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October 22

Polynomial degree vs order

Reading both Wikipedia and Wolfram I see that order and degree can be used interchangeably. But I was fairly certain that when I researched the same thing about 5 years ago the definitions were slightly different. In the next paragraph I do not state a fact, I am simply stating my previous (most likely erroneous) understanding.

I remember the "degree" being defined as the exponent of the highest order term, so a parabola would be 2nd degree. The order (and this is where I'm confused) was defined as the number of variables in the particular class of polynomial, so a general parabola would be 3rd order. A general 2nd degree polynomial of the form a*x^2 + b (no linear term), would be 2nd order.

Was it once defined like this or must I have been getting bad information?

196.211.175.2 (talk) 09:56, 22 October 2010 (UTC) Eon

The words degree and order are used interchangeably, according to degree of a polynomial. Bo Jacoby (talk) 12:37, 22 October 2010 (UTC).

I don't recall seeing the definition of order you described. But it's very possible that this definition is common in a particular place or you've seen it used by a particular author. -- Meni Rosenfeld (talk) 13:58, 22 October 2010 (UTC)

I've always used the word "degree" for things like the exponent in x³, and said that the degree of a polynomial is the exponent of the highest-degree (not highest-order) term. "Order", on the other hand, I use for differential operators; thus one can speak of a second-order differential equation. I think that's the way I was taugth to use those terms in school. But I know that in some eras the word "order" has been used the way I use the word "degree", and that some people are not fastidious. Michael Hardy (talk) 18:43, 22 October 2010 (UTC)

For a power series, order can refer to the smallest exponent that has a non-zero coefficient. Quantling (talk) 20:33, 26 October 2010 (UTC)

typesetting a long stream of numbers in Latex?

How do I typeset 114813069527425452423283320117768198402231770208869520047764273682576626139237031385665948631650626991844596463898746277344711896086305533142593135616665318539129989145312280000688779148240044871428926990063486244781615463646388363947317026040466353970904996558162398808944629605623311649536164221970332681344168908984458505602379484807914058900934776500429002716706625830522008132236281291761267883317206598995396418127021779858404042159853183251540889433902091920554957783589672039160081957216630582755380425583726015528348786419432054508915275783882625175435528800822842770817965453762184851149029376 so that it wraps and stuff? It just goes off the edge of the page now, so that only one line is visible... 84.153.179.167 (talk) 13:52, 22 October 2010 (UTC)

Insert \allowbreak in places where you want to allow it to break. You'd probably also need to mark the paragraph with \raggedright, as there will be no stretchable spaces on the lines.—Emil J. 14:14, 22 October 2010 (UTC)

If you want to allow a line break after every digit, and if you need to do this repeatedly, here's a macro:

\def\longnumber#1{\raggedright$\dolongnumber#1\stop}
\def\dolongnumber#1{\ifx\stop#1$\else#1\allowbreak\expandafter\dolongnumber\fi}

\longnumber{1148130695274254524232833201177...965453762184851149029376}

—Emil J. 14:28, 22 October 2010 (UTC)

Now I wonder: does anyone have a clue why the syntax highlighter coloured the first occurrence of "#1" black, the second one pink, and the other three blue?—Emil J. 15:19, 22 October 2010 (UTC)

I think it is confused by the (apparently) unpaired dollar signs. The anomaly changes somewhat if I replace them with \begin{math} and \end{math}, and goes away if I remove them completely. –Henning Makholm (talk) 15:40, 22 October 2010 (UTC)

I see. Thank you.—Emil J. 16:40, 22 October 2010 (UTC)

With hindsight, the \raggedright was a bad idea, as it spill over to subsequent paragraphs, and cannot be easily confined. Here is a better implementation:

\def\longnumber#1{$\dolongnumber#1\stop}
\def\dolongnumber#1{\ifx#1\stop$\else#1\allowbreak\hskip0pt plus.5pt\expandafter\dolongnumber\fi}

\longnumber{1148130695274254524232833201177...965453762184851149029376}

It inserts a slightly stretchable space after every digit to allow proper justification. Here's the same kind of thing with a bit of syntactic trickery:

\def\longnumber{\def\relax##1{##1\allowbreak\hskip0pt plus.5pt\relax}\relax}

$\longnumber1148130695274254524232833201177...965453762184851149029376$

—Emil J. 13:55, 25 October 2010 (UTC)

Redefining \relax is very much asking for trouble, even though it happens to work safely in the intended use case. It won't work with LaTeXy \begin{math}/\end{math}, nor for a primitive $$display$$. –Henning Makholm (talk) 15:38, 25 October 2010 (UTC)

That's why I called it a trickery, it's basically for fun, and it's designed to work only in the exact way I used above (a sequence of digits between single dollars). If that's a problem, use the previous solution (which however also relies on some assumptions on the token sequence in its argument, it won't process correctly stuff like \longnumber{123\bar{4}567}). If I were serious about it I'd also define a skip register for that "0pt plus .5pt".—Emil J. 15:57, 25 October 2010 (UTC)

if you had to brute-force a number but could be told some of the digits

If you had to brute-force a number but could be told some of the digits, would you rather have the n most significant or the n least significant digits - or wouldn't it make a difference? —Preceding unsigned comment added by 93.186.23.239 (talk) 14:07, 22 October 2010 (UTC)

If you do not know (or suspect) anything else about the number, it makes no difference. If you do, it depends on the other expected properties of the number.—Emil J. 14:37, 22 October 2010 (UTC)

all right, you're trying to factor a large composite number into its exactly two large random primes, and I'm telling you about the n most or least significant digits of the larger of them. Is the only reason you would prefer the former that the last digit is odd, so you already have slightly more information about it? Or would you have any other reason to prefer one over the other? Thank you.

You also know that it is coprime with 5 (if the digits are decimal digits). This is already a good reason (just oddness of primes essentially halves your search space if given the most significant digits), but I would also prefer the most significant digits because then the brute force requires me to sieve primes out of an interval, which is somewhat easier then to sieve them from the set of numbers with given ending (however, this is only relevant if you are really going to find the factors by brute force; if you are using a more sophisticated factoring algorithm, it may have quite different preferences as to which extra information on the factors is more useful, and it may well happen that it cannot use either of them).—Emil J. 15:37, 22 October 2010 (UTC)

I am struggling to see how you can say anything useful about the factorisation of the number if you are only told some of its digits and that it is a semiprime. For example, if you know the number is 1231x, it could be 13 x 947 = 12311, or 7 x 1759 = 12313 or 109 x 113 = 12317 or 97 x 127 = 12319. Gandalf61 (talk) 16:04, 22 October 2010 (UTC)

No, that wouldn't be an example of the kind the OP is describing. An example may be that you need to factor 12317 and know that one of the factors is 11x. –Henning Makholm (talk) 16:19, 22 October 2010 (UTC)

Yes (OP here, my IP has changed though) . If you were trying to factor 5725797461 would you rather know that the larger of them begins 75... or ends ...679 -- you only get to learn one of these things. Extrapolating, would you rather learn the first or the last hundred digits of a 1000-digit prime factor of an even larger composite number you were trying to factor into its exactly two large prime factors? 84.153.222.131 (talk) 18:03, 22 October 2010 (UTC)

Quotients

Resolved

I'm looking for a command in either Maple or Maxima to compute the quotient of a polynomial ring by a gradient ideal. I can do some simple examples by hand, but I don't know a general method. For example, let ƒ(x,y) = x² + y^k where k is a positive integer. I have

${\mathcal {A}}=\mathbb {R} [x,y]/\left\langle f_{x},f_{y}\right\rangle =\mathbb {R} [x,y]/\langle x,y^{k-1}\rangle \cong \mathbb {R} \langle 1,y,y^{2},\ldots ,y^{k-2}\rangle .$

Provided that the gradient vector field has an algebraically isolated zero at x = y = 0 the result will be a finite dimensional real vector space. Algebraically isolated zero means that the zero is isolated when x and y are considered as complex variables, and not just real. For example, can anyone calculate the vector space (called the local algebra) when ƒ(x,y) = x³ + xy^k?

${\mathcal {A}}=\mathbb {R} [x,y]/\left\langle f_{x},f_{y}\right\rangle =\mathbb {R} [x,y]/\langle 3x^{2}+y^{k},xy\rangle \cong \ldots \ ?$

More importantly, does anyone know a command in either Maple or Maxima to compute the quotient? — Fly by Night (talk) 15:10, 22 October 2010 (UTC)

I know what an ideal is, and I know what a gradient is, but what is a "gradient ideal"? Neither article so much as mentions the other word. –Henning Makholm (talk) 15:46, 22 October 2010 (UTC)

The thing in the denominator of my quotients above. For a function ƒ : Rⁿ → R the gradient ideal of ƒ is given by

$I_{f}:=\left\langle {\frac {\partial f}{\partial x_{1}}},\ldots ,{\frac {\partial f}{\partial x_{n}}}\right\rangle .$

It should really be thought of as an ideal in the ring of analytic function germs, but you can get away with thinking of R[x,y] instead. It's quite a common term. Google Scholar returns about 140 hits. — Fly by Night (talk) 15:54, 22 October 2010 (UTC)

It confused me that it looked as if "gradient ideal" was linked to something that ought to explain that particualar construction. Also, I'm not used to the angle bracket notation here; the texts I grew up with used round parentheses for "the ideal generated by such-and-such". –Henning Makholm (talk) 16:09, 22 October 2010 (UTC)

Ah, I see. Looking around on Google seems to show a mixture of rounded and angled brackets for the gradient ideal. Maybe it's not common in traditional ring theory, but the gradient ideal is an application of ring theory to another subject. So you'd have non-ring-theorists using rings and ideals. Maybe that would explain the non-familiar notation. — Fly by Night (talk) 17:59, 22 October 2010 (UTC)

In your example $\mathbb {R} [x,y]/\langle 3x^{2}+y^{k},xy\rangle$: When you quotient out (xy), all mixed terms drop out, so you have, R(1,x,y,x²,y²,x³,y³,...) left. Now quotient out (3x²+y^k). Then you can rewrite y^(k+1) as -3x²y which drops out because xy=0. Similarly, x³=x(-1/3)y^k=0. Then you're left with R(1,x,x²,y,y²,...,y^k), but x² is (-1/3)y^k, so your final space is R(1,x,y,y²,...,y^k).

This is hardly an example of a general method, of course. And the multiplicative structure of the quotient ring is lost when you write it simply as a vector space, but I suppose that is what you want? –Henning Makholm (talk) 16:09, 22 October 2010 (UTC)

It's a method to calculate the index at a singular point of a gradient vector field. You work out the quotient and then make a multiplication table of the spanning elements. You fill it in, modulo the gradient ideal. Then you find a linear map that takes the class of the determinant of the Hessian of ƒ to 1. The simplest one would send [det(Hess_ƒ)] to 1 and every other class to 0. This maps changes the k-by-k multiplication table into a k-by-k matrix. The signature of which is the index of the singular point of the gradient vector field. See here if your interested. (N.B. They use round brackets for ideals in this article.) — Fly by Night (talk) 18:12, 22 October 2010 (UTC)

You might want to try the SINGULAR CAS. I believe that it specialise in this sort of calculation.--Salix (talk): 19:44, 22 October 2010 (UTC)

I've just looked at the website, and it looks perfect. Thanks a lot Salix. I shall download it now. — Fly by Night (talk) 20:04, 22 October 2010 (UTC)

Scenario: Traveling to Pandora as seen in James Cameron's movie Avatar

I've been trying to figure out how long it would take to get to the fictional moon Pandora from Earth in the movie Avatar—at a specific faster-than-light speed—, but the unit conversions have been tedious. My problem is based on a few assumptions and definitions. I've laid them out as follows:

• Assumptions & Considerations:
  • Assume that fast than light travel is possible.
  • Either assume that time dilation does not occur or is not a concern.
  • Either assume that I'm not killed by the speed or that it is not a concern.
  • Assume their is no need for acceleration/deceleration time.
• Definitions:
  • The distance from Earth to Pandora is 4.37 light years.
  • One year is 31,557,600 seconds.
  • The speed of light is 186,282.397 miles per second.
• Givens:
  • The speed I'm traveling at is 5 trillion miles per nanosecond.
• Problems:
  • I would like the speed converted to miles per second—no rounding please.
  • I would like to know how long it would take me to get their—no rounding please.
  • I would like a real world comparison as to how quick that time is (example, 0.25 seconds is the blink of an eye).

