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January 29

how many points to determine a general parabola?

Surprising that neither parabola nor conic section gives this. How many points do you need to fix a parabola if you don't know its orientation? Is it one less than the number for a general conic? —Tamfang (talk) 07:09, 29 January 2008 (UTC)

Well, the formula for a conic section has six variables (with minor restrictions), and the parabolic case is a single additional equation. To specify the remaining five variables, you'd need five points, yes. Black Carrot (talk) 09:36, 29 January 2008 (UTC)

(after edit conflict) It is somewhat implicit, but from the general equation of a conic section

Ax² + Bxy + Cy² +Dx + Ey + F = 0

of a conic section with 6 parameters you have 5 degrees of freedom, since you can multiply all coefficients by a common multiplier so as to have A² + B² + C² = 1. It gives you a parabola if B² − 4AC = 0, which leaves 4 degrees of freedom. That means that you need, "in general", 4 points to fix it. I write "in general" because not all choices of 4 points will do (for example, you can't make a parabola go through the corners of a square), and in some cases two parabolas will go through 4 given points; for example, the set

{(0, −1), (−1, 0), ((1+√5)/2, (1+√5)/2), ((1−√5)/2, (1−√5)/2)}

can be fitted both to y = x² − 1 and x = y² − 1.

I haven't looked at how common it is that a given set will have multiple solutions, but suspect it will always be a finite number.  --Lambiam 10:07, 29 January 2008 (UTC)

Another way to get to same result is to define the parabola by two points that determine its directrix, a third non-colinear point that is its focus, and a fourth point on the parabola itself to select a unique curve from the family of parabolas that have the given directrix and focus. Incidentally, you can have a parabola that passes through the 4 corners of a square if you allow degenerate cases such as x²=1, when the "parabola" becomes two parallel lines. Gandalf61 (talk) 10:35, 29 January 2008 (UTC)

Yet another way to look at it: We all know that for a parabola of a known orientation we need 3 points. If we also need to find the orientation, that's another degree of freedom for which we need one more point. -- Meni Rosenfeld (talk) 13:33, 29 January 2008 (UTC)

January 30

grid coordinates from sequence number

This is a very elementary question but I am no good at math

I am writing a program and I need a function to determine the coordinates of an element in a table given its serial position (counting left to right, top to bottom) and the number of rows and columns. to visualise, I might have a grid like this.

1234
5678

I might want to look up the row and column of 6, which would be column 2 row 2. element 4 would have column 4, row 1. I have a feeling I should use mod somehow but I can't figure out how. Thanks! -- 71.86.121.200 (talk) 02:21, 30 January 2008 (UTC)

Numbering from 0 would make the relationship more obvious.

0123
4567

The number 6 is located in row 1, column 2 (the left and top are row/column 0). The table is 4 cells wide. The relationship between those numbers is (serial=row*width+column). To reverse that, row=serial/width (rounded down) and column=serial-row*width, a.k.a. serial mod width. If you insist on numbering from 1, you'll have to subtract 1 from serial first and then add 1 to the results afterward. --tcsetattr (talk / contribs) 03:06, 30 January 2008 (UTC)

Thanks! that looks right. Plugging that equation into my calculator (I told you im not good at math ;-) I see that row=(serial-column)/width which eliminates the need to round down. -- 71.86.121.200 (talk) 03:54, 30 January 2008 (UTC)

Finding the correct sequence for painting

The PVC figure like this one, requires manual panting.

http://www.entertainmentearth.com/images/%5CAUTOIMAGES%5CORG40601.jpg

Suppose it has 3 sections that require painting, A , B and C. And suppose each section has a painting failure rate. If a painting failure occurs, the whole thing would need to be discarded resulting in a loss of effort.

So the question is, in what sequence should the painting occurs? My gut feeling is that the section with the highest failure rate should be painted first, and then the second highest and so on until the section of the lowest failure rate is painted last.

Now if each section requires a different amount of effort, then this will change the problem. So what algorithm should be followed to find the sequence to minimize the loss of effort.

For a figure with only 3 section, the search is trivial and brute force can be used. But if a figure has 17 painting sections, each section with an effort rate and a failure rate then finding the optimum sequence is required.

Example:

sectionA {effort=5, failure=0.2}

sectionB {effort=1, failure=0.8}

sectionC {effort=20, failure=0.1}

Is this type of problem a hard problem to solve?

202.168.50.40 (talk) 02:30, 30 January 2008 (UTC)

Does it become clear only after the whole effort on painting one sector has been expended that this resulted in failure, or is that typically on the average detected more or less halfway?  --Lambiam 07:59, 30 January 2008 (UTC)

Never mind, I don't think that makes a difference.  --Lambiam 08:26, 30 January 2008 (UTC)

You can simply work out the average amount of wasted effort. With your numbers 0.2*5, 1*0.8, 20*0.1 = 1, 0.8, 2. Then you would do it in the order of LEAST wasted to MOST wasted (B, A, C), as you want to to minimal effort on something that fails.--58.111.143.164 (talk) 08:32, 30 January 2008 (UTC)

For a thourough analysis you can use a Decision tree, to calculate the Expected value. First choose the order you wish to use, say ABC.

Start -   A passes  - B passes  - C passes (effort 26, prob = .8 * .2 * .9)
      \            \            \ C fails  (effort 26 + E, prob = .8 * .2 * .1)
        \            \ B fails             (effort 6 +E, prob = .8 * .8)
          A fails                          (effort 5+E, prob = .2)

where E is the expected effort. This will give you a formula E=26*.8*.2*.9+(26+E)*.8*.2*.1+(6+E)*.8*.8+(5+E)*.2. You can solve this for E. Repeat for each combination and viola the order with the least expected effort is the one. If you have 17 steps then this will be too much to do by hand, but a computer should be able to cope. --Salix alba (talk) 09:18, 30 January 2008 (UTC)

If you have 17 steps, then there are 17! = 355687428096000 possible orderings of the sections. In general, the complexity of this method grows as the factorial function, which is going to become infeasible very quickly. I think the "brute force" method mentioned in the original post is what you're describing, and 202.168.50.40 wants a more efficient algorithm. —Bkell (talk) 12:18, 30 January 2008 (UTC)

Highest failure rate first would fail for Job1 having cost 1 and failure rate 0.1. And Job2 having cost 100 and failure rate 0.2. Because the effort put into job 1 is so much smaller it's less of a loss. Anon; you method would fail for Job1 having cost 20 and failure rate 0.8 while Job2 have cost 20 and failure rate 0.1. Here you would clearly want to do the unlikely job first, but this has a higher amount of wasted cycles. I think perhaps it's possible to solve this by comparing elements pairwise. Ie the sequence A.B has an expected wastage of 5*.2+(5+1)*.8=5.8, while B.A has 1*.8+(1+5)*.2=2 so B should go before A. Then compare A with C for 5*.2+(5+20)*0.1=3.5 or 20*0.1+(5+20)*0.2=7, so A should go before C. This sorting could in theory be done in O(n*log n) time. Taemyr (talk) 13:34, 30 January 2008 (UTC)

I believe I can provide a reasonably good argument that, to minimize the expected waste of effort, the pieces should be painted in descending order of failure rate per effort. To see why, consider dividing each task into arbitrarily many small subtasks, each requiring the same amount of effort and, for subtasks of the same original task, having the same probability of failure. Since the effort per subtask is constant, by the same argument as in the constant-effort case above, they should be done in descending order of probability of failure — but this is achieved exactly when the original tasks are ordered by decreasing probability of failure per effort.

