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April 26

A number that is itself

Hello,

I'm dealing with what I believe is an intractable problem. Here it is stated as succinctly as possible: A positive integer in its letter form can equal itself (given that a is 1, b is 2, and so on) when the sum of the letters are taken. What is the second one? Naturally, I want to be exhaustive in the answers I can give: does one write the number in traditional English as "TWENTY FIVE", "ONE HUNDRED AND THIRTY SIX", does one leave out "and", or does one resort to calling out each number within the greater string of numbers, e.g., "111" would be "ONE ONE ONE"? Besides that linguistic dimension of the problem, there is also the immense difficulty of finding the damn numbers! I'm trying to find these answers quickly and efficiently, but seeing how open-ended the problem is (but don't think there isn't an answer, because there definitely is), I'm wondering if anyone can provide a suitable approach to tackle this problem, even if I have to learn to write a program to do it. Thanks. I know this isn't your everyday sort of problem, and it probably will require a lot of thought. Another possibility that I don't know about is whether such a number would have a certain, known property, like an "honest number" or perfect number. If anyone knows of such numbers without having to go through the bricks and mortar of finding the answer, then please let me know that as well.Coitoergosum (talk) 02:02, 26 April 2010 (UTC)

The number of letters in the spelled-out form will increase linearly with the number of digits, while the actual number will increase exponentially. So if the two graphs ever cross, they will do it for fairly small n. You have to decide for yourself what spelling scheme to use, and then either use cleverness to solve for an answer, or use a program to exhaustively search for one. I get "two hundred and fifty one" and "two hundred and fifty nine" as solutions. Without the "ands", I get that there are no solutions. Maybe there are more solutions under other schemes, or in languages other than English. 69.228.170.24 (talk) 04:50, 26 April 2010 (UTC)

Thanks for that. I'm aware of the divergence that arises when one compares the additive sequence and the number sequence; however, neither of the answers you have given are the answer to the problem. I'm guessing that this problem is a little trickier than I had anticipated. There may be a solution involving the ordinal numbers (first, second, etc.) that I didn't think of before, so I'll have to try that. Again, thanks.
This isn't a resolved problem by any means. Another aspect to the problem is whether this is a known property of numbers, cf. "honest number"; it probably wouldn't be one found in formal mathematics, but it certainly can be construed as a linguistic property of numbers. If anyone knows about it, I'd like to know what this property is called.Coitoergosum (talk) 16:14, 26 April 2010 (UTC)

Well, this isn't a property of the number itself - it depends on which base you represent the number in. In binary you would be calculating the values of "one zero zero one" intead of "nine" etc. And it is language dependent. And it depends on your convention for valuing alphabetic strings (Why ignore spaces ? Why add letter values rather than multiplying or using a place value calculation ? Why not use ASCII character values instead of 1-26 ? etc.) There are so many arbitrary conventions here that it would't surprise me if this property has never been named or studied. Gandalf61 (talk) 16:31, 26 April 2010 (UTC)

While in principle I agree with your evaluation of the situation in general, the problem very explicitly and unambiguously states that one has to use letters when giving/finding the answer (a "space" isn't a "letter", either). The problem is also stated in English, so the answer must be one found in English. Different number bases came to mind, but I don't think the problem is a test of one's knowledge on that level (if it is, then it is extremely devious and nasty).Coitoergosum (talk) 20:41, 26 April 2010 (UTC)

Just assume that no number above 1000000 will have this property and use a brute force algorithm. You must have a specific encoding scheme decided before starting, but the question is ambiguous if you don't so that is not really something we can help with. Once it's clear how you are writing your numbers just start with 1 and go up, a reasonable program should reach 1 million in less than a minute. It's possible that there are tricks you can use to speed the computation up, but the problem is small enough that it's not really worth trying to find such tricks. The hard part might be proving the upper bound, ie. that no number above 1000000 has this property. Taemyr (talk) 17:03, 26 April 2010 (UTC)

Thanks. I've already thought this myself, but I have practically zero programming skills, so this will be an uphill battle.
You guys have been great; thanks for your help. I won't give up.Coitoergosum (talk) 20:41, 26 April 2010 (UTC)

It's actually a fairly decent task for introductory programming. Parsing a number into text is slightly tricky if you haven't done any programming, but will teach you the basic of string manipulation. Taemyr (talk) 20:29, 27 April 2010 (UTC)

apparent misprint:S

Hi all, My problem is little but very frustrating. I am reading from a book that has very little explanations and is not very rigorous too, and am stuck at a little point. I belive there is a misprint but then I can't make out what to substitute for the correct values: Consider the equation (in Z) $x_{1}+\cdots +x_{k-1}-x_{k}=-b$ (k,b are even). The book says that $x_{i}=a_{t};(1\leq i\leq {\frac {k}{2}}),x_{i}=a_{t+1}+1;({\frac {k}{2}}+1\leq i\leq k-1),x_{k}={\frac {k}{2}}a_{t}+({\frac {k}{2}}-1)(a_{t+1}+1)+b-1$ is a solution, which I fail to see. I believe the -1 at the end of the x_k term is an error. But reading onward it is evident that the -1 is very important in the subsequent discussion and in fact the value of x_k must be exactly what is proposed. So I feel the error must be in defining the other x_i terms. Subject to the additional contraints that x_i's don't belong to [a_t+1,a_t+1], a_t's are increasing and $a_{t+1}\neq a_{t}+1$ how should I modify the solution. i.e. the x_i's. Also is some modification actually needed or I am just plain missing something. Thanks-Shahab (talk) 05:35, 26 April 2010 (UTC)

I also get that the two sides are off by 1. However, without the context to know why these values were chosen it's impossible to say what the "correct" fix is. You could change any term you wanted by 1 to make the sum work, or even change lots of terms. 67.100.146.151 (talk) 05:58, 26 April 2010 (UTC)

Thanks, at least now I am assured that this is a misprint. The proper context will take too long to describe. Hence I gave the required constraints.

Let me rephrase my question now. Is there a fix where x_i (i is not k) is chosen as either a_t or a_t+1+1 and the only change that I have to do is to decide from where to where x_i takes either of these two values?-Shahab (talk) 06:17, 26 April 2010 (UTC)

Factorisation

Hello. I'm looking for information on the best way to factorise $ab(a^{2}-b^{2})+bc(b^{2}-c^{2})+ca(c^{2}-a^{2})$. I notice that swapping a and b negates the result, which suggests that the expression is divisible by (a-b), but I'm not really sure how to proceed from there. The people on IRC advised that considering a, b, and c as points in three-dimensional space and employing Lagrange multipliers could work, but I don't know much about that. Thanks for the help. —Anonymous Dissident^Talk 12:18, 26 April 2010 (UTC)

Well by symmetry (b-c) and (c-a) must then also be factors which since the original had fouth powers means you only have a linear expression in a, b and c left which must also be symmetric in them. Now I wonder what kind of expression is symmetric and linear sum of a, b and c? Dmcq (talk) 12:38, 26 April 2010 (UTC)

a + b + c. So inspection works best. Thanks. —Anonymous Dissident^Talk 13:12, 26 April 2010 (UTC)

Yes I think it is kind of funny that for two variables you've got two options a+b and a-b whereas for three you've only got one. In other circumstances one can have a+bω+cω² where ω is the cube root of 1 but then you'd need other terms like this to cancel out the ω's. Dmcq (talk) 13:55, 26 April 2010 (UTC)

Period (number)

The article says that "irrational algebraic numbers and functions are themselves expressible as integrals of rational functions over rational domains". Could someone handwave how algebraic numbers can be expressed that way? —Preceding unsigned comment added by 212.87.13.69 (talk) 16:54, 26 April 2010 (UTC)

The reference cited in the article looks very good. It gives the example: ${\sqrt {2}}=\int _{2x^{2}\leq 1}dx$. 69.228.170.24 (talk) 07:49, 27 April 2010 (UTC)

Generation of batting sequences

The game of cricket has a team, usually 11 in number, who in an innings bat in pairs, initially person 1 and person 2 together. Whoever is dismissed in a partnership is replaced by the next person in numerical order, this continuing until the 10th partnership is broken, when the innings is complete. If there were only 3 players there would be 2 possible partnership sequences, (1,2), (1,3) or (1,2), (2,3). A 4th player would double the possible number of sequences to these: (1,2), (1,3), (1,4) or (1,2), (1,3), (3,4) or (1,2), (2,3), (2,4) or (1,2), (2,3), (3,4). It's apparent that for n players there will be 2^(n-2) different sequences of length n-1. My problem - to get an algorithm to generate them. I feel that this should be fairly standard, but can't get one to work.→86.166.205.252 (talk) 20:32, 26 April 2010 (UTC)

I wrote it and tested it in Fortran:

       program CRICKET

       character*16  BIN
       integer*2     I,J,A,B,N

       print *,"Enter N (2-16):"
       read (*,*) N

       do I = 1, 2**(N-2)
         A = 1
         B = 2
         print *,"(",A,",",B,")"
         J = I-1
         call BINARY(J,  BIN) ! Returns 16 character binary string, 5 = "0000000000000101"
         do J = 19-N,16
           if (BIN(J:J) .eq. "0") B = B+1
           if (BIN(J:J) .eq. "1") then
             A = B
             B = A+1
           endif
           print *,"(",A,",",B,")"
         enddo
         print *," "
       enddo

       end

I didn't include the BINARY subroutine, which converts an integer to a 16 character binary string. Hopefully it's all self-explanatory. If not, let me know. StuRat (talk) 04:19, 27 April 2010 (UTC)

I think that problem may intended as an exercise in recursion. That's probably simpler than the explicit binary expansion. Is there some reason you want to use fortran? It's mostly for numerics, not so much for these combinatorial things. 69.228.170.24 (talk) 10:57, 27 April 2010 (UTC)

The OP didn't ask for Fortran, but I'm a Fortran programmer, so that's what I know best. StuRat (talk) 13:48, 27 April 2010 (UTC)

(OP) Thanks, StuRat, I checked that it worked. The problem was an exercise only in the sense that it occurred to me while watching cricket, bur afterwards I was unable to generate a complete list. If anyone feels inclined to derive a recursive method I'd be interested to see it.→86.155.208.113 (talk) 13:08, 27 April 2010 (UTC)

That's not too hard. For 2 players there is only one batting sequence: (1,2). For n > 2 players, you take each batting sequence for n-1 players, and use it to generate two new n player batting sequences as follows. If the final pair in the n-1 player sequence is (a,b), then one new sequence is created by appending (a,n) to that sequence and the other is created by appending (b,n). So from (1,2) you get the two 3 player sequences (1,2)(1,3) and (1,2)(2,3), then the four 4 players sequences (1,2)(1,3)(1,4), (1,2)(1,3)(3,4), (1,2)(2,3)(2,4), (1,2)(2,3)(3,4) and so on. Gandalf61 (talk) 13:20, 27 April 2010 (UTC)

OK, here's a recursive version, also in Fortran:

       program CRICKET2

       integer*2     N,A,B,I
       character*178 STRING  

       print *,"Enter N (>2):"
       read (*,*) N
       I = 2

       A = 1
       B = 2
       write (STRING,*) "(",A,",",B,")"

       if (N .ge. 2) call LOOP (N,  A,B,I,STRING  )

       end

 ******************************************
       subroutine LOOP (N,  A,B,I,STRING  )
  
       integer*2     N,A,B,I
       character*178 STRING

       integer*2     OLD_B,OLD_I
       character*178 OLD_STRING

       if (I .lt. N) then
         OLD_B      = B
         OLD_I      = I
         OLD_STRING = STRING

         I          = OLD_I+1
         B          = OLD_B+1       
         write (STRING(LEN_TRIM(STRING)+2:),*) "(",A,",",B,")"
         call LOOP (N,  A,B,I,STRING  )

         I          = OLD_I+1
         A          = OLD_B
         B          = A+1
         STRING     = OLD_STRING
         write (STRING(LEN_TRIM(STRING)+2:),*) "(",A,",",B,")"
         call LOOP (N,  A,B,I,STRING  )

       else
         print *,STRING(:LEN_TRIM(STRING))
       endif

       return
       end

It also prints out each sequence on one line instead of multiples, but I could have done that on the non-recursive version, too. StuRat (talk) 15:36, 27 April 2010 (UTC)

Here are the runs for 2-6:

Enter N (>2):
2
 ( 1, 2)

Enter N (>2):
3
 ( 1, 2)  ( 1, 3)
 ( 1, 2)  ( 2, 3)

Enter N (>2):
4
 ( 1, 2)  ( 1, 3)  ( 1, 4)
 ( 1, 2)  ( 1, 3)  ( 3, 4)
 ( 1, 2)  ( 2, 3)  ( 2, 4)
 ( 1, 2)  ( 2, 3)  ( 3, 4)

Enter N (>2):
5
 ( 1, 2)  ( 1, 3)  ( 1, 4)  ( 1, 5)
 ( 1, 2)  ( 1, 3)  ( 1, 4)  ( 4, 5)
 ( 1, 2)  ( 1, 3)  ( 3, 4)  ( 3, 5)
 ( 1, 2)  ( 1, 3)  ( 3, 4)  ( 4, 5)
 ( 1, 2)  ( 2, 3)  ( 2, 4)  ( 2, 5)
 ( 1, 2)  ( 2, 3)  ( 2, 4)  ( 4, 5)
 ( 1, 2)  ( 2, 3)  ( 3, 4)  ( 3, 5)
 ( 1, 2)  ( 2, 3)  ( 3, 4)  ( 4, 5)