--Melab±1 ☎ 19:02, 22 October 2010 (UTC)

Google is neat for unit conversions: 4.37 light years / 5e12 miles/ns, assuming short-scale trillions. –Henning Makholm (talk) 19:16, 22 October 2010 (UTC)

Thank you, but I need all three of my problems answered and I also do not know if Google's unit conversions use the definitions I have defined. --Melab±1 ☎ 19:48, 22 October 2010 (UTC)

Well, OK. 5 trillion is usually 5x10¹², and this occurs 10⁹ times per second, giving an exact speed of 5x10²¹ miles per second. At your definition, a light year is 5.87862537 x 10¹² miles (for a standardised definition, see light year). That makes a total distance of 2.56895929 × 10¹³ to a level of accuracy that's only true if Pandora was exactly 4.37 light years away. Divide the latter by the former, and you get 5.13791857 nanoseconds. That's less than the time it takes light to cover two metres, to give an example. That really short period of time is quite hard to quantify in everyday terms. Hope I've done this right, - Jarry1250 ^{[Who? Discuss.]} 20:23, 22 October 2010 (UTC)

I agree with the above calculations. Mine are... 5 trillion miles per nanosecond = 5*10^12 miles / (10^-9 sec) = 5*10^21 miles / sec = 5,000,000,000,000,000,000,000 miles / sec. 4.37 light years * 31,557,600 seconds / year * 186,282.397 miles / second = 5,878,625,371,567.2 miles * 4.37 = 25,689,592,873,748.664 miles. Distance = rate * time, so time = distance / rate (using the nonrelativistic approximation you seem to want). Here, travel time is 25,689,592,873,748.664 miles / (5*10^21 miles / sec) = 0.0000000051379185747497328 sec = 5.1379185747497328 nanoseconds. For scaling, about 5.14 nanoseconds in a second is the same ratio as about 2.5 minutes in a millenium. It should be noted that you're asking us to use wayyyyy too many significant figures in these calculations, and the speed is incredibly high. 67.158.43.41 (talk) 05:00, 23 October 2010 (UTC)

But faster than light travel isn't possible. Traveling exactly at the speed of light isn't possible either, unless you have zero mass or infinite energy: your mass increases with respect to your speed, so the faster you go the more energy you need to go faster. Close to the speed of light you would need almost infinite amounts of energy to accelerate your almost infinite mass. See this section of the Special Relativity article. Some theories hypothesise about anti-particles with negative mass travelling faster than the speed of light. Why don't you just assume the speed to be s, and the distance to by d? — Fly by Night (talk) 10:01, 23 October 2010 (UTC)

Generalization of ordinals

In ZFC, ordinals are equivalence classes of sets, since every set can be well-ordered in ZFC. In ZF+~AC, there are sets that cannot be well-ordered. Is there any way to organize these sets into equivalence classes that provide some notion of size similar to the ordinals? 99.237.180.170 (talk) 19:07, 22 October 2010 (UTC)

I'd quibble with the way you put it. There's no such thing as "sets in ZFC" or "sets in ZF+~AC"; there are only sets, and AC is either true of them, or it is not.

That said, yes, you can certainly get objects corresponding to cardinality, working in ZF alone. What you do is take equivalence classes up to the existence of a bijection. Unfortunately these are proper classes, so you use the Scott trick to cut them down to sets. --Trovatore (talk) 19:13, 22 October 2010 (UTC)

I think your terminology quibble is more Platonic than the mainstream mathematical culture makes it fashionable to be. But in that case, if you accept that the ZF axioms are true, then both of ZFC and ZF¬C will be consistent and thus have models. And the elements of such a model's universe are usually known as "sets" in that model, even though you may prefer to reserve the word for something more metaphysical. –Henning Makholm (talk) 19:31, 22 October 2010 (UTC)

Well, those objects are sets in the sense of the model, not in the sense of the theory. It's a critical distinction. --Trovatore (talk) 21:08, 22 October 2010 (UTC)

Like Henning Makholm, I stand by my original use of the word "sets." I don't see what you mean by "sets in the sense of the theory." To me, sets are objects described by the axioms of a set theory. I did consider using cardinality as you suggested, but I was hoping for something more specific. Maybe something with another axiom added to ZF¬C might work better for my purposes. Any more suggestions? 76.67.74.196 (talk) 22:04, 22 October 2010 (UTC)

The correct phrasing is ZFC proves that every set can be wellordered, not every set can be wellordered in ZFC. --Trovatore (talk) 23:17, 22 October 2010 (UTC)

I suspect what you mean is the correct phrasing is

"{Every set can be well-ordered} in ZFC

rather than

"Every set can be {well-ordered in ZFC}.

(The latter being meaningless.) Michael Hardy (talk) 16:57, 23 October 2010 (UTC)

No, I really don't. The mathematical standard is to talk like a Platonist whether you actually are one or not. This saves an immense amount of grief when it comes to things like expositions of the incompleteness theorems.

It's far cleaner in such cases to use the standard language, where a flat assertion of a statement is the same as the claim that the statement is true (disquotational truth) rather than provable. Then you can always make some sort of disclaimer afterwards about what interpretation you really have in mind. --Trovatore (talk) 17:49, 23 October 2010 (UTC)

This discussion is getting pretty far off topic. Returning to the original answer, doesn't the definition of the cardinal numbers depend on the ordinal numbers through cardinals like $\beth _{\omega }$, $\beth _{\omega +1}$, and $\beth _{2\omega }$? How can you name cardinals without the axiom of choice? Are the cardinals defined as equivalence classes, but they cannot be enumerated without the AC? 74.14.109.58 (talk) 20:01, 23 October 2010 (UTC)

No, without the AC there is no systematic enumeration of the cardinals; you cannot even always compare two cardinals to see whether one is larger than the other (the AC is equivalent to trichotomy between cardinals). Without AC there is no better general way of naming a cardinal than to exhibit a set (or many sets) of the cardinality you want to speak of. –Henning Makholm (talk) 20:12, 23 October 2010 (UTC)

A useful proxy in some contexts for such cardinalities is the notion of Borel reducibility of equivalence relations. If equivalence relation E is reducible to equivalence relation F, then the quotient space by E has cardinality less than or equal to the quotient space by F. See Borel equivalence relation for a bare start. An article I had ambitious plans for at one time but never got around to really developing. --Trovatore (talk) 20:17, 23 October 2010 (UTC)

Shoelace formula

Why does the shoelace formula work? --75.33.217.61 (talk) 19:59, 22 October 2010 (UTC)

Our article Shoelace formula explains (um, no it doesn't, but one easily sees that it is the case) how the formula can be expressed as a sum of cross products where each can be seen to be ±2 times the area of the triangle formed by the origin and two successive vertices.

The easiest case to convince onself of is when the polygon is convex and that the origin lies inside it. Draw lines from each vertex to the origin; this cuts it into a number of triangles that each corresponds to a cross product in the formula, and all of the cross products have the same sign.

For general polygons, the easiest approach may be to cut the polygon into triangles, and then show by induction on the number of triangles that the formula works and that the sum of the cross products is positive if the vertices are taken in clockwise (or is it anticlockwise? Never mind, one of them will turn out to work) order around the polygon. The base case for a single triangle is just a matter of calculation (the area of triangle ABC is ${\frac {|(B-A)\times (C-A)|}{2}}$, and the numerator can be rewritten using the distributive and anticommutative laws for the cross product, to give just the terms in the shoelace formula). In the induction case, add one triangle that shares one or two edges with the polygon you already have; the cross products corresponding to the shared edges will cancel each other out. –Henning Makholm (talk) 20:20, 22 October 2010 (UTC)

Reducing Transformation Arrays

Hi all, I'm looking for a way to reduce N 'transformation arrays' to a single transformation array without running through each individual transformation. Here's an example:

Given the alphabet, alpha = "abcd" and the transformation array, t1 = [4, 3, 2, 1]:

f(alpha, t1) = "dcba"

given t2 = [4, 2, 3, 1]

f(f(alpha, t1), t2) = "acbd"

So the first transformation we did was [4, 3, 2, 1] and the second was [4, 2, 3, 1] and the result was [1, 3, 2, 4]

I'm looking for a function, let's call it g(t1, ..., tN) to compute this. For instance:

g([4, 3, 2, 1], [4, 2, 3, 1]) = [1, 3, 2, 4]

Ideally I'd like for g(t1, ..., tN) to have a time complexity of O(n). If I actually do each of the transformations I can achieve O(n*m) where m is the length of the alphabet; thus, a time complexity of this, or worse, won't really help.

I'm not entirely sure if such a function can (does?) exist, but any ideas would be much appreciated!

Thanks!

24.147.249.232 (talk) 20:59, 22 October 2010 (UTC)

You cannot do better than O(nm); otherwise you don't have time to read all of the inputs!

(By the way, what you're referring to as a "transformation" is commonly known as a permutation) –Henning Makholm (talk) 21:06, 22 October 2010 (UTC)

Oh. Good point! I suppose a better question would be what's the optimal way to compute g(...)? I was hoping for a method where I could apply simple operations to each element, but that seems unlikely. The best method I've come up with, thus far, is (apologies for the poor pseudocode formatting!):

for(i in 0..m)

for(j in 0..n)

ndeck[j] = deck[permutations[i][j]]

deck = ndeck

Which is okay, but could be a lot faster if the last copy step didn't need to occur. Thus, perhaps my original question should have asked whether it's possible to do that innermost step in-place.

24.147.249.232 (talk) 22:16, 22 October 2010 (UTC)

Is this really the critical part of your entire program? Remember, premature optimization is the root of all evil. If it is critical, I would suggest unrolling the outer loop twice and shifting back and forth between two arrays:

for(i in 0,2,4,...,m-1)

for(j in 0..n)

decka[j] = deckb[permutations[i][j]]

for(j in 0..n)

deckb[j] = decka[permutations[i+1][j]]

Alternatively, you might also try to trace each element backwards through the permutations:

for(j in 0..n)

k = j

for(i in m..0)

k = permutations[i][k]

outj] = k

which seems to faster under an unit cost model, but is likely to have horrible cache locality. –Henning Makholm (talk) 23:00, 22 October 2010 (UTC)

Does the last copy step need to occur? It obviously depends on what goes on to be done with it and how modularised your code as a whole is, but if you've got space in memory for both deck and ndeck, can't you just carry on using ndeck once you're on the other side of the loop?

Thinking about it from another direction, would some recursion help? If you had a function to work out the end product of applying two re-orderings sequentially (even on a dummy "initial order" which you could then sub in your real data for at the end), you could just keep piling in pairs: t1 and t2, t3 and t4 ... up to tN-1, tN. Then you do it again to (the result from doing t1t2) and (the result from doing t3 and t4), etc. until you finally end up at the eventual answer. Obviously this is easiest to work for N a power of 2, but it's not difficult for some other N. Then again, that's probably only any practical use in speeding things up if you've got access to a parallel architecture where you can do all the initial pairs in parallel and then gradually bring them together. (So you use N/2 processes in your first go round, then N/4, then N/8 etc.) I'm rusty on all this, but it feels like it ought to be possible. --81.153.109.200 (talk) 10:36, 23 October 2010 (UTC)

GMAT scoring system

This page seems as good as any for my question, so here goes: On the math portion of the GMAT[1] I scored 50/95th percentile. Had I answered the last question correctly (it was not very hard; I had made my choice and was ready to press Next), was it probable that I might've raised my score? Thanks. 67.243.7.240 (talk) 21:31, 22 October 2010 (UTC)

I presume that by "50/95" you mean a score of 50 at the 95th percentile. You can see in the link you provided how an extra mark would increase the percentile score, assuming that you get one mark per correct answer. 92.24.178.5 (talk) 11:03, 23 October 2010 (UTC)

I don't know that one gets "one mark per correct answer"; thus my question hoping someone knows. 67.243.7.240 (talk) 23:48, 24 October 2010 (UTC)

Logic - deduction of axioms

Is it possible to deduce $\neg \neg p\Rightarrow p\,\forall \,p$ from $p\Rightarrow (q\Rightarrow p)\,\forall \,p,\,q$ and $(p\Rightarrow (q\Rightarrow r))\Rightarrow ((p\Rightarrow q)\Rightarrow (p\Rightarrow r))\,\forall \,p,\,q,\,r$? I'm studying on a logic theory course and I've been asked whether it is possible to deduce the third axiom we use from the first two? I strongly believe it isn't because otherwise we wouldn't need to take it as an axiom, but I'm not sure how to go about showing it. I know of most of the basic theorems by now (deduction, completeness etc), but I can't see any obvious way to get there. Could anyone please suggest anything? Thankyou, 62.3.246.201 (talk) 23:20, 22 October 2010 (UTC)

A typical way to prove that one axiom does not follow from the others, would be to exhibit a non-standard interpretation of the connectives that makes the two latter axioms (K and S, among friends) into tautologies but the first one into a non-tautology (while tautologies are still closed under whatever inference rules you have, such as modus ponens). This should be particularly easy because the two axioms you assume do not mention ¬ at all, so the only thing you need from an alternative truth table for ¬ is that it makes $\neg \neg p\rightarrow p$ into a non-tautology. –Henning Makholm (talk) 23:39, 22 October 2010 (UTC)

I'm sorry, I don't quite follow - what do you mean by 'exhibit a non-standard interpretation of the connectives'? Thankyou for the help :) 131.111.16.20 (talk) 11:53, 23 October 2010 (UTC)

Okay, let's take it more slowly then. A general method to show that some wff A cannot be proved from B, C, D is to define some set S of "special" well-formed formulas such that

1. B, C, D are all in S
2. A is not in S
3. For each rule of inference (such as modus ponens, which I suspect is actually the only one you have at the moment), if the rule allows you to infer a wff from wffs in S, then the wff you infer is also in S.