This proof sketch implicitly assumes that failures are detected as soon as they happen and that the failure rate is constant throughout each task, but as Lambiam mentions above, I don't think these assumptions should actually make any difference. It should be possible to construct a more rigorous proof by considering the integral of failure rate over effort, but I'm not going to start doing that just now. —Ilmari Karonen (talk) 07:25, 31 January 2008 (UTC)

Hmm. I had already given a counter example to this approach. There is two reasons that your argument fails. The easiest follows from the assumption that all work devoted to a task is lost when a failure occurs. This means that the task can not be divided into smaller tasks. Secondly the probability of failure during one of the smaller tasks would not be independent on the number of tasks the original is divided into. In my example above if the job with cost 100 and probability of failure 0.2 was divided into 100 tasks with a constant chance of failure such that the chance of failure during any of these 100 tasks was 0.2 the probability of failure in any single one of them would be less than 0.1. So these tasks should go after the single task that we divided the task with cost 1 and probability 0.1 into. Taemyr (talk) 06:51, 9 February 2008 (UTC)

Wow! I'm really impressed! I wonder how many anime otaku has actually considered the mathematics of painting anime PVC figures before they start painting their garage kits. I guess mathematics applies everywhere even where you would have least expected it. Thanks. Oh by the way, here is another nice figure http://www.tmpanime.com/images/medium/4560228201475_MED.jpg 122.107.141.142 (talk) 10:40, 31 January 2008 (UTC)

Polynomial Fitting

I know that you can put a polynomial through any finite set of points, no two of which are directly vertical. Can you do the same thing with a power series and an infinite set of points? Does it matter if there are accumulation points? Black Carrot (talk) 04:18, 30 January 2008 (UTC)

If you take the infinite set of points to be *every* point, then specifying the value at those points is simply specifying the function, and some functions do not have expressions as power series. If you take an infinite set of points including a limit point, then a function continuous on those points will have to have its value on the limit point to be the limit of the values converging to it. Convergent power series expansions tend to be continuous, so this gives some restrictions on accumulation points. If you take the points to be on the real line, then accumulation points and arbitrary values do not mix.

If you take an infinite set of points with no accumulation point, then I believe a variation on Weierstrass factorization provides a function with prescribed values on the points, and that has a convergent power series expansion. The discussion of this is in the section of (Rudin 1987, pp. 304–305) called "An interpolation problem", and the result itself is theorem 15.13.

• Rudin, Walter (1987), Real and complex analysis (3rd ed.), New York: McGraw-Hill Book Co., ISBN 978-0-07-054234-1, MR 0924157

The theorem even says you can prescribe finitely many values of the derivative at each point as well. JackSchmidt (talk) 05:51, 30 January 2008 (UTC)

integration and undesirable points

The purpose behinde this post is an attempt to expand integration space to include functions with undesirable points where the function could be undefined.I need help to understand if this post makes sense.The main question here, is this way could be more general than using the measurement set theory to expand the integration?

Assume the function F(x),

F:[a,b]→R 

We will put the interval[a,b] as acombination of subsets, UGk,

UGk=GNUGQUGQ̀....etc.(U,here stands for combination symbol) which stands for natural set,rational set ,irrational set,….etc.

Lets define the subsets,

өi={gk:F(xi)≥gk≥0},

xiЄ[a,b],

Let,(pi) be apartitoning of Gk in [a,b] and (pөi) apartitioning of(өi),

(Mөi=SUPөi), and (mөi=infөi),

(UFөi,pi)= Σi Mөi(xi-xi-1), (UFөi,pi)=upper darboux sum.

(LFөi,pi)= Σi mөi(xi-xi-1),(LFөi,pi)=lowe darboux sum.

(UFөi,pi,Pөi)=inf{UFөi:piPөi,(Pөi) partioning of(өi), (pi) partitioning of (Gk)},

(LFөi,pi,Pөi)=sup{LFөi:piPөi,(Pөi) partioning of(өi),(pi) partitioning (Gk)},

now, we put the integration in the form,

∫Fөi,over,Gk={0,UFөi}={0,LFөi}=the subsets,sөi,

∫F,over,[a,b]=Usөi

for example;

f(x):[0,1]→R,

f(x)=x , xЄ irrational numbers=Q̀, within[0,1],

f(x)=1,xЄ rational numbers=Q, within[0,1],

∫F,over,[o,1]={0,1\2}Q̀,U{0,1}Q={0,1\2}R,U,{1\2,1}Q ,(Q̀,Q AND,R,ARE SUFFIXES HERE AND U STANDS FOR COMBINATION symbol.)

this way enable us to exclude the undesired points from the integration where f(x)may be undefined. 88.116.163.226 (talk) 11:02, 30 January 2008 (UTC)husseinshimaljasim

What you are talking about seems to be similar to the difference between Riemann integration and Lebesgue integration. In Lebesgue integration, if two functions are identical everywhere except a set of measure zero (very roughly, a set of points that isn't too big), then their integrals will be equal. Thus, for example, a function which is continuous everywhere except x=0, where it is infinite, will have the same integral as the same function, but where at x=0 it takes the value 0, or similar. Since the rationals have measure zero over the reals, you could even have a function which is poorly-behaved over an infinite set of points (as long as it's a "small" infinite), and still be integrable. Confusing Manifestation(Say hi!) 04:37, 31 January 2008 (UTC)

well,i dont think so,in Lebesgue integration,the above example would be, μ(0,1\2)inQ`set+μ(0,1)inQset=1\2+0=1\2,iam suggesting to keep integration in form of combination of subsets regardless they were countable or not.wouldnot this be more genral?210.5.236.35 (talk) 14:36, 31 January 2008 (UTC)husseinshimaljasim

equation of natural numbers.

