about right from the complex number article), then I guess this space would indeed be ${\displaystyle \mathbb {C} \times \mathbb {R} }$. I just thought that the complex plane would be ${\displaystyle \mathbb {R} \times \mathbb {C} -\mathbb {R} }$ (reals in one axis, imaginary on the other). Made sense to me. ☢ Ҡieff 01:49, 23 January 2006 (UTC) Are you sure "extra real" means imaginary. To me extra-real means exactly ${\displaystyle (\mathbb {C} -\mathbb {R} )}$ , i.e., numbers Z such that Imag(Z) <> 0 (all numbers that are not real, like 5i and 2+3i ). deeptrivia (talk) 05:03, 23 January 2006 (UTC) Take the example (fixed now, actually): ${\displaystyle f(x)={\sqrt {r^{2}-x^{2}}}}$. If x > r, the function turns into an imaginary number, so you can imagine a third imaginary axis perpendicular to the x and y on the cartesian plane. That's what I meant... ☢ Ҡieff 05:11, 23 January 2006 (UTC) is that it? Two branchs of a hyperbola escaping in the imaginary direction, and a semicircle in the real direction. -lethe talk 05:26, 23 January 2006 (UTC) Yep, that was the space I meant! Where did you plot this? ☢ Ҡieff 06:46, 23 January 2006 (UTC) I'd thought of this before, a long time ago. I believe you can take your space to be R3, with your regular real-valued function remaining on the xy plane, and then having the complex values on the xz plane, since C is isomorphic to R2. Dysprosia 05:38, 23 January 2006 (UTC) ... though you wouldn't be restricted to xz. x is in R, so is the real part of f(x). The imaginary part of f(x) would just be mapped on the z axis. Example: ${\displaystyle f(x)=x+{\sqrt {r^{2}-x^{2}}}}$ wouldn't be restricted to two planes (xy & xz). ☢ Ҡieff 06:46, 23 January 2006 (UTC) You can choose different axes if you wish, but three should be enough. Dysprosia 22:55, 25 January 2006 (UTC) ## How do you memorize your multiplication table? Greetings: I was wondering how all of you Wikipedians rote your multiplication table when you were small? Is there any tricks or English nursery rhymes used to facilitate this learning proces? Regards, 129.97.252.63 03:06, 23 January 2006 (UTC) Well, whenever I forgot one I would figure it out by successive addition, which was a pain, so that taught me to remember them instead. =P —Keenan Pepper 03:11, 23 January 2006 (UTC) I don't recall how I memorized mine. There are a number of famous tricks, though. You might also consider using software - like this freeware - to learn them. Alas, I came around before the days of ubiquitous home PCs. --George 03:58, 23 January 2006 (UTC) There is no magic substitute for repetition. There are tricks for confirmation or reconstruction. For example, any product with an even number must be even, and the product of two odd numbers must be odd. Any product with 9 must give digits that sum to 9. (Thus 9×8 = 72, with 7+2 = 9.) Any product with 5 must end in 5 or 0; and the product with an even number halves the number and appends 0, or with an odd number appends 5. (Thus 6×5 = 30, 7×5 = 35.) To reconstruct, say, 3 times 7, count by threes, as in 3, 6, 9, 12, 15, 18, 21. Commutativity applies; thus 3×7 must equal 7×3, and we can count 7, 14, 21. For multiplying large numbers, it can be helpful to use an entirely different procedure such as that of Trachtenberg. --KSmrqT 05:32, 23 January 2006 (UTC) Since I learned them before PCs were common, I used flash cards. They are very helpful as you can remove cards you have already learned and concentrate on the rest. One pitfall to avoid is always memorizing them in the same order. When you do this, you may only be able to regurgitate them in the same order, which is frequently too slow to pass a test. StuRat 15:54, 23 January 2006 (UTC) We used songs like "she'll be coming around the mountain" for multiples of 8's. And "you are my sunshine" for 6's. I still remember them well :) Keepitrude 02:24, 24 January 2006 (UTC) ## Zero To whom it may concern, I’m sorry if I was rude “Ksmrq” and thank you “The Infidel”, for allowing me chance to post the question again. According to history, the Mayans did not create the number Zero (0), but the concept of Zero (0). This discovery, is very important because, by definition, the number Zero (0), not created, should not have a point of origin where infinity exists, but it does. Infinity, having no beginning and no end but at the same time, the number Zero (0), gives infinity a point of origin, a beginning. Examples: The number Zero (0), equal to infinity, is the point of origin, in which a number can go positive or negative, to infinity and back to Zero (0). I had time to sleep on it as to give an example as to how Zero (0) can be the Biggest number & the Smallest number. Here goes nothing. Example: 1. Zero (0) is the Biggest number when going from Zero (0) to -1, -2, -3, -4, > to Infinity. 2. Zero (0) is the Smallest number when going from Zero (0) to +1, +2, +3, +4, > to Infinity. I believe this sounds right. Based on this findings, I'm I correct to believe that: Zero (0 is equal to Infinity? I love waking up to a great tasting smell of coffee. Sincerely, Guadalupe Guerra, Jr. I'm really sorry, but you still haven't quite understood how this works. Please use the  link on your previous post to continue discussion. enochlau (talk) 09:54, 23 January 2006 (UTC) I'm sorry, but this is still just rambling. 37 is the smallest number when going from 37 to +Infinity and the biggest when going from 37 to -Infinity. Zero possesses special significance in cases such as the additive identity but is wholly unremarkable on the number line. — Lomn Talk 20:00, 23 January 2006 (UTC) You mix up definitions (ideas). Firstly, lets fix this: the numbers in your example are integers. There are two different concepts of "bigger than" for integers, which for natural numbers cannot be destinguished. One is the concept of "order", in the sense that one comes before two and six comes before seven. :The other is the concept of "magnitude" (see absolute value). This is "five sheep are more than four sheep". But with negative values possible, minus four comes before minus three. On the other hand, if have one billion dollars, this is a whole lot of money. If you have one billion dollars in debts, that's also an awful lot of money and in no respect small, althoug in mathematical notation it is "<" (spoken "less than") +1.000.000.000. The Infidel 20:25, 23 January 2006 (UTC) ## Area vs. Perimeter and Surface Area vs. Volume I was thinking about these questions, and have come up with answers for some, others in bold I have no answer to. In two dimensional space, which shape gives: • The largest perimeter to a constant area? • The perimeter could be infinite, as you could take a rectangle with the constant area, and repeatedly half the width and double the length. • The smallest perimeter to a constant area? • The isoperimetric inequality tells you that this must also be a circle -lethe talk 09:59, 23 January 2006 (UTC) • The largest area to a constant perimeter? • A circle • The smallest area to a constant perimeter? • A straight line with the length of the perimeter • Well, any open curve with the length of the perimeter. Or half the perimeter, if you look at it the right way. Black Carrot 01:27, 29 January 2006 (UTC) In three dimensional space, which shape gives: • The largest surface area (SA) to a constant volume? • Just like with the rectangle with constant area, this is unbounded -lethe talk 09:59, 23 January 2006 (UTC) • The smallest SA to a constant volume? • A sphere. • The largest volume to a constant SA? • sphere, by 3 dimensional isoperimetry -lethe talk 09:59, 23 January 2006 (UTC) • The smallest volume to a constant SA? • Once again, would be a straight line? • No, a straight line will have 0 SA. But a planar square will work. -- Meni Rosenfeld (talk) 10:09, 23 January 2006 (UTC) Thanks. ## dual spaces of (non) locally convex spaces So according to a couple places here, the Lp space for 0<p<1 has no nontrivial continuous functionals. Another example of a space