Then it's quite easy to show that no formal proof can ever contain a wff outside S, and in particular there's no formal proof that concludes A. (This ought to remind you of how soundness of the proof system was shown, where S was taken to be "all logically valid wffs").

I assume that by this point you have seen truth tables for ¬ and → and learned how to use them to determine whether a given formula is a tautology. My suggestion is then to invent your own crazy truth tables for the symbols, and then let S be all of those wffs that would have been "tautologies" under your crazy truth tables. –Henning Makholm (talk) 15:11, 23 October 2010 (UTC)

The axioms describe properties of the logical symbols $\neg$ and $\Rightarrow$. If you can find an interpretation that is true under two of the axioms, but not under a third, you have shown that the third does not follow from the other two. Try e.g. a three-valued logic. --Stephan Schulz (talk) 12:06, 23 October 2010 (UTC)

So just to be clear, there's no way to do this without going outside the realms of the 2-valued (just true/false) logic? I only ask because I'm being asked this question having been taught nothing but 2-valued logic on the course so far with no mention of other possibilities - basically we have covered basic propositional logic such as valuations, truth-tables (with only the values 0/1), the deduction, completeness, soundness, adequacy, Moor existence lemma, compactness and predictability theorems, the 3 axioms above and the deduction rule modus ponens. (I don't know how many of these things are course-specific and how many are widely known by these names so sorry if any of those sound unfamiliar!) I would usually happily read ahead on these things and I don't mean to sound glib or anything like that, it's just we've only been lectured on this material from scratch for the past 2 weeks (though I need to get my head around this for work due in just a week!), so I presume there must be some way to prove this using just 2-value logic - if that's what you meant Henning then please say, it's quite possible I misunderstood! Thankyou both for your patience, and if you can direct me to something I can read up on this via, rather than having to explain every step to a complete beginner (must be quite tiresome for you!) then I'd be very happy to do so, I'm just concerned I'm missing an easier way of doing this. Perhaps our lecturer expects us to deduce higher-valued logic for ourselves, it just seems a bit unlikely... Thanks again! 131.111.185.68 (talk) 21:01, 23 October 2010 (UTC)

The solution I have in mind does indeed stay within two values. It is a fairly standard technique, but by no means the only way to reach the goal, so there may or may not be hints in your text that point in another direction. (With the hints I've given so far, there are only 63 possible truth-table combinations to try out, most of which can be eliminated quickly, and all of which will yield to brute force. So you're not lost, though you might end up bored out of your skull unless you can think of a smarter way to home in on the right one).

The trouble with pointing to "something you can read up on" is that it is hard to know what will work for you better than the text you're already studying. Also it wouldn't do to simply point you to a complete solution; that way you would miss the learning experience of figuring it out yourself. (Except that it probably wouldn't work for exactly the same definitions your text is using. There are many slightly different ways to do formal logic, and they are all easily seen to be fairly unimportant variations of the same thing once you understand them in the first place. But until you get that basic understanding, it is best not to confuse oneself too much with alternative sources).

You syllabus so far sounds fairly standard, though the names "Moor existence lemma" and "predictability theorem" do not register with me. –Henning Makholm (talk) 22:01, 23 October 2010 (UTC)

Okay, I think I grasp the concept now - the 'not' and 'implies' symbols are literally just that - symbols. We can assign whatever meaning to them we want (we could say not(p) is always equal to 1, no matter what p, or not(p) has the same value as p, say): the reason I was finding it so hard was that I was retaining the meaning I'm used to for them, i.e. 0 implies 1 is true, 1 implies 0 is false, etc, rather than seeing we can define the values of p implies q, and not(p), to be whatever we want, and we just want to find a definition for which our 2 axioms and modus ponens are always tautologies (true no matter what values p,q,r take) and for which not(not(p)) implies p is -not- always true. So then, could we take implies under its normal meaning (always one except for 1 does not imply 0), given that we know this should make our 2 axioms and MP true (they do not involve the 'not' symbol so our definition of it is irrelevant with regards to these axioms), and take 'not' as not(p)=0 for p=0 or 1? So then not(not(1))=not(0)=0 which does not imply 1=p? 131.111.185.68 (talk) 01:41, 24 October 2010 (UTC)

Almost there, except 0→1 is 1. But 1→0 is 0. Nudge, wink. –Henning Makholm (talk) 02:28, 24 October 2010 (UTC)

Oh well, you can surely fix that one last mixup, so I'll pretend you that you've solved the problem, and feel free to point you to intuitionistic logic which explores what happens if one refuses to accept ¬¬p→p as an axiom. It turns out to be not quite as crazy as one would think at first; you may find it interesting. (Or it may make your head asplode, in which case don't sweat it. It usually takes several tries getting used to). –Henning Makholm (talk) 02:33, 24 October 2010 (UTC)

Ahh gotcha, so setting not(p)=1 and p=0, we get not(not(1))=1 which does not imply 0=p :) Thankyou so much for being so patient with me, I definitely owe you one! I will have a look at intuitionist logic the next time I'm feeling brave, all this is a bit of a brain-melter to say the least. Thanks again! Out of interest, where do the 'K' and 'S' ("among friends...") come from? Is there a list of standard axioms somewhere, each assigned a letter for conciseness or something like that? 131.111.185.68 (talk) 10:41, 24 October 2010 (UTC)

Hmm, after I've checked various sources, it may actually only be a small select circle of friends who call them that. I shouldn't have introduced them here. The reasoning I allude to is described in Combinatory logic#Logic, but has some fairly heavy prerequisites. If your brain is melting, you probably don't want to go there. –Henning Makholm (talk) 13:21, 24 October 2010 (UTC)

A simpler way to phrase the argument in this particular case would be: "Suppose we had a proof of ¬¬p→p. Then we would replace every subformula ¬φ anywhere in the proof with q, where q is any propositional variable that differs from p. Since none of the rules we use treat ¬φ specially, this substitution still leaves a valid proof. But that proof would conclude q→p which is manifestly not a tautology, so the assumed proof of ¬¬p→p cannot exist."

This may be what your teacher had envisioned. However, I couldn't find any hint for that version that wouldn't give it all away, and the method isn't as general. –Henning Makholm (talk) 13:44, 24 October 2010 (UTC)

October 23

the four-color theorem doesn't seem right to me

I can always come very close to producing a counter-example. What is the probability that the four-color theorem really is wrong, as it seems to me? If there is a simple, intuitive proof, maybe you could tell me it? 84.153.222.131 (talk) 15:40, 23 October 2010 (UTC)

Have your read our Four color theorem article? Basically, there is no known proof that is small enough to be checked by hand, but several independent computer-assisted proofs and at least one computer-verified formal proof have been constructed, and it is pretty implausible that they are all flawed.

I cannot imagine any meaningful sense in which one could assign a definite "probability" (other than 0 or 1) to the theorem being untrue. –Henning Makholm (talk) 16:00, 23 October 2010 (UTC)

See Bayesian inference for a paradigm in which assigning such probabilities makes sense. Robinh (talk) 16:10, 23 October 2010 (UTC)

I said "I cannot imagine any meaningful sense in which...". Pulling numbers out of one's ass and pretending that they have anything to do with the subject under discussion does not count. –Henning Makholm (talk) 16:53, 23 October 2010 (UTC)

Bayesian probability doesn't pretend to describe the subject under discussion, but rather our state of knowledge about the subject under discussion. This is of crucial importance in epistemology and decision theory. An example for someone making a decision based on his credence for the correctness of a proof is this. -- Meni Rosenfeld (talk) 07:29, 24 October 2010 (UTC)

I thought one pulled a prior out of thin air, not a posterior out of one's ass. Robinh (talk) 18:42, 24 October 2010 (UTC)

There is a relatively simple proof of the weaker Five color theorem. It does take several big steps to fully understand, though. If you want to give it a shot, here's a list of things you can look at:

1. Understand the translation of the map coloring problem into that of coloring graphs. This is explained in this section of the Four color theorem article. If you aren't already familiar with graph theory, you might want to play with those a little; a famous theorem that might help you get acquainted with graphs is the condition under which an Eulerian path exists (though that is not crucial to understanding the proof of the five color theorem).
2. Study the Euler characteristic for planar graphs. If a graph is planar, then we can count not only edges and vertices, but regions enclosed by the edges as well. It turns out that these three quantities satisfy a nice rule, V-E+F=2. The proof of this fact is outlined in that article.
3. Figure out why the Euler characteristic implies E ≤ 3V − 6 (assuming the planar graph has at least 3 edges). This can be algebraically derived from knowing that every edge touches at most two regions and every region touches at least 3 edges.
4. From this fact, prove that every planar graph must have some vertex of degree 5 or less.
5. If you are sufficiently well versed in proof by induction, a proof that every planar graph can be colored with 6 colors should follow naturally, knowing that last fact. Turning this into a proof that every planar graph can be colored with 5 colors requires a clever trick, which is explained in the article on the five color theorem. Why can't the same trick be used to prove the Four color theorem?

HTH. --COVIZAPIBETEFOKY (talk) 16:29, 23 October 2010 (UTC)

Actually, I have a vague memory of having heard an argument quantifying the sentiments of the IP supra. The argument, as far as I remember, went like this:

A graph is 4-colourable, iff 4 is a zero for its chromatic polynomial. Thus, the 4CT is equivalent to proving that the cromatic polynomial of any planar graph has a value strictly greater than zero at 4. However (the argument went), it is possible to prove (don't ask me how!) that there are zeros arbitrarily close to 4 for the set of all chromatic polynomials for planar graphs. One conclusion was very similar to the IP's: The 4CT is "almost false".

Thus, if you have found a reasonably "natural" family of "almost counterexamples", it might be interesting to check whether or not the set of zeroes of the chromatic polynomials of the graphs in this family has 4 as a limit point.

I'm sorry for not remembering more details. Possibly, some reference at or close to Chromatic polynomial#Chromatic roots might help. JoergenB (talk) 18:32, 29 October 2010 (UTC)

Help me impress my friend?

Hey I'm trying to pretend I know something about calculus to impress my friend. Technically that's why I want the answer to this question, it's not (my) homework.

So..

1/⅔ × sqrt(X) × constant = 3/2 X

What is the value of constant?? It's from a first year calculus course, I've already taken it, but I don't know where to start on this question. —Preceding unsigned comment added by 174.89.36.96 (talk) 16:07, 23 October 2010 (UTC)

This is not calculus. It is algebra. You need to isolate the X on the right. Multiply both sides by 2/3 and you'll have a very simple formula. However, the "constant" is not a constant. -- kainaw™ 16:28, 23 October 2010 (UTC)

(e/c) This isn't calculus, it's algebra. There is no differentiation or integration involved. Looking at your problem you have

${\frac {1}{2/3}}\times {\sqrt {x}}\times k={\frac {1}{2}}(3k{\sqrt {x}}).$

You claim that this is equal to 3x/2, i.e.

${\frac {1}{2}}(3k{\sqrt {x}})={\frac {3}{2}}x\iff k{\sqrt {x}}=x\iff k={\sqrt {x}}.$

But this doesn't really make sense. You see, x is usally though of as a variable. That means you would solve the equation for x and not k. (k was a constant after all, and was fixed.) A more likely solution would be

${\frac {1}{2}}(3k{\sqrt {x}})={\frac {3}{2}}x\iff x=0,k^{2}\ .$

If there's something you forget to mention then let us know. It will change the problem and hence change the answer. — Fly by Night (talk) 16:30, 23 October 2010 (UTC)

Ok thanks a lot guys, I asked if there are more details, but that seems to be the whole question. For now I guess this is resolved, I was hoping I was missing some advanced calculus technique that would let me find a genuine constant.

