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February 24

derivatives and second derivatives of rotation functions

hi, how do i find the derivative and second derivative of a function that rotates counter clockwise an angle theta? do i just take the derivative and second derivative of (-sin+cos, cos+sin)?

also, how might one prove that a harmonic function composed with a function that preserves dot product is still harmonic? thanks —Preceding unsigned comment added by 199.74.71.147 (talk) 03:33, 24 February 2008 (UTC)

For the first question I assume you mean rotation around the origin in the Euclidean plane, using the mapping

$(x,y)\mapsto (x\cos \theta -y\sin \theta ,x\sin \theta +y\cos \theta )\,.$

To get the n-th derivative with respect to the rotation angle θ, just work out

$(x,y)\mapsto {\frac {d^{n}}{{d\theta }^{n}}}(x\cos \theta -y\sin \theta ,x\sin \theta +y\cos \theta )\,,$

where a pair is differentiated coordinate-wise:

${\frac {d}{d\theta }}(X(\theta ),Y(\theta ))=({\frac {d}{d\theta }}X(\theta ),{\frac {d}{d\theta }}Y(\theta ))\,.$

Using matrix–vector notation, the mapping can be written as

${\begin{bmatrix}x\\y\end{bmatrix}}\mapsto M(\theta ){\begin{bmatrix}x\\y\end{bmatrix}}\,,$

where the 2×2 rotation matrix M(θ) is given by:

$M(\theta )={\begin{bmatrix}\cos \theta &-\sin \theta \\\sin \theta &\,\,\,\,\cos \theta \end{bmatrix}}\,.$

Here you have to find ^d⁄_dθM(θ), which can be done entry-wise:

${\frac {d}{d\theta }}M(\theta )={\begin{bmatrix}{\tfrac {d}{d\theta }}\cos \theta &-{\tfrac {d}{d\theta }}\sin \theta \\{\tfrac {d}{d\theta }}\sin \theta &\,\,\,\,{\tfrac {d}{d\theta }}\cos \theta \end{bmatrix}}\,.$

The second question is not clear. What does it mean for a function to preserve dot products? Normally you expect that to mean f(u·v) = f(u)·f(v), but the dot product turns two vectors into a scalar, so if u and v are vectors, on the left-hand side f is operating on a scalar and on the right-hand side on vectors. How can this be? Also, is it given in which order the harmonic function is composed with the dot-product preserving function? An example might help to clarify this.  --Lambiam 06:26, 24 February 2008 (UTC)

Probably means orthgonal, u·v = f(u)·f(v). With some very minor condition (possibly none), an orthogonal function is an orthogonal matrix. Then it is just asking to show that the laplacian is invariant under orthogonal coordinate changes, which is definitely a very common exercise to give (so likely to be his question). JackSchmidt (talk) 06:44, 24 February 2008 (UTC)

that probably is my question, but i am unfamiliar with most of the terminology you used as my professor is rather scatterbrained in lecture. could you please give me some pointers on where to start if i were to prove this? thanks! —Preceding unsigned comment added by 199.74.71.147 (talk) 08:04, 24 February 2008 (UTC)

Basically you use chain rule. If g:R^2->R is harmonic and f:R^2->R^2 preserves dot products, then f is actually given by an orthogonal matrix (which looks almost exactly like a rotation matrix). Define h:R^2->R by h(x) = g(f(x)). Write out what that means fairly explicitly in terms of the matrix entries of f (just label them f11, f12, f21, and f22; don't use cosines in my humble opinion). Then use chain rule to take the first and second derivatives. The derivatives will involve dot products of rows of f. f is orthogonal, so the dot products will be 0 for the mixed derivatives of g, and 1 for the repeated derivatives of g. In other words, lap(h) evaluated at x is equal to lap(g) evaluated at f(x). JackSchmidt (talk) 08:17, 24 February 2008 (UTC)

how would one avoid using cosines? and also, where do dot products come into the derivatives? I don't see them199.74.71.147 (talk) 20:22, 24 February 2008 (UTC)

(Sorry, I don't have an easy way to explain this; my comments have been meant for other more analytic types to help you out; same continues here). See directional derivative for the dot products. Applying f just changes the direction of a directional derivative. The laplacian is defined in terms of partials, but you get the same definition if you use directional derivatives in directions which are just a rotation of the previous ones. Simply writing it out and taking calc 1 derivatives suffices, it just takes a page. I recommend that method. JackSchmidt (talk) 20:47, 24 February 2008 (UTC)

can you explain a little more about how i get from taking the derivative to having dot products?

There is a dot product in the definition of a directional derivative, is that the one you're looking for? --Tango (talk) 11:44, 25 February 2008 (UTC)

Integral Powers of Trinomials

For expanding binomials, we have the binomial expansion which gives us all the info we need to know about expanding a binomial to (any) power. For example, if I want to find out what is the full expansion of $(x+y)^{25}$, I can do so. My question is, is there a similar expansion formula for $(x+y+z)^{25}$. Is there a shorthand way to write this expansion in terms of the coefficients of the expansion?A Real Kaiser (talk) 05:53, 24 February 2008 (UTC)

Memory of high school maths a bit hazy, but you can break $(x+y+z)^{25}$ into $((x+y)+z)^{25}$, which gives $\sum _{k=0}^{n}{C_{k}^{n}}(x+y)^{n-k}z^{k}$, and then expand $(x+y)^{n-k}$ binomially.

This gives something like $\sum _{k=0}^{n}\sum _{l=0}^{n-k}{C_{k}^{n}}{C_{l}^{n-k}}x^{l}y^{n-k-l}z^{k}$ --PalaceGuard008 (Talk) 06:19, 24 February 2008 (UTC)

You are looking for the Multinomial theorem. You are basically counting how many ways to rearrange "mississippi", or other such combinatorial questions. JackSchmidt (talk) 06:26, 24 February 2008 (UTC)

The coefficient of $x^{i}y^{j}z^{k}$ in the expansion of $(x+y+z)^{n}$ is

${n \choose i,j,k}={\frac {n!}{i!\,j!\,k!}}$

Pascal's pyramid contains the coefficients for the trinomial expansion - it bears the same relation to the trinomial expansion as Pascal's triangle does to the binomial expansion, but it has an extra dimension. So the coefficients of $(x+y+z)^{25}$ are the numbers in the 25th layer of Pascal's pyramid. Gandalf61 (talk) 11:26, 24 February 2008 (UTC)

Thanks guys, I had no idea that a generalization of Pascal's triangle existed.A Real Kaiser (talk) 01:29, 25 February 2008 (UTC)

Curiously enough, I independently discovered this relationship over a year ago. I found it a lot less obvious than Pascal's Triangle. A math-wiki (talk) 07:06, 27 February 2008 (UTC)

Vibrational modes of a drum

I'm interested in creating a few images for Wikipedia of the vibrations of a drum membrane.

It appears these vibrations are related to the Bessel functions - searching on google, I didn't find any precise descriptions of the phenomenon. Somes pages appeared helpful though (for example : Standing waves in a drum membrane...).

So, could anyone explain how it works, and what the equations describing the vibrations are ?

Thanks. -- Xedi (19:25, 24 February 2008 (UTC))

Well, I don't know your mathematical background (have you studied PDEs?), but the simplest thing to do is assume that the membrane is isotropic and homogeneous, and the waves are small enough that it responds linearly. Then you have to solve the wave equation (whose spatial part after separation of variables is the Helmholtz equation) by finding eigenfunctions of the Laplacian for whatever shape your drum is. The solutions for a circular drum do involve Bessel functions. See also Hearing the shape of a drum. —Keenan Pepper 23:13, 24 February 2008 (UTC)

If you want a textbook that discusses this stuff, try Elementary Applied Partial Differential Equations with Fourier Series and Boundary Value Problems by Richard Haberman. —Keenan Pepper 23:17, 24 February 2008 (UTC)

Thanks a lot, the article on the Helmholtz equation is pretty much what I needed. (I'll also have a look at the book) -- Xedi (10:08, 25 February 2008 (UTC))

February 25

1-1

let a,b positive integers.

Is

f(a,b)= 3^a + 10^b

1-1? —Preceding unsigned comment added by 71.97.4.194 (talk) 00:46, 25 February 2008 (UTC)

If you mean one-to-one, then yes, I think it is. Black Carrot (talk) 01:11, 25 February 2008 (UTC)

g(a,b) = 2^a + 3^b fails to be 1-1. g(3,1)=11=g(1,2). It seems like we could do something similar for f, but the numbers get too big. —Preceding unsigned comment added by 71.97.4.194 (talk) 01:25, 25 February 2008 (UTC)

If 3^a+10^b=3^c+10^d, then 3^a-3^c=10^d-10^b. Investigate the properties of differences between powers of 3 and differences between powers of 10. Does that help? —Bkell (talk) 04:48, 25 February 2008 (UTC)

Yes, I tried that, to no avail (yet). It did make me think of using the pigeonhole principle. *How*, I'm not quite sure. 192.91.253.52 (talk) 13:20, 25 February 2008 (UTC)

I was unable to solve the problem, though I approached it in the way Bkell suggested. If a>c then d>b, so I factored it as 3^c*(3^(a-c)-1) = 10^b*(10^(d-b)-1). You get congruences like a=c mod OrderMod( 3, 10^b ), and b=d mod OrderMod( 10, 3^c ) so if b or c is nonzero, this reduces the possible solutions, and in fact you can apply the same argument again with much larger moduli. However, induction does not quite work, because I could not rule out b and c being zero (especially not inductively). When the problem is changed to the g(a,b), in fact very early there is a solution with b or c being 0, g(2,0)=5=g(1,1). While one might decide the arguments are not allowed to be 0 for some reason, during the induction you have to allow that case. At any rate, there are no small solutions for f (0<= a,b,d <= 1000). JackSchmidt (talk) 15:43, 25 February 2008 (UTC)

The original problem states that the exponents are positive integers. —Bkell (talk) 16:33, 25 February 2008 (UTC)

My idea, while not fully fleshed out, is that a difference between powers of 3 is always going to be a string of 0's and 2's when written in base 3, and a difference between powers of 10 is always going to be a string of 0's and 9's when written in base 10. I think there's a contradiction lurking somewhere: it feels like it's going to involve a claim that in order to write a difference between powers of 3 as a sum of things that look like 9000…0 in base 10, you're going to have to use a particular 9000…0 twice. Sorry for the roughness of this idea; it's really just a hunch and needs a lot of work still. —Bkell (talk) 16:40, 25 February 2008 (UTC)

Consider this, if 3^a+10^b=3^c+10^d, then you can rewrite it in most cases as 3^f=((10^h-1)*10^g)/(3^j-1), which must be an integer. Is it possible for this to be an integer, though? (hint: look at 10^h-1 and 3^j-1 under the proper modulus). GromXXVII (talk) 19:33, 25 February 2008 (UTC)

Sure it can be an integer. For example, g=3, j=2 makes ((10^h-1)*10^g)/(3^j-1) an integer for every nonnegative h. JackSchmidt (talk) 19:42, 25 February 2008 (UTC)

I haven't come up with a good way to prove it yet, but I have another direction you might go in. If you write down a given power of 3, the question is whether you can subtract 1 from one digit, and add 1 to another, and get back a different power of 3. If you consider it base 3, the question is whether given a power of 101, you can subtract 1 from one digit and add 1 to another and get back another power of 101. Without carry, powers of 101 are pascal's triangle, which gives a clear enough pattern that it may be possible to prove it false. Black Carrot (talk) 20:50, 25 February 2008 (UTC)

I think I've got it. Take it mod 5, 4, and 20, in that order. The first shows that a=c(mod4), the second shows that that's only possible if either b or d is 1, and the third shows that that can't happen. It's one-to-one. Black Carrot (talk) 07:22, 26 February 2008 (UTC)

Unfortunately, simply looking mod finitely many numbers cannot prove the result. For instance, it is possible for f(a,b) = f(c,d) mod 4, mod 5, and mod 20 (simultaneously), without a=c, b=d: more explicitly, f(1,2) = f(9,10) mod 4, mod 5, and mod 20. JackSchmidt (talk) 16:45, 26 February 2008 (UTC)

Interesting problem. – b_jonas 10:59, 29 February 2008 (UTC)

Approximation by Smooth Functions

Basically, our teacher is trying to convince us that $C_{0}^{\infty }(\Omega )$ which is the set of all infinitely differentiable functions with compact support on a given set $\Omega$, is dense in $L^{2}(\Omega )$. Now, I know that one of the definitions of a set A being dense in another set B is that the closure of A must be B. Which is the same as saying that B contains A and all of its limit points. So he gave us a theorem in class saying that, for any $f\in L^{2}(\Omega )$, there exists a sequence $\phi _{n}\in C_{0}^{\infty }(\Omega )$ such that $\phi _{n}\rightarrow f$ in $L^{2}(\Omega )$. My question is how does this theorem prove that $C_{0}^{\infty }(\Omega )$ is dense in $L^{2}(\Omega )$? All we have done is shown that every square integrable function f is a limit of a sequence of smooth functions with compact support. We have shown that each f is a limit point but how do we know that those are all the limit points? What if there is a limit point of smooth functions with a compact support outside $L^{2}(\Omega )$?A Real Kaiser (talk) 04:52, 25 February 2008 (UTC)

$L^{2}(\Omega )$ is a closed set - it contains its own limits. Bo Jacoby (talk) 05:36, 25 February 2008 (UTC).