If ,N1,N2,s and k,represent 4 natural numbers,then why the equation, [(N1)^s]+2=[(N2)^k)],where,N1,N2,s,and k Є(N-SET) has only one solution in N-SET,where, N1=2,s=1,N2=2,k=2? i guess. 88.116.163.226 (talk) 11:19, 30 January 2008 (UTC)husseinshimaljasim

I don't think 2¹+2=2² is the only solution. Isn't 5²+2=3³ also a solution ? Unless maybe "N-SET" has some special meaning ?? Gandalf61 (talk) 11:29, 30 January 2008 (UTC)

There appear to be lots of trivial solutions, such as $40^{1}+2=42^{1}$. —Bkell (talk) 12:09, 30 January 2008 (UTC)

what i was thinking!?i must be needing some sleep.210.5.236.35 (talk) 15:56, 30 January 2008 (UTC)husseinshimaljasim

Capital-sigma notation

From the article Summation: Mathematical notation has a special representation for compactly representing summation of many similar terms: the summation symbol, a large upright capital Sigma. This is defined thus:

$\sum _{i=m}^{n}x_{i}=x_{m}+x_{m+1}+x_{m+2}+\cdots +x_{n-1}+x_{n}.$

The subscript gives the symbol for an index variable, i. Here, i represents the index of summation; m is the lower bound of summation, and n is the upper bound of summation. Here i = m under the summation symbol means that the index i starts out equal to m. Successive values of i are found by adding 1 to the previous value of i, stopping when i = n.

An example of the above would be? --Obsolete.fax (talk) 13:40, 30 January 2008 (UTC)

Isn't the equation an example in of itself? I suppose you could have say $n=\sum _{i=1}^{n}1$ - is that what you mean? -mattbuck 13:50, 30 January 2008 (UTC)

By definition, "example" is something specific, while the equation above is quite general. I think the OP is looking for something like:

$\sum _{k=2}^{6}k^{2}=2^{2}+3^{2}+4^{2}+5^{2}+6^{2}=90$

$\sum _{i=4}^{7}{\frac {1}{i}}={\frac {1}{4}}+{\frac {1}{5}}+{\frac {1}{6}}+{\frac {1}{7}}={\frac {319}{420}}$

-- Meni Rosenfeld (talk) 13:57, 30 January 2008 (UTC)

Spearmen correlation coefficient.

I have been reading on how to calculate correlation coefficient using spearman formula But I got confused on the way when ranking the scores of this nature. According to the procedure of ranking the arbitrary ranks of 64,64 and 64 of x if 9th 10th and 11th respectively. Now is it true that the ranks of each of the numbers above is 10? X Y 22 36 24 24 30 25 40 20 45 48 50 44 50 40 52 56 64 62 64 68 64 56 72 32 78 78 78 68 84 68 90 58 —Preceding unsigned comment added by Nkomali (talk • contribs) 14:31, 30 January 2008 (UTC)

Yes. See Spearman's rank correlation coefficient. If several values are tied, the rank of each of them is the average of the positions they occupy. -- Meni Rosenfeld (talk) 16:16, 30 January 2008 (UTC)

Dual Curve

Hello,

I've been looking at the article on dual curves, but it wasn't very informative.

Am I right when I say this is the construction for the dual curve :

For each point on the curve, draw the tangent. Then take the perpendicular to that tangent that passes through the origin. Labelling the point at the intersection of those two lines P, take the point H that is at distance 1/OP from the origin, that lies on the perpendicular line. The set of all such points H forms the dual curve.

If that's the case, doesn't that give rise to two different curves : one when we choose H to be the same side of the origin, one when H is the other side ?

Finally, why does this rather unlikely combination of constructions lead to a curve of any interest ? Seeing as the position of the origin has an importance for the resulting curve...

Thanks. --Xedi (talk) 23:31, 30 January 2008 (UTC)

I can't comment on your construction, but Duality_(projective_geometry)#Points_and_lines_in_the_plane might help clarify duality's goal. As it happens, given certain conditions, absolutely everything that can be proven in geometry has a corresponding dual proof that exchanges lines with points. For instance, every pair of points determines a line that crosses them, and every pair of lines (unless they're parallel) determines a point of intersection. Three lines determine a triangle, as do three points, etc. The dual of a curve is the corresponding curve with similar properties, where points have been exchanged with tangents. Black Carrot (talk) 00:43, 31 January 2008 (UTC)

Xedi does pick up an important point, which is not really represented in the article. We should really be working in the projective plane, so points and lines are represented by triples (x,y,z) subject to the equivalence relation (a x,a y,a z) ~ (x,y,z). The ambiguity happens when we project from the projective plane into R². Yes you can get different curves in R² but they will be projective equivalent. --Salix alba (talk) 08:44, 31 January 2008 (UTC)

Thanks ! I'll have to investigate projective geometry a bit further. -- Xedi (talk) 17:28, 31 January 2008 (UTC)

January 31

$\int$ operator

I have looked at the page on the operator $\int$ , but I don't really understand it. Would someone please explain the function of $\int$ in layman's terms? I think it means integral or something, but I don't know. Zrs 12 (talk) 02:43, 31 January 2008 (UTC)

Take a look at integral, but the easiest way to describe it might be "area under the curve". Or to give you a better idea of some of its uses, maybe it would be helpful to know that position is the integral of velocity, and velocity is the integral of acceleration? - Rainwarrior (talk) 04:07, 31 January 2008 (UTC)

I didn't find a page about the operator itself, could you provide a link? Anyway, the symbol is used for two related, but conceptually different things. The first is for the antiderivative, and the second is for the definite integral. I'll try to explain the second first:

The definite integral can be thought of as a summation process, resulting in the area under a curve, as stated by Rainwarrior. You have limits for integration, denoted by little symbols (say a and b) below and above the ∫. You divide the x-axis in the interval a..b into tiny pieces, and calculate the sum of the products of each tiny x-axis piece multiplied by the corresponding height of the curve. As you reduce the length of each piece of the x-axis, the sum will converge into a well-defined number, provided the function is "well-behaved". This gives you your area, which may have a physical interpretation as stated by Rainwarrior.

The antiderivative of a function f(x) is written as ∫f(x)dx, without the little symbols below and above the ∫. The reason why the same symbol is used, lies in the fundamental theorem of calculus - once you know the antiderivative of a function, say that ∫f(x)dx = F(x) + C, where C is a constant term, you can calculate the area under the curve by simple subtraction - the integral from a to b is F(b)-F(a). However, not all functions have closed-form antiderivatives. --NorwegianBlue ^talk 16:25, 31 January 2008 (UTC)

Note that area under the curve only really makes sense when you're talking about a curve defined on and into the real numbers. It's a good way to start thinking about integration, but be aware that integration can be generalised to many different spaces. -mattbuck 16:52, 31 January 2008 (UTC)

Thank you all for your input. I went to this [[1]] link and found it was fairly helpful as it provided graphs (there is also a link on this page to derivatives). However, I have one more question. What is the purpose for defining these? Why would anyone need to calculate the area under the curve or get a derivative of a function? Zrs 12 (talk) 20:27, 31 January 2008 (UTC)

To answer the side question... the integration symbol is a Unicode character, U+222B, which in UTF-8 e2 88 ab which leads to http://en.wikipedia.org/wiki/%E2%88%AB being the article about the symbol itself. (I suppose you could make a proper link by putting that single character inside double brackets, but I have bad luck inserting fancy characters into wikipedia articles) --tcsetattr (talk / contribs) 21:28, 31 January 2008 (UTC)

In response to User:Zrs 12 - Why would anyone need to calculate the area under the curve or get a derivative of a function? - the answer is Calculus and, in particular, Calculus#Applications. Just about every field in physics, and many other sciences, has a set of differential equations that describe some kind of behaviour, and without the ability to differentiate and integrate functions those equations can't be solved. Some examples are Maxwell's equations, which underpin electromagnetism and hence make your computer work, Newton's law of universal gravitation, which tells you how things fall down and how planets orbit each other, the Einstein field equations of General relativity, which do the same thing but better, and everything in Fluid dynamics#Equations of fluid dynamics and aerodynamics, which let ships sail and planes fly. Admittedly these are all more complicated than just an "area under the curve", but they all come from differentiation and integration.