Hopefully she's impressed —Preceding unsigned comment added by 69.165.137.230 (talk) 17:08, 23 October 2010 (UTC)

If she knows calculus, she will quickly spot your naive error: your assumption that "any complicated math is calculus." Once you are outed thusly, she will see through your charade and will probably be unimpressed by you (and will likely consider your stunt a poor indicator of character). To avoid this eventuality, you should educate yourself first by reading our articles on algebra and calculus, (and your textbooks) and work out as many practice problems as you can. Girls (or guys) who would be impressed by pseudo-calculus are of dubious quality. You should stick to actually knowing things; this will impress everybody, including those girls who actually know calculus. Nimur (talk) 18:03, 23 October 2010 (UTC)

Quite the opposite! If she knows calculus, and realizes he's only pretending to, she is far more likely to find him a hot boy-toy, ie physically lust for him, than otherwise. In fact, if he were a Fields-metal level genius, it would only hurt his chances... 84.153.221.42 (talk) 18:44, 23 October 2010 (UTC)

I beg to differ! Nimur (talk) 18:46, 23 October 2010 (UTC)

Sir, may I ask you, kindly to leave the inquiring gentleman's delusions intact? If indeed his friend can tell calculus from pseudocalculus, it would be discourteous to deny her this opportunity to form an accurate assessment of his moral character. –Henning Makholm (talk) 18:55, 23 October 2010 (UTC)

The idea that one might be able to use pseudo-calculus as an effective dating tool implies such a fundamental misunderstanding of dating interactions and the psychology of women that I don't see any real possibility that anything we say will un-delude this guy. Just give him a Darwin Award for trying, because if he keeps it up it's extremely unlikely he will ever reproduce. --Ludwigs2 19:06, 23 October 2010 (UTC)

Unless he becomes a sexual predator. 67.243.7.240 (talk) 23:53, 24 October 2010 (UTC)

Don't worry she doesn't know too much calculus. —Preceding unsigned comment added by 99.224.233.138 (talk) 19:21, 23 October 2010 (UTC)

See [2] for a joke about assuming a waitress doesn't know calculus. I think the whole business is a bad idea. Dmcq (talk) 22:07, 23 October 2010 (UTC)

secure hash

how to securely hash any two numbers each over 100 digits long into a number between 1 - 10 with equal distribution/probability. The easier the hash is to explain/follow, the better. this is not homework. thank you! 84.153.221.42 (talk) 18:41, 23 October 2010 (UTC)

How about modulo ten? There is no way to avoid hash collision with your scenario, so why try? Nimur (talk) 18:44, 23 October 2010 (UTC)

What do you mean? How do you combine the numbers? Anyway both my numbers are odd (they're large primes), so if you're adding them, you would only ever get 2,4,6,8, or 0 and never any of the other numbers. Do you have a better suggestion, that would use more of the digits of both numbers? (and 'depend' on both of them) 84.153.221.42 (talk) 18:55, 23 October 2010 (UTC)

(ec) Just add all the digits modulo ten. If you want something "more secure", you'll need to specify which kind of security property you're thinking about (since quite clearly the usual characteristics of a cryptographic hash functions are unachievable with an output that small). –Henning Makholm (talk) 19:02, 23 October 2010 (UTC)

(Is what you're looking for perhaps not a hash but a message authentication code? –Henning Makholm (talk) 19:05, 23 October 2010 (UTC)

Yes, I am looking for a message authentication code, but that article says it is sometimes called a keyed (cryptographic) hash function. This is the sense in which I used the word "hash". 84.153.221.42 (talk) 20:08, 23 October 2010 (UTC)

From a security standpoint, the recommendation would be something tried and true, such as HMAC-SHA1, and then use mod 10 (or whatever) to cut `it down to the size you want. If you insist on simpler calculations (always at the risk of accidentally introducing cryptographic weaknesses), you'll need to be much more explicit about which threats you want to secure against. –Henning Makholm (talk) 20:28, 23 October 2010 (UTC)

Well, I don't really want to secure against anything so much as make sure that there is an equal distribution.... 84.153.221.42 (talk) 20:38, 23 October 2010 (UTC)

To be clear, you want a hash function mapping into {1, ..., 10} which has a roughly uniform distribution for input primes in the ~100's of digits? How about... just take the first two (i.e. most significant) nonzero decimal digits mod 10? I'd be surprised if the distribution was much different from uniform. Of course you should try it out with a thousand or so test cases to see. (I'm guessing you just want something decent, not perfect, and an attacker isn't in the picture. If these assumptions are incorrect, please clarify the question.) 67.158.43.41 (talk) 04:44, 24 October 2010 (UTC)

I'm not sure what you mean by "securely" here. It's not possible to achieve any of the usual security properties of a cryptographic hash function if the function can only take 10 distinct values; in particular, even if the hash was a random oracle, it would only take on average 10 queries to find a preimage for any output. —Ilmari Karonen (talk) 19:01, 23 October 2010 (UTC)

By the way, is modulo 10 equally distributed?

rand() % x in c is not equally distributed, as it slightly favors smaller numbers over larger ones. Would it be equally distributed 1-10 when applied to the sum of the digits of two primes? 84.153.221.42 (talk) 20:09, 23 October 2010 (UTC)

This isn't answering your question, but a more important reason that rand()%x might not be uniformly distributed in C is that the pseudorandom number generator used in many C libraries does not produce uniformly distributed least-significant bits. —Bkell (talk) 06:18, 24 October 2010 (UTC)

also followup

is your recommendation the same if we need a number 1-10, 1-100, 1-500. 1-1000, or 1-5000? (assuming we're looking at primes several hundred digits long.) 84.153.221.42 (talk) 20:11, 23 October 2010 (UTC)

terminology

Resolved

add two things. The resulting sum...
multiply two things. The resulting product...
take modulo 5. The resulting _____?

what is the missing word? 84.153.221.42 (talk) 20:36, 23 October 2010 (UTC)

Remainder or residue. –Henning Makholm (talk) 20:42, 23 October 2010 (UTC)

Thanks!

can you find an n-digit prime?

I know for binary digits, you can -- encryption uses, for example, 512-bit primes. Can I ask you to find an exactly 100-digit prime (or 500, etc), expressed in decimal? 84.153.221.42 (talk) 22:20, 23 October 2010 (UTC)

Yes, you can ask. Whether anyone will bother to actually do it for you is another matter. If you google for prime number generator, you will find websites and free software that purport to do things like this for you. –Henning Makholm (talk) 22:32, 23 October 2010 (UTC)

There always exist primes with any desired number of digits. The known bounds on the prime-counting function imply that for any $n>e^{\ln 2+10}\approx 44053$ there is at least one prime between n and 2n, and this property also holds for smaller n (by inspection of tables of primes). –Henning Makholm (talk) 23:06, 23 October 2010 (UTC)

Also see Bertrand's postulate. --Sam^Talk 05:15, 24 October 2010 (UTC)

Ah, that's slicker than the pedestrian argument I used. Thanks. –Henning Makholm (talk) 13:24, 24 October 2010 (UTC)

The following 500 decimal digits number is almost certainly a prime:

9871615190774109714631411860249057511981491262347298846971208567428248575453148043285875030217607786541394670874
1765030099536818318173353655505066711178433481944388749902895601864112760848063528268606416757060899302271352925
5864263819518034812848659084214983995663078865233769085509471793209927625145995901711340039317987632029515022435
2487509247090084455951957030991069493698109678946205440148128949032618192971706584082407007804900529851119894058
1723948298045506883722180286057686368471530972548567

-- Meni Rosenfeld (talk) 07:04, 24 October 2010 (UTC)

Encryption software just tests randomly chosen numbers of the appropriate size for primality, and that can just as well be done with decimal digits. But even not knowing that, it follows from the fact that you can find primes for any number of binary digits that you can find them for any number of decimal digits. For example, 2³²⁹ through 2³³² all have 100 decimal digits, so if you can find a prime of 329 to 331 binary digits then you've found a prime of 100 decimal digits. -- BenRG (talk) 07:13, 24 October 2010 (UTC)

October 24

Differentiate a function

How would I compute ${\frac {d}{dx}}(x!)$, assuming that non-integer values of n! exist and can be computed using the gamma function/Stirling's approximation. 24.92.78.167 (talk) 02:56, 24 October 2010 (UTC)

Technically, non-integer values of n! do not exist, by definition. If we were to use n! as a shorthand for $\Gamma (n+1)$, which is equal for integer values of n, then, according to the page on the gamma function,

${\frac {d}{dx}}(x!)={\frac {d}{dx}}\Gamma (x+1)=x!\cdot \left(-\gamma +\sum _{k=1}^{x}{\frac {1}{k}}\right),$

where $\gamma$ is the Euler–Mascheroni constant. According to Stirling's approximation,

$x!\sim {\sqrt {2\pi x}}\left({\frac {x}{e}}\right)^{x}.$

${\frac {d}{dx}}({\sqrt {2\pi x}}\left({\frac {x}{e}}\right)^{x})={\sqrt {\frac {\pi }{2}}}e^{-x}x^{x-1/2}(2x\log(x)+1)$

Since Stirling's approximation is different then the gamma function and they are not even equal at integer values, it is no surprise that this is a very different formula. 76.67.73.90 (talk) 03:31, 24 October 2010 (UTC)

The formulas aren't that different. ${\sqrt {\frac {\pi }{2}}}e^{-x}x^{x-1/2}(2x\log(x)+1)={\sqrt {2\pi x}}e^{-x}x^{x}(\log x+1/2x)\approx x!(\log x+1/2x)$. It is well known that $-\gamma +\sum _{k=1}^{x}{\frac {1}{k}}\approx \log x$ and slightly less well known that it is $\approx \log x+1/2x$. -- Meni Rosenfeld (talk) 06:55, 24 October 2010 (UTC)

"76.67.73.90", you wrote:

${\frac {d}{dx}}(x!)={\frac {d}{dx}}\Gamma (x+1)=x!\cdot \left(-\gamma +\sum _{k=1}^{m}{\frac {1}{k}}\right),$

Can you explain what you mean by m? What is m? Michael Hardy (talk) 04:43, 24 October 2010 (UTC)

Looking at the linked page on the Gamma function, they meant x instead of m. It should be noted x is an integer in that formula (...since otherwise x! isn't defined). 67.158.43.41 (talk) 04:50, 24 October 2010 (UTC)

Remarks. It turns out that what is easier to write is the logarithmic derivative of $\Gamma (x)$, that is $\Psi (x):={\frac {\Gamma '(x)}{\Gamma (x)}},$ rather than $\Gamma '(x)$ itself. Actually, the construction (or one construction, at least) of the $\Gamma$ function starts from the construction of the digamma function $\Psi (x),$ as a solution of a simple linear functional equation. For this reason, the simplest formula for the derivative of $\Gamma (x)$ is $\Gamma '(x)=\Psi (x)\Gamma (x)$. As to the series expansion for $\Psi (x),$ you can find it here; the formulae given above by anonymous users are wrong or nonsensical for non-integer x. --pma 13:43, 27 October 2010 (UTC)

Finite subgroups of the multiplication group of the unit quaternions

I want a list of all finite subgroups of the multiplication group of the unit quaternions. --84.61.153.119 (talk) 08:30, 24 October 2010 (UTC)

You really should do your own homework. At least show some effort. If you don't know what a quaternion is, you may find the page quaternion group helpful. Jkasd 08:50, 24 October 2010 (UTC)

Also the list of small groups may be useful. --JohnBlackburne^words_deeds 11:37, 24 October 2010 (UTC)

I want a list of all finite subgroups of the group O(4). --84.61.153.119 (talk) 14:37, 24 October 2010 (UTC)

Does the position of the median change if a plot is distorted like this?

Let's say I have a large number of values, creating a one dimensional plot, and I figure out what the median is and in what places on the graph it occurs. Then I want to change the values so that the highest and lowest values remain the same, while the current median value is changed, and all values in-between are "scaled" to allow this to happen. If I calculate the median again, will I always find it in the same places on the graph as before? I think it should be but I lack the knowledge to verify this. Thanks. (Sorry, I can't express this in any other way than with words, but I can try to rephrase it if it's hard to understand) Obiha (talk) 11:12, 24 October 2010 (UTC)

The median for a finite number of discrete values is simply the middle value: more precisely if there are an odd number of values, n say, it's the (n + 1)⁄2 value, if there are an even number it's midway between the n⁄2 and n⁄2 + 1 values. To change the median you need to change just these values, but keeping them at their positions which may mean moving other values too. E.g. if the values are

1  5  9 13 17 21 25 29 33

the median is 17, the 5th element. To change the median to 21 you could change the numbers like so

1  6 11 16 21 23 25 28 31

The first and last numbers are unchanged but everything else has changed.--JohnBlackburne^words_deeds 11:47, 24 October 2010 (UTC)

What the OP asked is: He has an unsorted sequence of values, say

1 4 3 9 4 5 4

He finds their median 4, and then notes that it appears is position 2, 5 and 7. Then he applies a transformation which changes the median - I don't know if he means specifically a linear scaling, but for now assume that the transformed values are:

1 6 4 9 6 7 6

What he wants to know is - if he takes these new values, find their median and note where it appears, will he also get 2, 5 and 7? -- Meni Rosenfeld (talk) 12:04, 24 October 2010 (UTC)

[ec] Yes, if either there are an odd number of values or the middle two values are equal (otherwise it could depend on your definition of median, and your method for finding it). Let $x_{1},\ldots ,x_{n}$ be the values, f be the transformation you apply to each, $y_{i}=f(x_{i})$ be the transformed values, and ${\tilde {x}},{\tilde {y}}$ be the medians of x and y. A median is characterized by the fact that at most half the values are less than it and at most half the values are greater than it. So if f merely satisfies $a>{\tilde {x}}\Rightarrow f(a)>f({\tilde {x}})$ and $a<{\tilde {x}}\Rightarrow f(a)<f({\tilde {x}})$, it will follow that $f({\tilde {x}})$ is the median of y. From these conditions it will also follow that $a={\tilde {x}}\iff f(a)=f({\tilde {x}})$ which means that $x_{i}={\tilde {x}}\iff y_{i}={\tilde {y}}$, so the median will appear in the same places for both sets. -- Meni Rosenfeld (talk) 11:59, 24 October 2010 (UTC)

Venn Diagram Word Problems

I recently undertook a logic and problem solving course at a local community college as part of an IT diploma. We just commenced Venn diagrams and I seem to be having a lot of difficulty with them. Can anyone recommend a resource or site which contains questions on which to practice? I have used this site so far http://www.math.tamu.edu/~kahlig/venn/venn.html and I am looking for more?24.89.210.71 (talk) 12:16, 24 October 2010 (UTC)

Problem

Sir i have on problem plz solove it.

