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February 15

Simple question on exponents

With an expression like 4^3^2, does one consider this as (4^3)^2 or 4^(3^2)? Myles325a (talk) 00:32, 15 February 2008 (UTC)[reply]

In what context? TeX doesn't seem to like it at all: <math>x^y^z</math> yields an error message (Failed to parse (Conversion error. Server ("https://wikimedia.org/enwiki/api/rest_") reported: "Cannot get mml. TeX parse error: Double exponent: use braces to clarify"): {\displaystyle x^{y}^{z}} ).
There's a good argument for treating it as right-associative: if you wanted (x^y)^z you'd be better off writing it as x^(y*z) but if you want x^(y^z) there's no alternative. That's how it's interpreted in some programming languages, Perl for example, although the operator is spelled ** because ^ means something else. The double-asterisk comes from FORTRAN, which also treats it as right-associative. --tcsetattr (talk / contribs) 01:11, 15 February 2008 (UTC)[reply]
In TeX you typeset it as , which makes it obvious what is intended when you look at the source. But you have to know the convention (which, as you say, is quite standardly right-associative; there's no ambiguity whatsoever) to interpret the output. --Trovatore (talk) 01:25, 15 February 2008 (UTC)[reply]

OP myles325a here. Good Grief! Who said anything about programming languages? Think of a dude with a pen and paper, and the job of writing the example expression so it is unambiguous. What is the accepted convention for doing that? Does right-associativity mean you bracket from right to left? I clicked on right-associativity, and I have never seen a more obscurantist effort to make a simple thing complicated. Myles325a (talk) 04:18, 19 February 2008 (UTC)[reply]

With pen and paper you don't write it with ^. You write it as , or if you wanted the other meaning . Taemyr (talk) 11:23, 19 February 2008 (UTC)[reply]

OP myles325a here. I know that. I'm not concerned with how it is actually written. I am concerned with whether there is an explicit and formal convention for grouping exponents, however they be written. Are you saying the order in which the exponents are to be grouped is derived solely via the font size of the exponents as they are represented in the conventional superscript notation? Because I wouldn't mind seeing a site cite for that. Try printing out 5^4^3^2 like that and show it here. And I might be wrong here, but how is equivalent to one of the 2 possible values of x^y^z. Myles325a (talk) 23:44, 19 February 2008 (UTC)[reply]

Proof by contradiction

In the article about mathematical proofs, there is a proof that is irrational, this proof assumes that if a2 is even then a must also be even. My question is : can you prove that if (a2 is even ) then (a must also be even ) even if is rational ? --George (talk) 02:51, 15 February 2008 (UTC)[reply]

Sure, that's easy to prove. Assume a is odd. Then it is of the form , for some integer n. (I take this as the definition of an odd number; if you use a different definition the proof will be slightly different.) Then . If n is an integer, then is also an integer, so is odd by definition. But this contradicts the fact that is even, so the assumption that a was odd must have been false. —Keenan Pepper 03:58, 15 February 2008 (UTC)[reply]

I changed the question , can you take another look at it ? Thanks--George (talk) 04:01, 15 February 2008 (UTC)[reply]

Um, it seems like you just removed the phrase "for any integer a" from the hypothesis. This makes the statement false, because as far as I know even and odd apply only to integers. If , then is the even integer 2, but a itself is neither even nor odd because it is not an integer. Am I missing something? —Keenan Pepper 04:11, 15 February 2008 (UTC)[reply]

I don't know what's wrong, but it doesn't seem right. I'll think about it again, maybe I am missing something.--George (talk) 04:32, 15 February 2008 (UTC)[reply]

As User:Keenan Pepper writes, we assume you must be talking about integers. And note that any integer is either of the form 2b for some b or else of the form 2c+1 for some c, call them type-0 and type-1 integers, respectively, for what gets added in either case to the multiple of 2. Any type-1 integer 2c+1 will have square equal to 4c2+4c+1. But if we define d to be 2c2+2c, then we can see that the type-1 integer has a square of the form 2d+1, so the square, too, is type-1. So, because all type-1 integers have type-1 squares, no type-1 integers have type-0 squares. Thus if a2 is type-0—and therefore not type-1—neither can a itself be type-1; the only possible type for a is thus 0. (P.S. - feel free to replace type-0 with even and type-1 with odd. Furthermore, the proof made no mention at all of )—PaulTanenbaum (talk) 04:34, 15 February 2008 (UTC)[reply]

Urgent integration question!

This is for school. I need to find the area bounded by the graphs of f(y) = -y2 + 2y and g(y) = -y. I equated them and got the intersection points (0,0) and (3,-3). I'm supposed to find the area by integrating in terms of y, so based on what the graphs look like, I guess it should be ∫-30(-y2 + 2y - [-y]) = ∫-30(-y2 + 3y), which would then give me
(-y3/3 + 3y2/2)]-30. The upper limit yields only zero values, so this should be equal to -(-(-3)3/3 + 3(-3)2/2) = -(9 + 27/2) = -22.5. Aside from the disconcerting fact that this "area" value is negative (though I know that can happen), it seems far too large to be reasonable. What am I doing wrong? My utmost thanks for any help, anon.

The problem is with the integration limits. Maybe you got confused because the independent variable is called "y" here? Given your intersection points (y,f(y))=(0,0) and (3,-3), think again about what range of y you should integrate over. --mglg(talk) 06:16, 15 February 2008 (UTC)[reply]
Also, make sure you are subtracting the small function from the large one - if you do the opposite you will get a negative result. -- Meni Rosenfeld (talk) 09:15, 15 February 2008 (UTC)[reply]
It helps to draw a rough graph with these kinds of questions. Then you can see how to use the intersection points and what to integrate. --Tango (talk) 13:43, 15 February 2008 (UTC)[reply]

Number of solutions for a system of linear equations over finite fields.

How can we find out the number of solutions for a system of linear equations over finite fields. Are there any results giving the number of solutions in terms of the number of unknowns, "rank", etc.? 210.212.228.94 (talk) 05:58, 15 February 2008 (UTC)[reply]

I think Gaussian elimination still works. Black Carrot (talk) 07:56, 15 February 2008 (UTC)[reply]
Yeah, according to the article Gaussian elimination can be performed over any field. Black Carrot (talk) 08:16, 15 February 2008 (UTC)[reply]
Yes; but what about the number of solutions? Is there any result giving any bounds for it? ( in terms of the number of variables, rank, q and n if our field is GF(qn). 210.212.228.94 (talk) 10:23, 15 February 2008 (UTC)[reply]
Exactly the same results hold as for infinite fields (that's the beauty of linear algebra). If we have m equations in n unknowns, then this is the same as a matrix equation Ax=b where A is an m x n matrix, of rank r say and nullity n-r. If b is outside the image of A then there are no solutions, otherwise there is an (n-r)-dimensional space of them. If the ground field has q elements, then the solution space has qn-r. Algebraist 12:22, 15 February 2008 (UTC)[reply]

Sobolev Spaces

Okay, so in my graduate PDE class we are working with the Sobolev Space denoted by which consists of all square integrable functions (over the interval (0,1)) such that their first derivatives are also square integrable over (0,1). Since this is our first introduction to Sobolev Spaces, our teacher gave us a few functions to determine if they are in or not. My question is that, is it just enough to show that the first derivative of the function is square integrable or do we have to show that the function itself is also square integrable? Because I think that there is some theorem in analysis which says that if a function's derivative is square integrable, then the function itself is also square integrable. If not, then does anyone have any example of a function (could be multi-variable) whose derivative is square integrable but the function itself is not?A Real Kaiser (talk) 08:22, 15 February 2008 (UTC)[reply]

How about ? 134.173.92.17 (talk) 08:45, 15 February 2008 (UTC)[reply]
Erm, you probably wanted a bounded domain. Scratch that. 134.173.92.17 (talk) 08:55, 15 February 2008 (UTC)[reply]
You want what is called the Poincare inequality: , where is the average value of f on [0,1], and C is some constant independent of f.
Basically this says that for functions in a sobolev space on a bounded domain (finite volume), the sobolev norm up to the first derivative is equivalent to the norm of the first derivative. In an unbounded domain this is not true, as the previous example 1/sqrt(x) on [1,oo) shows. JackSchmidt (talk) 18:21, 15 February 2008 (UTC)[reply]
In case you actually needed a proof. If g is a continuous function on [0,1], then m(t)=g(0)+t*(g(1)-g(0)) has the property that f=g-m is square integrable if and only if g is, f'=g'-(g(1)-g(0)) is square integrable if and only if g' is, and f(0)=f(1)=0. Then where the first inequality is the fundamental theorem of calculus, and the second is cauchy-schwarz applied to f'*1 (you can use Hoelder inequality in general), so after squaring both sides and replacing the factors on the right with their maxes. Integrating over x=0..1 gives the result:
Note that one used that the volume of the domain was 1 really quite a few times, so for a more general domain there will be some constants. For the complete proof you might need some density arguments to say that it is ok to only have considered C^1 functions, and you need to handle switching from f to g if you want an explicit inequality. On wikipedia, I think this version is called the Friedrichs' inequality, but it and several others were called Poincare in our analysis class. JackSchmidt (talk) 18:59, 15 February 2008 (UTC)[reply]

Wow, thanks! This is exactly what I was looking for.A Real Kaiser (talk) 06:18, 16 February 2008 (UTC)[reply]

relation between A.M, Median and Mode

in acadamic books it is given that the relation between arithmatic mean, median and mode is mode = 3MEDIAN - 2 A.M Please explain this and explain any limitations for it because for some uni-model group of values this formula is not working out. F0r example

0, 0, 0, 2, 2, 3, 18

its mean is 12.5, meadian is 2, mode is 0 which do not satisfy the above relation: Please explain in detailKasiraoj (talk) 08:22, 15 February 2008 (UTC)[reply]

It is definitely not true in general; and I have never heard of it. What book is this and what exactly does it say? --Spoon! (talk) 09:06, 15 February 2008 (UTC)[reply]
You should have read the links provided to you earlier (and followed up there instead of posting a new question). This relation is just an empirical rule of thumb that has been found to approximately hold in some distributions commonly encountered in practice - as I understand, in particular those that resemble the normal distribution (and yet have some asymmetry). -- Meni Rosenfeld (talk) 09:12, 15 February 2008 (UTC)[reply]