Enter N (>2): 
6
 ( 1, 2)  ( 1, 3)  ( 1, 4)  ( 1, 5)  ( 1, 6)
 ( 1, 2)  ( 1, 3)  ( 1, 4)  ( 1, 5)  ( 5, 6)
 ( 1, 2)  ( 1, 3)  ( 1, 4)  ( 4, 5)  ( 4, 6)
 ( 1, 2)  ( 1, 3)  ( 1, 4)  ( 4, 5)  ( 5, 6)
 ( 1, 2)  ( 1, 3)  ( 3, 4)  ( 3, 5)  ( 3, 6)
 ( 1, 2)  ( 1, 3)  ( 3, 4)  ( 3, 5)  ( 5, 6)
 ( 1, 2)  ( 1, 3)  ( 3, 4)  ( 4, 5)  ( 4, 6)
 ( 1, 2)  ( 1, 3)  ( 3, 4)  ( 4, 5)  ( 5, 6)
 ( 1, 2)  ( 2, 3)  ( 2, 4)  ( 2, 5)  ( 2, 6)
 ( 1, 2)  ( 2, 3)  ( 2, 4)  ( 2, 5)  ( 5, 6)
 ( 1, 2)  ( 2, 3)  ( 2, 4)  ( 4, 5)  ( 4, 6)
 ( 1, 2)  ( 2, 3)  ( 2, 4)  ( 4, 5)  ( 5, 6)
 ( 1, 2)  ( 2, 3)  ( 3, 4)  ( 3, 5)  ( 3, 6)
 ( 1, 2)  ( 2, 3)  ( 3, 4)  ( 3, 5)  ( 5, 6)
 ( 1, 2)  ( 2, 3)  ( 3, 4)  ( 4, 5)  ( 4, 6)

StuRat (talk) 20:44, 27 April 2010 (UTC)

Here's a C++ metaprogram that generates the sequences at compile-time :)

template<int P1, int P2, typename Prev=void>
struct Pair : Prev {  enum { first = P1, second = P2 }; };

template<> struct Pair<1,2> { enum { first = 1, second = 2 }; };

template<int Max, typename P = Pair<1,2>, int Next = (P::first > P::second ? P::first : P::second)>
struct Seq : Seq<Max, Pair<P::first, Next+1 , P> >, Seq<Max, Pair<Next+1, P::second, P> > {};

template<int Max,typename P> struct Seq<Max, P, Max> { typedef P result; };

int main() 
{ 
  Seq<5>::result(); 
}
/* Output: sourceFile.cpp(13) : error C2385: ambiguous access of 'sequence'
        could be the 'sequence' in base 'Seq<5,Pair<1,5,Pair<1,4,Pair<1,3,Pair<1,2,void> > > >,5>'
        or could be the 'sequence' in base 'Seq<5,Pair<5,4,Pair<1,4,Pair<1,3,Pair<1,2,void> > > >,5>'
        or could be the 'sequence' in base 'Seq<5,Pair<4,5,Pair<4,3,Pair<1,3,Pair<1,2,void> > > >,5>'
        or could be the 'sequence' in base 'Seq<5,Pair<5,3,Pair<4,3,Pair<1,3,Pair<1,2,void> > > >,5>'
        or could be the 'sequence' in base 'Seq<5,Pair<3,5,Pair<3,4,Pair<3,2,Pair<1,2,void> > > >,5>'
        or could be the 'sequence' in base 'Seq<5,Pair<5,4,Pair<3,4,Pair<3,2,Pair<1,2,void> > > >,5>'
        or could be the 'sequence' in base 'Seq<5,Pair<4,5,Pair<4,2,Pair<3,2,Pair<1,2,void> > > >,5>'
        or could be the 'sequence' in base 'Seq<5,Pair<5,2,Pair<4,2,Pair<3,2,Pair<1,2,void> > > >,5>'
*/

End example. decltype (talk) 16:31, 30 April 2010 (UTC)

Ouch, decltype, that was brilliant but cruel ;-). I like Haskell for problems like this:

cricket [a,b] = [[(a,b)]]
cricket (a:b:c:ds) = [(a,b):es | es<-cricket (b:c:ds) ++ cricket (a:c:ds)]

main = mapM_ print $ cricket [1..4]

{- output:
[(1,2),(2,3),(3,4)]
[(1,2),(2,3),(2,4)]
[(1,2),(1,3),(3,4)]
[(1,2),(1,3),(1,4)]
-}

This is recursive: the "cricket [a,b]" equation handles the 2-element case, and "cricket (a:b:c:ds)" handles the recursive case (3 elements a,b,c followed by a possibly empty remainder of the list). The bracketed expression in the recursive case is a list comprehensions. The mapM_ in the "main" function runs "print" on each of the sublists, so they are shown one per line. [1..4] is shorthand for [1,2,3,4]. 69.228.170.24 (talk) 09:07, 2 May 2010 (UTC)

Work Word problem

hi the problem i have says "50 people build 10 bridges in 500 hours. If 15 people build 40 bridges, how many hours will they take to make them?" how do i solve this?? —Preceding unsigned comment added by Needlotsofhelp (talk • contribs) 21:38, 26 April 2010 (UTC)

Hint: If 50 people can build 10 bridges in 500 hours, then 200 people can build 40 bridges in 500 hours (assuming they do not work more efficiently, the more there are). Now it's a matter of simple division. —Preceding unsigned comment added by 76.230.149.101 (talk) 22:02, 26 April 2010 (UTC)

well my teacher told me about a way to find the answer using a fomrula but i dont remember it was it like multiply people by hours and then divide by stuff they built??? i'm not sure.... —Preceding unsigned comment added by Needlotsofhelp (talk • contribs) 22:20, 26 April 2010 (UTC)

Forget about the formula, you can do it yourself. To make it concrete, let's say you pay each man 1 dollar per hour. Building 10 bridges costs 500 hours of 50 men. How much is it? And 1 bridge? And 40 bridges? How much will you pay each man if they're 15 of them? --84.221.198.37 (talk) 22:57, 26 April 2010 (UTC)

Your teacher's method will have used the following formula to calculate the man-hours required to build each bridge:

${\frac {{\text{people}}\times {\text{hours}}}{\text{bridges}}}={\text{man hours per bridge}}$

Then we assume that the man hours per bridge is the same in both parts of the problem. So we have two ways of calculating the man hours per bridge figure, which must both give the same value. In other words, if it takes 15 people x hours to build 40 bridges then

${\frac {50\times 500}{10}}={\frac {15\times x}{40}}$

Now you can rearrange this equation to find the value of x. Gandalf61 (talk) 12:11, 27 April 2010 (UTC)

Uncountably Large Subfields of the Real Numbers?

Well? --128.62.32.123 (talk) 23:00, 26 April 2010 (UTC)

Well what? Dmcq (talk) 23:07, 26 April 2010 (UTC)

Do they exist? --128.62.32.123 (talk) 23:10, 26 April 2010 (UTC)

Yes, of course. R is one, for a start. If you want a proper one, then you could pick a transcendence basis of R over Q and take the field generated by some uncountable proper subset of it. Algebraist 23:21, 26 April 2010 (UTC)

Thank you. --128.62.32.123 (talk) 23:23, 26 April 2010 (UTC)

Wait, doesn't that mean you have to take a transcendence basis of infinite degree? Is it obvious that you can do that without getting all of R? 69.228.170.24 (talk) 06:14, 27 April 2010 (UTC)

It is, because of the algebraic independence of the bases. Different subsets of it generate different fields. --131.114.72.230 (talk) 09:08, 27 April 2010 (UTC)

Complicated Integration based on a Cylinder Cross-section

Is it possible to solve...

((ay^2/(b+c))-cy)*(1-(y^2/b^2))^0.5 dy from y=b to -b

Many thanks for any help. Here is a link to it on wolfram alpha if that helps to visualise the problem [1].

From Andrew McArthur --137.222.114.238 (talk) 23:00, 26 April 2010 (UTC)

Hint: Substitute in cos² θ for 1-(y^2/b^2). Do some arithmetic to figure out what θ is and don't forget to reset the bounds! 76.230.149.101 (talk) 02:25, 27 April 2010 (UTC)

If I'm reading the question right, the problem is to evaluate

$\int _{-b}^{b}\left({\frac {ay^{2}}{b+c}}-cy\right)\left(1-{\frac {y^{2}}{b^{2}}}\right)^{1/2}\,dy.$

I'd separate it thus:

$\int _{-b}^{b}{\frac {ay^{2}}{b+c}}\left(1-{\frac {y^{2}}{b^{2}}}\right)^{1/2}\,dy-\int _{-b}^{b}cy\left(1-{\frac {y^{2}}{b^{2}}}\right)^{1/2}\,dy.$

The second integral is that of an odd function over an interval symmetric about the origin; therefore the second integral is zero. The first integral is that of an even function over an interval symmetric about the origin; therefore its value is twice that of the integral over the half-interval:

$2\int _{0}^{b}{\frac {ay^{2}}{b+c}}\left(1-{\frac {y^{2}}{b^{2}}}\right)^{1/2}\,dy.$

Now let

${\begin{aligned}\sin \theta &={\frac {y}{b}}\\b\cos \theta \,d\theta &=dy\\\end{aligned}}$

etc. (And of course θ will go from 0 to π/2.) Michael Hardy (talk) 17:28, 27 April 2010 (UTC)

Redefining The Comparison of Cardinal Numbers

The standard definition for two sets X, Y of | X | ≤ | Y | is that there is an injection from X to Y. An alternative definition may be that there is a surjection from Y to X; under the assumption of the Axiom of Choice, this latter definition is equivalent. How does this definition behave if the axiom of choice isn't assumed? --128.62.32.123 (talk) 23:26, 26 April 2010 (UTC)

Note that even assuming AC the equivalence only holds for X non-empty (for e.g., there always exists an injection form the empty set to any set Y, but there exists a surjection from Y to the empty set only if Y itself is empty). Assuming X is not empty, an injection j:X→Y easily produces a surjection s:Y→X (no AC is needed). Indeed, say you have a bijection j':X→j(X), and say c∈X. The inverse of j' extends to a surjection s:Y→ X putting s(y)=c for y ∈ Y\j(X). --131.114.72.230 (talk) 08:29, 27 April 2010 (UTC)

Right, so I suppose that means the definition would have to be reworked to account for the special case of the empty set.

I already knew that the injective definition implies the surjective definition (I hadn't thought about the corner case of the empty set, but again, that's not really terribly important for my question; it's the infinite cardinals that are interesting); of course the interesting question is the reverse, and it seems doubtful that the reverse implication will always hold in non-AC settings, since the most direct way to get an injection from X to Y given a surjection from Y to X would be a right-inverse, and the ability to take a right-inverse for every surjective function is equivalent to AC. There may be another more clever way that I'm not thinking of, but like I say, doubtfully so.

The questions I'm more interested in are things such as trichotomy, how various potential definitions of infinitude using this notion of comparison behave (eg we may define infinitude as greater than or equal to every natural number, or greater than or equal to the set of all natural numbers, etc.), and so on. --128.62.32.123 (talk) 13:02, 27 April 2010 (UTC)

ZF does not prove (unless it is inconsistent) that a surjection from Y to X implies the existence of an injection from X to Y, and it does not prove that any two sets are comparable (using either definition of comparison). As for definitions of infinity, you can find a lot of information in our article on Dedekind-infinite sets.—Emil J. 15:47, 27 April 2010 (UTC)

Another important thing about your relation that ZF does not prove is that it's a partial order (antisymmetry may fail). Algebraist 17:20, 27 April 2010 (UTC)

Well, obviously, ZF proves that it is not antisymmetric, as two different sets can have the same cardinality. But this has nothing to do with failure of AC, this is inherent in the notion.—Emil J. 17:33, 27 April 2010 (UTC)

I meant as a relation on cardinals, of course. Algebraist 17:45, 27 April 2010 (UTC)

I see. However, if the OP is contemplating alternative definitions of comparison of cardinalities, there is nothing stopping them from redefining equality of cardinalities so that it fits. So let us be precise: you are saying that ZF does not prove the surjective variant of the Cantor–Bernstein theorem: if there is a surjection from X to Y and a surjection from Y to X, then there is a bijection from X to Y (whereas it does prove the original, injective version).—Emil J. 18:00, 27 April 2010 (UTC)

Ask user:Trovatore about this one. Michael Hardy (talk) 17:35, 27 April 2010 (UTC)

I notice that mathematicians seem to have difficulty with the notion of an open-ended question. So far, only one point has been brought up that I didn't specifically mention in trying to give some examples of what I'm looking for. --128.62.55.11 (talk) 00:24, 28 April 2010 (UTC)

Well, "128.62.55.11", I guess you like congratulating yourself based on the fact that you've figured out that we colored people like fried chicken. Here's a bit of advice: you might seem less like an idiot if you would abstain from shooting your mouth of before putting your brain in gear, as in your posting of 00:24, 28 April 2010. Michael Hardy (talk) 01:42, 28 April 2010 (UTC)

EmilJ gave a complete answer: without AC, ZF cannot prove that every pair of sets is comparable, under either definition. — Carl (CBM · talk) 00:28, 28 April 2010 (UTC)

That doesn't seem like a complete answer at all; it says very little. Michael Hardy (talk) 03:17, 28 April 2010 (UTC)

Thanks, people who made an attempt, I guess. I was hoping for more. I'd be very surprised if no one has ever studied this alternative definition in far more depth than the responses here seem to indicate. I'll assume the lack of further answers and insights indicate that no one else has anything substantial to add. I'll see if I can hunt down some real mathematicians who know more about this, since no one here seems to. --128.62.45.2 (talk) 18:00, 28 April 2010 (UTC)

Have you asked user:Trovatore this question? He's a mathematician with some expertise in this area. Michael Hardy (talk) 18:57, 28 April 2010 (UTC)

I'm sorry that we weren't up to your standards, and that some of us have areas of expertise other than set theory (which is what real mathematicians do). You can have your money back. -- Meni Rosenfeld (talk) 20:06, 28 April 2010 (UTC)

April 27

Squaring the wavefunction #2

Why is it that:

$|t|^{2}={\frac {1}{1+{\frac {V_{0}^{2}\sinh ^{2}(k_{1}a)}{4E(V_{0}-E)}}}}$

when:

$t={\frac {4k_{0}k_{1}e^{-ia(k_{0}-k_{1})}}{(k_{0}+k_{1})^{2}-e^{2iak_{1}}(k_{0}-k_{1})^{2}}}$

and k1 is imaginary? How do you calculate the absolute value of t anyhow? --99.237.234.104 (talk) 00:50, 25 April 2010 (UTC)

EDIT: I didn't realize the two expressions use different variables. The two equations came from the rectangular potential barrier article, and k0 and k1 are defined in terms of E and V0 there.