that's not locally convex is stated to have no nontrivial continuous functionals as well. I got to thinking that it's probably true in general that if a space isn't locally convex, then it has to have a trivial continuous dual space. So I tried to prove it. Proof: Suppose f is a linear functional on TVS V. Let U be an open set in space V. Assuming the underlying field is locally convex (since R and C are the only topological fields worth spit), I may choose a convex open set O in f(U) (an interval if the field is R or a disc if the field is C). Define p(x) to be |f(x)|. Then p–1(O) is a convex, balanced, and absorbing subset of U, provides a base, and hence V is a locally convex space. OK, so as a sanity check, for example, the Dirac delta functional should not be continuous on L1/2([–1,1]). I should be able to find a convergent sequence of functions fnf so that f(0) ≠ lim fn(0). In other words, it converges in the ||*||1/2 norm, but not pointwise. So I want someone to look at my proof and tell me it looks OK, and to help me find a sequence to show the failure of continuity. I know some examples of sequence of functions which converge to something in the uniform norm but not in the pointwise topology, but those examples, at least the ones I know, the two limits disagree on sets of measure zero, so those examples don't help with Lp spaces. -lethe talk 09:52, 23 January 2006 (UTC) I haven't dealt with TVS's for a long time. It looks okay except that I am too stupid to see why this is a basis of the topology right now. About your example: How do you even define the Dirac delta on a space that is not a subspace of C0? You need a different type of example to start with if you don't want the set of measure zero problem to come in, maybe try something in the dual space of ${\displaystyle L^{\infty }}$? Kusma (討論) 02:26, 25 January 2006 (UTC) I don't think you're too stupid, I think I am. What I mean is, since writing that, I've come to learn that in fact, my sets are not a base. I didn't update it, because I thought no one was looking at it, but I should have. OK, so I write that p–1(O) ⊆ U. If this were true, then for any open set U, I would have a convex subset, which is the definition of a base. Except of course it's not true. Consider basically f(x1,x2) = x1, and U the unit disc in R2. p–1(O) is not in U. I think I blundered. I've come to know that my conjecture is wrong, so we know my proof must be as well (it's like you say, those sets don't comprise a base). There are spaces which are not locally convex but still have nontrivial dual spaces. I'm still struggling with it, but when I've got it all internalized, I'm going to add a section to locally convex spaces about it. As far as the Dirac delta, I guess you're right. Members of Lp spaces are only defined up to almost everywhere, so the delta functional isn't well-defined. Is that your point? That was sloppy of me as well. (Won't L have the same problem though?) Anyway, thanks for your input. I'm struggling my way through all this. -lethe talk 02:42, 25 January 2006 (UTC) The (continuous) dual spaces of C0 and L are two different spaces of measures, with the dual space of L being measures that are slightly more regular (the Dirac delta is not regular enough), hence can be integrated against functions that are only defined almost everywhere. I can't remember the details, though. Kusma (討論) 04:39, 25 January 2006 (UTC) I guess there should be some list with three columns: functions dual space measure continuous functions of compact support distributions of order 0 Borel measures smooth functions of rapid decrease tempered distributions ? smooth functions of compact support distributions ? where each space of functions has a dual space of certain distributions, and each distribution gives a measure by the Riesz-Markov theorem. Where can we find a list like this? And then your point will be that for the right choice of distribution, it will be defined on equivalence classes of functions that are equal almost everywhere. Somehow, I think the Dirac delta distribution can still be defined over functions defined almost everywhere by using the Dirac measure, but then of course that's a different notion of almost everywhere. Hmm -lethe talk 07:07, 25 January 2006 (UTC) ## mathematics Hello there,my name is Fatima and i am an undergrad student i have two questions 1. whats the difference between Δx ,δx , dx (as written in differential equations) and dx ( as written partial differential equations) since they all signify small change in the variable under consideration ( which is x in this case) 2.whats iota. i know thats a strange question, but i know its the under root of -1, but whats its physical significance, and why do we treat imaginary parts of numbers and equations if its imaginary?? i hope i have been able to explain myself. Well, 1) Δx isn't a derivative, it usually represents a change in x. δx is usually the same, only for a smaller change. Sometimes it's used for a partial derivative. The difference between the 'd' in 'dx' in a diff equation and the 'd' in a partial diff equation is to signfify that it's a partial derivative (the derivative of only one variable of several). That's my experience, although someone might know a more formal definition. 2) Ok, the root of -1. Well.. it doesn't in itself have any physical significance. It's just an abstract mathematical object with a certain property, namely that it's the square root of -1. Now, this gains physical significance when you're modelling different problems. For instance, if you've got a second-order diff equation which you solve the characteristic equation for, the real roots will represent exponential solutions and the imaginary ones will represent periodic ones. And these periodic solutions to diff equations show up everywhere in nature, from the vibration of strings to the Schrödinger equation. --BluePlatypus 17:39, 23 January 2006 (UTC) And for your last question, I once wondered that myself a lot. The issue is really that "imaginary" is an unfortunate name. Imaginary numbers aren't imaginary, they're just as real (or unreal) as other numbers. The natural numbers are an abstract thing (five cows are real, five apples are real, 'five' isn't), negative numbers are more abstract (how do you have minus five cows?), and imaginary numbers are even more abstract. Natural numbers work well for representing everyday countable objects. Decimal numbers or fractions are needed once you're not dealing with whole objects. Negative numbers are useful for balance sheets of money. And imaginary numbers are really good for things that are representable by a differential equation. None of them are more or less "real", they just follow different rules. Remember grade school when you had to re-learn addition and multiplication with negative numbers? Complex numbers are just another extension, one which has turned out to be very useful (Which is why you learn it.) There are other extensions too, like Quaternions, which didn't turn out to be terribly useful, and aren't used as much. --BluePlatypus 18:53, 23 January 2006 (UTC) A protest is warranted by the characterization of quaternions as not terribly useful. They are widely used in many applications that need to manipulate 3D rotations and orientations. Examples include robotics, satellite control, guidance systems, six-degree-of-freedom input devices, and 3D computer graphics (including most games). Also, quaternions paved the way for today's wider concept of abstract algebra, where we can define an algebra of matrices, or choose formalisms like groups or rings or p-adic numbers or whatever we choose. They also live on in physics, disguised as quantum spin, and generalize within Clifford algebra as spin groups. Maxwell's equations for electromagnetism, one of the cornerstones of modern physics, are originally quaternion-based, though that dependency is often suppressed. Also, we have the notable theorem of Frobenius that there are exactly three real