Duh, I seem to have forgotten that little fact. Thanks!A Real Kaiser (talk) 05:52, 25 February 2008 (UTC)

The above comment obscures the point at hand. It is not enough to talk about a set being open or closed; you have to specify the topology you are talking about. Every set is closed in its own subspace topology. Here, the question is whether or not the closure of $C_{0}^{\infty }(\Omega )$ in the metric topology of $L^{2}(\Omega )$ is $L^{2}(\Omega )$. In this topology, it is impossible for the closure of $C_{0}^{\infty }(\Omega )$ to be bigger than $L^{2}(\Omega )$. Perhaps the above post was referring to the fact that $L^{2}(\Omega )$ is complete (is this true actually? ahh...), which would mean that however you embed $L^{2}(\Omega )$, the closure of $C_{0}^{\infty }(\Omega )$ will be exactly $L^{2}(\Omega )$. However, that would not be a requirement for saying that the one is dense in the other. If $A\subset B\subset C$, to say that A is dense in B is to say that the closure of A, in B's subspace topology, is all of B. It would not matter if B were not closed in C, because in the B subspace topology, the closure of A cannot exceed B. (Can someone with more experience please check my post for correctness? Thanks...) J Elliot (talk) 06:08, 25 February 2008 (UTC)

$L^{2}(\Omega )$ is a normed space, the norm being the integral of the absolute square. This norm defines the topology. In this topology the set $L^{2}(\Omega )$ is closed. A function outside $L^{2}(\Omega )$ has no well defined distance to a a function within $L^{2}(\Omega )$ , so it is not a limiting function of a sequence of functions in $L^{2}(\Omega ).$ Bo Jacoby (talk) 14:15, 25 February 2008 (UTC).

The sentence "In this topology [on a space X] the set X is closed" is one of the axioms of a topology, and thus is never a meaningful observation. As was noted above, the only meaningful thing that we can say in this direction (and indeed it is very meaningful) is that L²(Ω) is a normed space (and thus more generally a uniform space), and with respect to this structure L²(Ω) is complete. In particular, this implies that whenever any metric space V is isometrically embedded into L²(Ω) (we might take V=C^∞₀(Ω) with the induced square-integral norm), then any Cauchy sequence in V has a limit in L. Thus it is the completeness of L²(Ω), rather than the meaningless "closedness", which implies that the completion of C^∞₀(Ω) [with the L²-norm] is L²(Ω). This is not necessary to show that C^∞₀(Ω) is dense in L²(Ω), however. Tesseran (talk) 05:26, 27 February 2008 (UTC)

Kaiser, you should take some algebra classes next semester, because my analysis books had been getting nice and dusty before you started posting this semester. :)

BTW, you may seriously want to prove that if f_n are C^oo with compact support, and they form a cauchy sequence under the L^2 norm (so they "converge to something"), then there is some number C such that the L^2 norm of all f_n is bounded by C. In other words, if something is an L^2 limit of compact functions, then it is L^2. This should be easy to prove, other than the fact that it is posed randomly on the net. Once you do this, some of the previous posters talking about completeness should make more sense.

This idea is actually pretty important, even though some people might think my question is silly. The point of using the L^2 norm is to control the L^2 norms of limits. As long as everything in the sequence was L^2, and the sequence converges (is cauchy) according to the L^2 norm, then the limit is L^2. This is why you need to use the Sobolev norms in PDE. You want to take a sequence of approximate solutions to a PDE (or solutions to some approximately equal PDE), and know that the limit is still a solution. If you use L^2 for that, you will often fail, but if you use H^1 (L^2 of the function and its derivative), you will have much better luck, since the limit function at least has an L^2 derivative. JackSchmidt (talk) 15:57, 25 February 2008 (UTC)

I think, Kaiser, that your original post misses the point a bit. B containing A and its limits is nothing like sufficient for A being dense in B (consider $A=[0,1],B=\mathbb {R}$). Instead, what you want is that $\forall x\in B\;\exists \{y_{n}\}\subseteq A\;\lim _{n\rightarrow \infty }\|y_{n}-x\|=0$ (with some metric appropriate to B, which is where topology comes in). The theorem you were given establishes precisely this fact. --Tardis (talk) 16:08, 25 February 2008 (UTC)

Jack, you know what, you are absolutely correct. I am going for a degree in Applied Math and I incorrectly assumed (in my early childhood as a lower division Math student) that Applied Math and Pure Math were two disjoint sets. So I decided to stay away from Pure courses which is exactly why now I have huge holes in my understanding of real, complex, functional analysis, and algebra. I was already thinking about an algebra course next semester but it seems vital that I must take it. Well, lol, don't put those books away. This doesn't mean that I am going to stop asking questions. There are plenty more where these come from. I know that all of you don't have to help me but I really appreciate it. Thanks everyone! :)
Ok, so let me recap. L^2 is the space of all square integrable functions. It has a norm defined on it (the usual L^2 norm). It has an inner product defined on it and it is complete with respect to this norm. Therefore L^2 is a Hilbert space. Being complete means that any Cauchy sequence of L^2 functions converges to an L^2 function. So there are no limit points of L^2 that are outside L^2. Furthermore, if I want to show that the set of all smooth functions (members of $c^{\infty }$) is dense in L^2, then it will suffice to show that if I take any square integrable function, I can always find a sequence of smooth functions that converges to that square integrable function. This is the same as showing that any square integrable function can be approximated arbitrarily well by smooth functions. It is also not possible for a sequence of smooth functions to converge outside the L^2 space because smooth functions are a subset of square integrable functions and the set of all square integrable functions is closed. Is this correct, guys? I want to make sure that my line of reasoning is correct.A Real Kaiser (talk) 20:04, 25 February 2008 (UTC)

GMAT data sufficiency problem

A homeowner must pick between paint A, which costs $6 per liter, and paint B at $4.50 per liter. Paint B takes one-third longer to apply than paint A. If the homeowner must pay the cost of labor at $36 per hour, which of the two paints will be cheaper to apply?

(1) The ratio of the area covered by one liter of paint A to that covered by one liter of paint B is 4:3.

(2) Paint A will require 40 liters of paint and 100 hours of labor.

I first encountered the problem in Arco Math Workbook (2000). Kaplan's 2005 version is similar, but it's got the same problem and explanation, which I cannot agree with:

"Statements (1) and (2) taken together are not sufficient to answer the question. To make an intelligent decision, we need to know which requires more paint and how much more, how long each will take, and we need some info on their labor costs...Using both statements together, we still cannot find the labor costs."

It seems rather clear to me how to find the labor costs.

First, the two statement taken together mean that it'd take 40*4/3 or 160/3 L of paint B.

Second, statement (2) says that paint A is applied at a rate of 40L per 100 hours, or 2/5 L/hr. - but the original problem says it would take a third longer for paint B, which means that paint B is applied at a rate of 2/5 * 3/4 L/hr. or 3/10 L/hr.

This last conversion was the most difficult for me, but you can see it's just multiplying A's rate by 4/3 hr/L.

Thus, it will take (160/3) / (3/10) or 1600/9 hours to use the 160/3 L of paint B to cover the house.

Therefore we know how long it takes. Given the unit labor cost in the original problem, we can figure out how much labor costs for either paint. We also know how much paint we need and how much it costs. Why then would the problem be unsolvable so long as we have statements (1) and (2)?

Imagine Reason (talk) 07:26, 25 February 2008 (UTC)

(1) and (2) on their own aren't enough, you need to information in your first paragraph as well - is that what it means? --Tango (talk) 11:42, 25 February 2008 (UTC)

The first paragraph is a given, while the two statements are optional. I'm saying that if we have the two statements (the first paragraph is always there), we can solve the problem, but the review books say we can't even then. —Preceding unsigned comment added by Imagine Reason (talk • contribs) 17:14, 25 February 2008 (UTC)

Well, it's hard to comment on what the review book says without seeing it, but the problem you've posted is certainly solvable in the way you came up with. Black Carrot (talk) 06:32, 26 February 2008 (UTC)

Clarity

Working in dollars throughout...

Focusing on the given paragraph ONLY... We assume that equal amounts (L) of paint in litres are required no matter whether paint A or paint B is used and that if H hours are needed for paint A, then 4H/3 hours are needed for paint B.

Total cost of using paint A is 6L + 36H and the total cost of using paint B is 4.5L + 48H

Setting these equal to each other gives 6L + 36H = 4.5L + 48H and a solution of L = 8H

Therefore numerically, if L > 8H then paint B is cheaper and if L < 8H then paint A is cheaper.

Conclusion is that additional information is needed.

Incorporating ONLY statement 1 with the given paragraph... We need to make an assumption that the reduced volume of paint A needed (3/4 of the volume required by using paint B) results only in a cost saving on the paint, and NOT that less paint equals less hours of labour: after all, the same area needs covering and we know it takes a third longer with paint B. If this assumption is false and that less paint ALSO equals less labour hours then we have a slightly different problem.

Relative to the L litres required for paint A, paint B requires 4L/3 litres which now costs 6L (dollars).

Case 1.1 Less paint only required.

The expression for the cost of paint A remains 6L + 36H which in all cases is less than the cost for paint B of 6L + 48H.

Paint A is cheaper.

Case 1.2 Less paint and 3/4 of the previous amount of labour hours required (4/3 times the amount are now needed for B compared to A), i.e. relative to the H hours required for paint A, paint B requires 4/3 × 4/3 = 16/9 times the hours (cost = 36H × 16/9 = 64H).

The expression for the cost of paint A remains 6L + 36H which in all cases is less than the cost for paint B of 6L + 64H.

Paint A is even cheaper.

Statement 2 is irrelevant in determining which paint is cheaper to apply; it just gives the information to calculate the ACTUAL costs. All the problem requires though is a general 'which is cheaper', not by how much. So, as far as I can read it from the information provided, choosing paint A is cheaper in all cases if statement 1 is included with the given paragraph; statement 2 being irrelevant, as is the request for any additional information.

However, all this is blindingly obvious from the simple point that statement 1 means that the effective price of paint B is $6.00 per litre, identical to that of paint A. B takes longer to apply at an additional cost of $12 per hour for case 1.1 and $28 per hour for case 1.2 above based on the number of hours needed to apply paint A.

The only ambiguity is in calculating ACTUAL costs for paint B depending on case 1.1 or 1.2 above. Paint A costs $3840: Paint B either costs $5040 or $6640.

Please don't hesitate to contact me if you feel I have made any error in reading this situation! AirdishStraus (talk) 11:23, 26 February 2008 (UTC)

I agree that the wording is ambiguous, but I'm pretty sure it is saying that a gallon of paint B takes a third longer to apply than a gallon of paint A. Regardless, you're right--we only need the first statement to answer the question. I think that's actually what I got the first time I saw the problem, actually, but somehow I let it go. Now I've seen the same nonsensical explanation in a second book, I'm not happy. Imagine Reason (talk) 23:36, 26 February 2008 (UTC)

aquestion about closed surface

i am wondering about something,can we find or create aclosed surface where there all of the points inside that surface has adifferent distances to any of the all points of the surface?i mean here we cannot finde apoint inside the space of that surface has the same distance to at least 2points on the surface?this surface should not be for example like asphere because the center of the sphere is apoint that has asame distances to all of the points on the spherer`s surface.i hope my words is clear.thank you.Husseinshimaljasimdini (talk) 12:04, 25 February 2008 (UTC)

If I understand the question correctly, then I would think not, since you can just choose any two points on the surface and pick the midpoint between them. Well, if the surface isn't convex, you couldn't choose *any* two points, but there there will still be some points - just choose any line that passes through the interior of the region bounded by the surface and the intersection points of that line with the surface should do. I think as long as the surface does bound a region with a non-empty interior, there will be points in the interior equidistant to points on the surface. --Tango (talk) 13:41, 25 February 2008 (UTC)

Interesting question - I think the answer is no. I'll explain in two dimensions, you can easily expand the answer to three dimensions for surfaces..

Suppose I start my asymmetric ring at point A with distance r from the 'centre'.. then increasing r as I turn around the surface eg using polar coodinates (radius,angle) - the first thing I notice is that if I increase r with angle I cannot every decrease it since that would return r to a previously used value.. but to get back to r at angle=0 I must decrease.. Both can't be true so it's impossible..

Such a shape would be a spiral - spirals cannot be closed surfaces.. Did that make sense?83.100.158.211 (talk) 15:00, 25 February 2008 (UTC)

Even a spiral loses: take the midpoint of a small enough chord that it doesn't intersect the next layer (so as to avoid questions of containment). --Tardis (talk) 15:54, 25 February 2008 (UTC)

Didn't understand that - the spiral r=e^angle is single valued in r for all values of angle ?83.100.158.211 (talk) 15:59, 25 February 2008 (UTC)

The OP's request was for an open region $\Omega \in \mathbb {R} ^{n}\ni \forall x\in \Omega \,\not \!\exists (y,z)\subseteq \partial \Omega \;\;y\neq z\land \|y-x\|=\|z-x\|$. It's not enough for the "unique distances" clause to hold for one contained point. --Tardis (talk) 16:50, 25 February 2008 (UTC)

((For the spiral described above every point is at a unique distance, - am I wrong?))