Arithmetic mean = Sum of all items divided by number of items
Median = The number with the middle value
Mode = The number that appears most in the list
(I think you know that^)
The book should give the limitations for the equation. If not, the book should be wrong. Visit me at Ftbhrygvn (Talk|Contribs|Log|Userboxes) 03:51, 16 February 2008 (UTC)[reply]

Percentage Calculator

I'm looking for a calculator online that can perform this simple function: I have two (or more) values. I want the calculator to add up the values and tell me what the percentage of each value is compared to the total value. Basically, I'm trying to calculate vote totals from election websites that list vote amounts but not the percentages, and I know there's got to be an easier way than using a calculator and adding up the values and dividing manually. —Preceding unsigned comment added by Rc251 (talkcontribs)

You could try using a spreadsheet (Excel or similar). --Tango (talk) 16:01, 15 February 2008 (UTC)[reply]
This online spreadsheet is easy to learn and to use. --hydnjo talk 04:10, 16 February 2008 (UTC)[reply]

Or use google and just type in the calculation. For example '50/95*100' in google will return a 'google calculator' answer (http://www.google.co.uk/search?hl=en&q=50%2F95*100&btnG=Search&meta=) not a search result. ny156uk (talk) 10:30, 16 February 2008 (UTC)[reply]

which number is higher?

hi there, which number or fractions is higher? 3.5%(7/2), 3.4%(17/5), 4.3%(43/10), 3.1%(31/10), 1.8%(9/5), or 5.4%(27/5)? —Preceding unsigned comment added by Don Mustafa (talkcontribs) 15:41, 15 February 2008 (UTC)[reply]

You've already got them in decimal form, so you can easily which is the bigger number. Start by looking at the whole part (ie. the bit before the decimal point), and see which is larger. If there is one with a whole part larger than all the others, that number is larger. If there are two (or more) the same, then you look at the fractional parts (ie. the bits after the decimal points) of them and see which of them is larger, and then that number is the larger (if they have the same number of digits after the point, it's just like with whole numbers, if they have different numbers of digits, add 0's to the end of the shorter to make them the same length, and then it's just like whole numbers). Alternatively, you can write all the numbers in a column making sure to line up the decimal points above each other (and make each digit the same size so they all line up - squared paper might help), then start with the column of digits furthest left, and see which is larger (if a number doesn't have a digit in that column, you know it's smaller), if two have the same digit, compare the next digits until you find ones that are different, and the number with the large digit is the larger number. Hope that helps. --Tango (talk) 15:59, 15 February 2008 (UTC)[reply]
You got to be kidding me right?. Which number is larger? 3.5 , 3.4 , 4.3 , 3.1 , 1.8 or 5.4 . This seems to be such an easy question. 122.107.129.141 (talk) 23:05, 15 February 2008 (UTC)[reply]
Hi. Please don't make any assumptions about the OP's (what does OP mean anyway) age or mathematical abilities. Thanks. ~AH1(TCU) 02:58, 16 February 2008 (UTC)[reply]
It means Original Poster, and it's necessary to make assumptions to even provide an answer. Black Carrot (talk) 04:19, 16 February 2008 (UTC)[reply]

I got the numbers from [[1]] [[2]] and [[3]]. Please, have a look the deep red colored ones and calculate the percentages into numbers. —Preceding unsigned comment added by Don Mustafa (talkcontribs) 01:46, 18 February 2008 (UTC)[reply]

Please tell us what it is exactly that you want to find out. I understand you are trying to do some comparison involving minority populations in Toronto, but the questions you have asked so far are pretty much meaningless. -- Meni Rosenfeld (talk) 10:10, 18 February 2008 (UTC)[reply]

Look, All I want is you to look at these sites and look at the deep red ones and they have the numbers in decimals and at the bottom of the screen, bottom-left, they have the legends of the colours with numbers and the numbers that are high are deep red. All I want you to do is convert the decimals into the actual numbers like the ones at the bottom of the screen. —Preceding unsigned comment added by Don Mustafa (talkcontribs) 16:52, 18 February 2008 (UTC)[reply]

Can you not do this yourself? Oh, and please remember to sign your posts. -mattbuck (Talk) 16:56, 18 February 2008 (UTC)[reply]
It looks like they are different data: the colours code the actual number of persons, while the percentage (3.3, 0.5 etc.) give the same figures as percentage of population. So by combining the two data you might find an approximation of the population of the counties/districts/quarter(?), but probably this are data available elsewhere. If you are only interested in the data about the minority there is no computation to be done. Just notice that in a populous county there might be more Arabs even if they constitute a lesser percentage, and viceversa. Goochelaar (talk) 17:37, 18 February 2008 (UTC)[reply]

which number is higher? 2

Which number is higher? 23.5%, 26.9% or 20.1%? —Preceding unsigned comment added by Don Mustafa (talkcontribs) 15:49, 15 February 2008 (UTC)[reply]

Ignore the percent sign and just treat them as you would any other numbers (see above). --Tango (talk) 15:59, 15 February 2008 (UTC)[reply]
Is this a joke? Surely someone who can type with capitals and know that "2" is bigger than "1" can read numbers... --PalaceGuard008 (Talk) 02:50, 16 February 2008 (UTC)[reply]


February 16

Coordinate system

Hello friends. I have a cartesian coordinate system, where four points are given: (0|0), (0|1), (1|0), (1|1). You have now to move these points so that at the end these points are at (0|0), (1|3)(1|1), (3|0), (2|-1). But there's one rule given: One point can only be moved, if there is another point between its old place and its new place. Now my question: Is it possible to get this latter layout from the first one? --85.178.63.105 (talk) 10:30, 16 February 2008 (UTC)[reply]

Sounds similar to Hi-Q, but without removals. Here's a solution. I don't guarantee it's minimal. It was easy enough (working backwards from the goal to the starting position) that I wonder if there is some other constraint you forgot to mention. Each of my moves jumps over a point that is exactly halfway to the destination, which you didn't even ask for, so call that a bonus.
(0,1) -> (2,-1) -- the (1,0) peg is in the middle of the jump.
(1,1) -> (1,-1) -- the (1,0) peg is in the middle.
(1,0) -> (1,-2) -- the (1,-1) peg is in the middle.
(1,-2) -> (3,0) -- the (2,-1) peg is in the middle.
(1,-1) -> (5,1) -- the (3,0) peg is in the middle.
(5,1) -> (-1,-3) -- the (2,-1) peg is in the middle.
(-1,-3) -> (1,3) -- the (0,0) peg is in the middle.
--tcsetattr (talk / contribs) 10:57, 16 February 2008 (UTC)[reply]
Oh, I'm so much sorry I do not mean (1|3) but (1|1) at the end. Can you write the solution again for this way please? --85.178.63.105 (talk) 11:10, 16 February 2008 (UTC)[reply]
In that case, assuming the middle peg has to be halfway between the starting and ending positions, it's not hard to show that there's no solution. I'll withhold the details for the moment since this looks like the kind of problem that might be assigned as math homework. Try starting with the ending position and seeing where you can go from there to get a feel for what's going on. -- BenRG (talk) 19:09, 16 February 2008 (UTC)[reply]
No, this is not a homework or something like that. My friend told me if I want to puzzle over something and gave me this exercise for the weekend. Now I don't want to tell him that I failed to do this [and I want to get a beer for free]. So please can you give me please the solution? --85.178.32.123 (talk) 23:22, 16 February 2008 (UTC)[reply]
You want me to help you defraud your friend out of a beer? I'll give you a hint. Take a checkerboard with red and white squares and put some tokens on red squares. Now move the tokens around according to the rules of this game. You will never be able to move a token onto a white square (do you see why?). This means if your friend had told you to go from, say, (0|1), (0|-1), (1|0), (-1|0) to (0|0), (0|1), (1|0), (1|1), you would know it was impossible, because the former positions are all on the same color and the latter positions aren't. It's also impossible to do it in the other direction, because all of the moves in this game are reversible. That coloring argument doesn't work for the problem he actually gave you, but a slightly different coloring does work. To figure out that coloring, start with the pieces in the ending position of your friend's problem and try moving them around until you can see which squares are reachable and which aren't. -- BenRG (talk) 00:22, 17 February 2008 (UTC)[reply]
Something I'm not clear on: is this confined to the integer lattice? If not, it can still be solved. Black Carrot (talk) 02:37, 18 February 2008 (UTC)[reply]
You're right, it was never completely clear what the rules were. My comments above only apply if the rules say that you can only move a piece from (a|b) to (c|d) if there's a piece at (½(a+c) | ½(b+d)). In that case, of course, the pieces are confined to the lattice even if the rules don't explicitly say so. If you're allowed to move whenever there's a piece on the open segment (a|b)–(c|d), then the problem seems easily solvable even on the lattice: (1|0) → (−1|0) → (3|0) then (0|1) → (4|1) → (2|−1). -- BenRG (talk) 11:25, 18 February 2008 (UTC)[reply]


February 17

Functions with different arguments

Is there a standard notation to distinguish different arguments/formulations of a given function that is usually expressed as a single argument? For example:

If you are given , it could be either or ! Would you incorporate the constant as a coargument——e.g., or , or something else?
In the case where the function, itself, is an argument of another function, expressed either singularly or as a bivariable, it would seem perfectly proper to express it likewise:

 ~Kaimbridge~00:44, 17 February 2008 (UTC)[reply]

I really can't understand what this question is asking. If , then . That's the substitution property of equality. Introducing a system of notation that breaks this rule is an awful idea. If you want to have one definition and to have a different definition, then you're really talking about two different functions, and you need to give them two different names. —Keenan Pepper 01:19, 17 February 2008 (UTC)[reply]
I think Kaimbridge is working with some already awful notation commonly used in some subject, and trying to modify it to make more sense. If so, it would help to see a reference where this notation is used. -- Meni Rosenfeld (talk) 15:18, 17 February 2008 (UTC)[reply]
Using functions like that is an abuse of notation. It's quite common in science where the difference between a physical value and the function for calculating that value is rather blurred (hence the common shorthand for "position is determined by time" of "r=r(t)", the first r is a physical value, the second r is a function). The idea is that it's obvious from context what you mean, if it isn't, you need a better notation. In formal mathematics, you should give different functions different names, so you should have a and a or something. Two functions are equal if and only if they give the same result for any given input, which your two forms of clearly don't, so they aren't the same function, so they shouldn't have the same name. (You can get away with using the same name if you can distinguish the functions by number of variables, since that removes the ambiguity, but it's often best to distinguish them more clearly.) --Tango (talk) 15:51, 17 February 2008 (UTC)[reply]

Okay, let me give you the actual scenerio.
For Earth, is geographic latitude and is the reduced latitude. At the same time, is the spherical/geographical azimuth and is the transverse geographical colatitude (likewise, and are the elliptical equivalent, based on ). My particular question had to do with derivative. With respect to latitude relationship,

But in terms of to latitude conversion,

While the functions equate, the derivatives obviously don't! In the particular situation I am dealing with, the elliptical values are localized, with respect to :

So, .