For a simpler example, like User:Rainwarrior said above, differentiation can take you from distance to velocity, and from veloity to acceleration, and integration can take you in the opposite direction. Confusing Manifestation(Say hi!) 22:28, 31 January 2008 (UTC)

Integration is also important in probability theory, and thus in statistics and hence all the social sciences. Algebraist 14:52, 1 February 2008 (UTC)

The page intergral should be helpful. Ftbhrygvn (talk) 05:37, 5 February 2008 (UTC)

Determinants

I know that the determinant is a multiplicative map, meaning that det(AB)=det(A)det(B). I also know that det(A+B) is not necessarily the same as det(A)+det(B). Is there a special case where det(A+B)=det(A)+det(B)? What are the restrictions on matrices A and B in this case? A Real Kaiser (talk) 04:47, 31 January 2008 (UTC)

This is certainly not a general case, but if A and B share a nontrivial eigenspace with eigenvalue 0, then we have det(A+B)=0=0+0=det(A)+det(B). For example, this holds if A and B represent linear transformations with ker A $\cap$ ker B ≠ {0}.

Also, if A = -B and the dimension of your matrices is odd, then det(A) = -det(B) and you have det(A+B)=det(0)=0=det(A)+det(B). Probably these are not the kind of examples you were looking for. Tesseran (talk) 09:40, 31 January 2008 (UTC)

det(A) and det(B) are essentially the volumes of parallelpipeds with vectors corresponding to rows or columns of the matrix. So you need something so that the vector sum parallelpipeds have volume equal to the sum of the vector parallelpipeds. You can probably construct some with good enough use of algebra, but there doesn't seem to be an easy answer.--Fangz (talk) 16:06, 31 January 2008 (UTC)

What about if the matrices A and B are (real) orthogonal matrices (of any dimension nxn)? Actually, I am trying to prove that if A and B are real orthogonal matrices and if det(A)=-det(B), then A+B is singular. We can also use the fact that |det(A)|=|det(B)|=1. I was thinking about showing that det(A+B)=0 which would be really easy if det(A+B)=det(A)+det(B)=det(A)-det(A)=0. If the determinant is not additive in this case, then how can I prove the above proposition? A Real Kaiser (talk) 17:17, 31 January 2008 (UTC)

The fact that |det(A)| = |det(B)| = 1 shouldn't matter: if det(A+B) = 0 then det(λA+λB) = λⁿdet(A+B) = 0 for all scalars λ. —Ilmari Karonen (talk) 20:25, 31 January 2008 (UTC)

I don't think determinants are a good way of solving this problem. Remember that an orthogonal matrix has determinant +1 for a rotation, and -1 for a reflectionnon-rotation.--Fangz (talk) 01:01, 1 February 2008 (UTC)

While considering the geometric interpretation is often a good trick for linear algebra problems (at least as motivation of a proof, if not for the proof itself), what is the geometric interpretation of addition of matrices? I may be missing something obvious, but I can't see one... --Tango (talk) 22:28, 1 February 2008 (UTC)

What we want is, given arbitrary rotations (=det 1 orthog operators) A and B (A a rotation, B a -rotation), there exists a vector v s.t. A(v)=-B(v). In other words, we want a v s.t. v=-A^-1Bv. -A^-1B is just an arbitrary rotation/-rotation (depending on dimension), so we want that all rotations (n odd)/det -1 orthogonal maps (n even) to have an eigenvalue 1. The rest is fairly easy (using fact that orthogonal matrices are diagonable over C). Algebraist 23:03, 1 February 2008 (UTC)

I just wrote that as I thought it, and realised only at the end that it was in no way geometric. Oops. Looks like I'm incurably tainted with algebra. Algebraist 23:04, 1 February 2008 (UTC)

I'm not following you - I don't see how that proves the required theorem. For a start, the theorem says det(A)=-det(B), so one is a rotation and one a rotation+reflection, secondly, the theorem says it's over the reals, and you've explicitly mentioned doing it over C (in the real case, an eigenvector with eigenvalue one would correspond to an axis of rotation, which I seem to remember reading only exists for n=3, although it could be that there's more than one for n>3, I'm pretty sure there are none for n=2). Lastly, Av=Bv => A-B is singular, not A+B (this error compensates for the det=-1 vs det=1 error for odd numbers of dimensions, but n odd was not given in the theorem). --Tango (talk) 23:20, 1 February 2008 (UTC)

Sorry, I got slightly very confused in my typing. The argument works, I just typed it erroneously. And the fact that I'm using C is not a problem; I can give the details if you want. Algebraist 23:46, 1 February 2008 (UTC)

I'll take your word for it. The question that I'd like an answer to (because it's really bugging me now) is what the geometric interpretation of addition of matrices is. Matrix multiplication is just composition of linear maps, but what is addition? Any ideas? --Tango (talk) 00:15, 2 February 2008 (UTC)

Well, I suppose you'd have to base it on the geometric interpretation of vector addition, but it's not very nice in any case. Indeed I arrived at my proof by the thought process 'Matrix addition is horrible. Let's get multiplication involved.' Algebraist 00:19, 2 February 2008 (UTC)

Yeah, I guess so, but I still can't really see it. You're adding a vector to itself, but transforming each copy in some way... I can't see what significance that has. Getting rid of the addition by moving one term onto the other side of the equals sign is good trick, though. (Of course, it only works if the sum is singular, otherwise you just get a different addition.)--Tango (talk) 01:08, 2 February 2008 (UTC)

In response to Tango's question: matrix addition is just pointwise addition of linear maps. Let f and g be linear maps from a vector space V to a vector space W. Then we can define the linear map f + g from V to W by (f + g)(x) = f(x) + g(x). The coordinate representation of this is just matrix addition. I would hardly call it 'horrible'. -- Fropuff (talk) 00:42, 2 February 2008 (UTC)

That's just a restatement of the definition of matrix addition, really. Restated in the same terms, the question becomes: What is the geometric interpretation of pointwise addition of linear maps? A linear map is basically a transformation (stretching, squashing, twisting, rotating, reflecting, whatever) - what does it mean to add two transformations together? --Tango (talk) 01:08, 2 February 2008 (UTC)

(undent) Would you find it easier to geometrically interpret the mean of two vectors? The sum, after all, is simply twice the mean (and, in particular, the sum of two vectors is zero if and only if their mean is also zero). In terms of geometric operations, to apply the mean of two transformations to a figure, just apply each of the original transformations to the figure and then average them together — i.e. map each point in the original figure to the point halfway between its images under each original transformation. —Ilmari Karonen (talk) 01:34, 2 February 2008 (UTC)

Thank you! That's an excellent way to look at it. --Tango (talk) 13:46, 2 February 2008 (UTC)

Predicate logic

Express the following statement as a well formed formula in predicate logic:

"There is a natural number which is a divisor of no natural number except itself."