You've confused me now - I understood the OP asked for a closed surface (hence not a spiral) - but said that excluding that 'closed surface' clause a spiral would be such a shape (if only in 2d) I see that a spiral fails for a single unique point in 3D ie a spiral surface has more than one point with a given radius. Was that what your original point was saying?83.100.158.211 (talk) 17:12, 25 February 2008 (UTC)

cf open region with Closed manifold - they asked for closed manifold eg 'closed surface'83.100.158.211 (talk) 17:42, 25 February 2008 (UTC)

I interpreted the question as taking place in $\mathbb {R} ^{3}$ where the closed manifold (ie. closed surface) bounds an open region. --Tango (talk) 17:59, 25 February 2008 (UTC)

still a bit confused about the 'chord' explanation - the answer is a definate 'no' anyway -

another way to look at it is to note that the closed surface must be (at least) double valued (of r in angle) - to be closed (in 2D) {and gets worse in 3d ie infinite values with same r for (angle1,angle2) could go on about being able to draw a loop/ring on the surface of a 3D closed surface at any point }

- which of course means it cannot be single valued as requested. are we answering the same question??? at least we are getting same answer83.100.158.211 (talk) 18:19, 25 February 2008 (UTC)

What function are you saying is multi-valued, $r(\theta )$ or $\theta (r)$? I assume the latter, since that's what we need, but your terminology is a little confusing (to me, at least). The OP (in the 2D case - I'm pretty sure he's actually asking about the 3D case, since you can't have a closed surface in $\mathbb {R} ^{2}$) requires that $\theta (r)$ be multi-valued for some choice of origin within the interior of the region bounded by the surface. That's just a restatement of the question, though, it doesn't prove anything. --Tango (talk) 18:39, 25 February 2008 (UTC)

Yes I wrote "(of r in angle)" is should of course have read "doubled valued in angle for a given r" - my mistake that was very unclear.

$\theta (r)$ is multivalued for not true.. or single valued if the original construct was possible - I think I was showing that when $\theta (r)$=singlevalued the surface cannot be enclosed .

Clearly for a given radius r there must be only one value of theta (2D) or (combiantion) theta,mu (3D) that gives that radius ie if r=fn(theta) then the function theta=fn^-1(r) must have a single value of theta for that value of r eg for a ellipse there are 4 values of theta that give a specific r, so I would have said that (in the case of the ellipse) theta is multivalued.

and multivalued = not possible83.100.158.211 (talk) 19:51, 25 February 2008 (UTC)

My argument for it's impossibility was of this kind:

given that the rate of change of radius with angle is x

x cannot be zero (since this gives a 'flat' region - ie two points have same r)

so x is either positive or negative

x cannot change sign since this would require x=0 at some point.

So x is either always positive or negative.

BUT r=fn(x) = fn(x+2pi) (for a continuous surface) (a full rotation) (equation A)

equation A cannot be true since d fn(x)/dx is not zero and never changes sign.

Therefor assuming that the example function is single valued ie x=fn^-1(r) is single valued for r (PROBABLY DID STATE THIS THE OTHER WAY ROUND at some point sorry)

Then fn(x) <> fn(x+2pi) - it's not a closed ring.

This expands easily to 3D since for a closed 3D surface any plane section through it (through the 'centre') must be a closed ring.

So single valued in x=fn^-1(r) and 'closed ring' are mutually exclusive.

I've expanded a bit on the original explanation (including the differentials) - but it's the same un-proof. I guess I wrote single valued for fn(x) somewhere before when it should have been fn^-1(r) - that would account for any confusion you are getting.

(whether r=fn(angle) is multivalued (or not) (ie non-convex surfaces) doesn't affect this proof)83.100.158.211 (talk) 20:06, 25 February 2008 (UTC)

I think x=0 is allowed, but only instantaneously, and you still couldn't have a change of sign (it would have to be a point of inflexion), so your basic conclusion is correct. Dealing with the 3D case by taking the intersection with a plane is a good idea (although, pedantically, the intersection must be a union of closed rings, but that doesn't affect anything). --Tango (talk) 20:13, 25 February 2008 (UTC)

Yes, forgot about an inflexion -good point.83.100.158.211 (talk) 20:18, 25 February 2008 (UTC)

Geometry Help: Mysterious Equation

So, I'm here studying for a test and one of the formula's in my notes makes no sense to me. I have completly forgotten how to use it and I don't have any examples. The formula is V=BiT or V=BLT. Either or.. does anyone know what the formula stands for.. —Preceding unsigned comment added by 80.148.24.98 (talk) 21:26, 25 February 2008 (UTC)

I expect we're going to need some context. What's the section it's in about? Is there any other mention of V, B, i/L and T near it? --Tango (talk) 21:39, 25 February 2008 (UTC)

No, nothing at all.. We're on a unit about volume and surface area of rectangular prisms and pyramids..... —Preceding unsigned comment added by 80.148.24.98 (talk) 21:42, 25 February 2008 (UTC)

I think you'll have to ask your teacher. I don't see any way we can identify one formula in isolation. The volume of a prism is Volume=Base area * Length, that gives you V=BL, but I don't know what the T could be. --Tango (talk) 22:38, 25 February 2008 (UTC)

The volume of a pyramid is base*height/3, so maybe that extra factor has something to do with the third letter. On the other hand, the volume of a prism is also length*width*height. So yeah, your teacher's probably the only person who knows what she meant. Or one of your classmates. Black Carrot (talk) 22:42, 25 February 2008 (UTC)

T = tallness? —Bkell (talk) 23:48, 25 February 2008 (UTC)

What I meant here, I guess, was "Volume = Breadth × Length × Tallness", though "Tallness" is a strange word to use for that. Maybe BLT? ;-) —Bkell (talk) 23:53, 25 February 2008 (UTC)

Perhaps it's just bad handwriting? V=bh (volume = base area x height). Imagine Reason (talk) 23:49, 25 February 2008 (UTC)

February 26

Measure Theory

Let C be a semi-algebra on a given set $\Omega$ and $\mu :C\rightarrow [0,\infty ]$ such that $\mu (\emptyset )=0$ and $\mu$ is sigma additive.

We then define $\mu ^{*}(A)=\inf \sum _{k=1}^{\infty }\mu (D_{k})$ where $A\subset (\cup D_{k})$ and each $D_{k}\in C$.

I am trying to prove that $\mu ^{*}(E)=\mu ^{*}(B\cap E)+\mu ^{*}(B^{c}\cap E)\,\forall B\in C$ and $\forall E\in P(\Omega )$.

One inequality is easy to show because E is a subset of $(B\cap E)\cup (B^{C}\cap E)$ and $\mu ^{*}$ is monotonic.

But any hints as to how to show $\mu ^{*}(E)\geq \mu ^{*}(B\cap E)+\mu ^{*}(B^{c}\cap E)$ would be appreciated.

$\sum \mu (D_{k})=\sum (\mu (D_{k}\cap B)+\mu (D_{k}\cap B^{c}))=\sum \mu (D_{k}\cap B)+\sum \mu (D_{k}\cap B^{c})$.

Furthermore, $(\cup D_{k})\cap X=\cup (D_{k}\cap X)$. Does this help? -- Meni Rosenfeld (talk) 00:41, 26 February 2008 (UTC)

This fact is necessary for proving the existence of the Caratheodory extension. The proof of it is very ugly and technical. See [1] starting from page 3. 128.139.226.37 (talk) 10:02, 28 February 2008 (UTC)

Are you sure this is relevant? There, they are starting with an outer measure and proving some things about it. Here, we start with a measure and construct from it what will later turn out to be an outer measure. Unless I am missing something, what I have written is one line away from a complete proof of the statement. -- Meni Rosenfeld (talk) 10:15, 28 February 2008 (UTC)

Grammars

Given {a, ba, bba, bbba} bⁿa|nEN
I calculated that S→aS → bSa → bbSa → bbbSa would be the grammar.
I wanted to make sure that I am on the correct path. NanohaA'sYuri^{Talk, My master} 00:04, 26 February 2008 (UTC)

I don't think that's right, because if "a" is in the set and "S→aS" is a rule, then you can get "aa", which isn't in the set. Am I misunderstanding the question? —Keenan Pepper 05:06, 26 February 2008 (UTC)

One way of reading the rule "S→aS" is the following:

If σ ∈ L(S), then also aσ ∈ L(S) – where L(S) here denotes the language produced using S as the start symbol.

So if bba ∈ L(S), as desired, then you would also get abba ∈ L(S), which would be wrong.

There is a context-free grammar for the language {a, ba, bba, bbba, ...} = {bⁿa|n∈N} with one nonterminal symbol and two production rules. I think the assignment calls for a context-free grammar. In a context-free grammar all rules have a single nonterminal symbol on the left-hand side. What you wrote, "S→aS → bSa → bbSa → bbbSa", does not look like (production rules of) a grammar. Production rules contain only one arrow. It looks more like the begin of a derivation. However, no context-free grammar could have a derivation step of the form aS → bSa. If at some stage of a derivation the string starts with a terminal symbol, like "a" here, then it will start with "a" in all following stages.  --Lambiam 06:40, 26 February 2008 (UTC)

Completing the Square

I was just looking at the article on completing the square and saw in the derivation of the quadratic formula from the equation $ax^{2}+bx+c=0\,\!$ that $\displaystyle {x^{2}+{\frac {b}{a}}x+\left({\frac {b}{2a}}\right)^{2}=\left({\frac {b}{2a}}\right)^{2}-{\frac {c}{a}}\to \left(x+{\frac {b}{2a}}\right)^{2}={\frac {b^{2}-4ac}{4a^{2}}}}$. What I am wondering (although it is probably very simple) is where does the ${\frac {b}{a}}x$ go from the first equation to the next. Much appreciated, Zrs 12 (talk) 02:00, 26 February 2008 (UTC)

Try expanding out $\left(x+{\frac {b}{2a}}\right)^{2}$. Remember, when you expand something like $(L+R)^{2}$, the result is not $L^{2}+R^{2}$, but $L^{2}+2LR+R^{2}$ (the FOIL rule). —Bkell (talk) 02:34, 26 February 2008 (UTC)

Thanks Bkell, I understand it a little better now. Expanding it gives $x^{2}+{\frac {b}{a}}x+{\frac {b}{a}}$, right? Where does ${\frac {b}{a}}x$ go on the right side? Many thanks, Zrs 12 (talk) 03:04, 26 February 2008 (UTC) I got it. When expanded $\left(x+{\frac {b}{2a}}\right)^{2}=x^{2}+{\frac {b}{a}}x+{\frac {b^{2}}{4a^{2}}}$. That was easy enough. I just had to think and write it out. Thanks! Zrs 12 (talk) 03:25, 26 February 2008 (UTC)

Glad to help. —Bkell (talk) 04:23, 26 February 2008 (UTC)

Geometry Help

In a triangular prism with the width 3, and each diagonal side 3 and a length of 7, how would you find the surface area and volume?

--Devol4 (talk) 07:41, 26 February 2008 (UTC)

Triangular prism helps with the volume. Zain Ebrahim (talk) 11:09, 26 February 2008 (UTC)

Well, if I understand the question correctly, your prism can be interpreted geometrically as two equilateral triangles joined by rectangles. To find the volume, just multiply cross-sectional area (the triangle) by the length, and to find surface area, just find the area of all components and sum them. -mattbuck (Talk) 11:10, 26 February 2008 (UTC)

What I don't understand in Devol4's question is what exactly the width, the diagonal side, and the length means in such a prism. – b_jonas 10:52, 29 February 2008 (UTC)

egytian method

divide 16 by 9 —Preceding unsigned comment added by 41.205.189.9 (talk) 15:27, 26 February 2008 (UTC)

16/9. =) –King Bee ^{(τ • γ)} 15:35, 26 February 2008 (UTC)

I assume the OP means the "Egyptian method". Here's one explanation of the process. — Lomn 15:49, 26 February 2008 (UTC)

We do have an article on this, check out Egyptian fraction. After having done the calculations, you can check your work using these calculators. --NorwegianBlue ^talk 18:37, 26 February 2008 (UTC)

maths trick i just found out

1. think of a number and put it in your calculator [make sure you remember it]
2. add 5
3. multiply your answer by 6
4. divide your answer by 2
5. subtract your answer by 15
6. divide your answer by the number you first thought of

• answer is always = 3
• check source at [doxadeocollege.co.za] —Preceding unsigned comment added by 196.207.47.60 (talk) 16:15, 26 February 2008 (UTC)

That is true. It's not really a trick so much as obfuscation - simply writing it out on a piece of paper and you can see the following steps:

• $S_{2}:x\rightarrow x+5$
• $S_{3}:x+5\rightarrow 6(x+5)$
• $S_{4}:6(x+5)\rightarrow {\frac {6(x+5)}{2}}=3(x+5)$
• $S_{5}:3(x+5)\rightarrow 3(x+5)-15=3x+15-15=3x$
• $S_{6}:3x\rightarrow {\frac {3x}{x}}=3$

-mattbuck (Talk) 16:30, 26 February 2008 (UTC)

And what if the number I choose is 0? –King Bee ^{(τ • γ)} 16:36, 26 February 2008 (UTC)

Define 0/0:=3. That's afterall one reason 0/0 is indeterminant, because processes that result in 0/0 could in fact by any real number, or infinity (by appropiate choice of the process). 130.127.186.122 (talk) 17:15, 26 February 2008 (UTC)

How is "obfuscation" not a more specific word for "trick"? In fact, most actual magic tricks use nothing but clever obfuscation. It's so hard to make the hand quicker than the eye, but so easy to find gullible spectators. Take the classic "cup and ball" routine, for instance. You keep the mark's eyes on the cup and ball, and get caught clumsily trying to remove the ball, so that his attention will be on you when your friend picks his pocket. Black Carrot (talk) 17:32, 26 February 2008 (UTC)