associative division algebras: real numbers, complex numbers, and quaternions. A typical reason for stopping with complex numbers is because they are enough to solve any polynomial equation; that hardly justifies denigrating quaternions. --KSmrqT 01:23, 25 January 2006 (UTC) I believe Δx is used for finite changes in x, as in slope = Δy/Δx, while dx is typically used for the infinitely small divisions of x used in calculus. The square root of -1, i, is also used in some common problems in electronics dealing with reactance. However, it is called j there, since i is used in electronics to mean current. StuRat 18:55, 23 January 2006 (UTC) See this site for an excellent visualization of complex numbers: StuRat 19:01, 23 January 2006 (UTC) Δx vs. δx Hopefully this illustration helps with Δx and δx. Both tend to dx as the independent variable becomes infinitesimal. deeptrivia (talk) 19:24, 23 January 2006 (UTC) The names "real" and "imaginary" are not the best. Perhaps "visible" and "hidden" numbers would be a better way to describe the differences. Of course, in French hidden is "occult", and having "occult numbers" might attract a lot of weirdos. LOL. StuRat 19:39, 23 January 2006 (UTC) ;-) O Fatima, if your question is really about mathematics and not about lowely computations, then do not let yourself get befooled by the crude understanding and missleading talk of the engineers ;-) To be serious now: if you want not only the ability to correctly solve equations but to have a deeper understanding, it is wrong to think of the imaginary unit as "the root of -1", ${\displaystyle i={\sqrt {-1}}}$, as this leads to ${\displaystyle -1=({\sqrt {-1}})^{2}={\sqrt {(-1)^{2}}}={\sqrt {1}}=+1}$. I disagree. Note that it wasn't necessary for you to use i in your example, so it isn't a problem with i. This is just a feature of a non-function (where there is more than one value of f(x) for a given x). You can't just square a number, then take the square root, and expect to get the same answer, regardless of you definition of i. Instead, think of i as a (one of two) solutions (roots) of the equation ${\displaystyle x^{2}+1=0}$, which just happen not to be included in the set of real numbers but require an extension thereof, the complex numbers. In case you want a hint of the real beauty of complex numbers, look at Euler's formula and as it expands using Taylor series. The Infidel 20:59, 23 January 2006 (UTC) I'm wondering if there is a font/character problem with your first question. Usually the character used for partial differentials is a curly d, "∂". I'll assume that's what you meant. • The notation "Δx" is used consistently to mean a finite change in x, however small. So we could say that if y is a function of x, then Δyx gives a number, a slope, which approximates the rate at which y changes proportional to a given change in x. • The notation dx is often used to mean an infinitesimal change in x, a subtle concept which takes great care to formalize. Thus dy/dx is not an approximation to the proportion of change, but exact. In more sophisticated mathematical contexts the notation dx is used for a wider range of purposes, which we needn't explore today. • The notation δx is typically only seen in variational calculus. This is difficult to explain succinctly. For ordinary differential calculus, dy/dx indicates that a numeric quantity, y, varies with a change in another numeric quantity, x, by some definite proportion. But in variational calculus we essentially change all the values of a function proportional to an input change. The function itself varies. An example is where we begin with a collection of functions that give position in a plane in response to distance along a path. If we want to find which of those functions gives the shortest path from one fixed position to another fixed position, we use variational calculus, and the "δ" symbol denotes variation in this sense. • The notation ∂x is used in multivariable calculus, where a function like f(x,y) = x2+3y is a function of more than one variable. Thus ∂f/∂x describes the proportion by which f changes when x alone changes, holding y fixed. 1. Imaginary numbers are written using the letter lower-case " i ", not Greek iota (" ι "); the latter has no dot. By definition, this letter denotes a quantity that squares to −1. Where ordinary numbers allow us to step forwards and backwards along the real number line, imaginary numbers allow us to step to the side. More generally, complex numbers allow us to step at any angle and by any amount we wish. The need for such numbers arises naturally when we look for roots of polynomials. For example, the roots of x2−1, the values of x that cause the polynomial to evaluate to zero, are twofold in number: +1 and −1. The roots of x5−1 are fivefold in number, and step off a regular pentagon around zero. Such numbers are just as normal and physical as, say, the square root of 2; the name "imaginary" is a holdover from earlier times when they were poorly understood and treated with suspicion. --KSmrqT 01:47, 24 January 2006 (UTC) To illustrate the above x5−1 example, we get 5 roots: • +1 • +cos(72) + sin(72)i = +.309 + .951i • +cos(72) - sin(72)i = +.309 - .951i • -cos(36) + sin(36)i = -.809 + .588i • -cos(36) - sin(36)i = -.809 - .588i Now let's graph them on the complex plane:  ........................+i......................... .........................^......................... .......................+1|......................... .........................|.----O................... .....................----|.......\................. ................----.....|......................... .........O----...........|..........\.............. .........|...............|......................... .........|...............|.............\........... .........|...............|......................... .........|...............|................\........ .........|...............|......................... -R <-----+---------------+-------------------O-> +R ....-1...|...............|..................+1..... .........|...............|................/........ .........|...............|......................... .........|...............|............./........... .........|...............|......................... .........O----...........|........../.............. ................----.....|......................... .....................----|......./................. .........................|.----O................... .......................-1|......................... .........................v......................... ........................-i.........................  StuRat 10:15, 24 January 2006 (UTC) thanks every body :) u ppl helped me a lot fatima ## decrypt a message given f:Z26->Z26, f(x)=23x+10 (mod 26) is a bijection(one to one and onto) that it can be used as a subtitution cipher, then decrypt the message ZYCU was enctypted by using the function." Sounds like homework to me. --BluePlatypus 17:49, 23 January 2006 (UTC) Homework! Homework! Please try the problem out and you can always come again and write up where your actual difficulty is, or rather where your gettin held up. Then maybe we wikipedians can help you out. But never post the whole homework problem, that spoils the whole point of it! Right? -- Rohit 18:18, 23 January 2006 (UTC) I guessed the answer was CODE but it turns out that was a wrong guess. As a hint, the inverse of 23 with the modulus 26 is 17. – b_jonas 20:30, 23 January 2006 (UTC) # January 24 ## mathematical symbols What is the notation used to signify Infinity, please? A.Hortin It is somewhat like an "8" turned 90° on its side, like this: ${\displaystyle \infty }$Ҡieff 04:11, 24 January 2006 (UTC) It is a lemniscate. -lethe talk 04:12, 24 January 2006 (UTC) In case lethe's answer wasn't clear: Yes, ${\displaystyle \infty }$ is used to denote infinity, but only in those cases when we refer to "just infinity". When we want to distinguish different kinds of infinity, there is a different symbol for each kind. -- Meni