Here's what I've put:

$x,y\in \mathbb {N}$

$P(x,y):x|y\,$

$\exists x(\forall y((y=x)\leftrightarrow P(x,y)))$

I just wanted to see if it was valid. Damien Karras (talk) 20:43, 31 January 2008 (UTC)

(ec) It strikes me that you're assuming that every natural number is a divisor of itself. That's true, of course (at least, if you adopt the convention that zero divides zero, which I do), but it's not a purely logical truth. --Trovatore (talk) 20:50, 31 January 2008 (UTC)

(off-topic) Are there really people who have 0 not dividing 0? How bizarre. Algebraist 21:09, 31 January 2008 (UTC)

I don't think so. We have

$(\exists x(\forall y((y=x)\leftrightarrow P(x,y))))\Leftrightarrow$

$(\exists x(\forall y(((y=x)\rightarrow P(x,y))\land (P(x,y)\rightarrow (y=x)))))\Leftrightarrow$

$(\exists x((\forall y((y=x)\rightarrow P(x,y)))\land (\forall y(P(x,y)\rightarrow (y=x)))))\Leftrightarrow$

$(\exists x(P(x,x)\land (\forall y(P(x,y)\rightarrow (y=x)))))\Leftrightarrow$

$(\exists x(P(x,x)\land (\forall y((y=x)\lor (\lnot (P(x,y)))))))\Leftrightarrow$

$(\exists x(P(x,x)\land (\lnot (\exists y(\lnot ((y=x)\lor (\lnot (P(x,y)))))))))\Leftrightarrow$

$(\exists x(P(x,x)\land (\lnot (\exists y((y\neq x)\land P(x,y))))))$,

which is a faithful translation. There may be an hidden assumption underlying the way the phrase is formulated (i.e. the choice of the word 'except'), but it doesn't appear in the formula. Morana (talk) 21:36, 31 January 2008 (UTC)

I don't agree that your last equivalent is a faithful translation. "No natural number except itself" does not imply that "itself" is included; it simply fails to exclude it. --Trovatore (talk) 21:57, 31 January 2008 (UTC)

I don't know, your interpretation seems unusual. It tried reformulating the given sentence in all the ways I could think of but I cannot possibly understand it like you do. Can you justify this interpretation ? This is more an issue of grammar than logic. In any case this is an interesting illustration of the many ambiguities of the natural language. Morana (talk) 00:13, 1 February 2008 (UTC)

Seems straightforward to me. When I say "no natural number" unmodified, that means none at all. If I say "except" I'm weakening that claim. Just weakening it, not making any new claim.

Example: If I say "all eighteen-year-olds in Elbonia enter the military, except for the women", I am not wrong if it turns out that some women actually do join the military. --Trovatore (talk) 06:28, 1 February 2008 (UTC)

Did you notice the negation ? I can't make a poll right now but I argue that if you say "no eighteen-year-olds enter the military, except for the women", any normal (i.e. non-mathematician) person would think that you just claimed that women enter the military.

Do you have any kids ? I you do, tell them "you cannot take any candy, except for the blue ones", and tell me if they didn't jump on the blue ones. Morana (talk) 09:11, 1 February 2008 (UTC)

At the very least the formulation in natural language is ambiguous. If I say "Noone except idiots will buy that", I don't claim that idiots will buy that.  --Lambiam 10:56, 1 February 2008 (UTC)

I agree with Trovatores interpretation. "no eighteen-year-olds enter the military, except for the women" a normal person would draw the inference that there exists women entering since otherwise you could have made a stronger statement with easier form. However a valid interpretation would go, I know that no guys enter the military, but I don't know about the women. "you cannot take any candy, except for the blue ones" means that there is no rule saying the kids can't take the blue candy, so of course they take it. I write the statement as $\exists x(\forall y(P(x,y)\rightarrow (y=x)))$, using Trovatores interpretation of except. Taemyr (talk) 10:59, 1 February 2008 (UTC)

Regression analysis question

Given that X (X-bar) = 3, S_x = 3.2, Y (Y-bar) = 2, and S_1.7 = 3.2, and r = -0.7, is the least squares regression line Ŷ (y-hat) = 3.12-0.37x, or something else? I think I'm right (it seems right) but I want to check. 128.227.206.220 (talk) 22:22, 31 January 2008 (UTC)

It's not exactly right, as Y ≠ 3.12 - 0.37X. Further, X = 3, S_x = 3.2 gives the number of observations as the non-integral 3.2/3, and what is the meaning of S_1.7? Nor does mixing x, X, y and Y help. At least one of us is deeply confused.…86.132.166.138 (talk) 11:58, 1 February 2008 (UTC)

Infinite set containing itself

Is it possible for an infinite set to contain itself? For example, would {{{...{{{}}}...}}} be considered a valid set?

As an unrelated question, how many sets are there? I know it's an infinite number, but which infinite number? — Daniel 23:52, 31 January 2008 (UTC)

In standard set theory (the study of the von Neumann universe) it is not possible for a set to be an element of itself. There are alternative set theories in which it is -- see anti-foundation axiom. There are a proper class of sets (that is, there are more sets than can be measured by any transfinite cardinal number). --Trovatore (talk) 00:09, 1 February 2008 (UTC)

It is the possibility of a set containing itself that led to Russell's paradox, that in turn led to the formal axioms of set theory. Confusing Manifestation(Say hi!) 05:38, 1 February 2008 (UTC)

No, I don't really agree with you there. The Russell paradox was the fruit of conflating the notion of intensional class with the notion of extensional set. It's probably true that historically the antinomies led to the axiomatizations, but it is quite possible to have a non-axiomatic conception of set that avoids them. That non-axiomatic conception also avoids sets that are their own elements, but that isn't the reason it avoids the antinomies. --Trovatore (talk) 06:13, 1 February 2008 (UTC)