First off, it sounds better, and to me, a mathematical trick is some way of simplifying a result or theory. For instance, an easy way to calculate the multiplicity of a pole is to find the multiplicity of the zero at that point of the function's reciprocal - that I would consider a mathematical trick. -mattbuck (Talk) 18:38, 26 February 2008 (UTC)

Trig equation

(Note: this isn't homework, but I'd appreciate a hint rather than being told the answer) I'm having trouble with solving $\!\sin 2x+\sin ^{2}x=1$. We've not covered this in class, so the only things I can see ($\!\sin 2x=2\sin x\cos x=2\sin x\sin(x+{\frac {\pi }{2}})$ or $\!\sin 2x-\cos ^{2}+1$ or some combination thereof) don't seem to be awfully helpful. Any pointers – is there an identity I need that I don't have? Angus Lepper^{(T, C, D)} 20:58, 26 February 2008 (UTC)

Combine the double-angle identity $\sin 2x=2\sin x\cos x\,\!$ with the Pythagorean identity $\sin ^{2}x+\cos ^{2}x=1\,.$  --Lambiam 21:09, 26 February 2008 (UTC)

To 'idiot proof' the above... (almost full solution commented out by Meni Rosenfeld) AirdishStraus (talk) 22:28, 26 February 2008 (UTC)

Why would you give a full solution when the OP explicitly asked you not to? --Tango (talk) 22:41, 26 February 2008 (UTC)

Or call them an idiot? Black Carrot (talk) 23:04, 26 February 2008 (UTC)

Tango is right. I have commented out the solution. -- Meni Rosenfeld (talk) 10:15, 27 February 2008 (UTC)

I agree Tango is right. I got carried away doing the near full solution and had forgotten the 'hint only' request. My apologies. On a seperate note... 'idiot-proofing' something is a mere colloquialism for unambiguous clarification; I never used the word idiot to directly describe anyone and wouldn't ever use it in ANY case. Maybe Black Carrott hasn't come across this term before? AirdishStraus (talk) 11:15, 27 February 2008 (UTC)

I have no doubt that you didn't intend to insult the OP, but using the phrase "idiot-proof" in that way could very easily be misinterpreted, especially if someone isn't a native English speaker, so it's probably best to avoid it. --Tango (talk) 13:06, 27 February 2008 (UTC)

Triangle

The angle bisector of alpha w_α, the median of s_b and the altitude of c h_c of a acute-angled triangle ABC cross in one point, if w_α, the side BC and a circle around the point H_c, which goes through the corner A, have also a common point together.

How can one proof this statement? --85.178.22.17 (talk) 23:18, 26 February 2008 (UTC)

You need to do it one step at a time

1. First find the intersection of the "angle bisector of alpha w_α and the median of s_b" as a function of the parameters of a generalised triangle eg (0,0) (a,b) (c,d) as the points of the triangle.

2. Then find the coodrinates of the crossing point of "the median of s_b and the altitude of c h_c of a acute-angled triangle ABC"

Next you need to show that the two equations abtained above give the same point if the third premise is true.. ie that "w_α, the side BC and a circle around the point H_c, which goes through the corner A, have also a common point together."

3. I'm not sure what H_c is here - but if you do you should be able to generate some sort of parametric equation relating the vertexs of the triangle. {{ Sound's like Hc is the centre of the circle that goes through A and also passes through the intersection of (w_α and the side BC ) - this makes somewhere Hc along a line then}}

Then you need to show that if for instance the equation obtained in 1 is true at the same time as the equation obtained in 3. then equation 2 is also true.87.102.93.245 (talk) 11:43, 27 February 2008 (UTC)

HINT for the "angle bisector of alpha w_α, the median of s_b and the altitude of c h_c" to cross at one point the triangle must be at least iscocoles since we have three lines intersecting at one point and two of them "angle bisector of alpha w_α" "and the altitude of c h_c" already cross at A.. Because the median cannot pass through A, the crossing of the "angle bisector of alpha w_α" "and the altitude of c h_c" must give a line (ie they are the same line) and not just a single point. therefor ac=ab I hope that is enough to get you finished.87.102.93.245 (talk) 12:26, 27 February 2008 (UTC)

February 27

The washer method (disk integration)

I need to find the volume of a solid of revolution. The 2-dimensional area is bounded by y = √x, y = 0, and x = 4, and it is spun around the axis x = 6. I know that my outer radius is 6 - x, but since I'm integrating in terms of y (since the axis is parallel to the y-axis), I should write that as R(y) = 6 - y². But I can't figure out how to express the inner radius r(y) as a function, so that I can plug it into the formula V = π∫_a^b[R²(y) - r²(y)]dy! (I believe a and b = 0 and √6 respectively.) Can anyone offer any help? Thanks, anon. —Preceding unsigned comment added by 70.19.34.80 (talk) 01:10, 27 February 2008 (UTC)

Since the inner wall is a vertical line, the inner radius is constant with respect to y, and in this case is 2. Black Carrot (talk) 03:51, 27 February 2008 (UTC)

Also, if you're integrating with respect to y, it should run from 0 to 2. You should draw some diagrams if you haven't already. Black Carrot (talk) 03:54, 27 February 2008 (UTC)

Friend challenged me

IF A box of nutmeg ways 1 1/3 ounces and there is 20 scoops how much is each scoop? —Preceding unsigned comment added by Inutasha De Fallen (talk • contribs) 02:51, 27 February 2008 (UTC)

Are you asking fot the ounces per one scoop where you are told that there are 1.333... ounces per 20 scoops? --hydnjo talk 03:05, 27 February 2008 (UTC)

It has to end in as a fraction like 1/20 —Preceding unsigned comment added by Inutasha De Fallen (talk • contribs) 03:09, 27 February 2008 (UTC)

Well you know that 1 1/3 equals 4/3, right? And you know that dividing by 20 is the same as multiplying by 1/20, right? And how to reduce fractions, right? So, what more help do you need? --hydnjo talk 03:42, 27 February 2008 (UTC)

it is a pain in the ass... im not really understanding it it would be 0.065.. but thats not it! damn this is upseting —Preceding unsigned comment added by Inutasha De Fallen (talk • contribs) 03:45, 27 February 2008 (UTC)

You're close, but not there yet. Come on, I would expect a 12 year old to know this, so why a 14 year old like you don't? --antilived^{T | C | G} 05:01, 27 February 2008 (UTC)

Shiβe i Fell stupid...i Got it. do simple things ever Kick your ass? —Preceding unsigned comment added by Inutasha De Fallen (talk • contribs) 05:27, 27 February 2008 (UTC)

Nope. I've never made a mistkske in my life. Black Carrot (talk) 06:42, 27 February 2008 (UTC)

Eh? well im gonna have to raise the bullshit flag on that one. >.< —Preceding unsigned comment added by Inutasha De Fallen (talk • contribs) 07:05, 27 February 2008 (UTC)

Look carefully at Black Carrot's spelling, and see Sarcasm. —Keenan Pepper 13:07, 27 February 2008 (UTC)

Chain Rule and Higher Derivative

Let

$z=f(x,y)$

$x=g(s,t)$

$y=h(s,t)$

Then

${\frac {\partial z}{\partial t}}={\frac {\partial z}{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial z}{\partial y}}{\frac {\partial y}{\partial t}}=\left({\frac {\partial }{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial }{\partial y}}{\frac {\partial y}{\partial t}}\right)z$ (1)

${\frac {\partial ^{2}z}{\partial t^{2}}}={\frac {\partial }{\partial t}}{\frac {\partial z}{\partial t}}={\frac {\partial }{\partial t}}\left[\left({\frac {\partial }{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial }{\partial y}}{\frac {\partial y}{\partial t}}\right)z\right]=\left({\frac {\partial }{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial }{\partial y}}{\frac {\partial y}{\partial t}}\right){\frac {\partial z}{\partial t}}$ (2)

Replace (1) into (2)

${\frac {\partial ^{2}z}{\partial t^{2}}}=\left({\frac {\partial }{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial }{\partial y}}{\frac {\partial y}{\partial t}}\right)\left({\frac {\partial }{\partial x}}{\frac {\partial x}{\partial t}}+{\frac {\partial }{\partial y}}{\frac {\partial y}{\partial t}}\right)z$ (3)

${\frac {\partial ^{2}z}{\partial t^{2}}}=\left[{\frac {\partial ^{2}}{\partial x^{2}}}\left({\frac {\partial x}{\partial t}}\right)^{2}+{\frac {\partial ^{2}}{\partial x\partial y}}{\frac {\partial x\partial y}{\partial t^{2}}}+{\frac {\partial ^{2}}{\partial y\partial x}}{\frac {\partial y\partial x}{\partial t^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}\left({\frac {\partial y}{\partial t}}\right)^{2}\right]z$ (4)

${\frac {\partial ^{2}z}{\partial t^{2}}}={\frac {\partial ^{2}z}{\partial x^{2}}}\left({\frac {\partial x}{\partial t}}\right)^{2}+{\frac {\partial ^{2}z}{\partial x\partial y}}{\frac {\partial x\partial y}{\partial t^{2}}}+{\frac {\partial ^{2}z}{\partial y\partial x}}{\frac {\partial y\partial x}{\partial t^{2}}}+{\frac {\partial ^{2}z}{\partial y^{2}}}\left({\frac {\partial y}{\partial t}}\right)^{2}$ (5)

Is every step above correct? - Justin545 (talk) 07:07, 27 February 2008 (UTC)

No.

${\frac {\partial z}{\partial x}}{\frac {\partial x}{\partial t}}$

is not the same as

$\left({\frac {\partial }{\partial x}}{\frac {\partial x}{\partial t}}\right)z.$

For example, if z = x = t, the former evaluates to , and the latter to Think of ${\tfrac {\partial }{\partial x}}$ as an operator, and abbreviate it as D. Also abbreviate ${\tfrac {\partial x}{\partial t}}$ as U. Then in your first line of equations you replaced by .  --Lambiam 09:46, 27 February 2008 (UTC)

Reply to Lambiam: You are right indeed. According to your answer, does that mean (5) is also an incorrect result? What is the correct answer if (5) is incorrect? (i.e. what is ${\frac {\partial ^{2}z}{\partial t^{2}}}$ equal to?) Thanks! - Justin545 (talk) 12:15, 28 February 2008 (UTC)

If you take and , then , and the result should be 2. However, the right-hand side of (5) evaluates to 0. I think two terms have gone AWOL. Applying the product rule,

${\frac {\partial }{\partial t}}\left({\frac {\partial z}{\partial x}}\cdot {\frac {\partial x}{\partial t}}\right)={\frac {\partial }{\partial t}}{\frac {\partial z}{\partial x}}\cdot {\frac {\partial x}{\partial t}}+{\frac {\partial z}{\partial x}}\cdot {\frac {\partial }{\partial t}}{\frac {\partial x}{\partial t}}.$

The second term seems to be missing, and likewise with x replaced by y. Taking them together, the following should be added to the right-hand side of (5):

$+{\frac {\partial z}{\partial x}}\cdot {\frac {\partial ^{2}x}{{\partial t}^{2}}}+{\frac {\partial z}{\partial y}}\cdot {\frac {\partial ^{2}y}{{\partial t}^{2}}}.$

 --Lambiam 21:59, 28 February 2008 (UTC)

According to your reply

${\frac {\partial }{\partial t}}\left({\frac {\partial z}{\partial x}}\cdot {\frac {\partial x}{\partial t}}\right)={\frac {\partial }{\partial t}}{\frac {\partial z}{\partial x}}\cdot {\frac {\partial x}{\partial t}}+{\frac {\partial z}{\partial x}}\cdot {\frac {\partial }{\partial t}}{\frac {\partial x}{\partial t}}.$ (6)

Then

${\frac {\partial }{\partial t}}\left({\frac {\partial z}{\partial x}}\cdot {\frac {\partial x}{\partial t}}\right)={\frac {\partial }{\partial t}}{\frac {\partial z}{\partial x}}\cdot {\frac {\partial x}{\partial t}}+{\frac {\partial z}{\partial x}}\cdot {\frac {\partial ^{2}x}{\partial t^{2}}}$ (7)

But what does ${\tfrac {\partial }{\partial t}}{\tfrac {\partial z}{\partial x}}\cdot {\tfrac {\partial x}{\partial t}}$ evaluate to? Does it evaluate to 0 or does it evaluate to the rest of terms on the right-hand side of (5)? Thanks! - Justin545 (talk) 03:56, 29 February 2008 (UTC)

The latter (or, more precisely, half of these terms – there is also the same with x replaced by y).  --Lambiam 00:50, 1 March 2008 (UTC)

The dy/dx notation is not exactly a fraction, although often you can treat it as a fraction (chain rule, integration by substitution) --wj32 ^t/c 05:47, 28 February 2008 (UTC)

Reply to wj32^t/c: That is ture. I think (well, I'm just a layman) Leibniz notation is somewhat confusing to me. I didn't see (yes, I am not experienced enough) any formal introduction which states when can I treat it as a fraction and when can't I after I learned the derivative for many years. It seems to be a mystery to me! :) - Justin545 (talk) 12:15, 28 February 2008 (UTC)

Recursively defined sets

I want to generate the first several iterations of this set, but I feel I'm not doing it correctly; my notes are sparse on certain details:

Defintion of $C\,$.