If:

  7-3=124
  6+3=279
  5-2=63
  11+2=2613

then 15-3=? —Preceding unsigned comment added by Apsirkeel (talk • contribs) 12:58, 24 October 2010 (UTC)

I'll give you a hint:

7-3=12|4
6+3=27|9
5-2=6|3
11+2=26|13
15-3=??|??

-- Meni Rosenfeld (talk) 13:11, 24 October 2010 (UTC)

Choosing a pineapple at a supermarket

The pineapples, for example, at a supermarket are all the same price but vary in quality according to a Gaussian distribution. If I choose three pineapples at random from the pile, and then select the best of those three, what is the probability that the selected pineapple is within the top ten percent of quality of the whole pile?

How many pineapples do I need to compare to have at least a 50% chance of being in the top x percent of quality? Not a homework question, but a practical problem. Thanks 92.15.31.47 (talk) 16:57, 24 October 2010 (UTC)

ditto for wives, guys. how many past candidates should you first reject in order to have a sample base, and afterward pick the first one better than all of these? Assume that although normally distributed, the "value" is hidden to you, and you have no absolute way to compare candidates, but instead can only do a comparison function on candidates you have already rejected? You don't want to waste your time, so we're looking at the least number to have some high confidence (as with the number above) 84.153.247.182 (talk) 17:48, 24 October 2010 (UTC)

See Secretary problem. —Bkell (talk) 18:13, 24 October 2010 (UTC)

The answer to the first is straightforward, and the distribution doesn't matter. The chance of picking 1 that's in the top 10% is 0.1. The chance it's not is 0.9. If you pick three the chance all three are not in the top 10% is 0.9³ = 0.729. The probability at least one is in the top 10%, so that the best is in the top 10%, is therefore 0.271.

If you want to choose n so there's at least a 50% chance that the best of those is in the top 10% then you need n = 7, as 0.9⁷ = 0.478, so there's about a 52% change one of the seven will be in the top 10%.--JohnBlackburne^words_deeds 18:17, 24 October 2010 (UTC)

Nit-picking, but would the fact that you are not replacing the chosen pinapple make any difference? 92.28.246.6 (talk) 20:12, 24 October 2010 (UTC)

Yes, if the qualities are not independently, identically distributed (we may discard this difficulty here, I guess) Pallida  Mors 20:21, 24 October 2010 (UTC)

The questioner stated "top 10% of quality", not "top 10% of the population of pineapples." So, if quality goes from 0 to 99, he wants a quality of 90-99. If the distribution is skewed towards 99, more than 10% of the population will have the top 10% of quality. -- kainaw™ 23:00, 24 October 2010 (UTC)

"top x percent of quality". I normally read that as the corresponding percentile. However, your point can be answered, considering the nth order statistic distribution, computing $1-F^{n}\left(x*\right)$, the probability that the highest of the n pineapples extracted has a quality as good as the threshold x* [again, considering the iid assumption]. (Forgotten sign of Pallida  Mors 00:04, 25 October 2010 (UTC))

A relevant article is order statistic. Pallida  Mors 19:53, 24 October 2010 (UTC)

This is not the same as the secretary problem, or the wives problem suggested by 84.153.., since with the pineapples you can choose any of the pineapples you have previously inspected. Its more like a quality control (poor article) or Sampling (statistics) problem. 92.28.246.6 (talk) 20:03, 24 October 2010 (UTC)

Nitpicking: If the whole pile is N pineapples, and K of them are within the top ten percent of quality of the whole pile, and you buy n pineapples, then the probability that k of these are in the top ten percent of quality is

${\frac {{\binom {K}{k}}{\binom {N-K}{n-k}}}{\binom {N}{n}}}$

See hypergeometric distribution. Bo Jacoby (talk) 22:14, 24 October 2010 (UTC).

October 25

Roulette

Forgive my weakness in math. If I had $100 and I played roulette with the intent of walking away with as much money as possible, would I be better off placing all $100 on red/black and having it out in one spin, or incrementally wagering on red/black, say $5 at a time 20 times (so that the total wager is the same)? With a 1-to-1 payout, the possible results for my first option are $200 or $0. For the second, it'd be $105/$95, then $110/$100/$100/$90, and continue branching for 18 more levels. My gut says the most likely outcome would be something like $45 at the end, but I'm not sure how to do the calculations save for the brute force approach. The Masked Booby (talk) 02:12, 25 October 2010 (UTC)

You have to specify your goals more carefully. What do you mean by "as much money as possible"? Literally speaking, you could be saying that you want the maximum probability of breaking the house, even though that chance is going to be vanishingly small. Is that what you mean? That's probably not too hard to figure out, once we know how much money the house has, and which kind of roulette we're talking about (is there 0 and 00, or just 0)?. --Trovatore (talk) 02:55, 25 October 2010 (UTC)

Ah, never mind; I didn't read carefully enough. I didn't realize you were asking only about two discrete alternatives. --Trovatore (talk) 06:43, 25 October 2010 (UTC)

The House has some (small) advantage, since not every number is red or black. The expected value on a $1 bet is a $0.053 loss, in American roulette. Either strategy has an expected loss of 100*$0.053 = $5.30. The variance of the first strategy is much higher than the second strategy--you either win big or lose big, instead of winning small and losing small. The second strategy basically follows the multinomial distribution with p1=16/38, p2=16/38, p3=2/38. 67.158.43.41 (talk) 03:21, 25 October 2010 (UTC)

Make that the binomial distribution with p=16/38. The above is also correct, though if you bet "black" all the time, for winning total purposes, "red" and "neither black nor red" are the same outcome, so p2 and p3 can be combined into the implicit loss fraction of the binomial distribution. 67.158.43.41 (talk) 04:20, 25 October 2010 (UTC)

If there's no house advantage, then the distribution of the number of wins with 20 plays is distributed binomially with $n=20,\ p=0.5$, which can be seen here. With house advantage it's $p=8/19$ which can be seen here.

Of course, since the house does have an advantage, if you want to maximize your expected money, the correct answer is not to play. -- Meni Rosenfeld (talk) 07:33, 25 October 2010 (UTC)

To answer your question specifically, I think on average you'd be better off making one big bet, since making many smaller bets would give a higher expectation of losing to the house edge when the ball falls into the zero pocket(s).

The start of the articles Martingale (betting system) and Martingale (probability theory), describe a system where you bet on black or red only, and double your bet if you lose the previous bet. This system trades making a small profit with a small probability of losing all your money. The golden rule would seem to be: walk away when you are winning. If you enjoy gambling, then making a series of small bets will entertain you for longer.

I believe it is impossible to have any simple chance gambling system which makes any profit in the long run and on average, as the house always has an edge. This does not apply to poker, which involves skill, not just chance. You may also be interested in the Kelly criterion which is used more in racehorse betting etc. 92.24.186.217 (talk) 17:10, 26 October 2010 (UTC)

For every bet of $n on red/black, you expect to lose $0.053*n as above. If you spread these bets out over time into $n/3, $n/3, and $n/3, you still expect to lose 3*$0.053*(n/3) = $0.053*n. There is no difference in the expectation value of the different strategies. Put another way, if you implemented both strategies a million times, the average loss for both over all trials would be (virtually) the same. In a game with such well-defined odds, barring cheating or otherwise breaking the rules, we can confidently say there is no gambling system which makes a long-term profit, betting a fixed total, period. Every single one has the same expected loss, $0.053*n, where n is the total amount you decide beforehand to bet. Other games are more complicated to analyze, like Poker, depending on your rules. They might actually give a chance of winning long-term, but of course casinos try to avoid such a situation.

The betting system you mention is really quite interesting, but it just yields a massive variance in your profit (or loss). That is, you either win huge or you lose huge using that strategy. The expected value is the same as any other strategy--for every $1 you end up betting, you'll lose $0.053, on average.

Overall, the point is (a) if you just want to make money, the only sane thing to do is not to play; (b) if you enjoy gambling, but don't want to lose much money, make lots and lots of very small bets so your eventual loss takes forever; (c) if you enjoy gambling for the risk, find some help (my own opinion) or another hobby unless you're rich, 'cause you'll want to bet large and you'll just lose big. 67.158.43.41 (talk) 22:17, 27 October 2010 (UTC)

Mentally generating random numbers

Suppose I am playing a game like poker, and my strategy includes bluffing with a certain (fixed) probability each hand. Of course, humans are notoriously bad at generating "random" numbers themselves. The bluff (poker) article mentions using the colors of my hidden cards or the second hand on my watch as randomizing devices, but suppose I want to use some kind of pseudorandom number generator. Nearly all of the results about PRNGs are for computers that can easily remember and manipulate large numbers, which humans cannot do. Are there results for "human-implementable" PRNGs that use numbers with no more than, say, four decimal digits and computations that can easily be performed mentally? —Bkell (talk) 14:14, 25 October 2010 (UTC)

My initial thought was to do something like this:

• Choose prime numbers p, q.
• Choose m and n, small generators of the multiplicative groups modulo p and q, respectively (this will be easier if m and n are equal, more so if they are equal to 2)
• Choose thresholds $1\leq t<p,\ 1\leq u<q$ so that ${\frac {tu}{(p-1)(q-1)}}$ is equal to the probability of bluffing.
• Choose seeds $1\leq a_{0}<p,\ 1\leq b_{0}<q$.
• At each iteration, take $a_{i+1}\equiv ma_{i}{\pmod {p}},\,b_{i+1}\equiv nb_{i}{\pmod {q}}$. Bluff if $a_{i}\leq t,b_{i}\leq u$.

However, testing this gave a sequence with longer runs than random. Maybe some modification of this can give better results.

Another approach is to generate a sequence in advance and memorize it using some mnemonic. -- Meni Rosenfeld (talk) 15:19, 25 October 2010 (UTC)

I would just remember a long string of numbers. Or you could use some other more easily remembered pattern. E.g. if you've foolishly remembered π to 50 decimal places, or know the phone numbers of all your friends. You can extend the use of these by looking at them more than one way. For example you could alternate between bluffing if a digit is odd and bluffing when a digit is 5 or more. It even need not be numbers. A text, which might be easier to remember, could do the same job, though you need to be smarter at generating random numbers from it as the frequency and distribution of letters in most texts is far from random (or if it is random the text isn't easy to remember).--JohnBlackburne^words_deeds 15:29, 25 October 2010 (UTC)

complete the sequence

5+3+2 = 151012
9+2+4 = 183662
8+6+3 = 482466
5+4+5 = 202504
 Then
7+2+5 = ?  —Preceding unsigned comment added by 119.154.134.131 (talk) 17:42, 25 October 2010 (UTC) 

sorry, i have just solved it answer is 143542 —Preceding unsigned comment added by 119.154.134.131 (talk) 17:56, 25 October 2010 (UTC)

For anyone looking at this, the solution is, given a+b+c, a*b concatenated with a*c concatenated with b(a+c) reversed. So, 7*2=14, 7*5=35, and 2(7+5)=24, reversed is 42. Concatenated, it is 143542. -- kainaw™ 12:56, 27 October 2010 (UTC)

Ergodicity

I'm teaching myself a little ergodic theory. Looking at the article ergodicity, I have a couple of questions. First, in the second item in the formal definition, that A should be an E, yes?