The ultimate use and purpose for all of this is this equation:

or


Thus is azimuth in nature while is elliptical. Is it just the author's responsibility to effectively distinguish the two via some indexing——e.g., or ——or is there some established special notation?  ~Kaimbridge~16:22, 17 February 2008 (UTC)[reply]

Empty set

One of my fellow ask me a simple question about Empty Set.As we call set that thing if there are some thing in it,eg people,number etc but an empty set is call a "SET" becoz there are nothing in it according to text base definition and i am not mathematician so plz simplefy me. —Preceding unsigned comment added by Usmanzia1 (talkcontribs) 02:26, 17 February 2008 (UTC)[reply]

Yes, an empty set is a set with nothing in it. It is a collection of objects, which makes it a set - just that this collection of objects is empty. x42bn6 Talk Mess 13:55, 17 February 2008 (UTC)[reply]
Try reading our article, Empty set, it should help. It's a rather strange concept to get used to, but it's common throughout maths - the lack of something is an example of a something (The empty product equals 1, the empty sum equals 0, the empty union is the empty set, the empty intersection is the universe, etc. etc.) --Tango (talk) 14:37, 17 February 2008 (UTC)[reply]

and

What is the difference between and ? Thanks, Zrs 12 (talk) 03:21, 17 February 2008 (UTC)[reply]

When they are separate, there is no difference. However, if they are used in the same equation, such as "", it often denotes that if you use + in the first instance, you match it with - in the second. The previous example would generally be used to indicate two possibilities: "" and "". 134.173.92.17 (talk) 03:53, 17 February 2008 (UTC)[reply]
Also see Plus-minus sign#Minus-plus sign. --hydnjo talk 03:58, 17 February 2008 (UTC)[reply]
Thanks, Zrs 12 (talk) 05:17, 17 February 2008 (UTC)[reply]
We want to find the probability that the system is in a given state after a given number of time steps. The set of probabilities for each state after k time steps is given by the probability vector pk. The purpose of the formula is that it gives an expression for the probability vector after k time steps in terms of the initial state vector v and the stochastic matric P - so if we know v and P we can find the probability vector at any subsequent time. The "mathematical induction" part just means that we can derive the general formula for pk by looking at the formulae for p1, p2 etc. and then generalising the pattern that we see to k time steps. Can you see where the formulae that I give above for p1, p2 come from ? Can you see how they lead to a general formula for pk ? Gandalf61 (talk) 09:35, 12 February 2008 (UTC)[reply]

I am assuming the formulae that you gave above for p1, p2 came from Summation? If this is true then can the formula be put in Sigma notation format  ? --Obsolete.fax (talk) 05:28, 17 February 2008 (UTC)[reply]

I can't see any connection with summation at all. Gandalf61 (talk) 15:02, 21 February 2008 (UTC)[reply]
Then could you answer "Can you see where the formulae that I give above for p1, p2 come from ? Can you see how they lead to a general formula for pk?" I don't know. Please help, would really appreciate. --Obsolete.fax (talk) 18:23, 23 February 2008 (UTC)[reply]
Then go back and carefully read the explanations that Lambian gave here and that I gave here. They show how the general formula for pk is derived. If there is a step that you don't understand, then say which step you get stuck on. It is not possible to help you any further unless you say which part of the explanation you don't understand. Gandalf61 (talk) 11:09, 24 February 2008 (UTC)[reply]

Indefinite integral of the Gamma function

Is there any way to integrate (doesn't have to be closed form) Gamma(x)?

--wj32 t/c 10:34, 17 February 2008 (UTC)[reply]

Well, gamma is continuous, so yes, it has an antiderivative. The Integrator gives nothing, so I suspect the antiderivative can't be expressed in terms of functions anyone's bothered naming, but I could be wrong, and don't know anything like enough differential Galois theory to prove such a result. Algebraist 14:50, 17 February 2008 (UTC)[reply]

Factorials

How do you prove, or whats the mathematical proof of O!=1? I've asked this to several math teachers and they always answer someting like: "You just take it for granted", "It's just a rule", etc... —Preceding unsigned comment added by 201.167.101.193 (talk) 21:58, 17 February 2008 (UTC)[reply]

See Factorial#Definition. Basically, it's because it turns out that whenever you come across 0! in some setting, you want it to be equal to 1. For example, n! is the number of ways you can arrange n different objects in a line. How many ways can you arrange zero objects in a line? Well, one—there is one way to arrange zero objects, but that's it. —Bkell (talk) 22:12, 17 February 2008 (UTC)[reply]
You can arrange 0 objects? Without any objects, how can you do this? I'm confused. Zrs 12 (talk) 23:38, 18 February 2008 (UTC)[reply]
Okay, here I go. Are you watching? There, I did it. I arranged zero objects on my desk just now. I didn't have to do anything but stare at my desk, but I certainly did arrange zero objects on my desk. There they are, all zero of them, all in their correct places. (Note that I didn't have any choices to make while arranging these zero objects, so there was only one way to do it.) It sounds very strange to talk about arranging zero objects, but things like this are useful concepts in mathematics.
Sometimes similar statements are made about the empty set. For example, it is a true statement that "All elements of the empty set are even numbers"—if that weren't true, then you should be able to show me an element of the empty set which is not an even number in order to disprove the statement, but you can't. You can't show me any elements of the empty set at all, because it doesn't have any elements in the first place. Another way to say this is that the statement "All elements of the empty set are even numbers" is true because there are no exceptions. (Of course, it is also a true statement that "All elements of the empty set are odd numbers," or "All elements of the empty set are invisible pink unicorns.") —Bkell (talk) 14:43, 19 February 2008 (UTC)[reply]
Yeah, I get the concept. The wording is what doesn't work for me. By your logic, I would have to prove that they can't be arranged and you would have to prove that they could. Since neither can be proven, they are both correct? Anyway, I was just trying to bring attention to maybe getting a better wording. Zrs 12 (talk) 15:19, 19 February 2008 (UTC)[reply]
Well, if we're going to try to prove something, then we need to have rigorous definitions instead of vague terms like "arrange". If the rigorous definition of "arrange" is given in terms of permutations of the elements of a set, as is common, then it can be shown that there is exactly one way to arrange zero objects, because there is exactly one permutation of the elements of the empty set. This is true whether a permutation is defined as a sequence of elements listing each element exactly once (there is exactly one empty sequence) or as a bijection from a set to itself (there is exactly one function whose domain and codomain are the empty set). —Bkell (talk) 15:46, 19 February 2008 (UTC)[reply]
It's not something that can be proved, it's simply part of the definition of the function. -mattbuck (Talk) 22:50, 17 February 2008 (UTC)[reply]
Sure, but the reason it's defined that way is because it's the "right" definition to make. Defining 0! to be anything but 1 would break a lot of elegant formulas and identities. —Bkell (talk) 23:16, 17 February 2008 (UTC)[reply]
A better question than "why does 0! equal 1" is "why does 0! exist in the first place", and for that matter, "why does factorial exist in the first place?" It exists, by that I mean it is given its own name and used all over the place, because it is a convenient way of arranging information. 0! is included because in many applications the input winds up as zero in some special cases, and those special cases might as well be dealt with officially. It equals 1 because that's what you'd put as the output in each of those individual cases anyway. (Contrast this with division by zero, where the most convenient answer varies and no single definition would do the job.) At each step, the point is to summarize things as efficiently as possible. To do that, you introduce a structure, and make refinements of it. Only once you have a structure to describe can you start proving things about it. Black Carrot (talk) 23:26, 17 February 2008 (UTC)[reply]
When you're working with natural numbers, it's best to think of the factorial function as counting permutations. That is, n! is the number of distinct ways of arranging n different objects in a line. If you have zero objects, there's only one way to arrange them (don't do anything), so 0! = 1. More formally, if n! is the number of bijections from an n-element set to itself, then 0! = 1 because there is precisely one bijection from the empty set to itself. Michael Slone (talk) 23:39, 17 February 2008 (UTC)[reply]
Another way to think of it is by observing that n!=(n+1)!/(n+1) for all n>1, so extend that for n=0, and you get 0!=1.--Tango (talk) 23:45, 17 February 2008 (UTC)[reply]
And there's the fact that the Gamma function, a beautiful extension of the factorial, has a value of 1 at 1 (Gamma(t+1)=t! for t non-negative integer). Pallida  Mors 03:10, 18 February 2008 (UTC)[reply]

February 18

Cubic function

If you go here and enter as the equation ax3 + bx2 + cx + d = 0 and tell it to solve for x, you get a long expression as the solution. Can someone show algebraically how you can solve ax3 + bx2 + cx + d = 0 for x? Or link to a page that does? Thanks. 70.162.25.53 (talk) 03:39, 18 February 2008 (UTC)[reply]