I posted this question before, but didn't fully explain my question. Somebody at the science reference desk suggested I post it again, so here goes. --99.237.234.104 (talk) 04:04, 27 April 2010 (UTC)

I think the following formula may help

$|t|^{2}=t{\bar {t}}$

where ${\bar {t}}$ is complex conjugate of t,

${\bar {t}}={\frac {4k_{0}(-k_{1})e^{+ia(k_{0}+k_{1})}}{(k_{0}-k_{1})^{2}-e^{2iak_{1}}(k_{0}+k_{1})^{2}}}$

(Igny (talk) 04:15, 27 April 2010 (UTC))

Why is that true? Remember that k1 is imaginary, not real. --99.237.234.104 (talk) 20:19, 27 April 2010 (UTC)

integer solutions

Consider two natural numbers a and b, with a<b. Within the closed interval [b,b²-b] can there exist a solution within the natural numbers to the equation ax+by=bz? If not, why not? Thanks.-Shahab (talk) 05:08, 27 April 2010 (UTC)

Assume a=1. Then LHS ≥ 1×b+b×b = b(b+1), RHS ≤ b×(b²−b) = b²×(b−1)

For b=3, LHS ≥ 12 and RHS ≤ 18, so the possible solution is:

a=1, b=3, [b,b²−b]=[3,6], x=3, y=3, z=4,

which gives

ax+by = 1×3+3×3 = 12 = 3×4 = bz.

CiaPan (talk) 07:28, 27 April 2010 (UTC)

I understand you want x, y, z all in the interval [b,b²-b], and that a<b are natural numbers. If b=1 this interval is empty: so, no solution if (a,b)=(0,1). If b=2, the interval reduces to the singleton {2}, so the only possibility would be x=y=z=2, which is ok if a=0, otherwise, no solution if (a,b)=(1,2). If b≥3, you always have the solution x=b, y=b, z=a+b, for b≤a+b<2b≤b²-b. So the answer is: there is always a solution for all pairs (a,b) with the exception of (0,1) and (1,2). --131.114.72.230 (talk) 08:54, 27 April 2010 (UTC)

Degree by which data is partitioned

Resolved

Is there a term for the degree to which data is partitioned by a particular attribute? So, say I have a dataset of objects of different shape, size, colour, texture, etc. and I'm looking for the attribute which will best separate the data into evenly-sized subsets E.g. if 50% of them are red, this would be an excellent way to sub-divide my objects. So, "Is the object red?" would be the question of maximum (something) and I'm looking for the term to use in place of (something). Sorry for the dire description but, as you can tell, I'm no mathematician! --Frumpo (talk) 13:48, 27 April 2010 (UTC)

Just a guess, but maybe what you're looking for is information content. Basically you're dividing up the objects into different bins; the information content is a measure of a probability of a particular way of dividing up the objects out all possible ways of dividing them up. The labeling of the bins is arbitrary so that should be taken into account by multiplying the probability by the number of possible labelings. Just how you do this depends on the nature of the problem you have but the upshot should be that putting everything a single bin, or putting everything into a bit by itself results in a probability of 1 giving an information content of 0. Anything in between will have a probability of less than 1 so a positive information content. I don't want to get into a lot of details without knowing more about the specific problem and this is just a guess as I said.--RDBury (talk) 14:52, 27 April 2010 (UTC)

In computer programming it's frequently important to break data up into approximately equal "bins". This comes up when sorting or searching a list, for example. For this reason, many database management systems gather stats on data which can then be used to optimize those functions.

For example, let's say we want to search an alphabetically sorted list of names for "Clark". Without any further info, we might start right in the middle. If we know that names all start with A-Z, we might say that we should start the search 3/26ths (2.5/26ths, technically) of the way down the list, since C is the 3rd letter out of 26. If, however, we have stored the starting position of each letter, and we know we need to search the C's only, this can shorten the search considerably. If we know the starting position of CL's and CM's, then we can shorten the search even further.

This might suggest that the more partitioning we have, the better, but there is a limit. At some point, the partitioning info will take up more space than the data itself, and here we start hitting a level of diminishing returns. StuRat (talk) 16:25, 27 April 2010 (UTC)

The direct concept you're asking for, I think, is information gain. Finding ways to partition data to maximize the gain is a standard problem in machine learning. The articles cluster analysis and statistical classification might be helpful. 69.228.170.24 (talk) 19:17, 27 April 2010 (UTC)

Thank you. I think Information Gain is, indeed, the term that I need. My query was somewhat related to machine learning.--Frumpo (talk) 09:17, 28 April 2010 (UTC)

Non-associativity of substitution in calculus

I am looking for a reference for a particular strange phenomenon in the syntactics of calculus.

Define $F(a)=\displaystyle {\frac {xa^{2}}{2}}$, which might arise from an integral such as $\displaystyle \int _{0}^{a}xt\,dt.$

Now, in the usual syntactic notation of calculus,

$F(a){\Big |}_{a=x+1}=\displaystyle {\frac {x(x+1)^{2}}{2}}$

On the other hand,

$\left(F(a){\Big |}_{a=x}\right){\Big |}_{x=x+1}=F(x){\big |}_{x=x+1}=\displaystyle {\frac {(x+1)^{3}}{2}}$

So the "vertical bar" operator has a sort of non-associativity, which is obviously related to the concept of free variables. Is there any calculus text that actually examines the vertical bar operator rigorously?

Note: I am not asking why the expressions are not equal; that part is obvious. I am just asking for a reference. The example here is a minimal example of an issue that came up when I was thinking about the fundamental theorem of calculus and the more general formula for differentiation under the integral sign. — Carl (CBM · talk) 15:04, 27 April 2010 (UTC)

This would seem to fit in a logic text better than in a calculus text. And if I were to spend time on this in a calculus class, it would be time spent not teaching calculus. Michael Hardy (talk) 17:17, 27 April 2010 (UTC)

Sure, I wouldn't teach this in calc, but its resemblance to logic is only superficial. I started thinking about this when I was preparing material on the fundamental theorem, and realized I couldn't give a 1-second answer to why the fundamental theorem does not apply to this derivative: $\textstyle {\frac {d}{da}}\textstyle \int _{0}^{a}at\,dt$. The answer is that $\textstyle {\frac {\partial }{\partial a}}at$ is nonzero. — Carl (CBM · talk) 20:19, 27 April 2010 (UTC)

Why it should apply to that situation is another question one could raise. I've noticed non-mathematicians have an oddly negative way of asking some questions. If one asks how many prime numbers there are, they might say "Aren't there infinitely many?", as if the absence of a known reason why there aren't infinitely many justifies the conclusion that there are infinitely many. Michael Hardy (talk) 03:22, 28 April 2010 (UTC)

Yeah, but they're the ones we have to teach. — Carl (CBM · talk) 14:36, 28 April 2010 (UTC)

I think the problem here is just that "x=x+1" is a contradiction. If you assert a contradiction to be true, you aren't going to get a meaningful answer. Changing x to x+1 is more of a transformation than a substitution. It's not surprising that substitutions and transformations behave differently. --Tango (talk) 18:04, 27 April 2010 (UTC)

I'm sorry; the notation I used may not be as well known as I thought. In general, the notation $E{\Big |}_{a=\sigma }$ means to replace every occurrence of the letter a in the expression E with the string σ. As in, $\textstyle \int _{\sigma }^{\tau }f(x)\,dx=\left(\textstyle \int f(x)\,dx\right){\Big |}_{x=\sigma }^{\tau }$. The reason that the thing in my original post fails is obvious. But the usual story in calc I is that you can work syntactically in a completely naive way and you will still get the right answer. — Carl (CBM · talk) 20:19, 27 April 2010 (UTC)

The notation is standard, but I would only use it for substituting one thing for another. x->x+1 isn't really a substitution. It is clear what the notation is supposed to mean, but I don't think it is wise to use the notation like that. Transforming a variable is a very different operation to substituting one variable for another, so I wouldn't use the same notation for both. --Tango (talk) 00:53, 28 April 2010 (UTC)

This case is exactly what is meant by substitution in elementary logic, as Carl explained. I'm not sure what you mean when you insist it is not a substitution. Algebraist 01:08, 28 April 2010 (UTC)

How would you compute $\textstyle \int _{a+1}^{a+3}a\,da=(a^{2}/2){\Big |}_{a=a+1}^{a+3}$ without this type of substitution? — Carl (CBM · talk) 14:36, 28 April 2010 (UTC)

I wouldn't use the same symbol for a dummy variable as a real one. It would be much clearer if you wrote $\textstyle \int _{a+1}^{a+3}b\,db=(b^{2}/2){\Big |}_{b=a+1}^{a+3}$. --Tango (talk) 14:55, 28 April 2010 (UTC)

That sort of reminds me of a problem in automatic differentiation called "perturbation confusion".[2] I remember hearing that the book SICM uses Scheme code instead of traditional math notation partly to get around such problems, but I haven't read the book. 69.228.170.24 (talk) 19:53, 27 April 2010 (UTC)

Thanks! I had not thought to look into the CS literature on automatic differentiation, but now that you say it I can see how they will run into the same problems. — Carl (CBM · talk) 20:20, 27 April 2010 (UTC)

I would look into logic for this. Substitutions really is used for a lot in it. I am a bit puzzled as to why you find it surprising that substitutions is non assosiative though. Consider σ=a->b and ρ=b->a. Then, using prefix notation for substitution, σρ=a->b=σ, while ρσ=b->a. This is because, to take σρ we first apply ρ changing all b's to a's, then apply σ changing the a's to b's. Hence b's gets returned back to b's and a's gets changed to b's. Taemyr (talk) 07:15, 28 April 2010 (UTC)

I think that the bar operator is the same thing as the lambda calculus (except that your description of it doesn't make it clear what happens if you have a bar inside a bar, both of which substitute the same variable). The two equations you discuss look like this in the lambda notation (I took out the division by two):

$(\lambda a.xa^{2})(x+1)=x(x+1)^{2}$

$(\lambda x.(\lambda a.xa^{2})x)(x+1)=(\lambda a.(x+1)a^{2})(x+1)=(x+1)(x+1)^{2}$

(You can express the second case more simply as $(\lambda a.xa^{2})(x+1)$)

In any event, the important thing is that thing that comes between the $\lambda$ and the period is not a value. (For example, $(\lambda 2^{2}.6)(3)$ is an incoherent lambda expression, just like $6{\Big |}_{2^{2}=3}$ is incoherent.) Rather, it's the name of a variable, and its presence causes all of the occurrences of that variable "under" the lambda to become bound variables. To answer your actual question, I'm pretty sure that a rigorous treatment of the bar operator is the lambda calculus, which is extensively studied in programming languages, but I don't know if many mathematicians care that much. But it's an interesting area of study; the lambda calculus has only two constructs (pure lambda calculus does not include arithmetic or anything), and it's still capable of representing any possible computation. Paul (Stansifer) 18:56, 28 April 2010 (UTC)

Oops, my attempt at simplification was bogus, because it ignored the x in the inner expression. In fact, the whole point of the problem was that that x should be captured and shadowed by the outer lambda, not free at the top level. The alpha-renaming rule, however, tells us that we can safely rephrase it as $(\lambda z.(\lambda a.za^{2})x)(x+1)$, removing the confusing name clash. Paul (Stansifer) 09:42, 29 April 2010 (UTC)

I think the idea of de Bruijn indexes is that the confusion goes away if you replace the variables in the λ-expressions with them. 69.228.170.24 (talk) 05:20, 29 April 2010 (UTC)

April 28

Grade of Shade Cloth

Is the percentage of sunlight 'blocked' by a weave of material. How is this calculated using widths and gaps? --124.182.160.177 (talk) 08:59, 28 April 2010 (UTC)