Also, am I mistaken in thinking that the third item in the formal definition implies the other three?--130.195.2.100 (talk) 21:53, 25 October 2010 (UTC)

Nevermind on the second. I just looked closer and noticed that the article mentions they're equivalent definitions.--130.195.2.100 (talk) 21:55, 25 October 2010 (UTC)

You can read the source of the definitions you indicate at Google Books. Maybe a book would be better anyway. Thm. 1.5 says that, yup, you're right, A should be E. Apparently it was badly copied. I've updated the page. 67.158.43.41 (talk) 09:06, 26 October 2010 (UTC)

October 26

Monomial Ordering

Resolved

How and why does monomial ordering effect the quotient of a polynomial ring by an ideal? For example, take a polynomial ring over a field of characteristic zero, with variables x and y, and with the so-called degree reverse lexicographical monomial ordering. If we calculate

${\mathbf {K}}[x,y]/\!\left\langle x+x^{2},y^{2}\right\rangle \cong {\mathbf {K}}\!\left\langle 1,x,y,xy,y^{2},xy^{2}\right\rangle .$

However, if we use the so-called negative degree reverse lexicographical monomial ordering, then we have

${\mathbf {K}}[x,y]/\!\left\langle x+x^{2},y^{2}\right\rangle \cong {\mathbf {K}}\!\left\langle 1,x,y\right\rangle .$

I understand the first one, but I don't understand why terms are missing in the second one. I used SINGULAR to do the second calculation. More information on SINGULAR's monomial orderings can be found at this page. It seems that the second line involves the localized polynomial ring; whatever that might be. Any suggestions? — Fly by Night (talk) 13:04, 26 October 2010 (UTC)

How do you understand the first one? It seems to me that $y^{2}$ and $xy^{2}$ shouldn't be there, if it's an ordinary ideal with two generators, and an ordinary ring quotient. –Henning Makholm (talk) 13:21, 26 October 2010 (UTC)

(e/c) I don't understand either, as far as I can see the result should be $\mathbf {K} \langle 1,x,y,xy\rangle$. The quotient of a ring by an ideal depends only on the ring and the ideal, nothing else; monomial ordering only comes into the picture when looking for a basis.—Emil J. 13:26, 26 October 2010 (UTC)

There's a typo' in my post; that's all. It should have been y³ instead of y². I deleted this question, but Henning Makholm seems to have reinstated it. I removed the question because I found the answer. The two monomial orderings change the calculations. One gives you the whole polynomial ring as the numerator and the other gives you the localized polynomial ring. It's all in the link I posted. Anyway, thanks all the same. — Fly by Night (talk) 14:14, 26 October 2010 (UTC)

Maybe you already found a source for what's going on in which case you can ignore this but here goes anyway. The quotient ring K[x,y]/<x²+x,y²> has dimension 4 (the total multiplicity of the roots of these polynomials). In reverse lex order, the monomial basis is 1,x,y,xy. For the local version, you're no longer dealing with K[x,y], but its localization at the origin (in other words, its localization by the maximal ideal <x,y>). This ring consists of all rational functions which are defined at the origin. In this setting every polynomial with a non-zero constant term is a unit. The ideal <x²+x,y²> ⊂ K[x,y] extends to the local ring, but there it behaves a bit differently. In particular we can only see the properties of the ideal near the origin (this is where the term "localization" comes from). x²+x = x(1+x) but 1+x is a unit, so <x²+x,y²> = <x,y²>. That's why Loc_<x,y>K[x,y]/<x²+x,y²> only has dimension 2 (the multiplicity of the root at the origin). In negative reverse lex order the monomial basis of this quotient ring is 1,y. The monomial order chosen can affect the monomial basis of the quotient (and equivalently the Grobner basis of the ideal), but it won't change the dimension. However, global orderings (where higher degree terms are larger) only makes sense in the context of the global ring, while local orderings (where the smaller degree terms are larger) only makes sense in the context of the local ring, so SINGULAR deduces which one of those you want to be working in based on your choice of a global or local ordering. Rckrone (talk) 05:18, 27 October 2010 (UTC)

What a fantastic answer: superb! Thanks a lot Rckrone. — Fly by Night (talk) 08:49, 27 October 2010 (UTC)

Averaging points on a sphere

If I have a cluster of points on a sphere, is there some way of averaging them analytically? In other words, is there any method other than brute-force computation to find the point that minimises the distance between those points (in terms of angular displacement, obviously - the point should be on the surface of the sphere itself, so simply finding the euclidean distance wouldn't be much use)? If it helps, the particular case I'm looking at only involves a small section of a sphere, which should avoid the weird effects you'd get if you tried to average antipodal points. Thanks! Laïka 17:52, 26 October 2010 (UTC)

You can use vector cosine to easily calculate the angle from each axis to the point. Average the angles to get place a new point at an average angle to each axis. -- kainaw™ 17:55, 26 October 2010 (UTC)

Unless you care about which exact kind of average you find for widely separated points, an easy thing to do would be to average the points in 3D and normalize the resulting vector. If you try to average exact antipodes (so the 3D sum ends up being zero) it's not quite well-defined what one would want the operation to yield anyway. –Henning Makholm (talk) 18:35, 26 October 2010 (UTC)

Life might be easier if you convert to spherical coordinates first. Or, you can use Euler angles to give precedence to a particular rotation. Nimur (talk) 18:45, 26 October 2010 (UTC)

How does this problem become easier in spherical coordinates? –Henning Makholm (talk) 18:49, 26 October 2010 (UTC)

Normalization is free. Nimur (talk) 19:20, 26 October 2010 (UTC)

But vector addition is very expensive. –Henning Makholm (talk) 20:45, 26 October 2010 (UTC)

Ah, should have mentioned, I'm working in spherical co-ordinates anyway. Laïka 18:53, 26 October 2010 (UTC)

To compute the cartesian arithmetic mean in this case, it doesn't seem to suffice to compute the mean of the spherical angles--take a couple of points in the x-y plane, one at 5 degrees and the other at 355 degrees. Is there any way to avoid lots of conversions that are equivalent to just converting to cartesian anyway? 67.158.43.41 (talk) 22:17, 26 October 2010 (UTC)

There is quite a bit of work on how to do this correctly Directional statistics is the field where this is discussed and "Statistics of Directional Data" by Mardia and Jupp, is probably a good introduction.--Salix (talk): 19:08, 26 October 2010 (UTC)

prove that a + b will have the same value as b + a

How do you do that? What if b is negative. Then a + -b is the same as a - b. So if a = 5 and b = 3, it's easy: 5-3. But let's look at b + a. Now it's -3 + 5, then you have to start counting back from 3, (you count -2, -1, 0) and then go forward. (...1, 2, finishing the 5). So in this case b + a would be 3 + 2, which is -b + (|a|-|b|). And there are a bewildering number of other similar arrangements (both negative? only b negative, only a negative, etc). Then there's zero. After all is said and done, I'm not even sure if a+b = b+a at all. (Like if they are both zero -- is 0+0=0+0; when I try to verify that I just get divide-by-zeros, so maybe it's not a valid operation at all!!)

Please help. Is a+b = b+a for ALL cases??? How do you prove that??? Thanks a million!!! 84.153.187.197 (talk) 18:50, 26 October 2010 (UTC)

Addition is commutative, subtraction is not. I think that you are getting confused between the two. On a number line addition mean "First do this, then do that" . Positives move the counter to the right, negatives to the left. So A+B means count A then count B. If A is 5 and B = -3, A+B means start at zero and move 5 to the right then move 3 to the left. The answer is where you end up. For B+A it means move 3 to the left then 5 to the right, oh look you get the same answer. I don't know if it can be proved or if it just observed that addition is commutative. Theresa Knott | Hasten to trek 18:59, 26 October 2010 (UTC)

Of course it can be proved, but how to prove it depends heavily on what your basic definitions and axioms look like. (Such as, what is a negative number anyway?) With the definitons in Integer#Construction, commutativity of integer addition follows immediately from commutativity for addition of natural numbers. For natural numbers, then you can either say, well, it's obvious, or prove it from the Peano axioms using induction. –Henning Makholm (talk) 19:16, 26 October 2010 (UTC)

Proof by pictures?: For adding two positive integer, a + b, the picture is a dots on the left and b dots on the right. Count up all the dots to compute a + b. If I rotate the picture 180 degrees, now all the dots that were on the right are on the left, and vice-versa. So now the total number of dots represents b + a, but since I haven't erased any dots, or drawn any extras this count is the same as the count I got before.

One could extend this to the sum of a positive integer and a negative integer by drawing o's for the positive number and x's for the negative integer. Connect an o with an x using a line to "cancel" them against each other, and repeat until you've run out of o's or x's. The count of o's or x's that survive gives you the result of the sum. Rotate the picture 180 degrees and, again you get that a + b = b + a. Quantling (talk) 20:32, 26 October 2010 (UTC)

I "define" addition similarly. I imagine counting clearly distinct stones on a beach. For some total number of stones, we can clearly group the stones into two piles, call one pile, say, "8" and the other "3" depending on how many stones are in each pile. We may reconstruct the total pile by either creating an "8" pile and then setting a "3" pile next to it, or by reversing this order. There we have commutativity from a few physical properties--conservation of stones, and the ability to move them about in space from pile to pile using different timings, along with our ability to recognize piles and stones. These arguments can give the basic properties of addition of natural numbers, where integer properties follow as Henning Makholm mentioned. The more rigorous approach of Peano Arithmetic (or similar), I would say, relies on an implicit connection between the axioms and reality, where the user of those axioms is at all interested in them because they list a sort of minimal set of intuitively obvious properties that a physical definition of addition satisfies. For instance, I believe PA "mirrors" my intuitive definition of addition. If I didn't, I wouldn't be able to add, and I wouldn't be able to trust calculators, so I wouldn't be able to pay for both an item and shipping--which is a stupid place to be. There's not really much more you can do than accept some axioms as given. 67.158.43.41 (talk) 06:00, 27 October 2010 (UTC)

While mathematicians will address the commutativity directly, it is very reasonable and logical of you to consider the situation in a case by case basis. Just what are your "bewildering number of ... arrangements"? Well, if a < 0 then b < a or b = a or a < b < 0 or b = 0 or b > 0 for five arrangements. (And yes, I am picturing a number line as I write this.) If a = 0 then b < a or b = a or b < a for three arrangements. If a > 0 then b < 0 or b = 0 or 0 < b < a or b = a or b > a for five arrangemants. So, there are only thirteen arrangements, which is not all that many for you to check, case by case, on a number line to see that it really does work. Compare this to the proof of the four color theorem, where the initial proof required examining 1,936 separate cases. I think that you may be experiencing one of the wonders of mathematics. When an equality is proven, then it will hold true. There is a certain, never diminishing joy which occurs when the results are obtained through a previously unanticipated means, and that equality still holds. You knew that it would (as long as the logical basis for mathematics was sound) but it brings joy nonetheless. Enjoy! -- 124.157.254.112 (talk) 21:42, 26 October 2010 (UTC)

See Proofs involving the addition of natural numbers#Proof of commutativity.

—Wavelength (talk) 22:03, 26 October 2010 (UTC)

Note that OP asked for addition where at least one digit is negative, having no difficulty when only the natural numbers are considered. Taemyr (talk) 00:02, 28 October 2010 (UTC)

I disagree. It mentions 0+0 specifically, and 0 isn't negative. (0 is arguably not a natural.) For that case, 0+0 = 0+0 by hand-wavey rules of substitution I've never actually heard clearly articulated. In some sense, one 0 is indistinguishable from another in that we simply call them all "equal", whatever that means. By assumption the result of the two-argument addition function + is +(0, 0) = 0 (where +(0, 0) is using prefix instead of infix notation). We then have 0+0=0=0+0, so 0+0=0+0 follows from another assumption, the transitivity (and I suppose symmetry) of equality. One might wish to define a sort of structural equality, where two structurally identical formulas are always equal. That seems problematic for some types of definitions: like the function which returns the number of times it's been evaluated. Mathematical functions, at least, are static, though, so I'd be willing to accept such structural equality in some cases. This is a surprisingly interesting question! 67.158.43.41 (talk) 10:21, 28 October 2010 (UTC)

Linear transformation (map) problem, shortcut?

Given a basis of R^n B=[a,b,c . . . ], T(a), T(b), T(c) . . . , (where t is an unknown transformation from R^n to R^n+1) and u (a vector in R^n)

find: T(u), a basis for the kernal and a basis for the range.

I think I understand one way to solve this class of problem. T can be represented by a matrix, of known dimensions, I could do the multiplication T(a, b, c) with the entries of T as unknowns, and come up with a system of equations, and thus find a representation of T. Is this necessary? Is there a way to solve this class of problem without finding T explicitly? --hacky (talk) 21:33, 26 October 2010 (UTC)

A better way is to decompose u into a linear combination of your basis vectors. You can then apply linearity to compute T's effect on u. For instance, if you find

$u=\sum _{i=1}^{n}\alpha _{i}a_{i}\,\,\,{\text{where}}\,\,\,\alpha _{i}\in \mathbb {R}$

(where I've written your basis as B=[a1, a2, ..., an]), you have

$T(u)=T(\sum _{i=1}^{n}\alpha _{i}a_{i})=\sum _{i=1}^{n}T(\alpha _{i}a_{i})=\sum _{i=1}^{n}\alpha _{i}T(a_{i})$.