Rosenfeld (talk) 07:41, 24 January 2006 (UTC) Meni says true. He's been discussing elsewhere the need to distinguish between +∞, –∞ and unsigned ∞. Among the ordinals and cardinals, there's also alef ℵ, beth ℶ, omega ω, epsilon ε, and more. Many different kinds of infinities, many different symbols! -lethe talk 07:46, 24 January 2006 (UTC) ## Mathematical news Anyone knows a good source of news on the field of mathematics? Even better if they have a newsfeed ☢ Ҡieff 09:19, 24 January 2006 (UTC) Googling 'math news' or 'math news RSS feed' brings up quite a few promising looking results, amongst them, http://mathworld.wolfram.com and http://www.mathforge.net --Noodhoog 13:15, 24 January 2006 (UTC) I sometimes look at the news section of the AMS website (of course, they concentrate on AMS and US news). -- Jitse Niesen (talk) 12:00, 25 January 2006 (UTC) Perhaps we could have a news section at the mathematics portal here on Wikipedia? Fredrik Johansson - talk - contribs 14:58, 25 January 2006 (UTC) # January 25 (no questions today) this sucks ☢ Ҡieff # January 26 ## elementry math please define simple balance. daw Do you mean balance as in a checkbook or as in a see-saw or something else ? StuRat 02:33, 26 January 2006 (UTC) Perhaps equals on both sides of an equation. User:AlMac|(talk) 10:04, 27 January 2006 (UTC) ## Canonical Painlevé ODEs From Painlevé transcendents: In a landmark achievement, they found that up to certain transformations, every such equation [2nd order ODEs with Painlevé property] can be put into one of fifty canonical forms. The page then covers the 6 equations that do not have elementary solutions. What I would like to know is where I can see a list of all 50 forms, preferably with at least some information on the solutions to each. I can find nothing on Wiki and far too many pages on Google covering Painlevé property-related study that is far beyond what I want to know. TIA, Confusing Manifestation 16:41, 26 January 2006 (UTC) According to my notes, it is covered in Ordinary Differential Equations by Edward Lindsay Ince, originally published in 1926, reprinted by Dover in 1956. -- Jitse Niesen (talk) 17:18, 26 January 2006 (UTC) ## Sequence i am trying to find out the sequence to the following mathematic sequence: 11235813.....what is the proper name to this sequence? thank you Assuming you mean 1, 1, 2, 3, 5, 8, 13, ..., the sequence you're looking for is the Fibonacci sequence. Fredrik Johansson - talk - contribs 20:58, 26 January 2006 (UTC) You should search for it on the OEIS. --cesarb 16:19, 28 January 2006 (UTC) ## Support (geometric) I asked a question a while back on Talk:Support (mathematics) and haven't gotten a response. In short: is the usage of the term "support function" meaning "a function taking an N-directional vector and returning the (any) point P on the N-dimensional convex hull of an object for which the function N dot P is maxed" a valid definition to add to the Support (mathematics) page, or is this usage specific to geometric computer algorithms? As my math-ese is rough, some examples of (local-space) support functions for simple shapes: • sphere: support(V) = normal(V) * radius • convex hull: support(V) = the vertex of the hull which is farthest along vector V --Kyle Davis 23:06, 26 January 2006 (UTC) According to my EDM2, this thing is called a supporting functional. The space has to be locally convex and Hausdorff, and the set itself must be convex. Then a linear functional on C is called a supporting functional if sup f(C) = f(x) for some x in the boundary of C, and x itself is called a supporting point of C. So that's about all EDM2 has about it. I'm not sure if this agrees with your expecation. Sounds like it might, but this definition doesn't assume an inner product, so I don't know what normal would mean in that context. -lethe talk 23:53, 26 January 2006 (UTC) PS Of course, in the case that your space is an inner product space, then every linear functional is of the form f(x)=(x,n) for some n. -lethe talk 23:57, 26 January 2006 (UTC) Assuming I'm understanding what you wrote correctly, it's related to what I'm describing, but not quite a match. Specifically, the topology requirements agree, but the function itself isn't a mapping from vector space into linear space but is instead a mapping from an input vector (representing a direction in localspace for the shape) to an output vector (representing a location on the hull). In looking online, I'm finding many examples in source code, but nothing formal. --Kyle Davis 00:14, 27 January 2006 (UTC) It sounds like your map is related to my map in the following way: you assign to every vector the supporting point of the functional defined by the vector. -lethe talk 00:31, 27 January 2006 (UTC) That would make sense. So would it be preferable for me to put this at Support (mathematics), or create a new Supporting functional and put a note at the former regarding the terminology "support(ing) function"? --Kyle Davis 00:36, 27 January 2006 (UTC) I don't have a strong opinion about what name it should go under. If you know this concept as a "support" rather than a "supporting functional", then use that name. If you don't use the name "supporting functional", you should of course make a redirect. -lethe talk 10:23, 27 January 2006 (UTC) # January 27 ## Descriptive term Howdy! I'm in the middle of creating an article on "Pin art/Pinpressions, but I'm struggling to come up with a term for the toy's inability to create shapes with a concave surface in the plane perpendicular to the original "blank" plane. I figure there must be a term for this type of function, and that I'd be better off asking here than on the Language Desk. Any ideas? GeeJo (t) (c) 10:31, 27 January 2006 (UTC) The only thing I can think of to say is that the height must be a function of the position (because each pin can have only one height). â€”Keenan Pepper 23:38, 27 January 2006 (UTC) Yeah, a function or, more specifically, multidimensional function (because it's a contour plane rather than a curve) is the only thing that seems to describe it. What exactly are you trying to do with this word, phrase or description? Are you hoping for a noun? Verb? Something for laymen or jargonauts? Black Carrot 03:16, 28 January 2006 (UTC) Under some assumptions about continuous differentiability, if you want an intrinsic definition you could say that the normal to the surface must (be able to be defined continuously so as to) point upwards everywhere, never downwards, or equivalently that the signed Jacobian thingy dR / dx dy must be positive everywhere (where x, y are coÃ¶rdinates of the "original plane"). â€”Blotwell 03:20, 28 January 2006 (UTC) ## vector sums? This question no verb. â€”Keenan Pepper 23:39, 27 January 2006 (UTC) I would say see vector addition, but that redirect doesn't seem to have one good example in it, so here are four decent examples: Let's say vector A is 4 units long and vector B is 3 units long... • If they are pointed in exactly the same direction, then just do straight addition to get a vector in that direction with a unit length of 7. • If they are pointed in exactly the opposite direction, then subtract the smaller from the other. In this case, the unit length is 1 and it points in the same direction as the longer vector. • If the vectors are at a right angle, use the Pythagorean Theorem to find the length of the diagonal vector. In this case, we would get ${\displaystyle 3^{2}+4^{2}=D^{2}}$, so D = 5. This means the unit length is 5. To find the angle of the resultant vector, use trig: arctan(3/4) = 36.87 deg from the unit length 4 vector toward the unit length 3 vector. • If the vectors are at some other angle, it is necessary to find the X and Y components of each vector (each of which may be positive or negative), then add those together. At this point it has been reduced to the previous case of two vectors at a right angle and you can use the method above to combine them into a single resultant vector. StuRat 