Let's assume that the no light passes through the threads and that the same threads are used in both directions, and the same gaps are also present. Then we get a square cell, the distance between the start of each thread and the start of the next thread:

  / * *       \
 /  * *        G
Wc  * *       /
 \  * * * * * \ Wt
  \ * * * * * /

So, the width of a cell is the width of a thread plus the gap:

Wc = Wt + G

The percentage of light that gets through is:

100×G²/Wc²

The percentage of shade, then, is:

100 - (100×G²/Wc²)

Alternatively, you could find the percentage of shade by adding up the components:

100 × (Wt² + 2G×Wt)/Wc²

However, this doesn't account for "fuzz" outside the threads, so actually measuring the reduction in light would be better. Also note that it may lose "fuzz" over time, and a more threadbare cloth likely provides less shade. StuRat (talk) 12:27, 28 April 2010 (UTC)

You've missed the $/W_{c}^{2}$ in the last line. -- Meni Rosenfeld (talk) 13:48, 28 April 2010 (UTC)

Thanks, I added it now. StuRat (talk) 19:22, 28 April 2010 (UTC)

Cross ratio

I don't understand how the cross ratio answers when 4 points in projective line are projectivly equivalent. According to my book, "if z_1,z_2,z_3,z_4 and w_1,w_2,w_3,w_4 are two sets of 4 points, the are projectivly equivalent if their cross ratios are the same". The explanation is "since the cross ratio of z_1,z_2,z_3,z_4 is the image of the (unique) projective transformation carrying z_1 to 1, z_2 to infinity and z_3 to 0". How can the cross ratio, a scalar, be the image of a map on the projective line, a point defined by two coordinates? Thanks m uchly. —Preceding unsigned comment added by 122.109.239.224 (talk) 10:01, 28 April 2010 (UTC)

I think it must mean that the cross-ratio is the image of z_4 under this projective tranformation. From its description, the transformation is

$f:z\mapsto {\frac {(z_{2}-z_{1})(z_{3}-z)}{(z_{3}-z_{1})(z_{2}-z)}}$

This makes the image of z_4

$f(z_{4})={\frac {(z_{2}-z_{1})(z_{3}-z_{4})}{(z_{3}-z_{1})(z_{2}-z_{4})}}$

which fits the standard definition of cross-ratio up to a permutation of terms. Gandalf61 (talk) 10:31, 28 April 2010 (UTC)

Any triple of distinct points on the line is "equivalent" to any other, so the shape of a tuple of four points is just a matter of where you put the fourth point, and that is expressed by a scalar. Michael Hardy (talk) 03:28, 29 April 2010 (UTC)

Thanks but how exactly? I thought that may be the scalar represents the point in affine coordinates, but what exactly does it mean to multiply and divide two points in the projective line? Could you give a proof that the cross ratio is the image of z_4? There isn't a proof in Wikipedia, so I'm not sure. All I got is if z_3=az_1+bz_2, then if z_1 is sent to (x,0) and z_2 is sent to (0,y) (just looking at the transformation on affine space) z_3 is sent to (ax,by). Thanks muchly. —Preceding unsigned comment added by 122.109.239.224 (talk • contribs)

1. When you calculate a cross-ratio, you aren't multiplying and dividing points, you are multiplying and dividing distances between points (or, to be completely accurate, signed distances). If you are working in the real projective line, then you have four points P_1, P_2 etc. represented by four extended real numbers z_1, z_2 etc. (they are extended reals because we have to assign a value to the point at infinity), and the signed distances between pairs of points is z_2 − z_1 etc.
2. The projective transformation doesn't multiply or divide points either - it turns points into extended reals, multiplies and divides the reals, then turns the result back into another point (but the invertible mapping between points and extended reals is not usually spelled out).
3. A projective transformation is defined by the images of three points, not two. Given any two sets of three points {P_1, P_2, P_3} and {Q_1, Q_2, Q_3}, there is a unique projective transformation that maps P_1 to Q_1, P_2 to Q_2 and P_3 to Q_3 - we say that the group of projective transformation is "triply transitive" on the projective line.
4. The cross ratio of four points is an extended real number and a property of the set {P_1, P_2, P_3, P_4} - and if you want to think of it as a point, you can regard it as the image of P_4 under the projective transformation given above.
5. Projective transformations preserve the cross-ratio - indeed, you can define projective geometry as the study of the group of transformations that preserve the cross-ratio. Gandalf61 (talk) 10:30, 29 April 2010 (UTC)

Math Shortcuts

I couldn't think of a better title. I thought this article was very interesting considering I am a mathmetically challenged. Does anyone have any more to add to this list? --Reticuli88 (talk) 12:59, 28 April 2010 (UTC)

I suppose there's the fairly obvious dividing by powers of 10 rule: just shift the decimal to the left by the number of zeros:

123/10 = 12.3
123/100 = 1.23
123/1000 = .123
123/10000 = .0123 <- Add leading zero
123/100000 = .00123         "

StuRat (talk) 13:17, 28 April 2010 (UTC)

Mental multiplication of large numbers can be done as follows. First consider the rather contrived case of two numbers close to a round number N. Let's write the numbers as x = N+x' and y = N + y'. Then we have:

x y = (N+x')(N+y') = N^2 + (x'+y') N + x' y' =

N(N + x' + y') + x' y' =

N(x + y') + x' y'

Let's consider a numerical example. What is 974*986?. If we take N = 1000, then x' = -26 and y' = -14. Then x' y' is easily computed mentally like e.g. 2*13*2*7 = 4*91 = 4*(90+1) = 360 + 4 = 364. Computing x + y' is even easier, this is 974 - 14 = 960. So, the answer is 960*1000 + 364 =

960364.

Next, suppose that we have two numbers, x and y, that are close to two different round numbers, N and p N where p a small integer. Then with x = N + x' and y = p N + y', we have:

x y = (N + x')(p N + y') = p N^2 + N y' + p N x' + x'y' =

N(p N + y' + p x') + x' y' =

N(y+px') + x'y'

Example, let's compute 982*2113. We can then take N = 1000,

x' = -18, y'=113, p = 2. We have x' y' = -18*113 = -(20-2)*113= -2*(10-1)*113 =-2(1130 - 113) = -2*1017 = -2034

Then y + p x' = 2113 - 2*18 = 2077, so the product is

2077*1000 -2034 = 2074966

So, clearly a piece of cake to compute mentally this way. But this is still a somewhat contrived case. In general, you won't be able to find a round N and a pN such that subtracting these numbers from the numbers you want to multiply yields small easy to multiply numbers. However, you can iterate this procedure. With some practice you can learn to multiply numbers with ten digits or more this way in your head.

Note that from a computational POV, the above method is not efficient at all, it uses far more steps than the most efficient multiplication method. But those efficient methods are worthless for mental multiplication. The reason why mental arithmetic is difficult has nothing to do with a limited brain capacity. The brain is in fact far more powerful than the fastest supercomputers. Mental arithmetic is difficult because the brain doesn't have the correct software for doing computations. You have to work around that by hijacking other things your brain can do well. This is necessary to be able to do a computation at all; attempting to do computations efficiently is irrelevant.

Good mental arithmetic stategies work by breaking the problem down in many easy to compute substeps. This in such a way that you can mentally keep track of all the substeps. Count Iblis (talk) 16:50, 28 April 2010 (UTC)

A trick I use is similar to that. I memorized all squares up to about 50 and wanted to do more but didn't do much, although 2-digit squares are pretty easy to calculate any way. That takes time but if you want to be good at mental math you have to take time to memorize some things I assume. With the knowledge of all squares up to 50, I can do calculations like 38 * 46 easily. Note these are both 4 from the number 42, so we have (42 - 4) * (42 + 4) = 42^2 - 4^2 = 1764 - 16 = 1748. This uses the formula (x - y)(x + y) = x^2 - y^2. It works as long as you have the squares memorized and the difference between the numbers is even. If it's not an even difference, there's one little extra step. For example, 38 * 47 = 38 + 38 * 46 and then the difference is even so you can use the same method. StatisticsMan (talk) 14:55, 30 April 2010 (UTC)

April 29

area of the sphere above a plane

Suppose there is some sphere centred at the origin x^2 + y^2 + z^2 = r^2. Take an arbitrary plane/slice z=a between z=0 and z=r. Can I have some help with a general formula for the surface area of the sphere above a plane?

I've tried using a surface integral but I get a nasty dS formula ... and apparently using spherical coordinates won't give me the right answer. Help! John Riemann Soong (talk) 02:11, 29 April 2010 (UTC)

Archimedes answered this question. He found that if two parallel planes intersect a sphere, the surface area between them depends on the distance between them, but not on where they are otherwise. I'll see if I can remember or work out or find the details. Michael Hardy (talk) 03:30, 29 April 2010 (UTC)

The area of a slice of a sphere is equal to the area of the corresponding slice of the circumscribed cylinder of the sphere. Bo Jacoby (talk) 05:46, 29 April 2010 (UTC).

At an angle θ above the horizontal, the circle has a circumference of 2πrcosθ, so the area you want is $\int _{0}^{\alpha }2\pi r\cos \theta \,rd\theta$ where α = sin^-1(a/r). That evaluates to 2πr²sinα = 2πar. But I guess that's exactly what Bo Jacoby already said. Rckrone (talk) 07:43, 29 April 2010 (UTC)

Second limit comparisson test?

A textbook of a friend's briefly mentioned a theorem which I had never heard of before, and am therefore somewhat dubious of. Suppose $\left\{a_{n}\right\},\left\{b_{n}\right\}>0$, $\sum _{n=1}^{\infty }b_{n}$ converges, and the limit $\lim _{n\to \infty }{\frac {a_{n}}{b_{n}}}=0$, then $\sum _{n=1}^{\infty }a_{n}$ converges as well. I've taken a class on series, and have never run across this, nor did I see it in an internet search. However, it's been useful in answering a number of otherwise tricky series, so I have no idea why this wouldn't have been mentioned to me before. Either I had a bad teacher, or something's wrong with this book. Which is it? 173.179.59.66 (talk) 02:38, 29 April 2010 (UTC) Another possibility is that it's not so useful, and any series that I've tackled with the above "theorem" could have been handled by the regular limit comparisson test or something. If that's so, how would you have determined the convergence or divergence of a series like $\sum _{i=1}^{\infty }ln(i)/i^{3}$? —Preceding unsigned comment added by 173.179.59.66 (talk) 02:41, 29 April 2010 (UTC)

Under the hypotheses you state, for sufficiently large n, a_n must be less than b_n. That in itself is enough to get you the conclusion (combined with the fact that all the terms are nonnegative, of course).

So the test you state is just a little shortcut — you could always simply go through the fact that the a_n are eventually dominated by the b_n. It strikes me as being a little like l'Hospital's rule — a clever little gadget that's occasionally useful, but never a substitute for understanding what's going on. --Trovatore (talk) 03:28, 29 April 2010 (UTC)

Oh, by the way, you don't need the limit to be zero. It's enough that it be finite. If the limit is finite, then the a_n are eventually dominated by some constant multiple of the b_n, and that will still get you the conclusion. --Trovatore (talk) 03:37, 29 April 2010 (UTC)

And, it's not clear if you're assuming the a_n to be positive too; but in any case it's not needed, for a_n=O(b_n) implies that the series of a_n is absolutely convergent as explained above. --pma 11:12, 29 April 2010 (UTC)

Okay, thanks! 173.179.59.66 (talk) 14:01, 29 April 2010 (UTC)

1/49

Is 49 the smallest or somehow the simplest example of the phenomenon described at 49_(number)#Reciprocal? Lazily — Michael Hardy (talk) 03:59, 29 April 2010 (UTC)

Generalisation: If 10ⁿ is divisble by m then (10ⁿ−m)/m is an integer, and its reciprocal is

${\frac {m}{10^{n}-m}}=\sum _{k=1}^{\infty }{\frac {m^{k}}{10^{nk}}}$

49 is the case n=2, m=2. Taking n=1, m=2 we have

${\frac {1}{4}}=\sum _{k=1}^{\infty }{\frac {2^{k}}{10^{k}}}=0.2+0.04+0.008+0.0016+0.00032...=0.24999...$

For n=2, m=4 we have

${\frac {1}{24}}=\sum _{k=1}^{\infty }{\frac {4^{k}}{10^{2k}}}=0.04+0.0016+0.000064+0.00000256+0.0000001024...=0.0416666...$

For n=3, m=8 we have

${\frac {1}{124}}=\sum _{k=1}^{\infty }{\frac {8^{k}}{10^{3k}}}=0.008+0.000064+0.000000512+0.000000004096...=0.008064516...$

The pattern is clearest in the decimal expansion when m is small and n is large. For example, when n=3 and m=2 we have

${\frac {1}{499}}=0.002004008016032064128256513...$ Gandalf61 (talk) 09:22, 29 April 2010 (UTC)

But it's not clear that those other examples are examples of what I had in mind. If you write

${\frac {1}{49}}=0.02+0.0004+0.00000008+\cdots ,$

there's nothing that emerges from just staring at that expansion that suggests that it starts repeating after 42 digits, nor is it clear from any of the above how long the various repitends are. Michael Hardy (talk) 10:34, 29 April 2010 (UTC)

Drawing a Penis on a Ti-84

What function(s) will most easily create a penis-shaped graph (complete with 2 testes the sides) on a ti-84 graphing calculator between x=-10,10 and y=-10,10 ?