See cubic function, specifically the Cardano's method section. Be warned, it's not quite as nice as the quadratic formula... 134.173.92.17 (talk) —Preceding comment was added at 03:59, 18 February 2008 (UTC)[reply]
(ec) The Wikipedia article on cubic functions may be of your interest, or at least this subsection of it. Outside Wikipedia, this page from MathWorld is another interesting choice. Pallida  Mors 04:06, 18 February 2008 (UTC)[reply]
I intended a reference to the subsection Root finding in that article, but the link doesn't seem to work fine. Nevermind. Pallida  Mors 04:19, 18 February 2008 (UTC)[reply]

Logic

Does anyone know where I can find a whole bunch of "Knights and Knaves" problems, with answer/explanations/solutions? Thanks. (Joseph A. Spadaro (talk) 07:51, 18 February 2008 (UTC))[reply]

Here's a bunch of problems with a twist. These problems explore the consequences of adding a third type of person, the philosophers. For more traditional questions, a Google search turns up a few pages. —Bkell (talk) 12:40, 18 February 2008 (UTC)[reply]
Also, check out the books of Raymond Smullyan. He has tons of logic puzzles that all can enjoy. –King Bee (τγ) 13:48, 18 February 2008 (UTC)[reply]
I especially recommend "Alice in Puzzleland" by Raymond Smullyan. 129.67.157.6 (talk) —Preceding comment was added at 15:15, 18 February 2008 (UTC)[reply]

Hi. Thanks to all for the input. I guess I should have made my question a little clearer. So, let me re-phrase it. I am looking for Knights and Knaves problems ... that are accompanied by answers / solutions / explanations. Right now, I only want Knight/Knave questions ... and not just logic puzzles in general. Also, I am only looking for the "simple" type (knights and knaves only), where they do not introduce third-parties (spies, randoms, philosophers, etc.) into the equation. Also, I'd like to find this on the Internet --- as I don't have too much time to go out, order a book, wait for its delivery, etc. At some later point, I will probably go out and get some Smullyan books ... but right now, I don't have that time / luxury. Any suggestions? Any good websites out there? I can't seem to find any, with my searches. Many thanks! (Joseph A. Spadaro (talk) 16:33, 18 February 2008 (UTC))[reply]

The Wikipedia article on Knights and Knaves has a few problems with explanations and solutions. There are some pages with individual knights-and-knaves problems (from one site, [4], [5], [6], [7], [8], [9], [10], [11]). —Bkell (talk) 17:46, 18 February 2008 (UTC)[reply]
Smullyan's book entitled "What is the name of this book?" has a lot of Knights/Knaves puzzles complete with solutions. His books should be available at your local library, so you needn't purchase a thing. –King Bee (τγ) 19:21, 18 February 2008 (UTC)[reply]

help with aproof

i need help with how to prove that a number (a) is a zero divisor mod (n) iff gcd(a,n)>1?thank youHusseinshimaljasimdini (talk) 10:52, 18 February 2008 (UTC)[reply]

Think about n/gcd(a,n) - let's call this b. First of all, you can be sure that b is an integer - why ? Secondly, you can be sure that if gcd(a,n) > 1 then b mod(n) is not 0 - why ? Now think about the product ab - what is ab mod(n) ? How does this help you show that a is a zero divisor mod(n) ? Gandalf61 (talk) 12:09, 18 February 2008 (UTC)[reply]

Collecting Terms

i get how to do everything else but i seem to bomb out on collecting the terms forgive me if it is wrong
Simplify:
now im not sure but im pretty sure this is wrong
—Preceding unsigned comment added by 124.176.206.9 (talkcontribs) 12:28, 18 February 2008 (UTC)[reply]

You have one square root of 3, minus two square roots of 2, plus 2 square roots of 3, plus one square root of 2: so in the end how many square roots of 3 do you have, and how many square roots of 2? —Bkell (talk) 12:31, 18 February 2008 (UTC)[reply]
If it helps you, forget for a moment what means, and just think of it as a symbol, like a circle, perhaps. The same thing for : think of that as a triangle. So you have one circle, minus two triangles, plus two circles, plus one triangle. How many circles do you have, and how many triangles? Just be sure to write "" in your answer instead of a triangle, of course. —Bkell (talk) 12:37, 18 February 2008 (UTC)[reply]

so  ? do you just take the sign infront? so that 12a - 10b + 9c + 5b
=12 - 5b + 9c?

Think about this, this is just two applications of the Distributive property. If it's not clear why that works, just try going the other way. (Note that algebraic laws are reversible!)A math-wiki (talk) 12:56, 18 February 2008 (UTC)[reply]
You're right: the simplified answer is . Usually −1 times something is just written −something (for example, −1 × 5 = −5), and when "+−" occurs in a mathematical formula it's usually just written "−" [for example, 3 + (−2) = 3 − 2]. So you can write in a simpler way. —Bkell (talk) 13:56, 18 February 2008 (UTC)[reply]

geometry - spheres and unit cubes

seeing as there's quite a few mathematicians in here i thought i'd ask about a geometry problem i've been having. i think it's quite easy but i don't really have a clue.

imagine a unit cube. now imagine a point within this cube (call it p). i want to place a sphere with its centre at p. how would i work out the radius needed for this sphere so that the intersection volume between it and the cube equals a given value (call it v)? AlmostCrimes (talk) 14:21, 18 February 2008 (UTC)[reply]

If the sphere just intersected one side you could find the volume of the Spherical cap cut off, then subtract this from the volume of the sphere. If the sphere intersects more than one side then things get a little trickier. --Salix alba (talk) 15:07, 18 February 2008 (UTC)[reply]
[ec]It's actually not that simple. To do that you will first need some way to calculate the volume of intersection of a cube and a ball, which is pretty messy. Fortunately this latter problem is one I have worked on a while back, so I have thought of a few approaches. Since none of them is perfect, I can help you more only if you provide more details about your application (with special attention to details such as relative and absolute error tolerance, computational resources, platform on which you will do the calculations, etc.). If the question is purely out of curiosity I suggest you just leave it, though I can describe the general ideas. -- Meni Rosenfeld (talk) 15:10, 18 February 2008 (UTC)[reply]
salix alba - Yes sadly, it is quite possible the sphere intersects more than one side, for example if the point was at one of the corners of the cube. the basis of the problem is that I'm investigating various approaches to deciding whether a given colour in an RGB colour space (here represented as a cube like the following RGB#Representations) is "similar" to another colour. the natural approach would be to calculate the distance between the two colours (or points) and to then determine whether this is below a certain threshold. This technique suffers in that not all colours are treated evenly - colours in the centre of the colour space (so at [0.5, 0.5, 0.5], or grey) would be flagged as being similar to a higher percentage of the total colour space than a colour like white would. I was hoping to even this out by dynamically modifying the threshold (analogous to the radius) so that every colour has an equal proportion of the colour space flagged as being similar to it. As i say, this is mostly investigative and it isn't important that I use this exact technique. it's very possible i'm completely overcomplicating things.
for what it's worth though error tolerance isn't too important, neither are computational resources. I'm programming it in MATLAB and basically just seeing what works well. the most important factor is simplicity and implementation time really as I don't want to get too hung up on one technique. If any of your solutions apply I'd be interested in hearing them. I thought briefly about representing both the sphere and cube in a boolean matrix and physically determining the best sphere radius but I was kind of hoping there'd be a simple algebraic method. --80.4.203.142 (talk) 19:44, 18 February 2008 (UTC)[reply]
Well, I can see several problems with this approach. If you're going for a notion of similarity which matches that of a human observer, I don't think using a simple Euclidean metric as a basis will cut it. I don't know a lot about this, but at the very least you should give different weights to the components - Grayscale mentions weights used for a different purpose, but they might be appropriate here. This would then leave you dealing with ellipsoids which is even messier. Second, I'm not sure any sort of "Affirmative action" is necessary - white will have less colors labeled as similar to it, simply because there are less colors genuinely similar to it. You next have symmetry to consider - You mentioned a ball centered around one point, but when doing a comparison there are two points to consider, and the result will depend on which one you choose to calculate the threshold. If you do some symmetric calculation taking both into account, it will be difficult to guarantee that indeed every color has the same portion of the color space labeled similar to it.
As said, solving the problem in you original question is not impossible, but not simple to do with satisfying accuracy and efficiency. Since I gather that these are not crucial, I can offer a simple approximation method for the volume of intersection of the cube and a given ball. Just pick several (something like 100 or 1000) randomly chosen points in the cube, and check whether each is inside the ball. The proportion which are is an estimation for the volume of intersection. You can do this for a ball centered at one point and with radius equal to the distance to the other point, compare it to the volume you want to be labeled similar and make the decision accordingly.
I hope this helps. -- Meni Rosenfeld (talk) 22:25, 18 February 2008 (UTC)[reply]
it's true there are problems with the general approach and while I am aware there are colour spaces that lend themselves more easily to this kind of work I have been asked to look at this RGB cube method specifically. As for the "affirmative action", it seems to me it is necessary. A general distance approach is far better at finding areas that look grey than areas that look white when given the same threshold. Increasing the threshold for white increases the perceived accuracy, which is why I wanted to try this method. Your symmetry point is interesting. I actually hadn't considered that...
Because of the nature of the exercise it isn't important that every technique I try is successful but regardless you've brought me round to not trying the technique, which I believe won't be worth the effort. It helped to have this discussion even if the focus of it changed a little, for which I thank you. --80.4.203.142 (talk) 23:44, 18 February 2008 (UTC)[reply]

Math Riddle

While studying for my math test, I came across this problem:

I am thinking of some polynomial with positive integer coefficients. You can ask me for the value of p[x], where p is the polynomial and x is some positive integer as many times as you want. What is the minimum number of questions you need to ask in order to find out what the polynomial is?