02:02, 28 January 2006 (UTC) Our article for this seems to be Vector (spatial). It's linked at the bottom of vector space where noÃ¶ne will find it and noÃ¶ne who does will realize it's what they want. Should we have a dab link at the top of vector space? Should we change the redirect on vector addition? â€”Blotwell 03:15, 28 January 2006 (UTC) Sounds good to me ! StuRat 06:42, 28 January 2006 (UTC) Could be a real question : Vector (band), a 1980s new wave rock band, sums (the money they made), although it could be refactored to "Does Vector sum ?" --DLL 20:16, 30 January 2006 (UTC) ## Principal component analysis I've tried to read principal component analysis and I can't understand it at all. Can someone give me a one or two line summary in lay man's terms? (This is not homework -- it is a statistical method which comes up in some historical work I am doing but I'm having trouble figuring out exactly what it is.) The best I can come up with is that it looks like a very fancy way to average a lot of data points. Right? Wrong? I'm completely clueless here. --Fastfission 04:09, 28 January 2006 (UTC) Let me try...Take a dataset generated from a bunch of variables. Can you reasonably represent this dataset with fewer variables than the original? That is to say, can you choose a fewer number of new variables which would describe most of the details of the original data set? PCA answers this. It also tells you which new variables to use, and it also tells you how much these new variables contributes to describing the original data set. --HappyCamper 05:42, 28 January 2006 (UTC) Hmm. Maybe if I told you the context it would help more: a number of geneticists use PCA to compile gene frequency data (plotted on a map) of 82 genes into one massive "synthetic" map, which is a combinationof all of them (the point is to look for human ancestry trends and soforth). So... what they are doing then with the data is "simplifying it" into one coherent dataset that reflects the combined influence of all of the data? (Basically, I need to describe this in about one sentence in a paper I am writing which mentions this gene mapping project.) --Fastfission 13:30, 28 January 2006 (UTC) What if I said... "They compiled the map using principal component analysis, a common statistical technique which simplified the frequency data from the other 82 genes into one composite map." Would that be true enough? --Fastfission 15:23, 28 January 2006 (UTC) It seems quite okay to me, but it is a bit hard to say without knowing what data the map represents. Instead of "simplified", I'd use "combined" or "aggregated", though neither of these words indicates that you lose some data in the process. And I don't understand why you included the word "other" in your sentence. -- Jitse Niesen (talk) 16:00, 28 January 2006 (UTC) Thanks for the suggestions! Yeah, the "other" just crept in there somehow. --Fastfission 19:57, 28 January 2006 (UTC) When you have a bunch of data, it is frequently more than 2-dimensional. It can be 3, 4, 5, &c. dims. I think that the method is like this : you try to find the plane that is nearer to as many points as possible in that space (and then lesser significant planes). E.g., if your data draw a 3-dim cube, the plane intersects as many lines as possible and you get maybe an hexagon : only the two farthest points are left. --DLL 20:05, 30 January 2006 (UTC) # January 28 ## names for symbols used to record time From the mathematical perspective, does the colon (as it is used to write time) have a specific technical name - or is it merely called a colon as it would be in composition? Our school has a desire to be precise when we teach mathematical concepts, and time is currently in lesson plans for first and second grade. Thanks for your help! --216.63.217.100 20:34, 28 January 2006 (UTC) I think colon is fine. That's the only word I've ever heard used. —Keenan Pepper 21:24, 28 January 2006 (UTC) In English class, the lesson on colons always included timekeeping, so yeah, I figure that's the right word. Black Carrot 01:22, 29 January 2006 (UTC) # January 29 ## Lyapunov stability for state space models The article says: A state space model ${\displaystyle {\dot {\textbf {x}}}=A{\textbf {x}}}$ is asymptotically stable if ${\displaystyle A^{T}M+MA+N=0}$ has a solution where ${\displaystyle N=N^{T}>0}$ and ${\displaystyle M=M^{T}>0}$ (positive definite matrices). Does anyone have a proof for this? deeptrivia (talk) 23:37, 28 January 2006 (UTC) You might get better responses if you try Wikipedia:Reference desk/Mathematics. Grutness...wha? 00:42, 29 January 2006 (UTC) I think that ${\displaystyle V(x)=x^{T}Mx}$ is a Lyapunov function under the condition. By the way, I'd write the condition as ${\displaystyle A^{T}M+MA}$ is negative definite and M is positive definite (and symmetric). -- Jitse Niesen (talk) 02:01, 29 January 2006 (UTC) So the task is to prove that ${\displaystyle {\dot {V}}(x)}$ is negative definite, i.e., ${\displaystyle {\dot {V}}(x)={\dot {x}}^{T}Mx+x^{T}M{\dot {x}}}$ ${\displaystyle =x^{T}A^{T}Mx+x^{T}MA^{T}x}$ ${\displaystyle =x^{T}(A^{T}M+MA^{T})x}$ is negative definite. And that implies ${\displaystyle A^{T}M+MA^{T}}$ is negative definite. Is that what you mean? deeptrivia (talk) 02:47, 29 January 2006 (UTC) Yes: then ${\displaystyle {\dot {V}}(x)={\dot {x}}^{T}Mx+x^{T}M{\dot {x}}=x^{T}A^{T}Mx+x^{T}MAx=-x^{T}Nx}$, so ${\displaystyle V(X)>0}$ and ${\displaystyle {\dot {V}}(x)<0}$ away from 0. Okay...got it! Thanks! deeptrivia (talk) 02:49, 29 January 2006 (UTC) Can you demonstrate how to use this? Let's say we have the case of a damped harmonic oscillator (which we know is asymptotically stable.) We have: ${\displaystyle A={\begin{bmatrix}0&1\\{-M^{-1}K}&{-M^{-1}C}\\\end{bmatrix}}}$ M, K and C are matrices. How can we use this property to prove stability? Thanks! deeptrivia (talk) 03:00, 29 January 2006 (UTC) ## Beautiful this : eiπ + 1 = 0 is supposed to be the most beautiful (or one of the most beautiful)formula, no, not formula, theorem, but ...since I don't know any math, I don't find it beautiful because I do not understand it. lol. --Cosmic girl 05:16, 29 January 2006 (UTC) Its beauty stems from the fact that it relates seemingly unrelated constants like the base of natural logarithm, square root of negative one, ratio of circumference to diameter of a circle, multiplicative identity and additive identity. deeptrivia (talk) 05:20, 29 January 2006 (UTC) BTW, it's ${\displaystyle {e^{i\pi }}+1=0}$ deeptrivia (talk) 05:22, 29 January 2006 (UTC) It's a special case of Euler's formula, which is an extension of the exponential function to complex numbers. I highly recommend the book Visual Complex Analysis (ISBN 0-19-853446-9) if you can afford it. It has a geometric explanation of why the exponential function behaves so. —Keenan Pepper 07:07, 29 January 2006 (UTC) Beauty is in the mind, as well as the eye, of the beholder. We can look at a flower and see its beauty immediately. But it also has a beauty as a part of an ecosystem, using shape and color and nectar to attract and feed selected insects (bees) or mammals (bats), simultaneously spreading its pollen. This more sophisticated beauty only an educated mind can see. The beauty of eiπ+1 = 0 is not like the obvious color and shape of a flower, but more sophisticated, more ecological. It ties together diverse mathematical objects in a surprising and remarkable way. It links shapes (a circle, a line), constants, functions, and number systems. And it's useful. For those whose minds are trained to see, this deeper beauty of both nature and mathematics is as irresistible as the song of the sirens (but with a kinder reward). --KSmrqT 07:15, 29 January 2006 (UTC) ${\displaystyle \sin \left(x+y\right)=\sin x\cos y+\cos x\sin y}$ ${\displaystyle \sin \left(x-y\right)=\sin x\cos y-\cos x\sin y}$ ${\displaystyle \cos \left(x+y\right)=\cos x\cos y-\sin x\sin y}$ ${\displaystyle \cos \left(x-y\right)=\cos x\cos y+\sin x\sin y}$ Did you have to learn that by