Decomposing u can be done in a number of ways. You might be given u as a linear combination of basis vectors directly. You might be given u in another basis, in which case you could apply a change of basis transformation. 67.158.43.41 (talk) 22:31, 26 October 2010 (UTC)

Also, ${T(a_{i})}$ should itself be almost a basis for the range of T. However, it might not be linearly independent, so not really a true basis. I believe Gram-Schmidt Orthogonalization (or a slightly modified form, which can deal with intermediate 0's) should cull out the extra vectors while producing a nice orthonormal basis for the range. The 0's should give information on the null space as well.... 67.158.43.41 (talk) 03:46, 27 October 2010 (UTC)

There's a better way to find the matrix for T than how you described. Suppose M is the matrix for T. Then Ma = T(a), Mb = T(b), etc. So if [a,b,c,...] is the matrix with the elements of B as the columns, and [T(a), T(b),T(c),...] is the matrix with their images as the columns, M[a,b,c,...] = [T(a),T(b),T(c),...]. You can solve this by right multiplying by [a,b,c,...] inverse, or better yet, making the augmented matrix [a,b,c,...|T(a),T(b),T(c),...] and performing Gaussian elimination.--130.195.2.100 (talk) 01:29, 27 October 2010 (UTC)

physical constants as reported in Wikipedia

Why are Wikipedia physical constants displayed with parens around the last two digits? —Preceding unsigned comment added by J 0JFCfrmAyw59oVFk (talk • contribs) 22:09, 26 October 2010 (UTC)

They're reporting the measurement uncertainty in current measurements. For instance, 1.34(12) means the value is known to lie within 1.34-0.12 and 1.34+0.12. 67.158.43.41 (talk) 22:23, 26 October 2010 (UTC)

I haven't seen this in articles, but if it is enWP practice, then there should be a link to an explanation. Perhaps some template {{foo|1.34|12}} should yield "1.34(12)" or something.—msh210℠ 14:41, 27 October 2010 (UTC)

The physical constant article uses that notation extensively. 67.158.43.41 (talk) 22:47, 27 October 2010 (UTC)

As a layman, I've always assumed that 1.34(12) meant that the best estimate or measurement was 1.3412, but the last of these digits are not known to be accurate. Different from 1.34±0.12. Have I been wrong? -- Tom N (tcncv) talk/contrib 04:55, 28 October 2010 (UTC)

I'm afraid so. See Uncertainty#Measurements. It's also wrong to say that 1.34(12) means the value is known to lie within 1.34-0.12 and 1.34+0.12. The value 0.12 is an estimate of the standard error. Qwfp (talk) 06:48, 28 October 2010 (UTC)

I should have mentioned I was glossing over the uncertainty of the bounds. I figured the OP wouldn't mind the difference, but it's a good thing to mention that nothing is certain in experimental physics. 67.158.43.41 (talk) 10:05, 28 October 2010 (UTC)

October 27

Primitive roots modulo 5^n

Hi all,

Could anyone suggest an elegant (number theoretic) proof of the fact that both 2 and 3 are primitive roots modulo 5^n for all n, i.e. 2^i or 3^i both generate the multiplicative group of Z/((5^n)Z)? I've looked at inductive proof but it seems a bit messy and I'm hoping there's a nice clean/concise solution - I have no doubt the reference desk brains are far more capable of coming up with an elegant proof than I am ;) (or if there is a particularly concise inductive proof, I have nothing against induction!) Could anyone suggest anything, please?

Thankyou for the help! Delaypoems101 (talk) 00:02, 27 October 2010 (UTC)

$(\mathbb {Z} /5^{n}\mathbb {Z} )^{\times }$ is cyclic of order 4⋅5ⁿ⁻¹. If a = 2,3 were not a primitive root, its order would have to be a proper divisor of 4⋅5ⁿ⁻¹, or in other words, a would be a square mod 5ⁿ, or n > 1 and a would be a fifth power mod 5ⁿ. In the former case, a would also be a square mod 5, which it is not (the only squares are ±1). In the latter case, a would be a fifth power mod 25, which it also is not. (You can check that by simply enumerating all the 20 possibilities, or you can reverse the argument above: if a were a fifth power mod 25, then a⁴ ≡ 1 (mod 25), which does not actually hold.)—Emil J. 15:14, 27 October 2010 (UTC)

The moral of all this is that when p is an odd prime and n ≥ 2, then r is a primitive root modulo pⁿ if and only if r is a primitive root modulo p².—Emil J. 15:39, 27 October 2010 (UTC)

Standard error and standard deviation

In lit review today, the professor was going on about how poor the author's data was, and how we can infer this point from his use of standard error rather than standard deviation -- he explained that because standard error is equal to standard deviation/square root of the population (I think that's what he said), if the standard deviation appears too large for him to publish without everyone laughing at his research data, he can trick readers into viewing a smaller number by reporting standard error instead, and only the most erudite will catch on because most people haven't the foggiest notion of what any of this means anyway. I tried to read both articles above, but they a) both confuse me (the former much more than the latter) and b) they didn't really discuss either in terms of the other, which is what I'm really trying to get at. I guess that's my question...if you could please explain it without sigmas and thetas, 'cause then I might as well read the article again. Thanx! DRosenbach ^{(Talk | Contribs)} 02:08, 27 October 2010 (UTC)

Your formula for the standard error of the mean is right (standard deviation / sqrt(n), n = population size), so generally standard error of the mean is much smaller than standard deviation. The specifics really matter, though... it can be perfectly reasonable to report standard error of the mean instead of standard deviation. It would of course be possible to report one when the other was appropriate, since they measure quite different things.

The standard deviation measures how dispersed your sample data is. Standard error of the mean measures how accurate the average you compute from your sample data is. For instance, if you measure the heights of your classmates, you can get a standard deviation from that sample which says how diverse the heights are. You can then compute the average height of your class, but you don't know how accurate that average is. The standard error of the mean is a "reasonable" estimate of the accuracy of your class's average height when compared to that of, say, the average height of anyone in the world.

For the sake of the argument, suppose half of everyone is 2 feet tall and the other half is 8 feet tall. The standard deviation of any sample of those people's heights would be relatively large--say around 3 feet (a guesstimate), which should be roughly the same for each sample size. However, if you were to compute the average height from any sample, you'd get pretty close to 5 feet. In fact, as you took more and more people in your sample, you'd expect to get sample averages very very close to 5 feet--the error of the mean would start to vanish. In a crude way, that's why you have to divide by an expression involving the number of data points in your sample to compute the error of the mean.

So again, standard deviation = dispersion of your sample; standard error of the mean = accuracy of the average you calculate from your sample.

Note: I've dropped out a lot of details and technicalities that you clearly don't want to hear about, like that the above formula is an estimate for an infinitely large population where your samples are independent. But, I avoided almost all equations, sigmas, and thetas! 67.158.43.41 (talk) 05:41, 27 October 2010 (UTC)

Maybe a bit simpler (if not precisely correct): the standard deviation measures the variation expected from one individual item in the group, while the standard error measures the variation expected in the average of a group of size n of the items. For instance, you know that if you pick one person at random you have a reasonable chance of getting someone with a high IQ or a low IQ, but if you pick twenty people you have a much lower likelihood of the average IQ of the group being high or low (to get a very high or very low average IQ you'd need to pick 20 people all with high IQs or all with low IQs). Standard deviation is a descriptive statistic (it describes a population) and is usually reported but isn't often used in meaningful ways. Standard error is used extensively in hypothesis testing, to test if samples come from different populations (this is why sample size matters in hypothesis testing - the larger n is, the smaller standard error is, and the easier it is to see differences between the sample mean and the expected mean of the population). --Ludwigs2 06:12, 27 October 2010 (UTC)

1 = -1 ?

in my textbook, a puzzle (not homework)

we have learnt ${\sqrt {\frac {a}{b}}}={\frac {\sqrt {a}}{\sqrt {b}}}$ and ${\sqrt {a}}\times {\sqrt {a}}=a$

also $-1={\frac {1}{-1}}={\frac {-1}{1}}$

thus ${\sqrt {\frac {1}{-1}}}={\sqrt {\frac {-1}{1}}}$

multiply both sides by ${\sqrt {1}}{\sqrt {-1}}$ we have ${\sqrt {\frac {1}{-1}}}{\sqrt {1}}{\sqrt {-1}}={\sqrt {\frac {-1}{1}}}{\sqrt {1}}{\sqrt {-1}}$ therefore ${\sqrt {1}}{\sqrt {1}}={\sqrt {-1}}{\sqrt {-1}}$

hence $1=-1$

can you spot the wrong reasoning ??? --Umar1996 (talk) 15:14, 27 October 2010 (UTC)

Try Mathematical_fallacy#Power_and_root. -- SGBailey (talk) 15:20, 27 October 2010 (UTC)

Specifically, the rule ${\sqrt {\frac {a}{b}}}={\frac {\sqrt {a}}{\sqrt {b}}}$ works when b is positive, but not in general. -- Meni Rosenfeld (talk) 15:24, 27 October 2010 (UTC)

yes! ${\sqrt {-b}}$ is impossible if b is positive. --Umar1996 (talk) 11:55, 28 October 2010 (UTC)

Not impossible, but different. —Anonymous Dissident^Talk 12:05, 28 October 2010 (UTC)

how high do cryptographic functions scale (how big primes can you pick?)

take an algorithm like RSA, which uses big primes. Now, it is always big news when we generate a large prime, so, obviously there are limits to how many digits the primes can be. What is that limit? Can it work with 5000 (decimal) digit numbers, or 10,000 digit ones? Million digit ones? Where is the limit, before it becomes too slow to use realistically? 84.153.230.5 (talk) 16:13, 27 October 2010 (UTC)

In general it's not a really interesting question. The important question is, does it scale better than algorithms for cracking the cryptograms. And good encryption algorithms should be polynomial while requiring an exponential time to crack. So the answer is that, long before the algorithm becomes unusuable it will be essentially impossible to break the encryption. As to spesific limits, it will depend on application. Taemyr (talk) 23:57, 27 October 2010 (UTC)

Primality testing can be done in polynomial time in the number of digits in the input using the AKS algorithm. The polynomial exponent is (off the top of my head, so take it with a grain of salt) 6. That is, a million digit prime would take 1,000,000^6 = 10^36 operations to prove primality for. Probable primes are easier to check. The key step in RSA is modular exponentiation by squaring, which is shockingly fast. I recently had Python compute the largest known Mersenne prime, something in the neighborhood of 40 million binary digits, so, say, 10 million decimal ones, and take it modulo around 40 million. I didn't even combine the steps, which should have been much quicker. It completed several seconds later. Last I heard, RSA usually uses ~thousand digit primes, which, as Taemyr said, should be good enough.

Note: I've actually glossed over many details. I understand AKS is merely asymptotically polynomial, that certain primes take different amounts of time to prove primality for, that Mersenne primes are historically easier to find, that my example isn't representative, etc.; I don't believe the extra complication is worth it, for me or the OP. 67.158.43.41 (talk) 04:32, 28 October 2010 (UTC)

Was that several seconds for a single modulo operation? Keep in mind that while exponentiation by squaring is efficient, it still takes time proportional to the log of the exponent. I think in RSA decryption you use an exponent roughly the size of the primes. So for 40 million-bit primes, you'll need to multiply by 40 million the time it takes for a single operation, and several seconds quickly become several years. -- Meni Rosenfeld (talk) 08:41, 28 October 2010 (UTC)

If I recall RSA, you exponentiate by the encoding string, which can be on the order of the prime (product of the primes?). So you certainly have a good point--if the exponent was gigantic it would indeed take much, much longer than my admittedly nonrepresentative anecdote. If the OP is interested, perhaps they could try implementing RSA and see how quickly it works if they feed in some gigantic primes. I would be genuinely interested in the results. 67.158.43.41 (talk) 10:01, 28 October 2010 (UTC)

finding angles in 3D figures

Could you please give me an expression that gives the angle subtended by points placed symmetrically on a spherical surface. For instance, I understand that the angle between tetrahedrally-disposed covalent bonds on a carbon atom is ~109.5deg. How is this derived?

As a more general example suppose we had to place radio transmitters to produce optimum coverage over the Earth's surface (assumed spherical). For certain numbers e.g. 6, one of the many valid placements would be self-evident (e.g one at each pole and four at 90deg. intervals on the equator). But suppose we had a number such as 7 or 17? And would there be a unique configuration (disregarding "symmetry" equivalents) or could there be alternatives?

(Digression: It occurs to me that a physical analogue could be devised - with difficulty - to yield a solution: e.g. by placing the required number of equal repulsive electric charges confined within a thin spherical shell or surface).