Well, the set of all sets has cardinality greater than that of the real numbers $2^{\aleph _{0}}$, in fact it would be greater than that of the power set on the real numbers, $2^{2^{\aleph _{0}}}$. I suppose then you could take sets which are in the power set of the power set of the real numbers, so you get $2^{2^{2^{\aleph _{0}}}}$, etc etc etc. If you assume the continuum hypothesis is true, and that $2^{\aleph _{n}}=\aleph _{n+1}$ then the cardinality of the set of all sets would I suppose be $\aleph _{\aleph _{0}}$, though I expect it would be larger. -mattbuck 11:54, 2 February 2008 (UTC)

See Trovatore's answer: in conventional set theory (eg ZF), the class of all sets is not a set, and thus does not have a cardinality at all. Algebraist 14:28, 2 February 2008 (UTC)

Oh, and $\aleph _{\aleph _{0}}$ doesn't make much sense. I suspect you mean $\aleph _{\omega }$. Also, you're assuming (some cases of) the generalised continuum hypothesis, not just CH itself. Algebraist 15:18, 2 February 2008 (UTC)

Mmm, actually $\aleph _{\aleph _{n}}$ is a pretty common notation, though it's true I haven't seen it very often for n=0. You could write $\aleph _{\omega _{n}}$, but for whatever reason, people usually don't, in my experience. --Trovatore (talk) 01:12, 5 February 2008 (UTC)

To the OP: by the way, your notation {{{...{{{}}}...}}} is somewhat ambiguous. It could mean a set A whose only element is A, or it could mean (for example) a set A such that A={{A}} but A $\neq$ {A}. Both these (and much more) can exist in the absence of the axiom of foundation. Algebraist 15:23, 2 February 2008 (UTC)

I don't know anything about foundations (of mathematics), but is it really possible to have A = {{A}} but A ≠ {A}? In what sense can they be different? -- BenRG (talk) 14:02, 3 February 2008 (UTC)

The usual sense - that one has an element not in the other. I don't know enough about ZF sans foundation to support Algebraist's claim, but I can say that the scenario is not immediately inconsistent. A is different from {A} because A's unique element, {A}, is different from {A}'s unique element, A (yep, sounds circular. But again, I am only showing internal consistency here). -- Meni Rosenfeld (talk) 14:19, 3 February 2008 (UTC)

The most common anti-foundation axiom, due to Aczel ("every directed graph has a unique decoration") implies that if A={{A}}, then A={A}. Otherwise you would have two different decorations of the graph that has a top element and then just descends infinitely in a straight line.

However, a more ontologically expansive axiom, due to Boffa ("every injective decoration of a transitive subgraph of an extensional graph extends to an injective decoration of the whole graph", or something like that) implies that there is a pair A={B}, B={A}, but where A and B are different. This axiom is consistent with ZF. I did some work on this a long time ago but unfortunately never published it. --Trovatore (talk) 22:22, 3 February 2008 (UTC)

Um, of course I mean it's consistent with ZF minus the axiom of foundation. Obviously it contradicts the axiom of foundation. --Trovatore (talk) 18:35, 4 February 2008 (UTC)

February 1

The units of an interest rate

Interest is calculated as $P_{new}=P*(1+r)^{t}$, where $P_{new}$ is the new amount of money, $P$ is the principal, $r$ is the rate, and $t$ is the time elapsed. Since $P$ and $P_{new}$ are both measured in the same units, the term $(1+r)^{t}$ must be dimensionless. And since $t$ is an amount of time, $(1+r)$ must be raised to the power of $\mathbf {time} ^{-1}$.So a 2% annual interest rate be expressed as ${\sqrt[{\mathbf {year} }]{1.02}}$. This is a rather strange unit. How do you refer to this kind of unit, if at all? Rannovania (talk • contribs) 01:24, 1 February 2008 (UTC)

You are confused. t is not a measure of time but a measure of terms. If t is a measure of t then it would be in units of seconds. It is a measure of terms, ie how many terms. And terms are dimensionless value. 202.168.50.40 (talk) 03:14, 1 February 2008 (UTC)

It's probably better to think of the formula as $P_{new}=P*(1+r)^{(t/t_{0})}$, where $t_{0}$ is the unit of time used for expressing the interest rate. Then with a 2% annual interest rate (compound) the formula comes out as $P_{new}=P*(1.02)^{Y}$ where $Y$ is the time in years. HTH, Robinh (talk) 08:23, 1 February 2008 (UTC)

Quadratic inequalities

I know you are supposed to solve quadratic inequalities like you do quadratic equations, except that at the end, when you get the two solutions, you use an inequality sign instead of the equals sign. My question is, when do you have a solution like 1 < x < 2 and when do you have a solution like x < 1 OR x > 2? —Preceding unsigned comment added by 165.21.155.94 (talk) 05:06, 1 February 2008 (UTC)

That's a good question, and the very rough answer is that you do it either graphically, or with a couple of test points. So if you have, say, (x-a)(x-b) > 0, then you use the fact that it's concave up to show that the inequality holds on the left branch (x < a, say) and the right branch (x > b) but not on the little in-between interval. For high school mathematics that's about as far as you'd need to go, but for something more rigorous you'd look at the continuity of the function to ground your argument. Confusing Manifestation(Say hi!) 05:37, 1 February 2008 (UTC)

Take $e:=ax^{2}+bx+c$ as your equation, with solutions $x_{1}\leq x_{2}$

If $a>0$, then $e\geq 0\Leftrightarrow (x\leq x_{1})or(x\geq x2)$

If $a<0$, then $e\geq 0\Leftrightarrow x_{1}\leq x\leq x_{2}$

Morana (talk) 05:41, 1 February 2008 (UTC)

sin^2 x

Is it (sin x)^2 or sin (x^2) or something else? —Preceding unsigned comment added by 165.21.155.91 (talk) 05:11, 1 February 2008 (UTC)

sin²x means (sinx)². Strad (talk) 05:12, 1 February 2008 (UTC)

Thanks! —Preceding unsigned comment added by 165.21.155.91 (talk) 05:15, 1 February 2008 (UTC)

But be warned that sin⁻¹ x doesn't mean (sin x)⁻¹, it means arcsin x. -- BenRG (talk) 14:32, 1 February 2008 (UTC)

e to the power of

Given y-1=4, solving for y, you would say you "add one to both sides" or something similar. Same thing with subtraction, multiplication and division. But what about ln(y) = 4? Is there a word that describes "e to the power of both sides" as an action? My calculus teacher says "exponentiate", but admits he made that word up. —Preceding unsigned comment added by 70.240.254.243 (talk) 21:37, 1 February 2008 (UTC)

I would say "take exponentials of each side" or something similar. The inverse is "take logs of each side". --Tango (talk) 21:46, 1 February 2008 (UTC)

I'd be willing to bet that wikt:exponentiate was not made up by the teacher. --LarryMac | Talk 21:50, 1 February 2008 (UTC)

"Exponentiate both sides" is exactly what I'd say. I'd consider it standard usage. --Anonymous, 00:13 UTC, February 2, 2008.

The word is antilogarithm. Bo Jacoby (talk) 23:38, 1 February 2008 (UTC).