hart? That's about the same as having to divide 3639 by 3 when forced to use Roman numbers. That's a kind of cruel and unusual punishment. Now look at this: ${\displaystyle {\begin{matrix}\cos \left(x+y\right)+i\sin \left(x+y\right)&=&e^{i\left(x+y\right)}\\&=&e^{ix}e^{iy}\\&=&\left(\cos x+i\sin x\right)\left(\cos y+i\sin y\right)\\&=&\cos x\cos y-\sin x\sin y+i\left(\cos x\sin y+\cos y\sin x\right)\end{matrix}}}$ The Infidel 09:14, 29 January 2006 (UTC) User:Lethe plotted this. It's a nice way to see it visually: That's on my CxR space I proposed a few days ago. ☢ Ҡieff 12:24, 29 January 2006 (UTC) Thanks a lot guys!!! :D --Cosmic girl 20:22, 29 January 2006 (UTC) how d'y know I'm not a girl? (well, this is just an excursion into psychology ;-) The Infidel 19:47, 30 January 2006 (UTC) See the Mathematical beauty article. I say that it need images, as more than 50% of our brain is visual. Agree ? --DLL 19:57, 30 January 2006 (UTC) Hey anyone, why does the Image:Exponential corkscrew.jpg say that there are no links to it when there is obviously at least one link (above)? hydnjo talk 01:53, 31 January 2006 (UTC) I see two links, including the one above, in the "File links" section (not to be confused with "What links here"). --cesarb 14:28, 31 January 2006 (UTC) ## I don't understand general relativity theory ... and I don't expect this can be cured in just one go, but to start with a simple question: In a spacetime with Schwarzschild metric and Schwarzschild coordinates outside the Schwarzschild radius, how does a (small) spacetime vector (t, r, φ, θ) transform when the origin of the locale coordinate system is shifted from (0, R, 0, 0) to (0, R+ΔR, 0, 0) ? Is there an answer that doesn't need tensors, Christoffel symbols and the like? (Maybe the proof of the answer needs to involve these, but the transformation as such should be possible with vectors, matrices and calculus alone, I hope. The Infidel 11:11, 29 January 2006 (UTC) I'm not sure if this is really what you want to ask. First of all, the origin is usually at (0,0,0,0), not (0,R,0,0). Certainly that's the origin in Schwarzschild coordinates. So maybe you want to ask what happens when you shift the origin to (0,ΔR,0,0)? It's not a hard question to answer in principle, but it's a bit messy, since we're in polar coordinates and you lose the spherical symmetry if you change the origin. And of course, while the various components of the invariant distance may change, the total invariant distance itself must of course remain invariant. But let's see what we can say. I'm fooling around with it on paper, and doing it in spherical coordinates in 3d is too much paperwork for me, so I'm going to outline in 2d, and you can fill in the details if you want. I've drawn the triangle with the new origin, and from the law of sines, I can see that ${\displaystyle {\frac {r'}{\sin \phi }}={\frac {r}{\sin \phi '}}}$ and from the law of cosines I have ${\displaystyle r^{2}=r'^{2}+\Delta r^{2}-2r'\Delta r\cos \phi '.}$ Now solve those two equations for r and φ, take the derivatives, and stick them in the Schwarzschild metric and you're done. But you know, that actually still looks like a lot of work. Ugh. Maybe it can be done more easily, but who would want to? The whole point of coordinates is that they're arbitrary, and you should use them to take advantage of the symmetry. I don't think you'd even learn anything from this. -lethe talk 11:55, 29 January 2006 (UTC) Tank you. You are right, what I asked is not what I wanted to ask. I've got trapped with the "origin", and perhaps with more which I am not aware of. What I ment is, if an observer at (0, R, 0, 0)_S has a locale coordinate system with locale origin (0,0,0,0)_L1 at (0, R, 0, 0)_S and another observer at (0, R+ΔR, 0,0)_S with it's own locale coordinate system with locale origin (0,0,0,0)_L2 at (0, R+ΔR, 0,0)_S, how does a (small) spacetime vector (t, r, φ, θ)_L1 transform to (t', r', φ', θ')_L2, where _S denotes a vector in coordinates of the Schwarzschild metric, _L1 and _L2 in the coordinates of the observers respectivly ? (And I don't really care about φ and θ as this is probably only lengthy, messy writing down of formulas that don't contribute to the understaning). The Infidel 13:03, 29 January 2006 (UTC) OK, now I understand what the question is. It's a nice question, I'm sorry for misunderstanding before. Now, do you want your two observers to be free-falling? I don't know the answer right off the bat. You don't usually compare vectors in different tangent spaces, the best you can do is parallel transport the vector from one observer the other. And I don't think that's what you want. Is it? Maybe it is. And to parallel transport to a point that's infinitesimally close, you can use the Christoffel symbols to transform the vector to the adjacent space. -lethe talk 13:28, 29 January 2006 (UTC) Hm, I guess when I am able to put the question right that'll be a good deal of the answer. I think it is parallel transport that I'm asking for. Let me put it another way: if the observer L1 sees a spacetime vector with time length t and space length r (in the direction to the singularity) which "lengths" t' and r' would an observer L2 see? The observers are free-falling, but I'm not sure if this is ultimately necessary. The Infidel 14:00, 29 January 2006 (UTC) To compare two vectors, you have to parallel transport them so that they are next to each other. However, the result will depend on the path taken. For example, let me take two copies of a vector, both defined at a point. Let me put one vector on an airplane, and fly it around the world. When I bring it back to the lab, and compare it with its twin, I find that it is now pointing in a different direction. Just how it differs depends on the path the airplane took. This realization is the basic idea behind satellite tests of GR, e.g. Gravity Probe B. Note that you can't really do parallel transport without taking about Christoffel, etc. because that's what captures the path info. The other way to compare vectors at different points is to shine light on one of them, and then try to view it through a telescope, and try to se "what angle its sitting at", i.e. compare it to the local one. The problem now is that it takes a while for the light to get from here to there, and various paradoxes result, including various famous ones from special relativity. linas 16:49, 29 January 2006 (UTC) Thank you. I see that I have to take a deeper look at the manifolds complex of articles. But for now, let the path between the oserver at (0, R, 0, 0) and (0, R+ΔR, 0, 0) be ${\displaystyle p:[0,1]\mapsto \mathbb {R} ^{4};\;p(t)=(0,R+t\,\Delta R,0,0)}$. The Infidel 17:22, 29 January 2006 (UTC) So finding the parallel transport of a vector involves integrating the covariant derivative. In other words, solving some PDEs. That's why I said it's easier if you're only doing an infinitesimal parallel transport, since then "integrating" becomes just multiplying by dt. So let c(t) be a path, and v be a vector field. You would like to define the derivative of a vector field along the path as ${\displaystyle {\frac {dv}{dt}}=\lim _{dt\to 0}{\frac {v(c(t+dt))-v(c(t))}{dt}}}$ but you can't subtract vectors in different spaces. So let Γc be a linear operator which moves from one vector space to another along c', and define your derivative as ${\displaystyle \nabla _{\dot {c}}v=\lim _{dt\to 0}{\frac {\Gamma _{\dot {c}}(v(c(t+dt)))-v(c(t))}{dt}}}$ so Γ is a linear map that takes a vector in the tangent space at t+dt and returns a vector from the tangent space at t along c. Thus, for an infinitesimal change, we can see that ${\displaystyle v'=v+\nabla _{\dot {c}}v\,dt.}$ Changing to local coordinates, and using the linearity of Γ, I see that ${\displaystyle \nabla _{\dot {c}}v^{\mu }={\dot {c}}^{\nu }v^{\rho }\Gamma _{\nu \rho }^{\mu }+{\dot {c}}^{\nu }\partial _{\nu }v^{\mu }.