Wikipedia:Reference desk/Archives/Mathematics/2008 February 15

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is an archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

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)

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/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)

In TeX you typeset it as $x^{y^{z}}$, 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)

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)

With pen and paper you don't write it with ^. You write it as $x^{y^{z}}$, or if you wanted the other meaning $x^{yz}$. Taemyr (talk) 11:23, 19 February 2008 (UTC)

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 $x^{yz}$ equivalent to one of the 2 possible values of x^y^z. Myles325a (talk) 23:44, 19 February 2008 (UTC)

Proof by contradiction

In the article about mathematical proofs, there is a proof that ${\sqrt {2}}$ 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 ${\sqrt {2}}$ is rational ? --George (talk) 02:51, 15 February 2008 (UTC)

Sure, that's easy to prove. Assume a is odd. Then it is of the form $2n+1$, 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 $a^{2}=(2n+1)^{2}=4n^{2}+4n+1=2(2n^{2}+2n)+1$. If n is an integer, then $k=2n^{2}+2n$ is also an integer, so $a^{2}=2k+1$ is odd by definition. But this contradicts the fact that $a^{2}$ is even, so the assumption that a was odd must have been false. —Keenan Pepper 03:58, 15 February 2008 (UTC)

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

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 $a={\sqrt {2}}$, then $a^{2}$ 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)

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)

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 ${\sqrt {2}}$)—PaulTanenbaum (talk) 04:34, 15 February 2008 (UTC)

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)

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)

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)

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)

I think Gaussian elimination still works. Black Carrot (talk) 07:56, 15 February 2008 (UTC)

Yeah, according to the article Gaussian elimination can be performed over any field. Black Carrot (talk) 08:16, 15 February 2008 (UTC)

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)

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 q^n-r. Algebraist 12:22, 15 February 2008 (UTC)

Sobolev Spaces

Okay, so in my graduate PDE class we are working with the Sobolev Space denoted by $H^{1}(0,1)$ 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 $H^{1}$ 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)

How about ${\frac {1}{\sqrt {x}}}$? 134.173.92.17 (talk) 08:45, 15 February 2008 (UTC)

Erm, you probably wanted a bounded domain. Scratch that. 134.173.92.17 (talk) 08:55, 15 February 2008 (UTC)

You want what is called the Poincare inequality: $\int _{0}^{1}|f(t)-a|^{2}~dt\leq C\int _{0}^{1}|f'(t)|^{2}~dt$, where $a$ 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)

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 $|f(x)|\leq \int _{0}^{x}|f'(t)|\leq \left(\int _{0}^{x}|f'(t)|^{2}~dt\right)^{\tfrac {1}{2}}\cdot {\sqrt {x}}$ 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 $|f(x)|^{2}\leq \int _{0}^{1}|f(t)|^{2}~dt$ after squaring both sides and replacing the factors on the right with their maxes. Integrating over x=0..1 gives the result:

$\int _{0}^{1}|f(x)|^{2}~dx\leq \int _{0}^{1}|f'(x)|^{2}~dx$

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)

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

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)

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)

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)

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)

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 (talk • contribs)

You could try using a spreadsheet (Excel or similar). --Tango (talk) 16:01, 15 February 2008 (UTC)

This online spreadsheet is easy to learn and to use. --hydnjo talk 04:10, 16 February 2008 (UTC)

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)

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 (talk • contribs) 15:41, 15 February 2008 (UTC)

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)

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)

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)

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)

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 (talk • contribs) 01:46, 18 February 2008 (UTC)

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)

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 (talk • contribs) 16:52, 18 February 2008 (UTC)

Can you not do this yourself? Oh, and please remember to sign your posts. -mattbuck (Talk) 16:56, 18 February 2008 (UTC)

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)

which number is higher? 2

Which number is higher? 23.5%, 26.9% or 20.1%? —Preceding unsigned comment added by Don Mustafa (talk • contribs) 15:49, 15 February 2008 (UTC)

Ignore the percent sign and just treat them as you would any other numbers (see above). --Tango (talk) 15:59, 15 February 2008 (UTC)

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)