Thanks, Acceptable (talk) 04:49, 29 April 2010 (UTC)

Try y=sin(x)/x (Are you really sure you want to do this?) -- SGBailey (talk) 10:03, 29 April 2010 (UTC)

Well, that's one way to have fun with math. Here's the pic: [3].

Note that the portion of interest is approximately between x = -10 and x =10, but that the y value only ranges from (approximately) -0.25 to 1. To make it range from (approximately) -10 to 10, let's do some math.

We need to increase the range from 1.25 to 20, so we multiply by 20/1.25 or 16 to get y=16sin(x)/x. When we multiply this by 1 we get 16 and when we multiply by -0.25 we get -4. So, we now have a y range from (approximately) -4 to 16 and need to move this down by 6, or subtract 6 from the graph. This gives us y=(16sin(x)/x)-6.

Note that the lower y limit is only approximate, because the original graph didn't quite make it to -0.25. An interesting exercise might be to make the lower limit exactly y=-10. StuRat (talk) 14:11, 29 April 2010 (UTC)

You can also experiment with piecewise functions. Such as having semi circles, ellipses (sorry, you said functions, so you'll have to try two half ellipses) and two parallel lines. --Kvasir (talk) 15:53, 29 April 2010 (UTC)

Try $|\sin(x)/x|$, from x=-6 to +6 instead (that's the absolute value of the sinc function). Much better! HTH, Robinh (talk) 07:12, 30 April 2010 (UTC)

Perhaps you'll find one of these cubics erotic enough? They're not straightforward functions as such, I don't know about the Ti-84 but perhaps it'll be able to draw the graphs from the equations. Dmcq (talk) 09:48, 30 April 2010 (UTC)

Wow! these are wild! Never thought of these! Thx. --Kvasir (talk) 15:24, 30 April 2010 (UTC)

Jeez, xy² - x² + y² - 2x + 3 = 0 is downright pornographic. It should come in a plain brown wrapper. StuRat (talk) 22:40, 30 April 2010 (UTC)

I just had a look at the specs and it can't graph equations directly. It does seem to support polar mode graphing and I believe that should be good enough for your purpose with a little hacking, think about developing a polar plot of that sin(x)/x for instance should get halfway there. I'd go for something easy and pretty like a flower though. I'd have thought calculators would be far more powerful that that by now or perhaps they limit them so they are allowed into exams or so people can sell applications on them. Dmcq (talk) 10:20, 30 April 2010 (UTC)

That's because most of them are not true functions and do not past the vertical line test. Most people would not know how to graph Algebraic functions once out of highschool, beyond the typical circle, ellipse, hyperbola etc. --Kvasir (talk) 00:45, 1 May 2010 (UTC)

Changing the scale in a log log function

Say I have a function $\ln(y)=a+b\ln(x)$ and I change the scale of measurement of y (say from metres to kilometers), but dont change the scale of measurement of x, will the coefficient of the slope change, or just the intercept? —Preceding unsigned comment added by Damian Eldridge (talk • contribs) 08:57, 29 April 2010 (UTC)

Just the intercept. -- Meni Rosenfeld (talk) 09:14, 29 April 2010 (UTC)

You're going to change ln(y) to ln(cy) which is the same as ln(y)+ln(c). So you're just adding a constant to ln(y). 69.228.170.24 (talk) 09:18, 29 April 2010 (UTC)

Understanding Sphere Volume/Area Formulas

Does anyone know of a way to prove or at least conceptualize why the formulas for the volume and area of a sphere are what they are for someone who hasn't learned any calculus? It's ok if it's hand-wavy. I couldn't come up with anything any good. Rckrone (talk) 22:17, 29 April 2010 (UTC)

The volume can be found by using Cavalieri's principle. Last time I looked, the account there was fairly terse, but nonetheless covered the ideas. Michael Hardy (talk) 22:49, 29 April 2010 (UTC)

...and now I've looked again. The argument showing what the volume of a sphere is is all there, provided you know the volume of a cone is (1/3)×base×height. The argument showing that the volume of a cone is (1/3)×base×height is the part that's sketchy. Michael Hardy (talk) 22:52, 29 April 2010 (UTC)

(ec) The best I can think of is to do some soft calculus and essentially do a Method of exhaustion with of a stack of cylinders of varying radii. This is, of course, just Riemann sums for solids of revolution, but you can explain it without having to actually do the calculus or the integrals (the method of exhaustion was known to the ancient Greeks, long before calculus was invented). In fact I see someone has given you a more direct reference to this idea. -- Leland McInnes (talk) 22:55, 29 April 2010 (UTC)

Did you see the article on sphere?. Bo Jacoby (talk) 23:06, 29 April 2010 (UTC).

The way I remember the sphere volume formula is that I imagine a cylinder with radius r, with height r as well. Hence the volume of that cylinder is πr²*r = πr³. A sphere with the same radius r is just a little bit bigger than that cylinder, and hence the magic number 4/3.

As for surface area, just remember it's 4 times the area of the circle you'd see when you look at a sphere (same as the circular cross section of the sphere through its centre). Or imagine an open cube (no top and bottom) with length 2r thus fitting the sphere inside. Draw circle the same size of the sphere on each of the four sides. When you cut out the 4 circles, their combined area will be enough to cover the surface of the sphere.

Can't you tell i'm a visual person? --Kvasir (talk) 23:07, 29 April 2010 (UTC)

Integrating for volume

Hey all. Is it possible to extend integration to three dimensions? By this, I mean finding the volume of a three dimensional figure whose equation is known. Thanks --76.229.165.96 (talk) 23:45, 29 April 2010 (UTC)

Certainly. If you have an expression for the cross-sectional area at any value along one axis, all you need to do is integrate the area along that axis.

For example, imagine the equation $y=1/x$ with $x=1...\infty$. It's a curve that starts at {1,1} and gradually approaches y=0. Now imagine that curve rotated around the x axis, to make a trumpet shape. You now have a 3-D figure, of radius $r=1/x$, cross-sectional area of $\pi /x^{2}$. The volume is simply the area of the cross-section integrated over the length, i.e.:

$V=\int _{1}^{\infty }{\frac {\pi }{x^{2}}}dx=\pi$

Interestingly, for this shape, the surface area is

$A=\int _{1}^{\infty }{\frac {2\pi }{x}}dx=\left[2\pi \ln x\right]_{1}^{\infty }=\infty$

In other words, you can fill it with paint (due to finite volume), but you can't paint it (due to infinite surface area). ~Amatulić (talk) 00:19, 30 April 2010 (UTC)

In general, the concept is multiple integral. See in particular Multiple_integral#Computing_a_volume. Staecker (talk) 02:13, 30 April 2010 (UTC)

April 30

Need a tutorial on calculating moments

Would someone please point me to a layman's reference, or walk me through the procedure that was used to determine that the variance of the logistic distribution is $(\pi s)^{2}/3$ and the kurtosis is 6/5? The article on kurtosis, cumulant, and the like lost me, I'm afraid. I have a probability density function that's a nonlinear (but symmetrical) distortion of the logistic distribution, and I'd like to know its kurtosis and variance. If I know how to do it for the logistic distribution, I can figure it out for my version.

I can figure them out with a brute-force approach by generating thousands of values distributed according to my distribution, and then calculating the variance $\sigma ^{2}=\sum (x-{\bar {x}})^{2}/n$ and kurtosis as $a_{4}=\sum (x-{\bar {x}})^{4}/(n\sigma ^{4})$. That will give me an approximation. I need help figuring it out exactly, not numerically.

According to the moment (mathematics) article, I would find, say, the variance of the logistic distribution like this:

$\sigma ^{2}=\int _{-\infty }^{\infty }{\frac {x^{2}e^{-x/s}}{s(1+e^{-x/s})^{2}}}\,dx$

...but when I plug the indefinite integral into into Wolfram Alpha (which serves as a Mathematica front end), I get a result that involves a polylogarithm, which isn't exactly closed form. Wolfram Alpha won't solve the definite integral, and by inspection, substituting the -∞ and +∞ for x won't give a meaningful result. So where does $\sigma ^{2}=(\pi s)^{2}/3$ come from, as given in the logistic distribution article? ~Amatulić (talk) 04:52, 30 April 2010 (UTC)

[ec] Which part are you having trouble with - knowing what integrals to evaluate, or evaluating the integrals?

The former is fairly straightforward - the mean is $\mu =\int _{-\infty }^{\infty }f(x)x\ dx$, the variance is $\sigma ^{2}=\int _{-\infty }^{\infty }f(x)(x-\mu )^{2}\ dx$, the excess kurtosis is $\gamma _{2}={\frac {\mu _{4}}{\sigma ^{4}}}-3$ where $\mu _{4}=\int _{-\infty }^{\infty }f(x)(x-\mu )^{4}\ dx$, and so on.

The latter can be difficult to do by hand, but luckily you have the Wolfram Integrator Wolfram Alpha. -- Meni Rosenfeld (talk) 04:59, 30 April 2010 (UTC)

Looks like Alpha is still a bit quirky with these things. Giving the integral to Mathematica itself returns ${\frac {\pi ^{2}s^{2}}{3}}$ under the assumption $s>0$. -- Meni Rosenfeld (talk) 05:08, 30 April 2010 (UTC)

Ah, thanks. That explains my difficulty.

AAARGH! I had Mathematica at my previous employer, now I'm self-employed and must use Wolfram Alpha (can't afford Mathematica for personal use). I sure miss it.

The pdf for which I'm trying to find these moments is:

$f(x;\mu ,s)={\frac {a}{4s\cosh ^{2}\left[a\sinh ^{-1}\left({\frac {x-\mu }{2s}}\right)\right]{\sqrt {{\frac {(x-\mu )^{2}}{4s^{2}}}+1}}}}$

(for $\mu =0$ of course). If you have Mathematica handy, would you mind seeing if it gives a result for $\sigma ^{2}$ and $m_{4}$? ~Amatulić (talk) 05:16, 30 April 2010 (UTC)

Not at all. Unfortunately, I tried it and Mathematica is unable to evaluate these integrals.

If you do end up resorting to numerical calculation, numerical integration techniques will be much more accurate than the sampling method you mentioned. -- Meni Rosenfeld (talk) 06:04, 30 April 2010 (UTC)

It can be done for specific values of a (for example, $\mu =0,\ s=1,\ a=3$ gives $\sigma ^{2}={\frac {4\pi }{3{\sqrt {3}}}}-2,\ \mu _{4}=6-{\frac {8\pi }{3{\sqrt {3}}}}$), but for large a the expressions involve solutions to high-order polynomials. -- Meni Rosenfeld (talk) 14:26, 30 April 2010 (UTC)

Thanks. Actually my value of a is small, 1/tanh(1) or approximately 1.313. My experiments suggest that 1<a<2 always. ~Amatulić (talk) 17:49, 30 April 2010 (UTC)

Axiom is free and has a powerful integrator. I haven't tried it though. 69.228.170.24 (talk) 06:29, 30 April 2010 (UTC)

Thanks I'll give it a look. What I really would like is a symbolic rather than numeric solution as a function of s, but if I have to go with numeric that's how it must be, I suppose. ~Amatulić (talk) 17:49, 30 April 2010 (UTC)

I found a definite integral solver here. It doesn't give symbolic results, just numeric, but it was enough for me to determine that for a=1/tanh(1), $\sigma ^{2}=5.02793873736176s^{2}$. The odd thing is, the kurtosis came out negative (-4.25426), which the kurtosis article says means the peak is lower than a normal distribution and has thinner tails, but I know for a fact this distribution exhibits just the opposite character - taller thinner peak and fatter tails -- which is why I'm using it. ~Amatulić (talk) 05:07, 1 May 2010 (UTC)

The problem is not with the size of a per se, but with how complicated it is. 1/tanh(1) is too complicated to evaluate symbolically.

But if the value of a is fixed, then you have may have miscommunicated the problem. s is just a scaling factor, so the variance will be proportional to its square and the kurtosis doesn't depend on it. Finding the values numerically for a given a is easy.

The excess kurtosis is never less than -3. Your program gave a nonsensical result because the kurtosis is infinite - in fact, it is infinite for any a<2 (probably a=2, too). This is consistent with "taller peak, fatter tails", perhaps more so than you intended. The variance result is confirmed by Mathematica, though (the last 3 digits are off).

It sounds like you have a choice which distribution to use - perhaps you should choose one which is easier to work with. -- Meni Rosenfeld (talk) 19:50, 1 May 2010 (UTC)

Upon further investigation, it looks like the optimum value of a for my data is around 1.7.

Yes, my value of a is fixed once I determine the best fit to my data.

I was wondering if there was an elegant way to express $\sigma ^{2}$ in terms of a. I painstakingly solved $\sigma ^{2}$ for s=1 and a large number of values of a and, by inspection and experimentation, discovered five closed-form results in addition to the one you found (shown last):

Plots of variance and kurtosis as a function of a for f(x)=a/(4 cosh^2(a arcsinh(x/2)) sqrt(x^2/4+1))

Let $c=\pi /{\sqrt {3}}$

${\frac {\sigma ^{2}}{s^{2}}}={\begin{cases}-(20/3)c-2&a=4/5\\-(16/3)c-2&a=3/4\\(10/{\sqrt {3}})c-2&a=6/5\\(8/3)c-2&a=3/2\\{\sqrt {3}}c-2&a=2\\(4/3)c-2&a=3\end{cases}}$

(Corrected per reply comments below)

Those are just the ones I found. There may be more. For all I know, the whole relationship between sigma, a, and s might be a closed-form expression. That's one thing I was curious about.