Technically, maybe, not I've never heard anyone use the word (and I'm currently studying maths at Uni). --Tango (talk) 23:43, 1 February 2008 (UTC)

And it's a noun, so if you do use it you have to say "take the antilogarithm". --Anonymous, 00:13 UTC, February 2, 2008.

Our article on Exponentiation uses the verb "exponentiated". hydnjo talk 02:21, 2 February 2008 (UTC)

Both "to exponentiate" and "to take the antilog" are operations with an unspecified base. The number e plays a special role, in particular when the argument is complex. I don't know another name for the operation exp than exponential function. I'd abbreviate that in the obvious way when speaking out loud, like when tutoring, as "now take the exps of both sides" and expect to be understood.  --Lambiam 03:09, 2 February 2008 (UTC)

In higher level maths, the base is always e unless explicitly stated otherwise. I don't think you need to worry about the base. --Tango (talk) 13:45, 2 February 2008 (UTC)

To use exponentiate I think you’d have to say “exponentiate e to both sides”, because exponentating each side by e I believe would look like "ln(y)^e=4^e"

"take exponentials of each side" can also make sense though, because it’s commonly viewed as a function rather than an operation: "exp(ln(y))=exp(4)" which is "e^ln(y)=e^4". GromXXVII (talk) 12:20, 2 February 2008 (UTC)

That would be raising to the power e, not exponentiating, I don't think there is any ambiguity there. --Tango (talk) 13:45, 2 February 2008 (UTC)

Replying to Tango on antilogarithm. In the days when everything was done with published tables rather than calculators (not so long ago) there were tables of antilogarithms as well as logarithms and the phrases "take the log of both sides" and "take the antilog of both sides" were common currency. Both noun and verb forms were used. Terminology may well have changed now, I don't know. If you really want to know what geological era this was practiced, you will find that my age is displayed on my userpage. SpinningSpark 14:56, 2 February 2008 (UTC)

Using log tables to do arithmetic is very different to usings logs/exponentials in higher maths. (If nothing else, one uses base 10, the other base e and one uses numbers, the other algebraic expressions.) I wouldn't expect to use the same terminology in both. --Tango (talk) 17:17, 2 February 2008 (UTC)

Please do not be condescending. I know perfectly well the difference between arithmetic, pure, and applied mathematics. You are the one with a gap in your knowledge, which you highlighted yourself, I was merely politely giving you some information. You are wrong on just about every count. My mathematical tables (Knott, 1965, W&R Chambers) include both base 10 and base e. Likewise my slide-rule (a pre-computer analog calculating device) has scales for both log base 10 and log base e. In engineering calculations I have often (possibly more often) used base e for convenience. It is certainly not the case that mathematicians have a monopoly on base e while the rest of us grind out calcualtions in base 10. On the other hand there is no reason at all that algebraic expressions should not use log base 10 where appropriate; in engineering, many, in fact, do so. SpinningSpark 19:17, 2 February 2008 (UTC)

Ok, forget the difference in the bases then - the more important difference is that log tables are about the logs of numbers, not expressions. Exponentiating something generally refers to an expression (eg. "x"), and that is very different. As for engineers using base 10 - not when they're doing higher maths, they don't. Sure, decibels and things are defined in terms of base 10 logs, but that's not mathematics, that's just a convention. Base 10 might appear in the odd formula involving such units, but any time exponentials come up in relation to actual maths (eg. solving ODEs), they are base e. Oh, and don't complain about me being condescending and then do it yourself - I know what a slide rule is... I even own one, although I've no idea where I put it. --Tango (talk) 21:57, 2 February 2008 (UTC)

Just as you talk about the natural logarithm and the base ten logarithm, you may talk about the natural antilogarithm and the base ten antilogarithm. If ln(y) = 4, then take the natural antilogarithm on both sides and get y=antiln(4)=ln⁻¹(4)=exp(4)=e⁴. If sin(y)=a, then take the arcus-sine on both sides and get y=arcsin(a)=sin⁻¹(a). Bo Jacoby (talk) 21:11, 3 February 2008 (UTC).

February 2

Isn't there an interesting question out there about February 2?

Anyone? (extending and editing my previous remarks about February 2 ... —hydnjo talk 06:47, 4 February 2008 (UTC))

How about $(2^{(}2^{2}))/2='08$ or something. --hydnjo talk 07:18, 4 February 2008 (UTC)

Oh, ok then, - guess that'll have to do :-( --hydnjo talk 07:18, 4 February 2008 (UTC)

Well done87.102.90.249 (talk) 12:54, 4 February 2008 (UTC)

February 3

Distance

OK, so lets say I have points A,B,C on a plane forming the veritices of a triangle, with AB=1 and BC=1, both lines parallel to the axes. From the Pythagorean theorem, I know that AC=sqrt2. However, lets say in order to get from A to C, I instead started parallel to BC, walked, say 1/4 of the horizontal distance, then made a 90 degree turn and move forwards 1/4 of the horizontal distance, then turned 90 degrees again, etc. The distance of that path will be equal to AB+BC, because each of those little fragments in one direction corespond to a fragment of the line to which it is parallel. Using that logic, I could move, say, 1/32 of the distance before turning 90 degrees, etc, and the total distance would still be equal to AB+BC. In fact, any distance for each fragment will give AB+BC for the total distance! If I keep on making each incriment smaller and smaller, the total distance from A to C will always be 2. But, at the end of that sequence is the straight line, AC, which I know is not 2, but sqrt 2... An explanation? 70.156.60.236 (talk) 04:44, 3 February 2008 (UTC)

Wow, funny you should talk about that. We recently had a discussion about that in my analysis and my teacher gave us the same exact example. If you have a right triangle and one leg is A units long and the other leg is B units long. If you travel along the legs, the distance traveled is A+B and it will always be A+B unless you take a straight line along the hypotenuse in which case the distance will be sqrt{A^2+B^2}. So at which point, does the length change from A+B to ${\sqrt {(}}A^{2}+B^{2})$ and they are certainly not always equal? The answer has something to do with uniform convergence. I don't know how familiar with analysis you are but in a sequence of functions, if you don't have uniform convergence (as opposed to just convergence, which we shall now call pointwise convergence), then some of the most "obvious" results which we take for granted fall apart. For example, in calculus, if we do term by term integration or term by term differentiation of an infinite power series, we can only do that because that series converges uniformly. I will give you another example. Consider the sequence $\{x,x^{2},x^{3},x^{4},...,x^{n},...\}{\textrm {for}}x\in [0,1]$. This is a sequence of functions and all of them are continuous on the given interval. In fact, they are continuous everywhere. But if you take the limit as $n\rightarrow \infty$ then the limiting function becomes