$C\subseteq \Sigma$ where $\Sigma \,$ is the set of all strings over $\lbrace 1,2,3\rbrace \,$

Basis step:

$\lbrace 1,2\rbrace \subseteq C$

Recursive step:

$w\in C$ $\rightarrow$ $3\cdot w\in C$

$w_{1},w_{2}\in C$ $\rightarrow$ $w_{1}:w_{2}\in C$

$w\cdot 1\in C$ $\rightarrow$ $2\cdot w\in C$

The dot operator and the colon (:) operator represent character concatentation and string concatentation respectively.

It would follow, in my mind that the first few iterations of this set would go:

{1, 2}

{1, 2, 31, 11, 21}

{1, 2, 31, 11, 21, 32, 12, 22}

e.t.c

My main worry is this:

I assume w(1) and w(2) in the second part of the recursive step start at 1, 1 and go up 1, 2 then 2, 1 then 2, 2 and so on? Or do they have to be seperate values each time? Damien Karras (talk) 08:03, 27 February 2008 (UTC)

If I understand your last question correctly, the answer is that $w_{1}$ and $w_{2}$ don't have to be different. I'll guide you through the first iteration. You know that 1 ∈ C and 2 ∈ C. By rule 1, from 1 ∈ C it follows that 31 ∈ C, and from 2 ∈ C it follows that 32 ∈ C. We now apply rule 2: From 1 ∈ C and 1 ∈ C it follows that 11 ∈ C; from 1 ∈ C and 2 ∈ C it follows that 12 ∈ C; likewise 21 ∈ C and 22 ∈ C. Rule 3 doesn't give us anything new now. So the first iteration will be $\{1,2,31,32,11,12,21,22\}\subseteq C$. -- Meni Rosenfeld (talk) 09:51, 27 February 2008 (UTC)

(after edit conflict) I would interpret your rule #2 of the "recursive step" as being universally quantified over all w₁ and w₂ that are in C. So if 2 is a member of C_n in iteration n, then 22 is a member of the next generation C_n+1. Is that an answer to your question (which I do not fully understand)?

Because of rule #3, you can start the whole process by taking C₀ = {1}. Then the first few iterations give this:

C₀ = {1}

C₁ = {1, 31, 11, 2}

C₂ = {1, 31, 11, 2, 331, 311, 32, 131, 111, 12, 3131, 3111, 312, 1131, 1111, 112, 21, 231, 211, 22, 23}

 --Lambiam 10:10, 27 February 2008 (UTC)

I think what I meant my overall question to be is this: For each iteration, do I pick the value of w, w1 and w2 just once? So the first iteration would be: w = 1, w1 = 1 and w2 = 1 and then I would move to the next iteration.

OR do I consider all possible values of w, w1, w2 for each iteration? In which case I would do, for instance, step one for w = 1 and w = 2, step two would be done for w1 = 1, w1 = 2, w2 = 1, w2 = 2 e.t.c

Furthermore, are the new elements of the set generated on each step? Because if so that would mean that w1 and w2 take on even more values before the end of the total recursive step (unless of course the former point is true). Damien Karras (talk) 10:41, 27 February 2008 (UTC)

I am still confused by your questions, but I have a better idea what it is you are asking. To find $C_{n+1}$, you apply every possible rule to the items in $C_{n}$ and only them. So if you start with $C_{0}=\{1,2\}$, applying every possible rule to $C_{0}$ gives you $C_{1}=\{1,2,31,32,11,12,21,22\}$. You then apply every possible rule to $C_{1}$ to get $C_{2}$, and so on. In each step, you only apply rules to elements you have had in the previous step, not to those you have generated recently. -- Meni Rosenfeld (talk) 11:44, 27 February 2008 (UTC)

Thanks both of you! That's exactly the answer I was looking for, sorry the question was phrased poorly. Just to check will I be right in saying that the start of $C_{2}=\lbrace 1,2,31,32,11,12,21,22,331,332,311,312,321,322,3131,3132,3111,3112,3121...\rbrace \,$? (I won't write the whole thing out because it appears to be massive!) Damien Karras (talk) 12:27, 27 February 2008 (UTC)

This looks correct. Don't forget that $C_{2}$ also has elements such as 131, the concatenation of 1 and 31 which are both in $C_{1}$. -- Meni Rosenfeld (talk) 14:58, 27 February 2008 (UTC)

This is what I get starting from {1, 2}:

C₀ = {1, 2}

C₁ = {1, 2, 31, 32, 11, 12, 21, 22}

C₂ = {1, 2, 31, 32, 11, 12, 21, 22, 331, 332, 311, 312, 321, 322, 131, 132, 111, 112, 121, 122, 231, 232, 211, 212, 221, 222, 3131, 3132, 3111, 3112, 3121, 3122, 3231, 3232, 3211, 3212, 3221, 3222, 1131, 1132, 1111, 1112, 1121, 1122, 1231, 1232, 1211, 1212, 1221, 1222, 2131, 2132, 2111, 2112, 2121, 2122, 2231, 2232, 2211, 2212, 2221, 2222, 23}

 --Lambiam 23:22, 27 February 2008 (UTC)

modular algorithm

how can i solve this problem?

[ x1= a (mod 100) , a= 20 (mod 37) ]

[ x2= b (mod 100) , b= 15 (mod 37) ]

[ x3= c (mod 100) , c= 18 (mod 37) ]

must be ; x2= a.k + y (mod100)

and

x3= b.k + y (mod100)

i need find b and c.. thank you best regards.. Altan B. —Preceding unsigned comment added by 85.98.230.220 (talk) 10:20, 27 February 2008 (UTC)

x1 does not play a role. Eliminating x2 by combining with gives us Likewise eliminating x3 results in We can combine these to eliminate y and obtain or, equivalently,

b(k + 1) ≡ c + ak (mod 100).

Using the (mod 37) congruences given for a and b, and c, we rewrite these as and Substituting that in the equation brings us to:

(37β + 15)(k + 1) ≡ 37γ + 18 + (37α + 20)k (mod 100).

We now multiply both sides by the multiplicative inverse of 37 (mod 100), which is 73 (since This results in

(β + 95)(k + 1) ≡ γ + 14 +(α + 60)k (mod 100).

Bringing γ to one side,

γ ≡ (β + 95)(k + 1) + (99α + 40)k + 86 (mod 100).

Now pick any values for α, β, k, and a new variable n, and compute

γ = (β + 95)(k + 1) + (99α + 40)k + 86 + 100n,

b = 37β + 15,

c = 37γ + 18.

For example, choosing α = β = 0, k = 12, n = −18 gives the solution b = 15, c = 55.  --Lambiam 12:32, 27 February 2008 (UTC)

thank you for your cares, but this solution is not that i want.. 'cause i also know this solving system but, this system isnt exact solution. i will use this formula in the big machine ( this machine has 17 billions as modular ) as you know i cant expect(or according to your solution choosing) all value from my brain. thank you best regards Altan B. —Preceding unsigned comment added by 85.98.230.220 (talk) 14:51, 27 February 2008 (UTC)

Affedersiniz ama yukarıda yazdıklarınızı pek anlayamıyorum.  --Lambiam 19:22, 27 February 2008 (UTC)

bu formul uzerine makina yapacagım. her seferinde nasıl tek tek su rakam olsun bu rakam olsun diye secme yaptıracagım makinaya? bana net kesin formul lazım , cunku makinanın ilk mod alması 17 milyar uzerinden olacak. yani 37 moda inince ikincisinde her rakam icin 435 milyon alternatif olacak. kesin net formul olması illa ki sart. ilginiz icin simdiden tesekkurler Altan B —Preceding unsigned comment added by 85.98.230.220 (talk) 09:01, 28 February 2008 (UTC)

Verdiğim cevap, "b = 15, c = 55" , yazdığınız probleme çözümdür, ve bu kesin bir formüldür. Acaba yazdığınız problem sadece daha büyük bir gruptan tek bir örnek mi? Bu durumda, genel problem nedir?  --Lambiam 22:27, 28 February 2008 (UTC)

shapes and projections

I was wondering about something and I need abig help to correct my understanding to this issue.I am going to put my questions in two parts.part1:assume that we have ashape in two dimensions,like asphere,obviously we can determine the projections of that sphere on (x-y,y-z,z-x)planes, which they are going to be circles no matter how we rotate the sphere, the projections will be always circles.If we apply now the same thing on acone then we would have adifferent projections depend on how we rotate or fix the cone in two dimensions,we can get for example,acircle in(x-y)plane,atriangle in(y-z)plane and another triangle in(x-z)plane.we can determine the projections depend on the shape and its position or you can say the shape and the position vector between the origin and achosen point belong to that shape.Now my question is,can we determine the shape by looking at its projections?if yes,then what is the shape in two dimensions that has the projections,(circle,square and apoint)?p.s:you can improvise another projections.some times we can describe different shapes in 2 dimentions by looking at the projections,for example,assume we have 3identical circles as aprojections,one may say that the shape is atwo dimnensional sphere,or the shape could be 3 identical circles perpendicular to each other and sharing the same center. Part2:if we are able to say that the sphere in three dimensions has aprojection in two dimensions which is asphere too then for the cone in three dimensions what its projection in two dimensions going to be?IF THE ANSWER IS ACONE TOO NO MATTER HOW WE ROTATE OR CHANGE ITS POSITION IN 3 DIMENTIONS THEN IT SHOULD BE ASPHERE, because the sphere is the only shape that maintains its projections in all dimensions>2

Further discussion: Assume we have ashape in 3dimensions where its projections in 2dimensions as follows, (x,y,z)=s1,(x,y,w)=s2,(x,z,w)=s3&(y,z,w)=s4.The projections in one dimension would be,

[s1:(x,y)=a1,(x,z)=a2,(y,z)=a3],[s2:(x,y)=a1,(x,w)=a4,(y,w)=a5],[s3:(x,z)=a2,(x,w)=a4,(z,w)=a6],[s4:(y,z)=a3,(y,w)=a5,(z,w)=a6].We have 6 one dimensional projections .Now assume we have acone in 3 dimentions,we can finde some conditions to make that cone describable,for example ,if the con`s peak is located at the origin and the position vector between the peak and the center of the cone`s makes an angle=pi\4 with each of(x,y,z,w),then a1=a2=a3=a4=a5=a6 and each one of the six projections seems to be having ashape of sector of acircle.on the other hand if,a1=a2=a3=a4=a5=a6 ,shouldnot we get asphere in 3dimensions?my point here is,can we say that all different shapes in 3 dimensions with no preconditions have ashape of sphere???Husseinshimaljasimdini (talk) 12:05, 27 February 2008 (UTC)

Part 1 - you seem to have made a sort of mistake eg if I project a set of 3 circles orthogonal (at right angles) to each other onto the three planes I get a circle with a cross inside - not a filled circle as you might expect from the sphere.(Maybe this is worth thinking about?)

As for your question 'can we determine a shape from it's projections' - it might depend - the answer seems to be yes if we assume the shapes are all quadratics, ... but much more complicated shapes might project as for example 3 circles along one set of coordinate axis (suggesting a sphere) - but produce different shapes along a different set of coordinate axis.87.102.93.245 (talk) 12:39, 27 February 2008 (UTC)

Your question "(if yes),then what is the shape in two dimensions that has the projections,(circle,square and apoint)" - seems impossible - because if the shape projects onto a point it can only be a point or a line - therefor a circular or square projection is impossible...

(I removed a space from your question 2 since the text was going way off the screen..)87.102.93.245 (talk) 12:39, 27 February 2008 (UTC)

(edit conflict) I would guess in "3 identical circles perpendicular to each other and sharing the same center", he means *discs* not circles. Then the projections are the same as for a sphere. An obvious example of shapes that can't be distinguished by their projections are a sphere and a sphere containing another shape - you can't "see through" the sphere, so you'll never detect the contained shape. With suitable restrictions on the shapes (for example, 245's idea of restricting to quadratics) might work, but with general shapes there are all kinds of things that can go wrong. --Tango (talk) 13:21, 27 February 2008 (UTC)

In general, you cannot reconstruct the 3D-shape from three orthogonal 2D-projections. One shape that has three unit disks as projections is the union of these three disks:

{(x,y,z) | }.

Among the convex shapes, one solution is the unit ball {(x,y,z) | }. The largest solid object giving the same three projections is given by the intersection of three cylinders:

{(x,y,z) | }.

This is strictly larger than the unit ball: it contains the point which is outside the unit ball.  --Lambiam 13:18, 27 February 2008 (UTC)

Linear equations

Hello =], I'm just having some problems with a few linear equations and this text-book is terrible example wise. Ive had no trouble with the equations preceding it:
$5(y-2)=12$
$5y-10=12$ (Expand Brackets)
$5y=22$(Add 10 to both sides)
${\frac {5y}{5}}={\frac {22}{5}}$(Divide both sides by 5)
$y={\frac {22}{5}}$


Can the same method be applied to the equation below? As far my answers have been incorrect:
${\frac {x}{2}}+3=5$
${\frac {x}{2}}=2$ (Subtract 3 from both sides)
This looks finished but it is evidential that I have done something wrong, a point in the right direction will be appreciated.