If you look at the plot, you see that $\sigma ^{2}$ looks like it should be capable of being modeled somehow. The discontinuity at a=1 is where my distribution reduces to something Cauchy-like. I have no explanation for the behavior at a<1; in fact, between 0 and 1/2 there is only one value, a=1/4, that has a solution for variance.

While it's true that kurtosis can't be less than -3, mathematically a definite integral can have any value it wants. My red plot of kurtosis shows that it isn't defined at a=2, but also shows a smooth curve of negative values where a<2. Nevertheless, your comment above suggests that this may be an artifact of the online definite integral solver I used. Wolfram Alpha doesn't give me answers for a<2.

As to having a choice of distributions... I made this one up because it meets all the following requirements where other distributions fail:

• The pdf, cdf, and inverse cdf are all closed form, requiring no numerical methods. (The only numerical method is figuring out the relationship between variance, a, and s, once the value a is established, and I need to do that only once.)
• The distribution has a variance at the parameter a of interest to me.
• The pdf has a taller narrower peak and fatter tails than the normal distribution, while at the same time having a rounded peak, unlike, say, the stable pareto.
• The tails of the distribution bend outward when plotted on a log scale, crossing above the normal distribution tails past two standard deviations. The logistic distribution tails also cross the normal tails in that region, but plotted on a log scale they don't bend outward, they have a constant slope. I need the outward bend.
• The distribution fits my data! In fact it appears to universally fit several unrelated collections of samples.

I couldn't find anything else that met all those requirements.

One thing I want to do with this is modify the Black-Scholes option pricing formula, but for that I need kurtosis. On the other hand, the adjustment doesn't blow up with kurtosis values < -3. I'll try it and see (assuming those negative values I got are real results). ~Amatulić (talk) 04:50, 2 May 2010 (UTC)

It's unfortunate that you refuse to listen to both myself and your own common sense. Excess kurtosis is defined as ${\frac {\mu _{4}}{\sigma ^{4}}}-3$, where $\sigma ^{4}>0$ and $\mu _{4}$ is the integral of a non-negative quantity ($f(x)\geq 0,\ (x-\mu )^{4}\geq 0$ and thus $f(x)(x-\mu )^{4}\geq 0$). "Mathematically", the integral of a nonnegative quantity is nonnegative, and therefore the excess kurtosis cannot be less than -3. Any result that says the kurtosis is less than -3 is nonsense and is an artifact of the way it was calculated. As you may know, Mathematica can do arbitrary-precision arithmetic and control its error bounds, and I've used that to verify that the kurtosis is infinite for $a<2$. Also, it can be seen that $\cosh(a\sinh ^{-1}(x))$ grows like $x^{a}$ from which it can also be readily seen that the kurtosis is infinite.

And you can also see that the variance is reported negative for $a<1$, which I hope you believe me is also impossible. In fact it's, as you may have guessed, infinite. This includes $a=1/4$ for which, for whatever reason, a positive nonsensical result was reported. It may have something to do with the fact that for $a<1/2$, the distribution doesn't even have a mean.

And as I've explained, finding the value (for the variance, say) for any specific a is easy. There was no need to do anything "painstaking". However, the result for complicated a is unwieldy (as I said, solutions to high-order polynomials) which makes me doubt there is a nice closed form for general a, and you shouldn't look for one. It's naive to assume there's always a closed form in simple functions. There's a lot that can be done numerically (for example, using a root-finding algorithm to find the value of a which has a given variance), and I may help you with that if you explain what it is you're after (specifically, what did you intend to do with the closed form once you've found it).

By the way, your solution for the variance wasn't written correctly - the terms in the table multiply only the Pi term, not the -2. -- Meni Rosenfeld (talk) 08:57, 2 May 2010 (UTC)

Thanks, I corrected that above. That's what I meant. ~Amatulić (talk) 15:09, 2 May 2010 (UTC)

Note: It's not that surprising that the results came out negative. With a very specific algebraic interpretation, you can have results like 1+2+4+8...= -1 and 1+2+3+4...= -1/12. The former is roughly described as "If the series converges, you can do some algebraic manipulation to show that it must converge to -1". Likewise, numerical definite integrators over infinite domains can empirically determine the rate of convergence of the integrand to 0, do some algebraic tricks and reach an estimate for the integral. But if the integrand does not converge to 0 fast enough, and the integral is infinite, it is plausible that these tricks will lead to a negative result. But again, in our current context which is analytic, not algebraic, such a result makes no sense. -- Meni Rosenfeld (talk) 09:25, 2 May 2010 (UTC)

A simple example: Suppose you choose some large $x_{1}$ and find that $f(x_{1})=y_{1},\ f(ex_{1})=y_{2}$. If you assume the function is asymptotically of the form $ax^{b}$, you get $f(x)\approx y_{1}(x/x_{1})^{\ln(y_{2}/y_{1})}$ and thus $\int _{x_{1}}^{\infty }f(x)\approx {\frac {x_{1}y_{1}}{\ln(y_{1}/y_{2})-1}}$. If the function decreases sublinearly (and thus the integral diverges), you'll get $\ln(y_{1}/y_{2})<1$ and the result of this "approximation" will be negative. -- Meni Rosenfeld (talk) 10:47, 2 May 2010 (UTC)

You're right, that was a stupid comment on my part regarding negative kurtosis. I wrote that late last night before bed, then as I was falling asleep I realized that the integral is a positive function and can't have a negative result. I was hoping to delete that part this morning but I was too late!

And yes, I meant to exclude the 2 in the expressions for variance. Again, the only excuse I can offer is the late hour. I have corrected it above.

What was "painstaking" was solving the variance for many specific values of a to come up with that plot, in the hope of modeling the function somehow. What's also painstaking (for me) is to figure out by experimentation that, for example, 8.4791976 is $10\pi {\sqrt[{4}]{3}}-2$. The consistent pattern in these results suggested the existence of a closed form relationship. I don't really need to have a closed-form solution for variance, I just wanted the convenience of having a relationship that I could plug in a value for a and get a variance, in case I need to adjust a on the fly during processing that I might do in the future.

Thanks very much for the explanations. I have a much better understanding as a result of our dialogue. ~Amatulić (talk) 14:50, 2 May 2010 (UTC)

Ok, that's understandable. I'm glad everything is clearer to you now.

You can use the (numerical) values you've computed to fit a function of the form $\sigma ^{2}(a)\approx {\frac {b_{-1}}{a-1}}+b_{0}+b_{1}a+b_{2}a^{2}$ (with more terms if necessary), and use that to plug in different values of a. -- Meni Rosenfeld (talk) 15:51, 2 May 2010 (UTC)

Division Puzzle

Hi, I'm trying to solve this GeoCaching puzzle, and it's driving me mad. I need to find: "the lowest 10 digit integer which contains each of the digits 0 to 9 and which is exactly divisible by every number from 2 to 16 inclusive". I've tried working it out by testing each combination in order - I've got up to 1237546890 with no joy so far - is there a quicker way to solve this? The only digit I.m certain of is the zero, which has to be last for the number to be divisible by 10. Can anyone help me please? —Preceding unsigned comment added by 194.205.143.136 (talk) 05:42, 30 April 2010 (UTC)

Well, I thought the idea of a puzzle is you're supposed to do it yourself, but.... if you're doing it with a computer, use brute force, 10! is just 3.6 million. Some optimizations involving the least common multiple might speed things up. Do they say you are supposed to be able to do it by hand? 69.228.170.24 (talk) 06:38, 30 April 2010 (UTC)

(ec)

• If it contains all digits 0...9 then it is divisible by 3 and 9. It must end with zero to be divisible by 10. Then it also is divisible by 2, so by 6, too.
• 100=4×25, so the two last digits must form a number divisible by 4 for the whole number to be divisible by 4: with the last digit being 0 they may only be 20, 40, 60 or 80. Now the number is guaranteed to be divisible by 2, 3, 4, 5, 6, 9, 10, 12 and 15.
• 1000 is divisible by 8, so the 3−digits tail needs to be, too. That means the number of hundreds is even iff the 2−digits tail is divisible by 8, so the tail must be one of {1,3,5,7,9}{20,60} or one of {2,4,6,8}{40,80}, except repeating digits (440 and 880).
• 1/16=0.0625 → 10000 is divisible by 16 and no lower power of 10 is. So the last four digits must form a number divisible by 16. For each of sixteen possible 3−digits tails found before, check which of remaining 7 digits give 4−digits numbers divisible by 16. You can use modular arithmetic: for example 3−digits tail 760 ≡ 8 (mod 16), then 4−digits number will be "x760" ≡ 1000x + 8 ≡ 8x + 8 = 8(x + 1) (mod 16). Now x+1 must be even, so x is one of 1, 3, 5, 9 (7 got is already used). Similar way you'll find all possible 4−digits tail.
  Then your divisors set is {2..6, 8..10, 12, 15, 16}.
• For each 4−digit tail, there are 6 remainig digits, so you'll have to check only 6!=720 permutations of them (not combinations!) and test the whole number for divisibility by 7, 11 and 13 (14=2×7 will follow from 2 and 7). These three are all prime numbers, so they are coprime, so don't waste time to implement specific tests for divisibility by each them, just try dividing by 7×11×13=1001.

Good luck. CiaPan (talk) 08:14, 30 April 2010 (UTC)

You can also simply iterate through 10–digits multiples of LCM(2, 3, ..., 16) = 2⁴×3²×5×7×11×13 = 720720 and test which of them have all digits different. :) CiaPan (talk) 10:09, 30 April 2010 (UTC)

That's probably the most efficient approach for a computerised search, as it hits the lowest candidate first, and only needs to check about 3% of the search space. Gandalf61 (talk) 10:34, 30 April 2010 (UTC)

I certainly don't consider myself a math expert in any regard, but using CiaPan's advice, I used Excel to generate a list of all 10-digit multiples of 720,720, used the "text to columns" feature to split each number into ten columns, then used "sort columns" to eliminate numbers with duplicate digits until I found the answer. Good old left-brained fun, and it didn't take me all that long.... Kingsfold (talk) 13:36, 30 April 2010 (UTC)

Agreed, Gandalf. If we divide the lowest possible 10 digit number (assuming no leading zero allowed) by 720720 we get 1234567890/720720=1712.96. If we divide the largest possible number, we get 9876543210/720720 = 13703.72. So, we want to test all multiples of 720720 between 1713 and 13703. Here's a Fortran program to do just that:

 program mp    ! Math Problem.
 integer       I,J,N,C(10),CHECK
 character*11  STRING

 do I=1713,13703
   N = I*720720
   write (STRING,*) N
   STRING = STRING(2:) ! Needed since default format puts N 
                       ! in  STRING(2:11), 
                       ! not STRING(1:10).
   do J=1,10
     C(J) = 0
   enddo
   do J=1,10 ! Count number of each digit:
             ! C(1) = number of ones in
             ! current number, etc.
     if (STRING(J:J) .eq. "1") C( 1) = C( 1) + 1 
     if (STRING(J:J) .eq. "2") C( 2) = C( 2) + 1 
     if (STRING(J:J) .eq. "3") C( 3) = C( 3) + 1 
     if (STRING(J:J) .eq. "4") C( 4) = C( 4) + 1 
     if (STRING(J:J) .eq. "5") C( 5) = C( 5) + 1 
     if (STRING(J:J) .eq. "6") C( 6) = C( 6) + 1 
     if (STRING(J:J) .eq. "7") C( 7) = C( 7) + 1 
     if (STRING(J:J) .eq. "8") C( 8) = C( 8) + 1 
     if (STRING(J:J) .eq. "9") C( 9) = C( 9) + 1 
     if (STRING(J:J) .eq. "0") C(10) = C(10) + 1
   enddo
   CHECK = 0 ! Number of digits found
             ! exactly once in string.
   do J=1,10
     if (C(J) .eq. 1) CHECK = CHECK + 1
   enddo        
   if (CHECK .eq. 10) print *,STRING
 enddo

 end

It gives 8 solutions, but I don't want to give them away, as that would spoil all the fun. StuRat (talk) 13:49, 30 April 2010 (UTC)

This might be ticky, but the lowest 10-digit number to include all numerals between 0 and 9 is 1023456789, not 1234567890. (Right?) Kingsfold (talk) 15:43, 30 April 2010 (UTC)

That's true, but we also know that the number must end in a 0, since it has 10 as a factor. However, I modified the program to run with your lower limit. It doesn't make any diff in the outcome, but we might as well be throrough. StuRat (talk) 17:46, 30 April 2010 (UTC)

There are 8 solutions below 2³¹ and 58 above so I guess you have signed 32-bit integers. This is good enough to find the smallest solution but you would have missed the 4 that are also divisible by 17. PrimeHunter (talk) 14:12, 30 April 2010 (UTC)

Good catch. I was using the default INTEGER size, which is indeed a signed 32-bit integer. I've changed the program (below) to use a signed 64-bit INTEGER, and found all 66 matches. StuRat (talk) 17:46, 30 April 2010 (UTC)