}$ You should probably make sure you know how to derive that last expression. Post if you need help. OK, so now all we need is the Levi-Civita coefficients for the Schwarzschild metric and we're good to go. I'm going to calculate those coefficients and then the answer. You try it too, and we'll compare results. Is it clear enough for you to start calculating? -lethe talk 22:11, 29 January 2006 (UTC) That's great, thanks a lot. Now I need to digest this. This will take some time. The Infidel 19:37, 30 January 2006 (UTC) Buurrpp! I'm digesting. hydnjo talk 01:43, 31 January 2006 (UTC) If you want more details on the steps, go ahead and say so. I know I said I was going to do the calculation and get the answer myself, but I haven't done that yet. -lethe talk + 13:59, 31 January 2006 (UTC) Thanks again, but no hurry. Regrettably, I'm a bit short of time during the week and understanding is difficult when the mind is tired. But just to check the notations: ${\displaystyle \Gamma _{\nu \rho }^{\mu }}$ is the Christoffel symbol and ${\displaystyle \partial _{\nu }v^{\mu }}$ is the covariant derivative ${\displaystyle v^{\mu }\cdot \partial _{e_{\nu }}e^{\mu }}$ when ${\displaystyle e_{\nu },\,e^{\mu }}$ are the unit vectors of the local vector spaces? The Infidel 20:02, 1 February 2006 (UTC) yes, ${\displaystyle \Gamma _{\nu \rho }^{\mu }}$ are the Christoffel symbols. but ${\displaystyle \partial _{\nu }v^{\mu }}$ is not the covariant derivative of anything. ${\displaystyle v^{\mu }}$ is the μ-th component of a vector. A component of a vector is simply a scalar, and you can't covariant derivative scalars. You can only regular derivative scalars. so that's what ${\displaystyle \partial _{\nu }v^{\mu }}$ is the ν-th partial derivative of the μ-the component of v. That means that I'm taking my local basis to be ${\displaystyle e_{\mu }=\partial /\partial x^{\mu }=\partial _{\mu }}$. Note that even though ${\displaystyle \nabla _{\dot {c}}v^{\mu }}$ looks like the covariant derivative of the μ-th component of v, such a thing makes no sense, you can only covariant derivative vectors. This is a bad notation for the μ-th component of the covariant derivative. Sometimes I use parentheses like so ${\displaystyle (\nabla _{\dot {c}}v)^{\mu }}$ to make it clear. -lethe talk + 11:41, 2 February 2006 (UTC) # January 30 ## Such-and-such's Theorem I can remember once in a Statistics lecture the lecturer mentioning a "theorem" that went something like this: Any theorem in Statistics that is named after someone is named after someone other than the person who first discovered it. In other words, all eponymous statistical theorems are misnamed. He also pointed out that this theorem, which is itself named after someone who came up with it, falls victim to itself. Unfortunately, I have no idea who it was named after or who came up with it first. Any ideas? Confusing Manifestation 16:16, 30 January 2006 (UTC) I once read on USENET John Baez say: At least I can take some solace from Baez's First Law of Nomenclature: "Any law, effect, or theorem named after Professor X can safely be assumed to have been first discovered by Professor Y, for some value of Y not equal to X." In this case, it turns out the Lorentz force law was first discovered in 1889 by Heaviside! " He's being tongue in cheek, and he's not talking about statistics, so I'd be surprised if it's the same name that you heard in lecture. Maybe a lot of different people have made that observation and then named it after themselves (especially if they knew they weren't the first to make it). -lethe talk 22:24, 30 January 2006 (UTC) The book Új matematikai mozaik by (editor Hraskó András, Typotex kiadó, Budapest, 2002) also mentions this "theorem" in a footnote on page 399, in the writing of "A sündisznó megfésülése és egyéb gyakorlati problémák" by Szűcs András. It doesn't mention any name for it, so I think this must be part of the folklore. – b_jonas 13:50, 31 January 2006 (UTC) In Russia, it's called Arnold's theorem. Of course it applies to itself, so Arnold didn't invent it. Grue 22:02, 31 January 2006 (UTC) I thought theorems and formulas were named after the first person to discover them after Euler. linas 14:40, 1 February 2006 (UTC) # January 31 ## Recursive & Floor Functions Does anyone know a good source for advice on solving recursive functions? How about equations that use floor and ceiling functions? Black Carrot 02:57, 31 January 2006 (UTC) Get Concrete Mathematics by Ronald L. Graham, Donald E. Knuth, Oren Patashnik (Hungarian translation Konkrét Matematika. Műszaki könyvkiadó, Budapest, 1998). It's a great book. – b_jonas 13:45, 31 January 2006 (UTC) Are you using recursive function in the computer science sense of the term ? Or do you mean a function that is defined in terms of its previous values, such as the dyadic transformation xn+1=2xn - floor(2xn) ? Gandalf61 14:27, 31 January 2006 (UTC) The second one. I'm trying to find a way to either solve a particular equation for its fixed points, or iterate it to infinity starting from a low number (say 2), depending on which I figure out first. They should give more or less the same answer. ${\displaystyle x_{n+1}=\left\lceil {\frac {a}{\left\lfloor {\frac {a}{x_{n}}}\right\rfloor }}\right\rceil }$, where a equals a whole number constant. Black Carrot 20:34, 31 January 2006 (UTC) Well, as far as I know there's no easy way to deal with floor\ceiling functions, but you can extract a few info from just looking at it. For example, ${\displaystyle 0\neq |x_{n}|\leq |a|}$, because otherwise you'll have a division by zero. ☢ Ҡieff 21:39, 31 January 2006 (UTC) Black Carrot, do you want to find the fixed points of the particular equation that you gave? If yes, why? Generally, floor / ceiling functions are hard to handle, but that particular equation can be tackled with ad hoc methods. You can start by finding the fixed points for some value of a, e.g. ${\displaystyle a=6}$, and make sure you understand Kieff's remark. -- Jitse Niesen (talk) 22:02, 31 January 2006 (UTC) Let's call your function Ga, so xn+1=G(xn). Try to prove the following: 1. If x is an integer and a factor of a then x is a fixed point of Ga. 2. If x is not a factor of a then Ga(x)>x. 3. For any initial x in the range [1,a], iterating Ga(x) eventually reaches a factor of a. 4. For any initial x in the range [1,a], iterating Ga(x) eventually reaches the next factor of a that is equal to or greater than x (i.e. the iterations do not "overshoot" a fixed point). Gandalf61 10:40, 1 February 2006 (UTC) I have. That's what I designed it to do - to basically perform a search for the whole factors of a. The problem is that performing the entire search from beginning to end would be even less efficient than trial and error. Which is fine, since I designed it to be solved, not run through, but I'm having some trouble doing that. Does your quick and surprisingly detailed answer mean that someone's already failed at this, and written about it, or just that you're good with iterative functions? Black Carrot 13:10, 1 February 2006 (UTC) I haven't come across this particular function before, but I would be surprised if it were completely new. As you say, convergence to a factor of a is slow if a is large compared to the initial value of x. Specifically, if xn is an integer that is less than the square root of a but not a factor of a then xn+1 will be xn+1. Gandalf61 10:30, 2 February 2006 (UTC) Word. So, to get back to my question, does anyone know where I can find advice/tutorials/anything on recursive- and/or floor-related problem solving? I've looked in Wikipedia, and MathWorld, and anything else I can find, but there really doesn't seem to be much available. I'd appreciate any suggestions, but I probably won't follow up anything requiring spending$150 on a few paragraphs of applicable information (textbooks) or requiring library and journal access beyond the reach of a highschooler (me). Thanks. Black Carrot 20:00, 2 February 2006 (UTC)

As b_jonas said, if you want to understand recurrences and the floor and ceiling functions, Concrete Mathematics is the text to get.