Also something like this, is my answer correct?
$5x+4\leq 0$
$5x\leq -4$ (subtract 4 from both sides)

Then this is where i get lost do i divide or perform some other operation? —Preceding unsigned comment added by 58.165.62.73 (talk) 13:06, 27 February 2008 (UTC)

You are right as far as you've gone in part 2, but you haven't finished. I assume you want to know what x on its own is, and you have x/2. Double each side. As for part 3, yes, you can just divide through, but be careful - if you divide or multiply through by a negative number, you have to invert the inequality. -mattbuck (Talk) 13:13, 27 February 2008 (UTC)

Mattbuck, note than he only subtracted 4 from each side of the inequality. Visit me at Ftbhrygvn (Talk|Contribs|Log|Userboxes) 14:50, 27 February 2008 (UTC)

Yes, but the end result presumably asks for $x\leq {\frac {-4}{5}}$, which requires dividing through, and it's a note about multiplying by negative numbers generally. -mattbuck (Talk) 15:04, 27 February 2008 (UTC)

If you think about what these equations and expressions mean, you'll figure out how to solve these questions. Doing the same thing to both sides is just a simplification of what you're doing to two quantities--like other people have pointed out with multiplying/dividing by negative numbers, it's not enough. Imagine Reason (talk) 23:51, 27 February 2008 (UTC)

What you must remember about these types of problems is that 1.) your solving for a single variable (x in the above, and y above that) and 2.) That means isolating that variable so you get something like x=(whatever) or y=(whatever) etc. . To do that, you need to apply the inverse operations to those that act on the variable, take ${\frac {x}{2}}+3=5$ for example, for this problem, our unknown (variable) is x, and our equation says that, x is divided by 2 and then 3 is added to it to give us 5 as an "answer" our question however is what is x (our unknown) so we need to get the equation above to look like x=(whatever). First of what operation happens to x last? Remember the Order of operations? Try rereading the part in bold letters. It should be noted that when solving for a variable you apply the reverse of the order of operations to remove those terms from which ever side your variable is on to isolate the variable. A math-wiki (talk) 11:43, 1 March 2008 (UTC)

Sorting a list of points for "in order of closest points"

Hi, suppose I have a list of 3 dimensional points eg (x_n,y_n,z_n) in random order and I want to try to sort them so that closer points are nearer to each other in the list.

The only idea I had was to (after probably writing the numbers in binary form) would be to covert each set of numbers into a single 'string' in which the most significant digits are group together

eg in decimal (1234,6789,-5500) would convert to ( (1,6,-5) , (2,7,-5) , (3,8,0) , (4,9,0) ) and then sorting that in order from left to right... This would put points with 1000<x<=1999 close together in the list..

This works a bit, (but still gives more priority to x than y etc - working in binary reduces this effect a bit)

Can anyone suggest any improvements to this method, or a better method? Note that I only require the list be 'improved' so that nearer points be nearer in the list - it doesn't have to give a perfect sort in perfect order or nearness (that seems impossible anyway) (You can assume I don't need any help with the sort in order of linear magnitude algorhytm)

Any ideas welcome. Just the concept would do.. Thanks87.102.93.245 (talk) 16:58, 27 February 2008 (UTC)

Imagine that your points are uniformly distributed within a cube. How are you going to "walk" around the cube and visit each point? One way is to walk along y=0, z=0 from x=0 to x=N and then back along y=1 z=0 from x=N to x=0 etc. After a while you have visited all of the z=0 plane in a passably efficient manner, but the rest of z hasn't been started. so x=0, y=0, z=1 which is close to x=0, y=0, z=0 is N^2 points away from the that coord. In other words, unless the data has some known clustering that you can take advantage of, there isn't going to be a simple solution to your question. -- SGBailey (talk) 17:56, 27 February 2008 (UTC)

Is your goal to be able to quickly find the nearest points to a given point? If so, see kd-tree. --Trovatore (talk) 18:02, 27 February 2008 (UTC)

right thanks - I already had a sort of octree (if I used binary number) - this 'kd-tree' looks like a possible improvement (though a little more analysis of the data is require - it should give a better list)Thanks.87.102.93.245 (talk) 18:37, 27 February 2008 (UTC)

If you really need a single ordering rather than a data structure, you could try sorting them in their ordering along a Space-filling curve. For 3d variants of the Hilbert curve each comparison of the sorting algorithm can be done by finding the most significant bit at which any of the coordinates of the two compared points differ. —David Eppstein (talk) 18:24, 27 February 2008 (UTC)

Thanks, looks like a good possibilty. I've often wondered if I'd ever find a use for those 'peanut curve' things . Today was that day. I'm not convinced that the hilbert curve is better (than a bsp tree) in terms of points that are just on either side of 0.5 (ie halfway accross) - probably averages out to similarily as good (please correct me if you are more expert than me - which is likely..) Still very interesting though.87.102.93.245 (talk) 18:44, 27 February 2008 (UTC)

Assuming you use in-order depth-first traversal, a bsp tree does not tell you by itself in what order the children of a node should be put. If you partition by cutting up the cube at each level in eight equally sized cubes, it is possible to order the children at each node in such a way that the in-order traversal is the same as traversal along a 3D Hilbert curve, which is the best you can get with a uniform recursive method. But other orderings are not dramatically worse, at most a factor 2 or so in the path length, which is O(n^2/3).  --Lambiam 20:25, 27 February 2008 (UTC)

This sounds very much like the Travelling Salesman Problem, which has been extensively studied. -- BenRG (talk) 16:04, 28 February 2008 (UTC)

It's only vaguely similar - the difference here amoungst others is that the distances are the square root of sums of squares rather than a linear sum.. given that I read that computer times for the solution are measured in years, I'll make do with an approximation!87.102.84.112 (talk) 17:44, 28 February 2008 (UTC)

If the distance metric used is the Euclidean metric, then finding the optimal solution is known as the Euclidean Travelling Salesman Problem, which is a special case of the general TSP, but lso hard to solve. (Christos H. Papadimitriou. "The Euclidean travelling salesman problem is NP-complete". TCS 4:237–244, 1977) The essential difference with TSP is that the questioner does not ask for the optimal solution, but only for improvements to a method. In terms of order of magnitude, the O(n^2/3) bound cannot be improved upon, as can be seen by considering N×N×N points placed in a completely regular cubic grid pattern.  --Lambiam 22:53, 28 February 2008 (UTC)

Actually what I asked for was very vague, if I was being more specific to the problem I'd have to say that I wanted samples of points in order to be as close together as possible (L2 space), in fact what I really wanted was to group the set of points into n sets of m points so that the volume of the sets was minimised (I suppose some sort of 'sum of squares of volumes measure' would be applicable here..) further more I prefer that the volumes are close to cubic (perhaps some sum of distances^4? measure would work)..

The kdtree (or similar methods) look good enough, and I could work with the octree/peano/hilbert method too with a little extra algorhythm on the end.. I wouldn't try a exhaustive search. Thanks to everyone who replied. Feel free to give more ideas/information. Cheers.87.102.38.45 (talk) 12:02, 29 February 2008 (UTC)

Have a look at Cluster analysis. For a distance measure between two non-empty sets of points you may consider the Hausdorff metric, which is a proper metric, instead of the wishy-washy measures invented by social scientists.  --Lambiam 00:58, 1 March 2008 (UTC)

Rounding conundrum?

I disagree with former teaching colleagues over the correct answer to a rounding problem I posed a few years ago to some 13-year-old pupils, and for my own personal satisfaction I would be interested in people's views to the following simple question:

What is 0.00999 rounded to 2 Significant Figures (2 S.F.)?

The two answers to consider, from myself as head of department and a few of my maths teachers, are 0.010 and 0.0100 (I'm not revealing who claimed which! Yet!). Answer PLUS REASON please if you think you know for sure which is the correct answer! Thanks! AirdishStraus (talk) 18:18, 27 February 2008 (UTC)

It's clearly 0.010, at least according to the convention in our article. The two significant figures are 1 and the subsequent 0. -- Meni Rosenfeld (talk) 18:38, 27 February 2008 (UTC)

As far as I remember from Physics and Chemistry, it should be 0.010. The first two zeros (the one before the decimal and the one right after the decimal) are just place holders. They appear in all the numbers being operated on and I was told that they are not considered a significant value in a problem like this. So we have the one and then the second zero to have two significant digits.A Real Kaiser (talk) 18:48, 27 February 2008 (UTC)

Mmm... obviously the number 0.010 has two significant figures and the number 0.0100 has three, but the issue, I believe, is in the actual process of taking 0.00999 and applying the 'rules'...

'Chop' the number off after the second significant digit and round if the third digit is 5 or greater; write down as your answer the digits you now have up the the 'chop off line'.

So... 0.00999 $\rightarrow$ 0.0099|9 $\rightarrow$ 0.0100| $\rightarrow$ 0.0100

Where, in any article does it say that the resultant answer is further truncated to 2 significant figures? This would mean that a small percentage of all decimals would have a different rule: anything from 0.00995 to 0.00999999... etc. for example for those numbers starting in the third places of decimals. AirdishStraus (talk) 19:47, 27 February 2008 (UTC)

Now I see the problem - which is that significant figures are stupid. For a number that starts with 1, a given number of significant figures represent less accuracy than for a number that starts with 9. For the number in your example, we are faced with the dilemma of treating it as starting with 9, which it does, or starting with 1, like its rounded value. I say the rounded value wins, since we measure the significant figures for it, not for the real value. The quirk is that 0.00999 rounded to 2 sf is 0.010, and rounded to 3 sf it's 0.00999 which is two orders of magnitude more accurate. Again, sfs are stupid. It makes much more sense to discuss relative error, and sfs were never meant to deal with such subtleties. -- Meni Rosenfeld (talk) 20:18, 27 February 2008 (UTC)

I'd think of it this way: the "numbers with two significant figures" are ..., 0.0097, 0.0098, 0.0099, 0.010, 0.011, 0.012, ... and rounding to two significant figures consists in choosing the one of those that (considered as an exact real number) is closest to your number, in this case 0.010. A rounding algorithm, like the one you gave, which produces two significant figures in most cases and three in a few boundary cases I would describe as buggy. Think of it also from the perspective of someone seeing the answer 0.0100; they'll most likely assume that it was selected from the list ..., 0.00998, 0.00999, 0.0100, 0.0101, ... and not from the list ..., 0.0098, 0.0099, 0.0100, 0.010, 0.011, ... (which doesn't even make mathematical sense, since two of the entries are the same). Not that there's anything particularly sensible about the standard convention, it's just standard. As Meni said, the whole concept is rather badly broken, and if you really care about accuracy you have to do a proper error analysis and give explicit error bars. -- BenRG (talk) 20:24, 27 February 2008 (UTC)

(outdent) Just a thought. While strictly speaking, rounding to 2 sig figs does give 0.010, a partial way to sidestep the noted discontinuity of accuracy is to look at the unrounded value and note at which place (tenths, hundredths, etc.) the number loses reliability, then round so as to remove all digits at and past that place. So if the third "9" is not reliable, quote the result as 0.0100, and so on.

The above posting has a good point which I reflected in my advice: rounding should not comply to rote symbolic manipulation rules of digit strings but should reflect the precision and reliability in obtaining the value to be rounded.

This discussion also relates to Benford's law. Baccyak4H (Yak!) 20:38, 27 February 2008 (UTC)

Guys... thanks for the contributions (others still welcome). I'm the one who believes 0.0100 is the correct answer as opposed to the rest of my (now ex-) department who favoured 0.010. The reasons are that it follows the given rule with no deviation (i.e. no 'special cases') and that it gives an indication of the accuracy of the original number. 0.0100 declared as having been rounded to 2sf means something that HAD TO BE originally between the bounds of 0.00995 and 0.009999999.... etc. whereas 0.010 could have come from between bounds of 0.00995 and 0.0105 Once again, many thanks for the various inputs. AirdishStraus (talk) 22:06, 27 February 2008 (UTC)

The correct answer is definitely 0.010. If you round to 2sf, you have to have 2sf at the end, otherwise it's nonsense. 0.0100 clearly has 3sf. Your reason for rounding to 0.0100 sounds to me like simply "3sf is more precise than 2sf", which is obviously true, but irrelevant, since you asked for 2sf. --Tango (talk) 22:30, 27 February 2008 (UTC)

You're probably right, rounding in such a way as to accurately reflect the precision of the measurement is best, but that only works if people know how to interpret the data. If I see "0.010" I know that means the real value is between 0.0095 and 0.0105. Conventions have to be followed whether they are good conventions or not, because otherwise people have no idea how to interpret what you say (you could devote a paragraph at the beginning of your paper to explaining the obscure rounding system you've used, but your readers would probably prefer you just to use the standard method). —Preceding unsigned comment added by Tango (talk • contribs) 22:35, 27 February 2008 (UTC)

HI! my rule here would be to convert to a mantissa/exponent form eg 9.99 x10^-3 then convert the mantissa.. giving 1.0 x10^-3 1.0 x10^-2.. this method is always consistent... 87.102.84.112 (talk) 11:19, 28 February 2008 (UTC) see Significant_figures#Scientific_notation:

particular, the potential ambiguity about the significance of trailing zeros is eliminated

Also consider the reverse - how many sig figs in 0.0100 - the answer looks like 3 87.102.84.112 (talk) 11:21, 28 February 2008 (UTC)

I think you mean that 9.99 x10^-3 gives 10.0 x10^-3 and not 1.0 x10^-3 as you may have written in error. This, however, has the digits 1 0 0 supporting the 0.0100 version of the answer. I know that 0.0100 appears to be 3sf but the conundrum/ambiguity is in applying the commonly supported algorithm that I described above (don't forget that this rounding topic is aimed at 12- to 13-year-olds): 0.00999 $\rightarrow$ 0.0099|9 $\rightarrow$ 0.0100| $\rightarrow$ 0.0100

All text books in my possession (a considerable number!) avoid the issue completely and have no examples or questions of this nature whatsoever; nor does any Wiki article on rounding or sig figs. I'm still at home to any good-natured contributions or suggestions! AirdishStraus (talk) 11:38, 28 February 2008 (UTC)

NO! (but I made a typing error -see above) 2 sig figs means 2 digits in the mantissa - so 9.99 x10^-3 expressed to 2sig figs gives 1.0 x10^-2 to 2 sig fig..?87.102.84.112 (talk) 11:59, 28 February 2008 (UTC)

If you want an algorhythm that 'works' then how about this

0.0999 to 2sf

STEP 1

0.099|9 (ignore zeros on left, separate after 2 digits) ::gives 0.1 (this is rounding)

That is 0.1

STEP 2

Write the answer from STEP 1 to 2 significant figures

This is 0.10

This will give consistent and the answers will always be to the same number of significant figures as requested.