$f(x)={\begin{cases}0&x\in [0,1)\\1&x=1\end{cases}}$

So, what happens? Each function in the sequence is continuous but the limit breaks up. The reason is uniform convergence. This sequence does NOT converge uniformly and that is why even if each terms is continuous, the limiting function is not guaranteed to be continuous.A Real Kaiser (talk) 05:27, 3 February 2008 (UTC)

While I agree with your description of uniform convergence, I struggle to see how it applies to this question. I can't see a sequence of functions, I can only see a sequence of numbers {2*n/n}, which ought to converge to 2, no sqrt(2) - there's no "x" or similar in there for the convergence to be uniform or not with respect to... I should probably have paid more attention in my Analysis in Many Variables lectures, but maybe you can point out what I'm missing? --Tango (talk) 16:36, 3 February 2008 (UTC)

Everything that you computed is right, you just need to go the last step and also believe it. You are currently confused by an assumption that you took for granted, but it is false. The limit of the lengths of a series of curves is not equal to the length of the limit of a series of curves. If you had made the analysis in the hope to find the length of AC, than your analysis is wrong, because you made this incorrect assumption. The length of a curve is defined by integrating along the curve in the direction of the curve, but the direction of the curve changes when you switch from the staircase like approximations to the the straight line limit. Another view is if you look at the definition in Arc length. Your approximation staircases have points which are not on the straight line, but the definition requires all points to be on the line.

btw.: Very good question. Clever people have reasoned about what would happen if length were defined by a metric that works like your analysis does. The result is called Manhattan distance. Thorbadil (talk) 18:32, 3 February 2008 (UTC)

It might provide perspective to imagine other examples. Say, for instance, that instead of walking parallel to the axes you walk in curliques that get smaller and smaller, or back and forth along the line itself, or in a hairpin-turn zigzag that covers dozens of times the length of the line while never going far from it or backtracking. In general, you can get these paths be of whatever length you want while never getting more than an arbitrarily small distance from the diagonal line. With one exception: you can never produce a path with less length than the actual diagonal distance, and you can never achieve that minimum except with a perfectly straight line. So, it seems reasonable to think of these other paths as extremely intricate detours. Another bit of perspective might be gained by studying more pathological examples, like the Koch curve, Peano curve, and Weierstrass function. Black Carrot (talk) 20:09, 3 February 2008 (UTC)

By the way, I just noticed that the introduction of Koch curve makes the exact assumption that started this. It's important to notice the difference: The Koch curve is the actual limit curve, and any description of its length has to keep track of that. It is the set of all points that remain part of the simpler curves from some step onwards, and the limit points thereof. You can justify that it has infinite length by letting curves of known length hug it more and more tightly. They must have length less than the Koch curve, since they're smoothed-out versions of it, but can be made arbitrarily long. These approximation curves can be most easily taken as the curves that defined the Koch curve in the first place, since they certainly hug it tightly. The curve you mention works in the opposite way. Each point of the diagonal line is a point that was fixed at some point in the process (where the stepped line meets it) and the limit points thereof. However, there's no reason to think the stepped line is shorter than the diagonal. Since it's rougher, it could only be longer, meaning that you could prove the length of the diagonal at most 2, but not at least 2. Black Carrot (talk) 20:20, 3 February 2008 (UTC)

February 4

HEY IT`S ME THE PHYSICS MAGAZINE GUY THIS TIME WITH AN ALGEBRA QUESTION

From an Algebra Magazine. 1.Find the consecutive integers whose product is 156. Find two consecutive integers whose product is 156. Find two consecutive even integers whose product is 288. Find two consecutive odd integers whose product is 783.

I also have some word problems Michelle is 5 years younger than Cindy.Janet is years less than twice Cindy`s age.Kenny is 10 years old.The ratio of Janet`s age to Cindy`s age equals the ratio of Michelle`s age to Kenny`s age.How old are Cindy,Michelle and Janet.

A rectangular poster has an area of 190 square inches.The height of the poster is 1 inch less than twice it`s width.Find the dimensions of the poster.

Daryl earns twice as much per hour as Andy,and John earns 6 dollars more per hour than Andy.June earns 16 dollars per hour.The ratio of John`s hourly earnings to Andy`s hourly earnings is the same as the ratio of John`s hourly earnings to Andy`s hourly earnings is the same as the ratio of Daryl`s hourly earnings to June`s hourly earnings.How much does Daryl earn per hour.

A small rocket is launched upward from ground level.The height of the pocket from the ground is given by the quadratic equation h=-16t2 plus 144t where h is the height of the rocket in feet and t is the number of seconds since the rocket was launched.How many seconds will it take for the rocket to return to the ground.

given by the quadratic equation h0 —Preceding unsigned comment added by Yeats30 (talk • contribs) 00:13, 4 February 2008 (UTC)

Try taking the square root of the numbers. --Salix alba (talk) 00:33, 4 February 2008 (UTC)

Well I'd strongly encourage you to try them yourself - for example with no. 1 make the first number x and the second number y, then x*y = 156 (from "the product is 156") & y = x + 1 (from "the numbers are consecutive") and you can solve them like any pair of simultaneous equations...the others all work similarily - decide what the variables are and make equations with them, then try to solve those equations. So for the second question you might represent michelle, cindy, janet & kenny's ages by the letters m, c, j & k respectively. Then "michelle is 5 years younger than cindy" becomes m = c - 5, and so on and so forth. It's not that hard once you get equations. Trimethylxanthine (talk) 00:47, 4 February 2008 (UTC)

Complex Base

It's possible to write the set of complex numbers in decimal-like form using a complex base, like 2i, and the appropriate number of digits, in that case 0, 1, 2, and 3. Given a base, how do you determine whether it will produce a consistent number system in that way? Black Carrot (talk) 00:52, 4 February 2008 (UTC)

Yeah, it is. See quater-imaginary base — Kieff | Talk 01:17, 4 February 2008 (UTC)

I'm not sure you actually read my question. Black Carrot (talk) 07:34, 4 February 2008 (UTC)

Sorry, you're right. I read it as two questions ("is it possible, if then how to determine"). My bad. :( — Kieff | Talk 08:24, 4 February 2008 (UTC)

Not sure if I understood - but if the base is a+ib where a and b are integers then the new number system will be workable - ? (it's just solving linear equations of order 2)87.102.90.249 (talk) 12:57, 4 February 2008 (UTC)

solution of Euclide

i visted clay mathematics institute web site and i read this, Euclid gave the complete solution for that equation,(x^2+y^2=z^2).my question is how could we give such solution?by picking up an arbitrary values? or by finding operators, like we put,y=(a+s),x=(b+s),then z=(b+a+2s),where,a,b are constants and s,is variable.then we fixed a,and ,b,and start chossing differents value of ,s? is that how we find the solutions?or the general way is different?209.8.244.39 (talk) 12:01, 4 February 2008 (UTC)o`neil

You may want to take a look at Pythagorean triple. If you then have a more specific question we'll be happy to help. -- Meni Rosenfeld (talk) 12:06, 4 February 2008 (UTC)