I am unable to figure out the answer. Since it doesnt tell you how many terms or the order of the polynomial, it's hard to find a starting spot. Can anyone help? 99.240.177.206 (talk) 18:37, 18 February 2008 (UTC)[reply]

I remember that one. The answer is 2. Does that help? Black Carrot (talk) 18:45, 18 February 2008 (UTC)[reply]
It helps me, at least. Nice problem. Algebraist 19:45, 18 February 2008 (UTC)[reply]
If I understand Black Carrot correctly, the value you choose for x in your second question depends upon the answer you receive to your first question - yes ? Gandalf61 (talk) 20:06, 18 February 2008 (UTC)[reply]
It would have to, polynomials aren't completely determined by their values at two points. There must be some clever trick based on the answer to the first questions... I haven't worked it out yet, but give me some time... --Tango (talk) 20:08, 18 February 2008 (UTC)[reply]
You don't just have 2 points, you have 2 points plus the knowledge that the coefficients are positive integers. I just got it! Oh that is clever. --tcsetattr (talk / contribs) 20:40, 18 February 2008 (UTC)[reply]
...which was the point I was trying subtly (perhaps too subtly) to make. Answer to first question gives you information about coefficients which you use in second question. Gandalf61 (talk) 22:42, 18 February 2008 (UTC)[reply]
Haha, wow, that's pretty cool. —Bkell (talk) 21:23, 18 February 2008 (UTC)[reply]

Hm..suppose if one chose p(0) and the answer turned out to be 4 or 5. Obviously, this would indicate the the constant at the end is either a 4 or 5. Where would you go from there?Acceptable (talk) 20:39, 18 February 2008 (UTC)[reply]

Asking for p(0) is a bad start (aside from the fact that it wasn't technically allowed by the problem statement). --tcsetattr (talk / contribs) 20:55, 18 February 2008 (UTC)[reply]
Funnily enough, 0 (if you take it to be positive) is the only first question that doesn't work. Suppose you got f(0)=a (say), and used n for your second question. If you got f(n)=n^2+a, then f could be nx+a or x^2+a. Algebraist 21:50, 18 February 2008 (UTC)[reply]

Finally, I think I've got it! That shouldn't have taken me that long... --Tango (talk) 22:29, 18 February 2008 (UTC)[reply]

Just in case the OP needs another hint - how would you solve this if you also knew that all coefficients are less than 10? -- Meni Rosenfeld (talk) 22:32, 18 February 2008 (UTC)[reply]
It finally clicked when I twigged that it was only positive coefficients. -mattbuck (Talk) 23:02, 18 February 2008 (UTC)[reply]

I think the smallest number of questions you need to ask is (the degree of the polynomial + 1).--George (talk) 22:43, 18 February 2008 (UTC)[reply]

I wonder if there's a finite bound for more general polynomials, like with positive rational coefficients. Black Carrot (talk) 22:47, 18 February 2008 (UTC)[reply]
To George - that'd be right for polynomials with no restrictions on the coefficients, but you can take advantage of their positive integerosity to jump straight to the answer. Black Carrot (talk) 22:50, 18 February 2008 (UTC)[reply]
Do you have any specific technique for doing this or is this just a general statement ?--George (talk) 22:56, 18 February 2008 (UTC)[reply]
To the OP's riddle? Yeah, I've got the answer. Give it a little while, and reflect on Meni's hint. BTW, for a polynomial in n variables, with positive integer variables and coefficients, it could be done within n+1 questions. Black Carrot (talk) 23:00, 18 February 2008 (UTC)[reply]
I got it ,thanks. --George (talk) 23:53, 18 February 2008 (UTC)[reply]
I think 2 questions ought to work for real non-negative coefficients - that they're integers I wouldn't think was strictly necessary. -mattbuck (Talk) 23:02, 18 February 2008 (UTC)[reply]
Really? It doesn't seem like that would be enough. That is, for any two inputs you choose, it seems like there would be at least one pair of polynomials with the same output. Black Carrot (talk) 23:06, 18 February 2008 (UTC)[reply]
If you choose any two points, there are an infinite number of polynomials that go through them. But, restricted to non-negative coefficients, if you choose the 2nd point based on the first one, it ought to be possible to find the order. -mattbuck (Talk) 23:12, 18 February 2008 (UTC)[reply]
Thinking about Black Carrot's extended question, I can see that if you only know that all (non-zero) coefficients are real and positive, then with two questions you can find the maximum degree of the polynomial - say it is m (so at most m+1 non-zero coefficients). But don't you still need a further m-1 questions to find the values of the coefficients (so that altogether you have m+1 simultaneous equations) ? Key difference from the positive integer case is that there is no longer a lower bound on the magnitude of a non-zero coefficient. Gandalf61 (talk) 00:00, 19 February 2008 (UTC)[reply]
Generally with polynomials, rational coefficients and integer coefficients work in exactly the same way (you just multiply through by the lowest common multiple of all the denominators), so I imagine the case of positive rationals is doable (you may end up needing to additional fact that the polynomial is monic in order to get a unique answer, I'd have to try). The case of positive reals probably isn't, though. The key fact that p(1)<10 (or whatever) gives you is that 10 doesn't divide any of the coefficients, in the real case that fact is no longer true (it's not true for rationals, either, but you can probably fudge it). I haven't actually tried either case, this is just intuition and could well turn out to be wrong. --Tango (talk) 12:50, 19 February 2008 (UTC)[reply]
But how do you determine the lowest common multiple (or even any common multiple) of the denomoinators ? I think the positive rational coefficients case is the same as the positive real coefficients case - it only takes a finite number of questions, you don't know in advance how many, but after two questions you know how many further questions you need. Gandalf61 (talk) 15:08, 19 February 2008 (UTC)[reply]
That comment was just to demonstrate why polynomials over the integers and polynomials over the rationals are closely related. Obviously, you can't directly convert in this case, but there might be some trick that allows you to use the same basic method. I haven't looked for one, so I don't know. --Tango (talk) 16:44, 19 February 2008 (UTC)[reply]
For those still struggling, I suggest using p(1) as the first question. It's not necessary, but thinking about what I could learn from p(1) was what led me to the solution. And don't forget the coefficients are all positive! --tcsetattr (talk / contribs) 23:21, 18 February 2008 (UTC)[reply]
Sorry, I still am not getting it. The question does not state how many terms there are. So suppose I realize that none of the coefficients are greater than 10, I still dont know anything about how many terms or how large the coeff of each term is.

Let's take an example: suppose the equation was f(x)= 4x^3 + 3x^2 + 7x + 6. If I plug in p(1), i get p(1)=20. What do I ask next? Thanks. 99.240.177.206 (talk) 04:29, 19 February 2008 (UTC)[reply]

OK, that's enough. Full spoiler: suppose you had chosen that function and I asked you for p(1) and you said p(1)=20. I know the polynomial is A+Bx+Cx^2+Dx^3+... so what do I get when x=1? A+B+C+D+... in other words the sum of the coefficients. I don't know how many coefficients there are, but I know they are all positive integers and their sum is 20. If a set of positive numbers adds up to 20, then none of them individually can be greater than 20. The next question I'd ask is for p(100). I could ask for p(21) instead - any number greater than 20 will do - but I'm choosing 100 because it might make the procedure more obvious.
Your answer will be p(100) = 4(100^3) + 3(100^2) + 7(100) + 6 = 4030706. See how the coefficients 4,3,7,6 revealed themselves? If I had asked for p(21) you would have said 38520 which would make the procedure a little more mysterious-looking, but the secret is I would then convert 38520 into base 21, and the result would be: 437621!
Now we can go back to the first step and see how p(1) doesn't have to be used. If I asked for p(2) first, you'd say 64. In this version, 64 is not the sum of the coefficients, but it's still an upper bound on them individually. If any one of your coefficients was greater than 64, it would contribute a term of (something greater than 64)*(some power of 2) to your answer, so your answer would have been greater than 64 also. Knowing this upper bound, I could ask for p(65), get the answer 1111636, and convert it to base 65, where it is: 4376. --tcsetattr (talk / contribs) 04:44, 19 February 2008 (UTC)[reply]
And of course, my point was that if all coefficients are less than 10, you don't need two questions - you just ask what is . -- Meni Rosenfeld (talk) 09:49, 19 February 2008 (UTC)[reply]

This is nice. Ask for and the problem is solved in one question for all the cases where the coefficients are less than a million. You do not have to require that the integer coefficients are positive, just that they are not negative. What if the integer coefficients may be negative too, how do you solve that? Bo Jacoby (talk) 15:33, 19 February 2008 (UTC).[reply]

If the coefficients can be arbitrary integers, then infinitely many questions are required (consider the case when every question gets the answer '0'). If they're arbitrary positive reals, then finitely many questions suffice (as Gandalf pointed out above) but I'm not sure if you can get a fixed finite bound or not. Algebraist 16:25, 19 February 2008 (UTC)[reply]
You can't have infinitely many questions all giving 0, since the polynomial will always be of finite degree (otherwise it's not a polynomial). --Tango (talk) 16:44, 19 February 2008 (UTC)[reply]
Well, I suppose there's the 0 polynomial... it would take an infinite number of questions to spot that. Any non-0 polynomial would be worked out with finite (but unbounded) steps, though, I'd think. --Tango (talk) 16:52, 19 February 2008 (UTC)[reply]
No. However (finitely) many times I give you the answer zero, you won't know if it's the zero polynomial or just some other poly that happens to have roots at every point you've asked about. Algebraist 18:35, 19 February 2008 (UTC)[reply]
That's not quite what I said. I said that any non-zero polynomial could be worked out in finitely many steps, not that finitely many steps is enough to tell if a polynomial is non-zero. There are only finitely many roots to any polynomial (at most, it's degree), so after one more question than that (which is still a finite number), you'll know it's non-zero. Whether you can work out exactly what it is or not, I'm not sure - repeated roots would be an issue, but you would know it's non-zero. --Tango (talk) 20:39, 19 February 2008 (UTC)[reply]
That (while true) is not much use to solve the problem as stated. Sure, for any polynomial there is a question-asking strategy that proves the poly is non-zero in finitely many steps, but what we want is a strategy that for any polynomial will prove it is non-zero in finitely many steps. We can't base our strategy on an answer we don't know. Algebraist 22:34, 19 February 2008 (UTC)[reply]
How about asking for p(0), p(1), p(-1), p(2), p(-2), etc. until you get a non-zero answer? If the polynomial is non-zero, you will get a non-zero answer after a finite number of questions. If it is the 0-polynomial, then you'll never get an answer, but you're asking for a proof that it's non-zero, not a proof that it is zero. --Tango (talk) 12:08, 20 February 2008 (UTC)[reply]

Set of convergence of a sequence of measurable functions

This is an exercise from Lang's Real and Functional Analysis, Chapter VI: Let be a sequence of measurable functions. Show that the set of points for which converges is a measurable set.