Whether or not a 12 year old can be expected to perform a 2 step process isn't something I can know. But I hope that the answer will always be yes. Clearly they must have a perfect understanding of rounding numbers before they can begin to do sig figs reliably.87.102.84.112 (talk) 12:19, 28 February 2008 (UTC)

I see what you're driving at... you are saying that after the 'chop off' to however many sf you choose, you should always ignore all trailing zeros and then write whatever decimal that gives you to your required degree of accuracy. So, as you imply... 0.00999 becomes 0.01 which then becomes 0.010 written to 2sf, because after the first step all trailing zeros are ignored. If only that rule existed in print somewhere!!! It still doesn't indicate the initial bounds for the original number in the same way though as my own 'logical' method does... but I guess you can't have everything! AirdishStraus (talk) 14:48, 28 February 2008 (UTC)

I agree that I too can't find examples for numbers that round up more than one digit.. there's also (if you're teaching this) the question of sematics in grammar - "round this number to 2 significant digits' can be different from 'write this number rounded so that your answer has 2 sig. digits' as you have shown. ie I could even write "round this number to 2 s.ds and write the answer obtained to 4 s.ds" (though that would be wrong for other reasons - ie inferred accuracy). If I was a teacher (which I'm not) I'd try to approach this topic having first covered scientific notation.. For-warned is for-armed.87.102.84.112 (talk) 16:16, 28 February 2008 (UTC)

It indicates bounds on the original number, just not as precisely as you get when using 3sf, which is obvious. Your method gives you 3sf, so of course it's more precise, it's just not what you're asking for. --Tango (talk) 14:59, 28 February 2008 (UTC)

I see the ambiguity in this case, it has to do really with where to draw the line as to when a number has moved it's leading significant digit to another place value. There is no clear 'right way' either, I much prefer notation like .00999±.00005 to sig. figs. A math-wiki (talk) 12:03, 1 March 2008 (UTC)

February 28

Lebesgue measure of a complement?

No doubt this is trivial, but I don't have handy access to a measure theory text and web searching has availed me nothing. Is there a tidy expression for the Lebesgue measure μ(Rⁿ \ S) of the complement of an arbitrary measurable subset S of Rⁿ in terms of the measure μ(S) of S? My guess is that if μ(S) = c is finite, then μ(Rⁿ \ S) = ∞, and if not then there is nothing in general we can say about μ(Rⁿ \ S).—PaulTanenbaum (talk) 01:03, 28 February 2008 (UTC)

That's correct. The addivity formula for a measure continues to hold for sets of infinite measure, so you get µ(S)+µ(Rⁿ\S) = µ(Rⁿ) = ∞, which amounts to what you guessed (two elements of [0,∞] sum to ∞ iff one of them is ∞). You can also derive the same result using finite measures only by writing the Lebesgue measure as a limit of finite measures (e.g. measures supported on larger and larger closed balls). Bikasuishin (talk) 01:32, 28 February 2008 (UTC)

Finding the length of a vector

here's my question:

suppose u=[1, -2]. determine 3u.

so i do, i get [3, -6].

but then the question asks

find the length of the vector.

huh? i have no idea how to do this; none of my notes tell me how. i thought about finding its magnitude √(x²+y²), but you end up with √45. i checked the answer at the back of my textbook and it says 3√5. can anyone help me here? --24.109.218.172 (talk) 01:51, 28 February 2008 (UTC)

oh wait... i think i figured it out. √45 equals about 6.708; 3√5 equals about 6.708. does anyone know how they got 3√5?--24.109.218.172 (talk) 01:52, 28 February 2008 (UTC)

3u is three times the length of u. So |3u|=3|u|=3√(1²+(-2)1²)=3√5. Notice that 3√5=√(3^2*5)=√45. Taemyr (talk) 01:57, 28 February 2008 (UTC)

ahh i see thanks man.24.109.218.172 (talk) 02:59, 28 February 2008 (UTC)

A calculator is generally unhelpful in learning math. http://upload.wikimedia.org/math/5/e/5/5e5ea894cc317ec4bb22f46beec9d791.png —Preceding unsigned comment added by Imagine Reason (talk • contribs) 03:01, 28 February 2008 (UTC)

If you wanted to write ${\sqrt {xy}}={\sqrt {x}}{\sqrt {y}}$, all you needed was to take a look at Help:Formula. And no, I don't think I agree. A calculator can be extremely helpful if one uses it properly. -- Meni Rosenfeld (talk) 09:21, 28 February 2008 (UTC)

√45=√(9x5)=√9√5=3√5 should make it totally clear.. (another √2√2=√(2x2)=2) or √72=√2x36=√2√36=6√2 .. it helps if you spot that the value in the root has a factor that is a square eg 4,9,16 etc is a factor87.102.84.112 (talk) 10:27, 28 February 2008 (UTC)

point to point reflection formula

What is the formula for generating the reflection curve for a point to point reflection (not formula for ellipse)?

I don't understand. The curve in question is exactly (one of) the ellipse(s) with those two points as foci. -- Meni Rosenfeld (talk) 09:30, 28 February 2008 (UTC)

My error. I was thinking of 2 parabolas facing each other rather than an ellipse. The formula for an ellipse is the correct formula.

Are you asking how to prove/find that the 'foci' reflect onto each other for an ellipse? If so you're going to need the slope of the ellipse at a given point, which is given by the differential of the equation of the ellipse, and then be able to work out what is the reflected ray - if so ellipse, reflection, Ellipse/Proofs, dot product might be useful - ask for more if I was along the right lines —Preceding unsigned comment added by 87.102.84.112 (talk) 10:21, 28 February 2008 (UTC)

Linear equations Part II

Question 1

I think Ive managed to do the previous question:
${\frac {3a}{2}}-4={\frac {1}{2}}$
${\frac {3a}{2+1}}-4+4={\frac {1}{2}}$ (<---Plus 4 to the top of the fraction and 1 to the bottom)
${\frac {3a}{3}}={\frac {9}{3}}$
$a=3$

OK, not entirely sure where you got ${\frac {3a}{2+1}}$ from - the easiest way to do this would be to multiply through by 2, thus eliminating the fraction, then move all the constants to the right hand side and divide through by the coefficient of a (3). You got the right answer though. -mattbuck (Talk) 13:14, 28 February 2008 (UTC)

The next one proved a little more difficult:
${\frac {2}{3}}-{\frac {5b}{6}}={\frac {3}{2}}$
${\frac {2\times 2}{3\times 2}}-{\frac {5b}{6}}={\frac {3\times 3}{2\times 3}}$ (lowest common denominator)
${\frac {4}{6}}-{\frac {5b}{6}}={\frac {9}{6}}$
And Something Goes next.. im not sure what

You are correct so far. Multiply through by 6 and it should become obvious. -mattbuck (Talk) 13:14, 28 February 2008 (UTC)

Question 2

Im not sure how to do the questions like these because the numbers are different on both sides. $7x+4=4x-5$

What you want to do is get all the terms in x onto one side, and all the constant terms onto the other. If you do that, it should be obvious. -mattbuck (Talk) 13:14, 28 February 2008 (UTC)

(edit conflict) I don't understand your method for the first question. It gets the right answer, but I can't see how... There is a standard method for solving linear equations:

1. Use addition and subtraction to get all constant terms on one side of the equals sign and all terms with an x (or whatever your variable is) in on the other.
2. Combine the terms so you have one term on each side (this is just basic addition and subtraction).
3. Divide both sides by the coefficient of x
4. Read off the answer

That method will work for all the questions you've mentioned. --Tango (talk) 13:18, 28 February 2008 (UTC)

Recursively defined sets, Part II

I'm having trouble trying to determine some of the language in the following problem:

Consider the following set C, a subset of the set of all bitstrings:

Basis step:

$\lambda \in C$ where $\lambda$ is the empty string.

Recursive step:

$\omega \in C\rightarrow S_{1}(\omega )\in C$

$\omega \in C\rightarrow S_{2}(\omega )\in C$

$\omega \in C\rightarrow S_{3}(\omega )\in C$

Where...

$S_{1}:C\rightarrow C$ given by $S_{1}(\omega )=1\cdot (\omega \cdot 0)$

$S_{2}:C\rightarrow C$ given by $S_{2}(\omega )=1\cdot (\omega \cdot 1)$

$S_{3}:C\rightarrow C$ given by $S_{3}(\omega )=0\cdot (\omega :\omega )$

(Note $\cdot$ and $:$ are character and string concatentation respectively.)

My question is as follows:

Evaluate $\omega _{1}=(S_{3}\circ S_{3}\circ S_{3})(0).$

Now what does this mean? Do I just generate the set using just $S_{3}$ on three seperate steps and concatenate?:

$S_{3}=\lbrace 0\rbrace \,$

$S_{3}=\lbrace 0,000\rbrace \,$

$S_{3}=\lbrace 0,000,0000000\rbrace \,$

$w_{1}=0\cdot 000\cdot 0000000\cdot 0$

Then I suppose that's wrong. I'm fairly certain I have to concat the (0) at the end. Trouble is there is no mention of the $\cdot$ operator and there is no indication of what $\circ$ means. Last time I checked it was the function composition operator. Help start me off with this one! Damien Karras (talk) 13:16, 28 February 2008 (UTC)

It is function composition. S₃ is a function, you just apply that function three times. If it's easier, think of it as $\omega _{1}=S_{3}(S_{3}(S_{3}(0)))$, it means the same thing. --Tango (talk) 13:21, 28 February 2008 (UTC)

Apologies, my brain is useless - the bracketed function notation confused me. Am I right I declaring, therfore that

$w_{1}=0000~0000~0000~000?\,$

Damien Karras (talk) 15:19, 28 February 2008 (UTC)

That's what I got. --Tango (talk) 16:43, 28 February 2008 (UTC)

Second order, linear ODE with non-constant coefficients

${\frac {d^{2}x}{dt^{2}}}-at{\frac {dx}{dt}}-ax=0$

$a$ is a constant. Is there a closed-form solution for $x(t)$? —Keenan Pepper 16:34, 28 February 2008 (UTC)

Depending on what you're willing to accept as "closed-form", yes, there is. $x_{0}(t)=\exp(at^{2}/2)$ is a solution, so you can reduce the problem to a first order ODE by looking for solutions of the form $x(t)=x_{0}(t)z(t))$. The solutions turn out to be:

$x(t)=e^{at^{2}/2}\left(\lambda +\mu \cdot \mathop {\rm {erf}} {\Big (}{\sqrt {\frac {a}{2}}}t{\Big )}\right)$

where erf is the error function. Bikasuishin (talk) 16:54, 28 February 2008 (UTC)

That's not solved for $x(t)$, because $x$ appears in the left-hand right-hand side. Thanks for that one solution though; now I can apply reduction of order myself. —Keenan Pepper 17:51, 28 February 2008 (UTC)

Sorry, the x on the right-hand side was a mistake on my part. Fixed. Bikasuishin (talk) 18:13, 28 February 2008 (UTC)

Sinusoidal tourism model

I'm looking for a website that would contain information on how the amount of tourists in Japan (or another country) varies periodically throughout the year.Tuesday42 (talk) 16:46, 28 February 2008 (UTC)

mmmh some data here http://www.tra.australia.com/content/documents/Visitor%20Profile/Japan_visitor_profile_07.pdf page 7 which you can read and find out that they come from "Source: Australian Bureau of Statistics, Overseas Arrivals and Departures Data [Cat. 3401.0]" - so this might be a good source. I suppose (that was visitors from japan if that matters). Found using search for "tourism figures <country> month" or similar.87.102.84.112 (talk) 17:48, 28 February 2008 (UTC)

May as well get the actual publication, from here. Unfortunately a quick look on Japan's Statistics Bureau site didn't reveal any tourism data (the ABS gets their figures as administrative data from the Department of Immigration and Citizenship, so perhaps the Japanese equivalent of that could be of use). Confusing Manifestation(Say hi!) 22:25, 28 February 2008 (UTC)

That doesn't seem to be a sinusoidal model, which is what I really need. I was hoping that amount of tourists in a given area would vary sinusoidally due to seasonal weather. In retrospect, I should have suggested a tropical area.Tuesday42 (talk) 02:00, 1 March 2008 (UTC)

Maybe not tropical - if it's sunny all the year round - suggest somewhere that only has good weather in the summer - how about Cleethorpes?87.102.79.228 (talk) 09:42, 1 March 2008 (UTC)

Why do they teach complicated mathematics to middle and highschool students?