However, maybe you need to explain more clearly what you're trying to do. Your recurrence has a fixed point when xn is a factor of a. So are you in fact trying to find an efficient integer factorization algorithm? Gdr 17:55, 5 February 2006 (UTC)
Yup. And I'll try out the book, if it's that good. Is it okay if I move this down? It's about to get archived. Black Carrot 01:56, 7 February 2006 (UTC)

## Number of Bases for a Linear Code

If a linear code has dimension k,then derive the formula for the number of bases of that linear code.

Assume ${\displaystyle p}$ dimensions, where ${\displaystyle p>18.62}$. According to the Hungert Theorem, ${\displaystyle Q={\sqrt {x\zeta }}}$, so you can then prove by induction (remembering that ${\displaystyle e^{i\pi }+1=0}$ and ${\displaystyle 1+1=2}$) a statement which, when solved for ${\displaystyle \zeta }$, will give you the calorific cost of doing your own homework. — QuantumEleven | (talk) 14:30, 31 January 2006 (UTC)
That theorem wins the Intarnet. --Deville 04:06, 4 February 2006 (UTC)

## Area of a concave polygon

I know there are many ways of calculating the area of a concave polygon. I am writing a computer program. For a computer, using X,Y coordinate pairs, is there an algorithm that is considered best or is it just whatever feels best for the programmer? Right now, I am going to use the method of peeling off triangles and adding their areas until I use up all the points. --Kainaw (talk) 17:36, 31 January 2006 (UTC)

What about Pick's theorem? —Keenan Pepper 18:17, 31 January 2006 (UTC)
That is cool. I just have to work on an entirely different algorithm to calculate the number of points on the border and then the number of points contained inside the polygon. I don't know which is more difficult. --Kainaw (talk) 18:22, 31 January 2006 (UTC)
If the vertices in order have the coordinates ${\displaystyle ((x_{0},y_{0}),(x_{1},y_{1}),{}}$${\displaystyle \dots ,(x_{n-1},y_{n-1}),(x_{0},y_{0}))}$, then the area of the polygon is half of the absolute value of ${\displaystyle x_{0}y_{1}-x_{1}y_{0}+{}}$${\displaystyle x_{1}y_{2}-x_{2}y_{1}+{}}$${\displaystyle \dots +x_{n-2}y_{n-1}-x_{n-1}y_{n-2}+{}}$${\displaystyle x_{n-1}y_{0}-x_{0}y_{n-1}}$. Don't forget about the signs and the last two terms. See Cross product. – b_jonas 20:09, 31 January 2006 (UTC)
Thanks. this looks like it will calculate quicker. Now, I'm trying to figure out how to take a cross product. The cross product page explains the theory, but ignores a direct calculation - such as one you can type into a computer that only accepts things like addition, subtraction, multiplication... not sets and ranges. --Kainaw (talk) 20:21, 31 January 2006 (UTC)
The cross product of the three-dimensional vectors ${\displaystyle (x_{0},y_{0},z_{0})}$ and ${\displaystyle (x_{1},y_{1},z_{1})}$ is the vector ${\displaystyle (y_{0}z_{1}-z_{0}y_{1},z_{0}x_{1}-x_{0}z_{1},x_{0}y_{1}-y_{0}x_{1})}$. The magnitude of this vector gives twice the area of the triangle with vertices ${\displaystyle (0,0,0)}$, ${\displaystyle (x_{0},y_{0},z_{0})}$ and ${\displaystyle (x_{1},y_{1},z_{1})}$. In the case when the vectors are in the plane ${\displaystyle z=0}$, this degenerates to the unit vector ${\displaystyle (0,0,1)}$ times the outer product ${\displaystyle x_{0}y_{1}-x_{1}y_{0}}$; so that outer product is twice the signed area of the triangle. If you add these signed areas, you get the area of the polygon. – b_jonas 22:48, 31 January 2006 (UTC)
Unless I did something wrong, this is producing double the area. I just divided it by 2 and it is working fine. Thanks. --Kainaw (talk) 20:54, 31 January 2006 (UTC)
Argh, indeed. You must divide by two. Sorry. – b_jonas 22:48, 31 January 2006 (UTC)
A typical fast method is secretly an application of the generalized Stokes' theorem. Make a trapezoid with each edge and its projection onto a chosen fixed axis. Add up all the signed areas, these being easy to calculate. Note that the polygon must be "simple" (which here is a technical term). --KSmrqT 00:53, 1 February 2006 (UTC)

## translate this coordinate expression into a coordinate-free expression

When doing linear algebra or tensor analysis, if you write an expression in terms of its components in some basis (or local coordinates), but the total expression is invariant under coordinate changes (all the component indices are contracted à la Einstein), then there should be a coordinate-free way of writing the expression.

For example, given a bilinear form B with components Bab, and two vectors with components va and wb, the expression Bab(va,wb) has all its component indices contracted. This is the component expression for the coordinate-free expression B(v,w). Writing v = vaea and Bab=B(ea,eb) lets you change from the coordinate-free to component expresssion and back.

Now, with a nondegenerate bilinear form on the vector space V, and a basis {ea}a, I can form the coordinate expression Babeaeb, where Bab are the components of the inverse of the bilinear form B–1 (defined by B–1vw)=B(v,w) with σv a member of dual space V* given by σv(u) = B(v,u). In components, BabBbcac). I'm trying to figure out what the coordinate-free version of this expression is. Clearly it is a member of the tensor space VV. The weird thing is that, unlike the above example, it doesn't seem to be the result of a map; it has no free variables. Edit: Actually, I guess that's not so weird. The same comment applies to the volume form of an inner product space. -lethe talk + 23:55, 31 January 2006 (UTC)

OK, nevermind. I figured it out. The answer is of course that it is the coordinate-free expression for B–1 itself. -lethe talk + 00:49, 1 February 2006 (UTC)