Thanks. —Preceding unsigned comment added by Paul Renshaw (talk • contribs) 20:11, 27 October 2010 (UTC)

You can generate a tetrahedra, so the distribution needed, by placing points at non-adjacent vertices of a cube. The angle is then the angle between any two of these, such as between the vectors to (1, 1, 1) and (1, -1, -1) from the origin. Similar solutions exist I think for other regular polyhedra such as the octahedron that you describe. On the more general problem there is information at circle packing which links to Tammes problem and Thomson problem.--JohnBlackburne^words_deeds 20:20, 27 October 2010 (UTC)

There are only a few platonic solids, that is, regular (i.e. "symmetric", by what I think you mean) shapes that fit on the surface of a sphere. These each have well-known angles. The number of vertexes in these shapes determines the allowable solutions to your question, 4, 8, 6, 20, and 12. Apparently, if you place any other number of repelling charges on a sphere, they'll end up in a configuration slightly different from perfect symmetry. A special case is <= 3 points, since they don't form a polyhedron in that case, and the platonic solids argument doesn't apply. Clearly 1 point doesn't provide symmetry; 2 points do if they're placed on opposite sides of the sphere; 3 points do not, since they determine a plane which, from symmetry, must divide the sphere in half--which isn't a spherically symmetric configuration. 67.158.43.41 (talk) 22:44, 27 October 2010 (UTC)

Clustering algorithm

Is an efficient algorithm known for the kind of problem of which this is an example:

${\begin{matrix}5&3&8&17&2\\12&6&4&2&6\\8&8&3&4&11\\12&13&15&12&20\\9&6&3&1&8\end{matrix}}$

The figures show the distribution of population in a 5 by 5 block of 1 km squares, the total being 198. The aim is to cluster the squares into six groups each of size 198/6 = 33, or as close to this as possible in terms of minimising total absolute deviation. The groups can be of any shape, but no component square may be attached at just one or more of its corners.

The general problem will have a rectangular array, probably with some zero elements.→86.155.186.37 (talk) 22:17, 27 October 2010 (UTC)

I'm sorry, I do not know an algorithm. The general case feels quite hard to me. Thinking a little more, your problem is harder than the partition problem, a known NP-Complete problem. (Proof: take a 1D array of inputs; create a 2D array with 0's off the diagonal, stringing the 1D array along the diagonal. Cluster the 2D array into two groups. The clustering must consider cases where the first cluster takes the lower triangle and the second takes the upper triangle. From here, you can get all possible subsets of the diagonal split between the two clusters by choosing which cluster to put each diagonal element in. This is precisely the set of subsets considered by the partition problem, and you're considering them with respect to the same fitness function. Any other clustering also divides the diagonal into two sets, so is redundant. This instance of your problem is thus precisely equivalent to the full partition problem.) So, short of giving up, I would just throw a greedy algorithm at it and hope for the best. The article I linked is perhaps more hopeful: "The pseudo-polynomial time dynamic programming solution for the subset sum problem applies to the partition problem as well, and gives an exact answer in polynomial time when the size of the given integers is bounded." Perhaps you could use a variant of that solution and execute it in reasonable time. Good luck! 67.158.43.41 (talk) 04:19, 28 October 2010 (UTC)

This may be generalized into an NP problem, so an absolutely correct solution would be exhaustive to run - it would be a brute force check of every possible combination. However, a heuristic may be used. One that comes to mind would be to initially divide this into two subsets of equal size. It is similar to the partition problem mentioned above, but it is a bit easier because the limitations on which items may be included once another number is included. Once you have two equal size sets, divide each one into two equal size sets. If the heuristic fails to produce four nice sets, try rotating the entire set to see if your algorithm grabs a better group (since it will likely be biased to look in one direction for new numbers, rotating the set will make it technically look in a different direction). -- kainaw™ 12:39, 28 October 2010 (UTC)

Obscure 3D shape that folds

In a kids mathematics magazine I used to get back in the early 2000's, there was a free gift of a card net of an obscure geometric shape to fold 'n' glue yourself

the shape had a really long name which i think ended in "hedron" (narrows it down! :-\) which i cannot remember, the magazine may even have been forced to make up the name themselves but it sounded authentic (inside the magazine the name was explained - which brings to mind lots of uses of "hex" and "dec", but my knowledge of shapes since may have distorted this memory.

To add to the challenge I cannot remember the name of the magazine itself, but it had a sister publication of "Young Scientist" (try googling for that lol - its not the first one), which had a version for older children called "SciTec" or "SciTech", and there was an equivalent older kids version of the maths magazine, whose name also escapes me. the publisher of all four of these was Educational publishing, under the name of which i can find at least three different publishing houses, and the relevant one seems to have gone under, since it's website domain is up for sale the editor of the young scientist magazine (at least) was Jenny Smith

every issue of each of the mags came with some kind of science or maths toy, such as 3d glasses. I want to find the name and net of the shape so i can make another one - it folded much like a Paper fortune teller, but on both sides, and looked to be made of cubes with triangular "hinges"

I have looked for the shape on wikipedia - i wouldn't be surprised if there was at least a stub for it, but I am at a loss for how to search with no clue of the name I have googled several combinations of the magazine names to find a publishers website - but as I said it has been taken down replaced with a for sale http://www.educationalpublishing.com/ (.co.uk has a similar notics under a different agency I'm not sure which is correct)

The end goal is finding the shape, not the magazine, but if anyone gets as far as the magazine I'd be interested anyway ☭ мдснєтє тдлк^ЅТЦФФ 00:11, 28 October 2010 (UTC)

Hexaflexagon? -- 119.31.121.84 (talk) 00:45, 28 October 2010 (UTC)

Study our article polyhedron. Bo Jacoby (talk) 09:27, 28 October 2010 (UTC).

This site has something that sounds similar, which it calls a "kaleidocycle" or "flexahedron". the wub "?!" 12:08, 28 October 2010 (UTC)

Another Logic query - uncountability

Hi all, me again! I asked a question on axiomatic deductions a few days ago, but I was curious about something I couldn't figure out on my own. Won't need as thorough an answer this time, I don't actually -need- to do a question on this, I'm just curious :)

In the lecture course I attend we always assume (for propositional logic) that our primitive propositions are countable, and therefore so are all possible propositions. However, I'm asked here what happens when we allow uncountability, and I haven't really got a clue!

We call a set of propositions S independent if for all s in S, we can not deduce s from S-{s} (i.e. s is required to deduce s from S). We call 2 proposition sets S, T equivalent if for every s in S, we can deduce s from T, and for every t in T, we can deduce t from S. I've shown that for every proposition set (countable), there exists an independent set equivalent to it, but now what happens if we allow our initial propositions to be uncountable? Is there still always an independent equivalent subset?

By the nature of the question, I expect there to be a counterexample rather than a proof. Is there any way we can map this into the reals or [0,1] perhaps, then use a more well-known set of results to derive the result? I know you can consider a topological space on the valuations as a Stone space (though I don't know much about stone spaces), hence the name of the compactness theorem, as we reach a conclusion on a compact T.S. However, is there anything we can do in this case, any map into R we can use to help us for example?

Thankyou (time and time again!), 62.3.246.201 (talk) 00:39, 28 October 2010 (UTC)

I'm not terribly well versed in logic, or set theory for that matter, but I believe the uncountable case follows from Zorn's lemma. Take P as the set of all subsets of S, an uncountable set of propositions. Take E as the set of all elements of P equivalent to S, which is clearly nonempty, since it contains s. Partially order E by reverse inclusion. Zorn's lemma provides us with at least one maximal element M, which, from our reverse ordering, is actually minimal under standard inclusion. Any subset M' in E of M would be comparable to M under our partial order, so, since M is minimal, M' must be M. That is, no proper subset of M is equivalent to S, therefore, to M. Take M-{m} for any m, which is now not equivalent to M, so m cannot be deduced from M-{m}. Thus M is independent and equivalent to S.

I wouldn't be at all surprised if this question is (in a more technical sense than I've used) equivalent to Zorn's Lemma, which is itself equivalent to the Axiom of Choice and a number of other things, under the proper set theory. 67.158.43.41 (talk) 04:05, 28 October 2010 (UTC)

Zorn's lemma fails here: the relevant poset is not chain-complete. This can be seen even in the countable case: take S={p₁,p₁∧p₂,p₁∧p₂∧p₃,...}, and let E_n be S with the first n elements removed. Then each E_n is equivalent to S, and they form a chain in E, but the only set contained in all of them is the empty set, which is not in E. Algebraist 09:56, 28 October 2010 (UTC)

Integral calculus

Given ${\frac {d\sec x}{dx}}=\sec x\tan x$ and $\sec x={\frac {\sec ^{2}x+\sec x\tan x}{\sec x+\tan x}}$, find $\int \sec xdx$. What is the link between the first two identities and the question ? ?--115.178.29.142 (talk) 01:05, 28 October 2010 (UTC)

Hint: How is the numerator of the second identity related to its denominator? More hint; you should try yourself before reading past this word: It's the derivative. 67.158.43.41 (talk) 03:15, 28 October 2010 (UTC)

Unbelievable that I didn't notice that myself. Thanks. 115.178.29.142 (talk) 03:34, 28 October 2010 (UTC)

Tough probability question

If four cards are chosen at random from a standard deck (with all jokers removed, and face cards assigned value as follows: Jack = 11, Queen = 12, King = 13), what is the probability that using the elementary functions of addition, subtraction, multiplication, and division, they cannot be made to equal 24. In other words, for four cards with values a, b, c, and d (not necessarily distinct), what is the probability that using no elementary functions nor arrangements that a ₪ b ♥ c ♯ d (where ₪, ♥, and ♯ are elementary functions, also not necessarily distinct) will be 24. You cannot use the same card more than once, for example you can't say a ♦ a ₪ b ♥ c ♯ d, and you must use all the cards (so a ₪ b ♥ c does not work either). Thanks. 24.92.78.167 (talk) 02:26, 28 October 2010 (UTC)

How are parentheses handled? Aces are 1, I hope? A standard deck has 4 copies of each card, so there are 13^4 = 28561 possible distinct ordered four-card hands. There are four functions you listed. If parentheses are ignored and evaluation is assumed to be (((a ₪ b) ♥ c) ♯ d) there are only 4^3 = 64 choices for each (₪, ♥, ♯), leading to just 64*28561 = 1827904 possible combinations. If parentheses can be supplied arbitrarily, there are a few more possibilities:

((a ₪ b) ♥ c) ♯ d

(a ₪ b) ♥ (c ♯ d)

(a ₪ (b ♥ c)) ♯ d

a ₪ ((b ♥ c) ♯ d)

a ₪ (b ♥ (c ♯ d))

leading to 28561 * 64 * 5 = 9139520 possibilities to compute. These numbers are quite small for computers, so you could write a script to simply compute the answer. I'd be delighted to hear an analytic solution, but I suspect it's a horrifically complicated case-based analysis (or nearly equivalent to what I just wrote). There are a few optimizations. First, multiplication and addition are associative, so many of the extra parenthesizations are actually equivalent. Similarly multiplication and addition are commutative, so many hand rearrangements are equivalent. You could actually cut the run time by an ~eighth by running those (₪, ♥, ♯) made of only multiplication and addition on the set of unordered distinct four-card hands, which is of size (13 multichoose 4) = (13+4-1 choose 4) = 1820. Some combinations can be eliminated by inspection as well; for instance, 4 aces can't possibly get large enough with the operations you've given, and neither can anything with all 2's or under. You also can stop evaluation of a given (a, b, c, d) when you've found a combination that totals 24. I doubt these optimizations are worthwhile in your case, though, considering a script should be able to brute force the problem in a few seconds. 67.158.43.41 (talk) 03:43, 28 October 2010 (UTC)

Pairs of antihomologous points lie on a circle

Could someone please review this section of the Homothetic center article? I believe it to be wrong and there is a discussion on the article's talk page, but would like an expert opinion. If confirmed incorrect, is anything from that section salvageable? Thanks. -- Tom N (tcncv) talk/contrib 04:30, 28 October 2010 (UTC)

Fox calculus on [F', F']

I'm looking at words w in the free group, F, of rank n, with basis x_i, whose images in the abelianisation of F := F^o are trivial. I've modified the domains of the Fox calculus derivatives to be ZF^o, rather than ZF, and I'm trying to show that the only such words w which have zero (modified) Fox derivative with respect to all the basis elements of F are the members of [F ',F '], i.e. commutators of commutators. Unfortunately, I'm not having much luck.

I've tried working in the simple case where we can write $w=ux_{i}u'x_{i}^{-1}u''$, where w only contains x_i (and hence x_i^-1) once. Then the derivative of w with respect to x_i is $u-uu'$ which equals zero if and only if the image of $u'$ is trivial in the abelianisation, which forces the image of $uu''$ to also be trivial. I'm tempted to try some sort of induction argument, but even then, I'm not sure how to get a foothold in the general case, since the Fox derivative is so heavily dependent upon where the x_i 's appear in the word. Any help would be great. (I've considered that the statement I'm trying to prove isn't actually true, but I'm using it to try to flesh out the proof from a textbook, and I believe it boils down to the above situation). Icthyos (talk) 11:40, 28 October 2010 (UTC)

Differential equation

Hi, can anyone see whether there is any closed-form solution to the following?

d^2y/dx^2 = k*y*sqrt(1 + (dy/dx)^2)

k is an arbitrary constant.