Presumably this involves finding some collection of obviously measurable sets that can be unioned/intersected together to end up with the set in question, but I haven't been able to find such a collection. The only information available is the basic properties of σ-algebras and measurable functions.

[As a side issue, does anyone have experience with or thoughts on this book? I can't say I find it easy going, but struggling with it seems to be good practice and the more time I spend with it the more I appreciate it. I got it originally for the development of integration, which is the best I've seen anywhere--just beautiful! If only there were a solutions manual...  — merge 23:33, 18 February 2008 (UTC)][reply]

One way of doing this is to first show that supn{fn} is measurable. Dually the inf of countably many measurable functions is measurable, so the lim sup and the lim inf of the fn are both measurable. Then the set of x for which fn(x) converges is the set of points at which two measurable functions are equal, and is thus measurable. Slight care might be needed in dealing with infinity. Algebraist 23:51, 18 February 2008 (UTC)[reply]
Thanks! This idea had not occurred to me. However as it turns out the properties of lim sup and lim inf you refer to are dealt with in the next exercise! So I'm thinking there's a way to do this one that doesn't use them. Also, I forgot to mention that the take values in a Banach space, so one can't take the sup directly.  — merge 12:52, 19 February 2008 (UTC)[reply]
In that case, you could use the fact that fn(x) converges exactly if fn(x) is Cauchy. The latter statement is just
So the set of x for which fn(x) converges is just
where is the open ball of radius epsilon about 0 (I'm assuming this is measurable) and all the intersections and unions are countable (in particular we can restrict to positive rational epsilon). Algebraist 16:22, 19 February 2008 (UTC)[reply]
Wonderful. Someone else suggested the same idea in conversation and it works perfectly. I think this is "the" solution. Thank you!  — merge 21:23, 19 February 2008 (UTC)[reply]

Googol

Why is one googol not ? The Wikipedia article states it as being . Why is this? Thanks, Zrs 12 (talk) 23:50, 18 February 2008 (UTC)[reply]

One googol is by definition 10100, as correctly stated in the article. The article also correctly states that 70! is approximately 10100.0784. I don't see any source of confusion here. Algebraist 23:55, 18 February 2008 (UTC)[reply]
Perhaps OP did not understand Order of magnitude, and just skipped over those words to read "One googol is the same as 70!." --PalaceGuard008 (Talk) 23:57, 18 February 2008 (UTC)[reply]

Haha! How dumb of me! Gah, I must start reading more carefully. Zrs 12 (talk) 00:38, 19 February 2008 (UTC)[reply]

February 19

Singular Value Decomposition & Hermitian Matrices

This is a question posed in my class and the teacher himself could not figure it out either. The question is that given the singular value decomposition of a mxm square matrix where the * represents the complex conjugate transpose of a matrix, is there anything that we can say about the eigenvalue decomposition (diagonalization) of the 2mx2m Hermitian matrix ? Can B's eigenvalue decomposition be written in terms of and  ? I have also tried a few numerical examples on Matlab and it appears to me that the two are completely unrelated. Any help would be appreciated.A Real Kaiser (talk) 06:10, 19 February 2008 (UTC)[reply]

If , then . As is self-adjoint, we get as well. Now compute:
Moving the unitaries (one can check that those block matrices on the left and right of your are unitary) to the right gives the singular value decomposition. You have the absolute values of the eigenvalues at this point, but I'm not sure exactly what you mean by eigenvalue decomposition. Do you mean diagonalizing B? J Elliot (talk) 07:13, 19 February 2008 (UTC)[reply]
The eigenvalue decomposition goes similarly. For motivation, find the eigenvalue decomposition of [0,1;1,0] (taking A to be the 1x1 identity matrix). A=USV*, so A*=VSU*, AV=US, A*U=VS, so [0,A*;A,0][V;U] = [VS;US], so [V;U] is an "eigenvector" with eigenvalues S, and similarly [0,A*;A,0][V;-U]=[-VS;US], so [V;-U] is an "eigenvector" with eigenvalues -S. Since our matrix is hermitian, we want our eigenbasis to be unitary, so we'll divide the eigenvectors by their overall norm, 1/sqrt(2). Putting it all together gives:
So the eigenvalues of B are plus or minus the singular values of A. JackSchmidt (talk) 15:53, 19 February 2008 (UTC)[reply]

Thanks guys, that makes a lot more sense. But I still have a follow-up question. You have shown that [V;U] and [-V;U] are eigenvectors with respect to S and -S but how do we know that those are the only eigenvalues? What if S^2 of -2S^5 also turn out to be eigenvalues of our matrix B? How can we conclude that S and -S are the ONLY eigenvalues?A Real Kaiser (talk) 23:53, 19 February 2008 (UTC)[reply]

Sorry, I spoke too informally. U and V are actually m x m matrices, and S is a diagonal matrix with m values. [V,V;U,-U] has full rank (because it is unitary), so is actually 2m independent eigenvectors for B, and S,-S gives the 2m eigenvalues. The informal language was just to indicate how block matrices can simplify things. JackSchmidt (talk) 00:58, 20 February 2008 (UTC)[reply]

fitting a conic

I want to fit a general parabola (of unknown size and orientation) roughly to a given set of points in the plane. My first thought was to seek the coefficients that minimize the sum of the squares of ; to make the problem more linear, I then sought to settle for a general conic, ; but then it occurs to me that this penalizes those curves that go near the origin.

My next idea is to consider the family of cones tangent to some plane ; I'm not sure what to minimize.

Anyone know a better way? —Tamfang (talk) 06:24, 19 February 2008 (UTC)[reply]

Minimize the sum of squares of using some other normalizing condition than (which excludes curves through the origin). Bo Jacoby (talk) 07:50, 19 February 2008 (UTC).[reply]
You could try an iterative approach. Define for brevity
and
Given estimates for the coefficients A, B, etcetera, you can determine the distance ei of each point (xi,yi) to the curve determined by . If we give weight wi to the term Fi2 in a weighted sum of squares, we want the weighted square to come out like ei2, which suggests setting
as the weights for a next iteration.
Instead of determining the values ei exactly, which is computationally expensive, you can approximate it by using the linear approximation
Applied to the point (xi,yi), we write this as
The least value for for which the right-hand side can vanish, which provides an estimate for ei2, is then given by
So for the weights for the next iteration, you can use then
Since the value being inverted can become arbitrarily small and even vanish, you must exercise caution not to make this numerically unstable, and put limits on the size of the weights.  --Lambiam 08:49, 21 February 2008 (UTC)[reply]

Rationalising Surds

I was going over some of my notes and i tried this one but my answer isnt the same as in the book.. im not sure what im doing wrong
Rationalise the denominator



Kingpomba (talk) 11:26, 19 February 2008 (UTC)[reply]

Looks fine to me and quick calculator check confirms your answer. Could also be written as or . What does your book say ? Gandalf61 (talk) 11:47, 19 February 2008 (UTC)[reply]

it says: (hmm i got a different answer on paper and i understand how the 1/2 thing works i guess typing it out on wikipedia helped me solve it, well cheers anyway =] Kingpomba (talk) 11:57, 19 February 2008 (UTC).[reply]

Are you sure that's what it says? There should be a minus sign in front, shouldn't there? --Tango (talk) 12:54, 19 February 2008 (UTC)[reply]
Here's a quick sanity check: 5 < 7, so sqrt(5) < sqrt(7) (by the properties of the square root), and hence sqrt(5) - sqrt(7) < 0. Thus the denominator of the original fraction is negative, meaning the whole fraction is negative. Confusing Manifestation(Say hi!) 23:47, 19 February 2008 (UTC)[reply]

February 20

zeros of a periodic function

Is there any way to find an which solves the equation

for any arbitrary and ? I know that some value of x will make both and equal to an odd multiple of but can't think of a way to find this value. I know that the answer may lie in recasting the equation in the form

and in this case know that some value of x makes this product of cosine terms equal to negative one, but again can't think of how to find this x value. Can this be solved analytically or must I resort to numerical calculation (which would be very unsatisfying). Thank you very much for your assistance. —Preceding unsigned comment added by 128.223.131.21 (talk) 00:21, 20 February 2008 (UTC)[reply]

If the combined function is periodic, then a/b is a rational number, so after rescaling x, you can assume a and b are relatively prime integers. Then I believe x=pi is a solution if both a,b are odd, and there is no solution if one of a or b is even. JackSchmidt (talk) 01:16, 20 February 2008 (UTC)[reply]
To explain JackSchmidt's answer a bit: since (for ) , the only way to have is to have . So , so and the simplest form of must have both numbers odd. When that condition is satisfied, there is some smallest integer ; k is the common denominator for a and b. Any with integer l solves the equation. --Tardis (talk) 16:20, 20 February 2008 (UTC)[reply]

Two questions about the above-named 1959 Walt Disney film. (Question 1) In the film, a character incorrectly states the value of pi. Does anyone know how a Disney film that was designed to be a mathematics educational film for children (and presumably had math consultants working on it) could possibly contain such an error? How would that error slip by unnoticed? (Question 2) With regard to that error, someone stated (in a blog) that it was not really an error, since "the value of pi has changed so much since 1959". That cannot possibly be true, can it? I am sure that we have become enlightened about the accurate values of many digits of pi over the course of thousands of years of history ... but not since 1959, correct? Any ideas? Thanks. (Joseph A. Spadaro (talk) 04:10, 20 February 2008 (UTC))[reply]