Yes, I found 66 10-digit solutions with Python (which supports arbitrary-precision arithmetic), and I was just wondering where the other 58 had gone ! Gandalf61 (talk) 14:19, 30 April 2010 (UTC)

PARI/GP is also arbitrary precision. This computes the 66:

forstep(n=0,10^10,720720,if(vecsort(Col(Str(n)))==Col("0123456789"),print(n)))

PrimeHunter (talk) 14:24, 30 April 2010 (UTC)

Here's a strange thing. If you add 17 and 19 to the list of divisors then there are no 10-digit solutions using digits 0-9 once each. But if you relax that requirement and just look for pandigital solutions (using digits 0-9 at least once each) there are just two 11-digit solutions - and the larger is exactly twice the smaller ! Gandalf61 (talk) 14:32, 30 April 2010 (UTC)

OK, I updated the program to find all 66 values by using a longint (INTEGER*8, or 8 byte integer, in Fortran terminology). I also lowered the starting point so we will now find any matches between 1023456789 and 1234567890 (there aren't any), and I added a line count to the print:

 program mp    ! Math Problem.
 integer*8     I,J,K,N,C(10),CHECK
 character*11  STRING

 K = 0 ! Number of solutions found.
 do I=1421,13703
   N = I*720720
   write (STRING,*) N
   STRING = STRING(2:) ! Needed since default format puts N 
                       ! in  STRING(2:11), 
                       ! not STRING(1:10).
   do J=1,10
     C(J) = 0
   enddo
   do J=1,10 ! Count number of each digit:
             ! C(1) = number of ones in
             ! current number, etc.
     if (STRING(J:J) .eq. "1") C( 1) = C( 1) + 1 
     if (STRING(J:J) .eq. "2") C( 2) = C( 2) + 1 
     if (STRING(J:J) .eq. "3") C( 3) = C( 3) + 1 
     if (STRING(J:J) .eq. "4") C( 4) = C( 4) + 1 
     if (STRING(J:J) .eq. "5") C( 5) = C( 5) + 1 
     if (STRING(J:J) .eq. "6") C( 6) = C( 6) + 1 
     if (STRING(J:J) .eq. "7") C( 7) = C( 7) + 1 
     if (STRING(J:J) .eq. "8") C( 8) = C( 8) + 1 
     if (STRING(J:J) .eq. "9") C( 9) = C( 9) + 1 
     if (STRING(J:J) .eq. "0") C(10) = C(10) + 1
   enddo
   CHECK = 0 ! Number of digits found
             ! exactly once in string.
   do J=1,10
     if (C(J) .eq. 1) CHECK = CHECK + 1
   enddo        
   if (CHECK .eq. 10) THEN
     K = K + 1 ! Increment number of solutions found.
     print *,K," = ",STRING
   endif
 enddo

 end

StuRat (talk) 17:38, 30 April 2010 (UTC)

Thanks so much for all the replies to my query. Most of it was over my head to be honest, but I was able to pick out some hints to help my 'brute force' attempt to crack it. Finally solved it by starting with 720720*1713 - then just kept adding 720720 until I got 10 different digits! Took me about 10 minutes in total - versus the hours already spent!

Also, in reply to the first answer, I realize the idea of a puzzle is to 'do it myself' - I wasn't looking for (and didn't receive) an answer - just some idea on a quicker way to solve it. Also the puzzle makes no reference to 'solving it by hand' - previous logs state it has been solved by excel and programs like the one above - but I wouldn't know where to start, hence the brute force method.

Thanks again, Justin 194.205.143.136 (talk) 13:00, 1 May 2010 (UTC)

I enjoyed coding in the J (programming language) the lowest 10 digit integer which contains each of the digits 0 to 9 and which is exactly divisible by every number from 2 to 16 inclusive, and the number of solutions

  ({.,#) (#~(n=(\:])"_1&.((10#10)&#:))) (*[:i.[:<.(n=.9876543210x)&%) *./2+i.15
1274953680 66

I don't know any shortcut, but brute force works. Bo Jacoby (talk) 13:00, 1 May 2010 (UTC).

Quick query about eigenvalues of a positive real matrix

Hi all - I'm working my way through some old exam papers and I came upon this quick problem which I can't for the life of me seem to solve: could anyone give me a hand?

If A is a n*n matrix and all entries of A are real and positive, do all eigenvalues of A have positive real part? I know that since all entries are real the determinant (polynomial) will have only real coefficients so for any complex eigenvalue root, its conjugate must be an eigenvalue too - then I tried looking at the trace as the sum of the eigenvalues, and arguing that summing over all the eigenvalues will end up picking up only the real parts of every eigenvalue, since every complex root has a conjugate eigenvalue, so the trace will be the sum of the real parts of the eigenvalues - and certainly, since all entries >= 0, the trace >= 0, but that only tells me about the overall sum of the eigenvalues being positive - for all I know, there could be a negative eigenvalue and just an even larger positive eigenvalue which makes the trace positive despite a negative eigenvalue, and I can't see any way to 'fix' the matrix so that we can definitely obtain only that negative-real-part eigenvalue without any of those larger positive-real-part eigenvalues necessarily being there: or who knows, perhaps it's in fact false and I simply can't think of a counterexample!

Your thoughts would be much appreciated, 131.111.29.210 (talk) 15:18, 30 April 2010 (UTC)

${\begin{pmatrix}1&2\\2&1\end{pmatrix}}$ has eigenvalues 3 and −1, the latter corresponding to the eigenvector (1,−1).—Emil J. 15:35, 30 April 2010 (UTC)

And the way you should have come up with this example is that the determinant is the product of all eigenvalues, so if the determinant is negative there's no way all eigenvalues are conjugate pairs and positive reals. -- Meni Rosenfeld (talk) 15:41, 30 April 2010 (UTC)

Wonderful, thankyou both for the counterexample and the explanation :) —Preceding unsigned comment added by 82.6.96.22 (talk) 17:27, 30 April 2010 (UTC)

What is true is that if a nxn matrix has all nonnegative entries, it has at least a real nonnegative eigenvalue, and the correspondig eigenvector has nonnegative coordinates. It's a theorem by Frobenius; in fact quite an immediate consequence of the Brouwer fixed point theorem. --pma 18:43, 30 April 2010 (UTC)

Even distribution on a sphere

Is there any any algorithm to describe the distribution of n points on a sphere such that the distance between points is maximized? For instance, if there were n entities of a certain length, with a charge (the same) at one end, and all joined at the other end, how would they orient themselves to minimize the potential energy? For certain cases, it's trivial (2 = opposite, 3 = triangle), and it should be nice and symmetric for the Platonic solids (4 = tetrahedron, 6 = cube, 8 = icosahedron, 12 = icosahedron, 20 = dodecahedron). For 5, I assume it would be a double tetrahedon, with three oriented along the horizontal plane in a triangle, and the other two directly up and down. Is there a general algorithm or pattern? Perhaps there is not a unique solution? — Knowledge Seeker দ 21:20, 30 April 2010 (UTC)

Have you read Wikipedia:Reference desk/Archives/Mathematics/2007 August 29#Distributing points on a sphere? Algebraist 21:30, 30 April 2010 (UTC)

No! And I apologize; I tried searching the archives before asking, but I don't know how I didn't find that entry. I will read it now, thank you. — Knowledge Seeker দ 21:42, 30 April 2010 (UTC)

It helps to have been here long enough to remember it. Algebraist 21:57, 30 April 2010 (UTC)

If you want to minimize the potential energy, then the problem is not trivial, see here for an analogous problem which is discussed here and the outline of a solution is given here. Count Iblis (talk) 22:23, 30 April 2010 (UTC)

There's an article about that, one of two mentioned at the end of the circle packing article. I notice thats not referenced from packing problem, I'll go and add it. Dmcq (talk) 22:27, 30 April 2010 (UTC)

I think Coxeter's 12 Geometric Essays book addresses this problem. A Dover reprint bears the title The Beauty of Geometry, if I recall correctly. Michael Hardy (talk) 20:54, 1 May 2010 (UTC)

Sqrt

What is the antiderivative of square root. 76.199.144.250 (talk) 22:14, 30 April 2010 (UTC)

You can use the same rule as with any power of x. --Tango (talk) 22:21, 30 April 2010 (UTC)

See the first entry of List of integrals of rational functions, with n=1/2. Buddy431 (talk) 22:25, 30 April 2010 (UTC)

Product rule on Banach spaces

Let U, A and B be Banach spaces, L(A,B) the vector space of linear maps from A to B, let

$M:U\to L(A,B)$

$v:U\to A$

be differentiable functions.

I'm wandering whether some version of the product rule holds like

$D(M(x)v(x))=DM(x)v(x)+M(x)Dv(x)$

and in that case what would be the meaning of the product DM(x)v(x).

Can anybody help me?--Pokipsy76 (talk) 22:31, 30 April 2010 (UTC)

Yes, you just have to use the composition rule. The map $\scriptstyle U\to L(A,B)\times A$ taking $x$ to the pair $((M(x),v(x))$ is differentiable because its components are; and you are composing it with the evaluation map

$\scriptstyle L(A,B)\times A\to B$, taking the pair (T,x) to Tx.

The evaluation map is bilinear and continuous (of norm 1: essentially by definition of the operator norm: $\scriptstyle \|Tx\|\leq \|T\|\|x\|$).

As a general fact, any bounded bilinear map $u$ on Banach spaces is infinitely differentiable, and the differential is given immediately by the expansion: $\scriptstyle u(x+h,y+k)=u(x,y)+u(h,y)+(x,k)+u(h,k)$. Since $\scriptstyle u(h,k)=o(\|(h,k)\|$ you have $\scriptstyle Du(x,y)=u(\cdot ,y)+u(x,\cdot )$. As a consequence, the differential of the composed map $\scriptstyle x\mapsto M(x)v(x)$ is the linear map $\scriptstyle h\mapsto (DM(x)h)[v(x)]+M(x)[Dv(x)h]$, which is what you wrote (recall that for any $h$ in U , M(x) and DM(x)h are in L(A,B), while v(x) and Dv(x)h are in A, so that the meaning of the evaluations $\scriptstyle (DM(x)h)[v(x)]$ and $\scriptstyle M(x)[Dv(x)h]$ is clear).

Also, remember that DM(x), formally (that is, according with the general definition of differential) is an element of $\scriptstyle L(U,\,L(A,B))$. The latter Banach space is naturally isometrically isomorphic with $\scriptstyle L^{2}(U\times A,B)),$ the space of bounded bilinear maps. So if we wish we may identify DM(x) with the corresponding bilinear map $\scriptstyle U\times A\to B$, which is somehow a simpler object, and write in consequence e.g. $\scriptstyle DM(x)[h,v(x)]$ or $\scriptstyle DM(x)\cdot h\cdot v(x)$.--pma 08:49, 1 May 2010 (UTC)

Thank you, very illuminating. I still have some problem i guessing the nature of the isometry about DM, I don't understand if this isometry depends on some particular choice of the norm. I've just posted another question here which is related to this point.--Pokipsy76 (talk) 14:05, 1 May 2010 (UTC)

May 1

Gene DNA GC content against whole genome DNA GC content.

Hi guys, I have a simple question concerning GC (Guanine-Cytosine) content in DNA sequences. Suppose I have a bacterial genome, with a total GC content of 60%. And one of the genes in the genome is 1000 base pairs long, and has a GC content of 70%. I have been trying to do a bootstrap analysis to see if the 70% gene GC content value is significant, but I'm not sure about my steps:

- I create 100 random sequences, each 1000bp long, with each position in every sequence having a 60% chance to be G or C, and 40% to be A or T.

- For every sequence I calculate the resulting GC content, and add all values in an array.

- I calculate the Mean and StDev in the array.

- If the GC content of my gene is higher than Mean + StDev*4.47 or lower than Mean - StDev*4.47 i have a 95% probability chance of significance.

- If the GC content of my gene is higher than Mean + StDev*10 or lower than Mean - StDev*10 i have a 99% probability chance of significance.

I feel I'm wrong somewhere, and this produces the same results as if I had simply used the genome GC content for Mean and 1 for StDev. Any help would be much appreciated. PervyPirate (talk) 10:54, 1 May 2010 (UTC)

Are the base pairs of a genetic sequence independent of one another ? If not, it's a bit like saying "the average number of Ford logos per average car on the road is 0.3, yet my car has 4, is that significant ?". In the case of DNA, while they may initially be random, the selection process may tend to favor certain (non-random) combos. StuRat (talk) 11:23, 1 May 2010 (UTC)

I didn't want to complicate things, but yes, let's assume that we are talking about synonymous positions (the ones that have no effect on the encoded amino acid, and therefore they are independent from one another as well as there is no selective pressure on them). PervyPirate (talk) 11:35, 1 May 2010 (UTC)

Help in Solving Tensor Equation

Lets say we have a tensor equation $2A_{\alpha \beta \gamma }=B_{\gamma }^{\lambda }C_{\alpha \beta \lambda }+B_{\beta }^{\mu }C_{\alpha \gamma \mu }$. How do we solve it? Telling me how to find the tensor $D_{\beta }^{\alpha }$ or $D_{\beta }^{\alpha }$ given $E_{\beta \gamma }=D_{\beta }^{\alpha }F_{\alpha \gamma }$ or $E_{\beta \gamma }=D_{\beta }^{\alpha }F_{\alpha \gamma }$ respectively would be great help too. P.S. This is not a homework problem. Thanks in advance.The Successor of Physics 13:35, 1 May 2010 (UTC)

Operator norm and isomorphisms

Let A, B be Banach spaces and L(A,B) be the vector space of the linear operators between A and B.