Pi is an irrational number meaning that its' decimal representation will have and infinite amount of digits. Given the fact that we can never calculate an infinite amount of digits it is the case that new digits have been calculated since 1959. Of course this does not mean the the value of pi has changed, just the precision with which one can write it. Perhaps this is what 'someone' was reffering too? Any errors in the calculation of digits of pi since 1959 would, i suspect, occur at least several thousand (being conservative here) digits out so it would in fact be an error. I have no input on how the error actually occured. —Preceding unsigned comment added by 74.242.224.136 (talk) 04:55, 20 February 2008 (UTC)[reply]
The error could very well have been caused by something as simple as the voice actor getting lost while reading the digits. —Bkell (talk) 05:00, 20 February 2008 (UTC)[reply]

Thanks. Let me follow-up. I should have made this clearer. (Though, remember, we are dealing with a Disney children's video.) First ... the value of pi was recited to, perhaps, 15 or 20 decimal places ... certainly, no more than 20. So, since 1959, nothing has "changed" in that early chunk of the number - correct? Second ... regarding the voice actor flub. Certainly, that is a possibility. I guess my point was ... is not that an "inexcusable" error (that would call for a retake ... or, at the very least, a dubbed edit in post-production), given that it is an education video, and a math one to boot ( ... as opposed to, say, some character misstating the digits of pi on an inconsequential TV episode of Seinfeld) ...? Thanks. (Joseph A. Spadaro (talk) 05:16, 20 February 2008 (UTC))[reply]

Pi was known to 527 digits by the end of the 19th century, and to 2037 digits by 1949 (a decade earlier than the film was made). Our article on Donald in Mathmagic Land says that the bird only recites "the first few digits" of pi. The error was not due to pi not being known accurately enough in 1959. MrRedact (talk) 05:25, 20 February 2008 (UTC)[reply]
Probably the producers didn't consider it important. If the voice actor flubbed, either no one noticed or no one thought it was worth it to rerecord the line just to be pedantic. The character reciting pi is basically a throw-away gag anyway. They got the first few digits right (out to maybe eight or ten places, if I remember correctly), and that's certainly more than anyone really needs. The purpose of the film is not to teach children the digits of pi but to introduce them to some general mathematical concepts. —Bkell (talk) 06:12, 20 February 2008 (UTC)[reply]

Yes, I agree with all that is said above. However, as an education video --- I am not sure that being "pedantic" is a bad thing ... quite the contrary, I'd assume. I will also assume that Disney's multi-zillion dollar fortune could easily afford any "expensive" retakes. Finally, mathematicians are nothing, if not precise. I can't imagine a mathematician letting that glaring error slide. A film producer, yes ... a mathematician, nah. (Joseph A. Spadaro (talk) 06:56, 20 February 2008 (UTC))[reply]

I agree completely. This is just a symptom of a much more general problem. Very few people care about spreading misinformation, and very many then wonder how come people are so misinformed. -- Meni Rosenfeld (talk) 10:45, 20 February 2008 (UTC)[reply]
What a succinct -- yet woefully accurate -- synopsis, Meni. Bravo! (Joseph A. Spadaro (talk) 16:20, 20 February 2008 (UTC))[reply]
As Bertrand Russell appears to have written: 'PEDANT—A man who likes his statements to be true.' Algebraist 10:55, 21 February 2008 (UTC)[reply]

Predicate logic

I have to show, using a transformational proof (logical equivalence and rules of inference) that the following two premises are a contradiction:


1.
2.


Now these two statements look quite similar because they were actually two more complex statements I tried to "whittle down" into a similar form in order to see where I could take it. They almost contradict one another apart from the part in 1. versus the part in 2.

Can you offer me a suggestion for manipulating these statements further to show a clear contradiction? My first line of thought would be to show that the conjunction, , of the two statements is false but I get bogged down in symbols. Perhaps there is a more elegant way using inference rules, I'm not sure. Damien Karras (talk) 13:39, 20 February 2008 (UTC)[reply]

Start by consulting De Morgan's laws, including the ones for quantifiers, if you're not familiar with them. Then you should be able to make use of the fact that . Also, it may not be strictly necessary, but consider what it means to say that . --Tardis (talk) 16:34, 20 February 2008 (UTC)[reply]
The rwo premises are contradictory if (and only if) (2) implies the negation of (1). You will have to use somehow, directly or indirectly, a rule that says that implies . If you have a logic in which the domain of predicate P can be empty, so that this implication does not hold, while the domain of the predicates Q and R, from which variable x assumes its values, is not necessarily empty, the two premises are not contradictory: take for Q the constant always-true predicate and for R the always-false predicate; the premises then simplify to and . I don't know the precise rules of the you are allowed to use, but the logical connectives and , as well as existential quantification, are all "monotonic" with respect to implication, so that
is a consequence of
 --Lambiam 09:58, 21 February 2008 (UTC)[reply]
Thanks for the response both, I will look into what you have said. In the meantime let it be known that P, Q and R share the same domain. Damien Karras (talk) 13:55, 21 February 2008 (UTC)[reply]

There are no countably infinite σ-algebras

Hoping someone can help me spot the holes in my argument.  ;)

Suppose is a countably infinite σ-algebra over the set X. If we can show that contains an infinite sequence of sets or , then the sequence of differences or will be a countably infinite collection of pairwise disjoint sets , and the map will be an injection from the power set of N into .

Let be partially ordered under inclusion. We may assume that contains no infinite ascending chain , for if it does this gives us what we wanted.

The set has a maximal element F1 by Zorn's lemma. It follows that the only elements of are subsets of F1 and their complements; in particular this set contains infinitely many subsets of F1. The collection of these subsets cannot contain an infinite ascending chain, so has a maximal element F2. It follows that the only elements of are subsets of F2 and their complements relative to F1. Proceeding in this way yields an infinite descending chain .  — merge 16:36, 20 February 2008 (UTC)[reply]

I think you have confused maximal element with greatest element. -- Meni Rosenfeld (talk) 19:46, 20 February 2008 (UTC)[reply]

Meni's right -- your last "it follows that" is wrong. The result is right, though. Hint: You don't need Zorn's lemma; the hypothesis that the sigma-algebra is countably infinite already gives you a bijection between the sigma-algebra and the natural numbers. Do induction along that. --Trovatore (talk) 20:08, 20 February 2008 (UTC)[reply]

I had confused myself, although not about that--sorry! I'll definitely try what you said, but I've also tried to fix the above--any better?  — merge 21:08, 20 February 2008 (UTC)[reply]

Revised version at σ-algebra redux below.  — merge 13:26, 21 February 2008 (UTC)[reply]

Summing 1^1 + 2^2 +3^3 ..... + 1000^1000

I'm trying to improve my mathematics skills by doing the problems at Project Euler, but have hit a stumbling block with problem 48: (http://projecteuler.net/index.php?section=problems&id=48

I need to find the last 10 digits of 1^1 + 2^2 +3^3 ..... + 1000^1000

I've read the wikipedia page on Geometric_progression , but I want to do rather than (i.e. I don't have a constant value r)

I'm not so much looking for an answer, just what techniques or terminology I should be reading up on. Pointers welcomed!

Lordelph (talk) 16:56, 20 February 2008 (UTC)[reply]

The key point is that you're looking for the last 10 digits, not the whole number (which is enormous). Think about which numbers contribute what to the last ten digits. --Tango (talk) 17:14, 20 February 2008 (UTC)[reply]
I don't think there is a mathematical shortcut for this. But Project Euler is as much about programming as it is about mathematics, so you could take a brute force approach. To calculate the last 10 digits of 999^999, for example, start with 999; multiply it by 999; just keep the last 10 digits; repeat another 997 times and you are done. You only need to work with interim results of up to 13 digits, so its not even a bignum problem. I guess you end up doing around half a million integer multiplications altogether, but I don't think that will take very long. In the time it's taken me to write this, someone has probably already programmed it. Gandalf61 (talk) 17:27, 20 February 2008 (UTC)[reply]
Actually even the ultimate brute force approach of computing the whole 3001-digit sum will be more than fast enough on a modern computer. My 2004 laptop took a negligible fraction of a second to evaluate sum [n^n | n <- [1..1000]] in GHCi. They should've gone up to 1000000^1000000. -- BenRG (talk) 19:24, 20 February 2008 (UTC)[reply]
I've seen similar problems where there has been a shortcut, I can't spot one for this problem, though. It's a trivial programming exercise, though, so I can't see why they would have set it. I can see a few possible ways to optimise the code, but there's really no need (for example, 999^999=(990+9)^999, expand that out using the binomial expansion and each term will have a 990^n in it, which has all 0's for the last ten digits for all but the first 10 terms, so you can ignore all but those terms - probably not much quicker for 999^999 (you need to calculate the coefficients, which won't help), but could be for 999999^999999). --Tango (talk) 19:52, 20 February 2008 (UTC)[reply]
Many thanks for pointers, problem solved Lordelph (talk) 20:37, 20 February 2008 (UTC)[reply]
On a slow computational platform, this can be somewhat sped up as follows. The recursive algorithms for exponentiation by squaring works in any ring, including the ring Z/nZ of modulo arithmetic. So in computing nn, all computations can be done modulo 1010. And if n is a multiple of 10, nn reduces to 0 modulo 1010, so 10% of the terms can be skipped. —Preceding unsigned comment added by Lambiam (talkcontribs) 11:13, 21 February 2008 (UTC)[reply]

Resistance

These are more questions of physics than maths but considering the overlap between the two, and how this ref desk is much more specific than the science one, I'll ask them here.

I know that the formula for the overall resistance of two resistors in parallel is . So if you have two fifty ohm resistors in parallel, the total resistance is 25 ohms. What I don't understand is how the total resistance can be less than that of either resistor; it doesn't make any sense to me.

Also, if you have the below setup but R1 is just a wire, how would you calculate the total resistance of the circuit, ignoring internal resistance?