The natural norm for L(A,B) is the operator norm:

$|T|_{L(A,B)}:=\sup _{|x|_{A}=1}|Tx|_{B}$

now my problem is the following: there can be isomorphisms between these operator spaces, for example

$L(A,L(B,C))\cong L(B,L(A,C))$

and if we think the space in the first way rather than the second isomorphic way the definition of the operator norm changes.

My question is: is the operator norm independent on these isomorphism or do we get different operator norm if we look the space in a different way?

In other word is it true that

$|T|_{L(A,L(B,C))}:=\sup _{|x|_{A}=1}|Tx|_{L(B,C)}=\sup _{|y|_{B}=1}|Ty|_{L(A,C)}=:|T|_{L(B,L(A,C))}$

?

Thank you for your help.--Pokipsy76 (talk) 14:21, 1 May 2010 (UTC)

I think I've the answer. The equality holds just because if we explicit the meanings of the terms we have

$|T|_{L(A,L(B,C))}:=\sup _{|x|_{A}=1}|Tx|_{L(B,C)}=\sup _{|x|_{A}=1}\sup _{|y|_{B}=1}|Txy|_{C}=\sup _{|y|_{B}=1,|x|_{A}=1}|Txy|_{C}$

and the same happens for $|T|_{L(B,L(A,C))}$.--Pokipsy76 (talk) 16:14, 1 May 2010 (UTC)

Correct, and note that both spaces are also isometrically and linearly isomorphic with the space of bounded bilinear maps, L²(A×B, C). The norm of a bilinear map u:A×B→C is by definition

$\|u\|_{L^{2}(A\times B,C)}:=\sup _{|x|_{A}\leq 1,\,|y|_{B}\leq 1}\|u(x,y)\|_{C}$,

and of course does depend on the choice of the norms on the spaces A B C. But the other equivalent norms on A, B resp. C, would produce an equivalent norm on L²(A×B, C).--pma 17:45, 1 May 2010 (UTC)

games that do not end

What games do not have moves that progress through time, i.e., what games are not temporal? 71.100.1.71 (talk) 16:44, 1 May 2010 (UTC)

I don't know if this is what you mean, but some games consist of the players each making a single move all simultaneously. Those games don't progress through time in any meaningful sense since they happen at a single moment. Prisoner's dilemma is one example of such a game. Rckrone (talk) 16:55, 1 May 2010 (UTC)

These are known as Normal-form games. -- Meni Rosenfeld (talk) 20:01, 1 May 2010 (UTC)

There are games that can get into draw states, where no player can win. Chess is one. Conway's Game of Life can also get into repeating loops. StuRat (talk) 17:07, 1 May 2010 (UTC)

Many social games do not progress in any meaningful way. They are just there to maintain society. Dmcq (talk) 19:15, 1 May 2010 (UTC)

Do you mean games like rock, paper, scissors? 71.100.1.71 (talk) 09:11, 2 May 2010 (UTC)

No I mean like the banter between James Bond and Miss Moneypenny. Though I guess playing a game of rock scissors paper every day with someone would be roughly equivalent. Dmcq (talk) 15:49, 2 May 2010 (UTC)

Asians, Indians, and Eastern Europeans

Why are Asians, Indians, and Eastern Europeans so much better at math (in modern times, at least)? --75.33.219.230 (talk) 17:35, 1 May 2010 (UTC)

See Race and Intelligence. By the way Eastern Europe isn't part of Asia. Dmcq (talk) 19:08, 1 May 2010 (UTC)

That's why I listed Asians and Eastern Europeans separately. --75.33.219.230 (talk) 19:36, 1 May 2010 (UTC)

Assuming that the assertion in your question is indeed true, I think the reason is mostly cultural—how math is taught and how much emphasis is placed on math education, whether math is commonly perceived as a hard subject, whether great mathematicians and their achievements are seen as a source of national pride, existence of lucrative career possibilities not requiring much math, ... --98.114.146.58 (talk) 08:22, 2 May 2010 (UTC)

If you're talking about research in mathematics, France is probably the top country, in terms of Fields medallists. If not in absolute terms, per capita certainly. The US and Russia are also high.

On the other hand, if you're talking about the ability of schoolchildren, Eastern Europe no longer ranks highly. Since the fall of the Soviet Union, their education systems have gone down the tubes. France isn't great either. Have a look at the PISA results (for 15-year olds), ranking countries in the OECD as well as some others: [4]. Japan is tops in math, South Korea second, but then come New Zealand, Finland, Australia, Canada, Switzerland, the UK... Two Canadian provinces, Quebec and Alberta, would rank above South Korea if they were countries. The Czechs, Russians, Hungarians and Americans are all below average, and Latvia near the bottom. France is slightly above average. I would suspect that India wouldn't have ranked high if they had participated. 82.124.97.111 (talk) 16:16, 2 May 2010 (UTC)

Sorry, the latest results from 2006 (in which more countries participated) have the following top ten in math: Taiwan, Finland, Hong Kong, South Korea, Netherlands, Switzerland, Canada, Macao, Liechtenstein, Japan. Estonia 14th, Czech Republic 16th, Slovenia 20th, France 23rd, Poland 24th, the UK 25th, Slovakia 26th, Hungary 27th, Latvia 30th, Lithuania 31st, Russia 34th, the US 35th. Many Eastern European countries are ranked lower as well. Clearly, it's possible for Western countries to match Asia, since five of the top ten are Western. We don't need to try and match Japan. Let's copy whatever Finland is doing instead.

Derivatives

Hey guys, I'm having trouble with some derivatives; maybe someone could help me out? I don't get how ${\frac {d}{dx}}a^{x}=\ln(a)a^{x}$ or how ${\frac {d}{dx}}\ln(x)={\frac {1}{x}}$. Could someone show me a proof? What I'm not getting is where the natural logarithm is coming from. Thanks! PS: In the former, why is it that ${\frac {d}{dx}}a^{x}=\ln(a)a^{x}$, and not ${\frac {d}{dx}}a^{x}=(x)a^{x-1}$? 76.229.195.134 (talk) 19:49, 1 May 2010 (UTC)

${\frac {d}{dx}}a^{x}=\ln(a)a^{x}$ follows from $a^{x}=e^{xlna}$ and the chain rule. ${\frac {d}{dx}}\ln(x)={\frac {1}{x}}$ follows from the derivative of $e^{x}$ and the inverse function theorem. The former is not $xa^{x-1}$ because that rule is for when x is the base, not the exponent. -- Meni Rosenfeld (talk) 20:09, 1 May 2010 (UTC)

The erroneous proposed derivative xa^x−1 would imply that the slope is negative when x is negative, and obviously that's not what you see when you look at the graph. If you go back to the definition of the derivative, you've got

${\begin{aligned}{d \over dx}a^{x}&=\lim _{h\to 0}{a^{x+h}-a^{x} \over h}\\[10pt]&=\lim _{h\to 0}a^{x}{a^{h}-1 \over h}\\[10pt]&=a^{x}\lim _{h\to 0}{a^{h}-1 \over h}{\text{ (since }}a^{x}{\text{ does not depend on }}h)\\[10pt]&=\left(a^{x}\cdot {\text{constant}}\right){\text{ (since the limit does not depend on }}x).\end{aligned}}$

so next you need to know what the "constant" is. To be continued...... Michael Hardy (talk) 20:48, 1 May 2010 (UTC)

Note that the derivative of a function with respect to x looks at how the function changes when we change x. In that regard a^x behaves very differently from xⁿ because x is the variable we care about. It is true that ${\frac {d}{da}}a^{x}=xa^{x-1}$, but that's something different. Rckrone (talk) 05:59, 2 May 2010 (UTC)

long differential equations problem

This was a problem in my homework a couple of days ago.

"Use the method of variation of parameters to determine the general solution of the given differential equation: $y^{iv}+2y''+y=\sin(t)$"

The characteristic equation is $k^{4}+2k^{2}+1=0$. That has roots of {i, i, -i, -i} so the general solution of the corresponding homogeneous equation is $y=c_{1}\cos(t)+c_{2}\sin(t)+c_{3}t\cos(t)+c_{4}t\sin(t)$.

To find the particular solution, I have to evaluate five Wronskians:

$\left|{\begin{array}{cccc}\cos(t)&\sin(t)&t\cos(t)&t\sin(t)\\-\sin(t)&\cos(t)&\cos(t)-t\sin(t)&\sin(t)+t\cos(t)\\-\cos(t)&-\sin(t)&-2\sin(t)-t\cos(t)&2\cos(t)-t\sin(t)\\\sin(t)&-\cos(t)&t\sin(t)-3\cos(t)&-3\sin(t)-t\cos(t)\end{array}}\right|$

$\left|{\begin{array}{cccc}0&\sin(t)&t\cos(t)&t\sin(t)\\0&\cos(t)&\cos(t)-t\sin(t)&\sin(t)+t\cos(t)\\0&-\sin(t)&-2\sin(t)-t\cos(t)&2\cos(t)-t\sin(t)\\1&-\cos(t)&t\sin(t)-3\cos(t)&-3\sin(t)-t\cos(t)\end{array}}\right|$

$\left|{\begin{array}{cccc}\cos(t)&0&t\cos(t)&t\sin(t)\\-\sin(t)&0&\cos(t)-t\sin(t)&\sin(t)+t\cos(t)\\-\cos(t)&0&-2\sin(t)-t\cos(t)&2\cos(t)-t\sin(t)\\\sin(t)&1&t\sin(t)-3\cos(t)&-3\sin(t)-t\cos(t)\end{array}}\right|$

and so forth. How can I quickly calculate these by hand? I also need some help evaluating the resulting integrals. I know the problem can be done pretty easily using the method of undetermined coefficients but the problem says to use the method of variation of parameters. —Preceding unsigned comment added by Metroman (talk • contribs) 21:37, 1 May 2010 (UTC)

This assignment looks like torture to me. I would suggest simplifying things as follows. You can write the diff. equation as:

(D^2 + 1)^2 y = sin(t)

This means that if you first solve the equation:

(D^2 + 1) y_1 = sin(t)

then the solution of

(D^2 + 1)y = y_1(t)

Will also be a solution of the original diff. equation. You can use variation of constants here too, so this should be allowed. You can simplify things further by replacing sin(t) by exp(it) and taking the imaginary part of the solution. Also, you can factor D^2 + 1 = (D+i)(D-i) and solve each of the two equations in two steps. Count Iblis (talk) 22:43, 1 May 2010 (UTC)

As to the issue of computing by hand the determinants: note that, in the first matrix, if you sum the first row into the third, and the second row into the fourth, you get a matrix with a null 2x2 block, so its determinant is easily computed as the product of the determinants of the two principal 2x2 blocks (at a first glance: isn't -2?). Analogous simplifications work with the other matrices. --pma 07:07, 2 May 2010 (UTC)

Romanian army / vehicle volume

In What Am I Doing in New Jersey?, comedian George Carlin mentioned, "People who don't know how wide their cars are...'Well, I don't know if I can fit in there!' - You can get the f*****g Romanian Army in there!"

I have been unable to find data on the approximate volume of the largest commercially-available cars, nor the average body capacity of Romanian military personnel. I wonder if anyone could help prove or disprove his hypothesis?  Chzz  ►  23:53, 1 May 2010 (UTC)

The hypothesis is that the Romanian army will fit in there? I think it would depend on where. Rckrone (talk) 05:51, 2 May 2010 (UTC)

As far as I can ascertain, it is the space in which one could comfortably fit a car.  Chzz  ►  06:05, 2 May 2010 (UTC)

check Hyperbole. --pma 08:09, 2 May 2010 (UTC)

As an ex-resident of the Manchester (UK) area, the corresponding expression I'm familiar with is "You could get a Ferranti transformer through there". As it says in the article: "High voltage power transformers became an important product for Ferranti; some of the largest types weighed over a hundred tons". Moving the largest by road required a low-loader with motive power fore and aft, and skilled driving to pass through built-up areas.→81.131.162.203 (talk) 13:44, 2 May 2010 (UTC)

May 2

Information theory terminology

The entropy of a probability distribution $p$ over a discrete random variable $X$ is $\sum _{i=1}^{n}{-p(x_{i})\log p(x_{i})}$. I'm trying to find out what I should call one term of this sum. For example, what is $-p(x_{i})\log p(x_{i})$ called for $x_{i}$? It's not $x_{i}$'s surprisal, because that's $-\log p(x_{i})$. Thanks. --Bkkbrad (talk) 01:10, 2 May 2010 (UTC)

The entropy contribution from $x_{i}$, or the mean surprisal of $x_{i}$ ? Bo Jacoby (talk) 06:02, 2 May 2010 (UTC).

Winning games

Where can I find a list of just the winning games in TicTacToe, preferably with each move displayed in the format of the game? 71.100.1.71 (talk) 15:30, 2 May 2010 (UTC)

The Tic-tac-toe article states that "When winning combinations are considered, there are 255,168 possible games." hydnjo (talk) 15:52, 2 May 2010 (UTC)