A diagram of three resistors, two in parallel, which are in series with the other

Many thanks, 92.1.70.98 (talk) 18:07, 20 February 2008 (UTC)[reply]

It's actually very simple intuitively. Resistance measures how hard it is for electrons to move from one place to another. If there are two resistors in parallel, the electrons have 2 ways to choose from, thus travel is easier. The analogy is a mass of people trying to cross a narrow corridor - they will have a much easier time if there are two parallel corridors.
Any resistor network can be solved using Ohm's law () and Kirchhoff's current law (the sum of currents at every point is 0). This one is simpler because it's just two resistors in a series, where the second is composed of two resistors in parallel. Thus the equivalent resistance is .
There are many Wikipedia articles from which you can learn more about this - Resistor might be a good start. -- Meni Rosenfeld (talk) 18:35, 20 February 2008 (UTC)[reply]
Well, Meni beat me to it. But here it is. First of all, the total resistance decreases when you are talking about two resistors in PARALLEL not in series. In series, you get exactly what you think that you should get. The total resistance is just the sum. In parallel, you can think of it like this, you have two branches, so the current going in splits up in the branch so the energy loss in each branch is smaller than what it should have been. But then the current adds up again when the branches meet. In series, the entire current has to go through both of the resistors, so the loss is greater. As for the picture, first you calculate the combined resistance of R1 ad R2 as resistors in parallel (call it R12) which gives you and then you simply treat R3 as a resistor in series with the combined resistor R12 and you get that the total resistance of the circuit is .A Real Kaiser (talk) 18:39, 20 February 2008 (UTC)[reply]
OK the explanation for why the resistance is less in parallel is helpful but I think you both missed the bit about R1 being just a wire, not a resistor (it was the best picture I could find lying around on WIkipedia). Using the equation for resistors in parallel, it would make the resistance of the parallel section 0 (assuming that a wire has no resistance). I would interpret this as meaning that all current goes through the wire and none goes through the resistor. Is that right? Thanks 92.1.70.98 (talk) 18:45, 20 February 2008 (UTC)[reply]
That would be an idealised situation. Nothing really has 0 resistance, so R1 would be very small, but it would be non-zero. This would mean that the total resistance over R1 and R2 was almost 0, since R2's contribution would be negligible. -mattbuck (Talk) 19:24, 20 February 2008 (UTC)[reply]
Still, in the limit where , there will be no current going through R2 (this has nothing to do with interpretation, it's just a calculation). Assuming we want the current to go through R2, the extra wire is a short circuit. -- Meni Rosenfeld (talk) 19:32, 20 February 2008 (UTC)[reply]

Random calculus question

A friend of mine asked me to help him solve the following question, and I'm not a math major by any means. So, any help would be appreciated. :) Zidel333 (talk) 19:40, 20 February 2008 (UTC)[reply]

  • [1+1/N] ^ (N+1/2), prove that it is always increasing from zero to infinity.
It will help to try to prove something that is true: that function is always decreasing for . Its limit at large N is e, as can be seen immediately from that article. The normal procedure is to evaluate the function's derivative and show that it is negative for all positive N. However, the algebra involved is non-trivial; perhaps some information is missing? --Tardis (talk) 20:07, 20 February 2008 (UTC)[reply]
I'm sorry, that's all the info I have from my friend. If he tells me more, I'll add it to the discussion. Thanks for your help so far, far better than what I could have come up with. :) Zidel333 (talk) 20:12, 20 February 2008 (UTC)[reply]
As it's "N" not "x", I would assume N is ranging only over the integers, so taking the derivative isn't necessary. You can either show that f(n+1)-f(n)<0 or f(n+1)/f(n)<1 for all n. The latter is probably easier. --Tango (talk) 22:22, 20 February 2008 (UTC)[reply]
Hint: Perhaps rewriting the sequence as exp[(N+1/2)*log(1+1/N)] can help you (sorry, I'm not comfortable with LaTeX). --Taraborn (talk) 10:33, 21 February 2008 (UTC)[reply]

Sum

How would you work out, for j and k ranging over the integers, the sum of (j+k)^-2? Black Carrot (talk) 20:36, 20 February 2008 (UTC)[reply]

Well, the term for j=k=0 is going to be a problem :-) --Trovatore (talk) 20:50, 20 February 2008 (UTC)[reply]
But assuming you really meant the positive integers, the sum diverges. For a fixed j, the sum over all k>0 of (j+k)−2 is Ω(1/j). --Trovatore (talk) 20:53, 20 February 2008 (UTC)[reply]
(Edit conflict) Assuming you mean positive integers, since all integers doesn't make sense, as Trovatore points out, observe that for a fixed j, and varying k, j+k varies over all the integers, so
Now, the inner sum is clearly greater than 0, and summing an infinite number of non-zero constants gives you infinity (or, strictly speaking "doesn't converge"). So, unless I've missed something, the series doesn't converge. --Tango (talk) 21:00, 20 February 2008 (UTC)[reply]
Correction: For fixed j, varying k, j+k varies over all integers greater than j, I was thinking of k varying over all integers, rather than just positive ones... it doesn't change the conclusion, though - the series still diverges. --Tango (talk) 22:19, 20 February 2008 (UTC)[reply]
The statement that "summing an infinite number of non-zero constants gives you infinity" is very often not true, even assuming you're limiting yourself to positive constants. For example, . See the Taylor series for ex or tan x as a couple other examples. MrRedact (talk) 01:50, 21 February 2008 (UTC)[reply]
None of those examples are constants. --Tango (talk) 12:08, 21 February 2008 (UTC)[reply]

Integral Question

Hi I've got an integral question which i have the answer for but i don't understand why it works.
The question is: If f(x) is continuous on [0; ¼], show that:
I've got it into this form:
The next step i have written is:
But i don't understand how to get to that step. It doesnt help that my lecturer makes the odd mistake now and again. Can somebody explain how to get between these steps and how to complete the rest of the question. Thanks for any help. 212.140.139.225 (talk) 23:00, 20 February 2008 (UTC)[reply]

In the one-but-last formula, write f(x) instead of just f. I think the last formula should be something like
which is justified by the first mean value theorem for integration.  --Lambiam 11:39, 21 February 2008 (UTC)[reply]

February 21

Quick question about log convexity

I've read the article on power series, and it mentions that for more than one variable, the domain of convergence is a log-convex set instead of an interval.

But what is a log-convex set here ? I mean, how do we take the log of a set ? Surely the set can contain negative values for example...

Thanks. -- Xedi (talk) 02:13, 21 February 2008 (UTC)[reply]

Sorry, I have no idea what they could mean by log-convexity of a set. The multi-variable stuff was added in June 2004, and the log-convex comment was made in Oct 2004. It does not appear to have been touched since then. I'll ask the author about it.
The domain of convergence is a union of polydisks, but I think perhaps it need not be a polydisk. Some reasonably elementary notes are at Adam Coffman's website, specifically his research notes on Notes on series in several variables. It includes the polydisk result, as well as the continuity, and differentiability of functions defined by a series, and includes the nice "recentering" result familiar for one complex variable allowing analytic continuation. JackSchmidt (talk) 04:25, 21 February 2008 (UTC)[reply]

Natural Logs

I'm being asked to evaluate natural logs and I'm getting very confused. One problem I'm having particular trouble with is lne(-x/3)

Could anyone please help explain to me how to solve problems like these and how to determine how to get started?

Thanks, anon. —Preceding unsigned comment added by 66.76.125.76 (talk) 03:53, 21 February 2008 (UTC)[reply]

What is lne(-x/3) ?
Is it Log[ Exp[-x/3] ] ?
or is it Log[-x/3] where Log is Log based e.
202.168.50.40 (talk) 04:34, 21 February 2008 (UTC)[reply]
ln is usually a shorthand for loge --PalaceGuard008 (Talk) 05:07, 21 February 2008 (UTC)[reply]
  • y = log(-x/3)
  • ey = -x/3
  • x = -3ey

--wj32 t/c 05:16, 21 February 2008 (UTC)[reply]

If the question is about , what can you say in general about ?  --Lambiam 12:05, 21 February 2008 (UTC)[reply]

is math the same as logic?

Is math the same as logic, or why not.

In other words, is there math that isn't logical, but just evaluates truths, for example "experimenting" with numbers to find out "facts" but not using logic to show that these must be necessarily true. After all, the other sciences don't rely on their findings to be NECESSARILLY true, just that they happen to be true... so which is math? —Preceding unsigned comment added by 79.122.42.134 (talk) 10:12, 21 February 2008 (UTC)[reply]

You might be interested in logicism and experimental mathematics. Algebraist 10:43, 21 February 2008 (UTC)[reply]
You might also be interested in David Hilbert's project to formalize mathematics.--droptone (talk) 12:45, 21 February 2008 (UTC)[reply]

Moment

Suppose you have a ladder leaning against a wall, which makes an angle θ with the ground, and a man, who can be treated as a particle, stands a distance l up the ladder. Now his weight will obviously be acting downwards and so working to work out the moment produced, you would have to resolve his weight vector to find the component of it that acts perpendicular to the ladder.

My question is, if you used Pythagoras or trig to determine his perpendicular distance from the ladder's point of contact with the ground to the line of action of the weight vector and then multiplied that by his weight, would that give you the correct value of the moment? Cheers in advance 92.3.49.42 (talk) 12:45, 21 February 2008 (UTC)[reply]

Unless I've misunderstood, yes. What I tend to do is extend the "arrow" of the force such that the perpendicular goes through the point you are trying to resolve against, then it's a simple case of force times distance from point. x42bn6 Talk Mess 13:02, 21 February 2008 (UTC)[reply]

σ-algebra redux

A revised version of There are no countably infinite σ-algebras, hopefully closer to being correct?

Lemma. If is an infinite algebra over a set X, partially ordered by inclusion, then contains an infinite ascending chain or an infinite descending chain .

Proof: Let . If has a chain with no upper bound, then that gives us what we wanted. Otherwise has a maximal element b1 by Zorn's lemma. If and then so by maximality, and therefore . That means there are infinitely many subsets of b1 in .

Let be that collection of subsets, minus b1 itself. Then has a maximal element b2, and so again contains infinitely many subsets of b2. Inductively we obtain an infinite descending chain .

Corollary. Every infinite algebra contains an infinite collection of pairwise disjoint sets.

We just take the sequence of differences of elements of the chain.

Corollary. There are no countably infinite σ-algebras.

An infinite σ-algebra is an infinite algebra, and by the above has an infinite subcollection of pairwise disjoint sets. The map is then an injection from the power set of N into the σ-algebra.

 — merge 13:58, 21 February 2008 (UTC)[reply]