There's no doubt that mathematics is important to humanity. Engineering depends on it, along with statistics, and other fields. But why do they teach complicated math to the average highschool student who may or may not intend to go into those fields? If such a person decides to become a writer, historian, or go into business, he would never need something like geometry, calculus or pre-calc. So why? 64.236.121.129 (talk) 19:18, 28 February 2008 (UTC)

Full-on calculus is pretty rare for the average high school student. Geometry, algebra, and the like, however, are inextricably tied to everyday life. Those who understand the subject are better equipped, then, to intelligently handle situations where math arises. Additionally, mathematical topics tend to reinforce concepts like critical thinking and interpretation of data which arise in even more wide-ranging fields. Personally, the branch of mathematics I'd like to see get more of a push at the high school level is probability (and to a lesser-though-related degree, statistics). The lotteries wouldn't like me very much if I ever pulled that off. Finally, it helps develop the student as a well-rounded capable individual. Similar ideas could be put forth that engineers don't need writing classes, but having met engineers who hold to such an opinion, I can firmly state that there's a world of difference in the career potentials of engineers who express themselves well and those who can't. — Lomn 19:40, 28 February 2008 (UTC)

I can't really agree that Geometry is used in everyday life. Geo isn't exactly about knowing that a triangle has three sides, it's complicated proofs, which no average person ever uses. And yea, I guess the average HS student doesn't need to take Calc, but my point was, HS math teaches convoluted maths to students who typically won't ever use it. In theory I guess math should push critical thinking, but you don't actually have to know what you are doing in math to do well. You can get by just by following patterns and memorizing formulas without knowing what the point is. 64.236.121.129 (talk) 20:42, 28 February 2008 (UTC)

If you can pass the exam just be memorising things (which is often the case), it's a bad exam. That's true (and quite common) in many subjects, not just maths. It's a serious problem with modern education. We didn't do many (if any) complicated proofs in Geometry before A-level - it was mostly calculation based, which is what's useful in everyday life. --Tango (talk) 20:48, 28 February 2008 (UTC)

In the UK, calculus isn't taught until A-level, so you have to explicitly choose to take A-level maths to learn it. Geometry and basic algebra are very useful in the real world - being able to do basic right-angled trig could come in handy when working out how to lay out a room, for example. Algebra can be used to work out if a special offer is really a good deal. I agree with Lomn that there should be more probability (and, as he says, to a lesser degree, stats) - probability is very counter intuitive and without a proper understanding of it people make quite serious mistakes. --Tango (talk) 20:18, 28 February 2008 (UTC)

Of all the mathematics which I had learn in high school, I hated geometry and statistics the most. It's not that I cannot do geometry but that I just can't see myself solving real world problem in geometry and that real world engineers uses vectors to solve problems rather than using geometry.

As for statistics, it's useful but just not taught right to me in highschool. Furthermore as I didn't have computers in highschool, doing a statistics problem which I made up myself is excruciating painful (because adding lots of numbers together is very painful without a computer). Hence my dislike for statistics. 202.168.50.40 (talk) 21:19, 28 February 2008 (UTC)

Vectors are part of geometry, so I'm not quite sure what you mean. The difference between using vectors and doing geometry in the way it's taught at lower levels is pretty much just one of notation - the actual maths is the same. I've also never enjoyed studying statistics... I'm not sure why not. I enjoy playing around in Excel analysing data, but studying it formally has always seemed boring to me - probably because stats at school generally involved calculating means, etc. of tables of random numbers, which, as you say, is excruciating. --Tango (talk) 21:49, 28 February 2008 (UTC)

To be fair, when I say that I'd like to see the average student work more with statistics, it's really that I'd like the average student to better understand the general theory and limitations of statistics -- I hate doing the work, too. But I think it'd be great if more people understood why a random poll of 1000 Americans is meaningful (within America) as well as the caveats that accompany it. — Lomn 22:21, 28 February 2008 (UTC)

I'm rather confirming your statement here, but a random poll of 1000 Americans is meaningful? Don't you have to choose people from appropriate demographics to get statistically significant results from such a small sample? --Tango (talk) 22:28, 28 February 2008 (UTC)

Yes and no. I don't recall the exact number, but a fully random sample of approximately 1000 people is sufficient to give around ±5% results with 95% confidence for the US population. A table for populations up to 100,000 provides a pretty good benchmark for the estimate. Theoretically, one shouldn't adjust for demographics. In practice, as Black Carrot notes, some sort of adjustment has to be made, though I'm no expert on how this is actually applied. — Lomn 00:00, 29 February 2008 (UTC)

In terms of policy, I think the "may or may not" bit of the original post is also critical. Public school is not a trade school. It's supposed to prepare you for any occupation or hobby you could want. (Side effect of freedom of choice.) And, given the dropout and change-of-major rates at most colleges, you don't know what you want yet, even if you think you do. My sister is a software engineer right now, at a huge company. Until she took a mandatory computer programming class in high school, she wanted to be a vet. You have be given a taste of everything, by force if necessary, or you'll lose a lot of golden opportunities. I'm not saying you've been introduced to the subject particularly well, but you should at least recognize that you need to be introduced to the subject. Also, be careful what you assume doesn't use math. You mention business. Statistics and finance are the lifeblood of business. Not everyone in business knows that, but that's their problem. Black Carrot (talk) 22:38, 28 February 2008 (UTC)

On the sample thing - for a "random" sample, 1000 would be plenty. Since it's impossible to get a random sample, you have to counter the inherent biases by breaking it into demographics. Black Carrot (talk) 22:41, 28 February 2008 (UTC)

There's no doubt that maths gets complex. But complicated is a different matter. To me, that word suggests making it more complex than it needs to be. Surely the job of any teacher, and a maths teacher in particular, is to explain concepts in ways the students can understand, by de-complexifying them and expressing them in simpler ways appropriate for their age group. Once they've grasped that idea, then the teacher might proceed to more complex stuff, on the assumption that the students are up to speed. If that's actually not the case, then sure, what comes next will seem mystifying and incomprehensible. That's why depending on just what's presented in the class room is generally not enough, and doing homework (and not necessarily limiting yourself to the exercises that are set by the teacher) is important. A music student does most of their learning not at the weekly lesson with the teacher, but while at home practising by themselves, finding out what works and what doesn't, and trying out different approaches in an effort to find some of their own solutions. If they haven't mastered playing a scale error-free, they won't master a harder piece. Same with maths and other disciplines. -- JackofOz (talk) 23:01, 28 February 2008 (UTC)

I believe that teaching different ways of thought is helpful to any person. Understanding mathematics, functions, and programming bettered my thought significantly, and I can think of things in more than one perspective, kind of. Mac Davis (talk) 23:04, 28 February 2008 (UTC)

You could say..why teach history if they are going to be a plumber with no interest in the past? We teach for many reasons: To increase the intellect of the individuals that are being taught, to potentially inspire them into an avenue of work/development that will lead them to have a fulfilling life, to give people a basic knowledge of things that are 'important' to society, to provide a good entry to a subject that the individual has never looked at in detail, to teach frameworks/ideas, to promote thinking and asking questions, to challenge the brain/mind of individuals. How would you ever know you want to do maths without being given a branch into it? Yes the stuff isn't everyday-useful but really what is? Do i need to be able to understand verbs, adjectives and nouns? No. In my entire life post-schooling I have never had anybody ask me to identify what they are, and I doubt they are 'hugely important' to my understanding of sentences/books/whatever. You specialise more the further into education you go, but up to the 'opt out' level of schooling we really should try to give students a starting-block knowledge (at various levels of depth) for the main subject areas that people value. As the tv show Numb3rs introduction says "We all use maths everyday, to predict weather, to tell time. Maths is more than formulas or equations; it's logic, it's rationality, it's using your mind to solve the biggest mysteries we know" - A good grounding in maths/math-theories can help people be logical/approach things in a more structure way. Much like the formal-sciences can help with things like that - it's not so much the fine-grain detail of what you learn, but the frameworks, structures and boundaries that you learn to operate in and around that are important. ny156uk (talk) 23:23, 28 February 2008 (UTC)

The world is getting complicated, and you can't trust what you don't understand. That's why America is so behind in math and science - they don't care and they don't trust. Would anyone complain about English classics being taught? I don't think so. The anti-intellectualism in America is horrifying. Imagine Reason (talk) 00:44, 29 February 2008 (UTC)

Why learn math? Because solving problems is fun and joy and satisfaction. Most math courses teach some theory first and then presents you with exercises to check that you understand the theory. In my opinion that is a mistake. It would be better first to present some problems, which you can understand but not solve, and then teach the tools which enables you to solve the problems. Most problems have two solutions, a elementary brute force method, and an elegant shortcut. You do not appreciate the shortcut unless you know the elementary method. Bo Jacoby (talk) 01:52, 29 February 2008 (UTC).

New York City's new elementary math curricula for the past few years starts off each lesson with some very interesting problems. I don't think it goes as far as unsolvable questions, but it's awfully good. Imagine Reason (talk) 05:56, 29 February 2008 (UTC)

I could also turn this question backwards: why do they teach history in high schools? – b_jonas 10:43, 29 February 2008 (UTC)

All human knowledge is interesting. In high school I learnt, for example, how to identify a glaciated landscape; how to say "blood sucking, bandy-legged vulture" in German; and several stanzas of The Rime of the Ancient Mariner. I don't think that information will ever give me any pratical benefit, yet I don't regret learning those things. Gandalf61 (talk) 12:55, 29 February 2008 (UTC)

I agree, it's always best to motivate a theory before teaching it. Sometimes, it's fun to learn a bit of maths just for the sake of it, but usually you want to learn it in order to do something useful, so you should be told that useful thing before you start. --Tango (talk) 13:42, 29 February 2008 (UTC)

One thing I wish highschool math teachers would do is tell students what the maths they are teaching is actually used for. In my entire time in HS, or even middleschool, they never told us what these maths they were teaching us, were for. I remember a student asked a teacher what are we going to use these complicated algebra theories she was teaching us for, and she said "absolutely nothing". I had to learn on my own, what maths was for. Physics, engineering, etc.

I think the comment the ideology that, "we have to force everything on students whether they like it or not, just so they have a choice of what they want to do later in life" is not quite "right", in the sense that it's a good idea. There are basics to math that everyone should learn, sure. But the more complicated maths should be left to the future engineers, and such. Of course it isn't up to me, so things remain as they are. But yes, I agree that seems to be one of the reasons why they teach maths to students.

The critical thinking part definitely is a reason why they teach maths, I just don't think it works. Even in college, they make computer science students learn complicated calculus theories that they will probably never use. You don't need calculus to program in C++ (I think. I didn't get very far in my Comp Sci studies, so I don't know what the advanced classes encompass). So the only reason to learn calc for a comp sci student is for the critical thinking skills. But like I said before, you can ace a calc class just by memorizing formulas and recognizing patterns. "Oh this is the equation that we learned in lesson 5. I'll just follow the pattern they taught in lesson 5 to solve this equation". No critical thinking involved.

I remember at the end of one of our calc classes, my prof asked one student what a Derivative was. And the student just kept saying the derivative was the derivative. He had no idea a derivative was just the slope at a given point on a function. He just knew how to find the derivative, and solve the equation. He did well in the class, but he had no idea what the point of it was though. Another time another professor came into our calc class and asked us what the point of a math degree was. The only answer the students could give was "to teach math to other people". Amazing. 64.236.121.129 (talk) 14:26, 29 February 2008 (UTC)

What you (OP) have observed is a symptom of a much larger and I think more dangerous problem with the education system in the United States of America. The reason math has become something the "people never use" is because it's taught in such a way to be unless to all but those who have the most aptitude for it. If it were taught correctly (as even I could do), I am quite certain most people would use at least Algebra 1 level math on a daily or near daily basis. The only real question is why has the problem been allowed to persist when it has surely been noticed by many people over the years. The answer to that question is very intriguing indeed. The way mathematics is currently taught (at the HS level) is in a purely utilitarian form. Mathematics is best learned by posing the question, and then developing an answer based on what one already knows. That requires one to be able to think freely and critically about a problem. There is a quote on my brother's myspace page that reads as follows; "The most dangerous man, to any government, is the man who is able to think things out for himself, without regard to the prevailing superstitions and taboos. Almost inevitably he comes to the conclusion that the government he lives under is dishonest, insane and intolerable, and so, if he is romantic, he tries to change it. And even if he is not romantic personally he is very apt to spread discontempt among those who are." --- H.L. Mencken. To actually understand the concepts in mathematics one must be able to think things out for them selves, but someone who can do that is dangerous (in the eyes of those in power). A math-wiki (talk) 12:33, 1 March 2008 (UTC)

Mandelbrot set of Matrices

If you plot the Mandelbrot set with matrices instead of complex numbers, can you get interesting shapes with more than two dimensions? — DanielLC 23:18, 28 February 2008 (UTC)

I guess you don't get anything really different. Indeed, if you consider the iterates of 0 under a Mandelbrot-like transform $M\mapsto M^{2}+C$, all the matrices you get are polynomials in C. Assuming without loss of generality that C is upper-triangular (see Jordan normal form), you find that the asymptotic behaviour of that sequence is mostly given by the elements of the diagonal, i.e. the eigenvalues of C. The nilpotent part may matter at the border, but it is of little concern in general when looking at asymptotic dynamics. So I think it's mostly true that a matrix belongs to the "matrix Mandelbrot set" when all of its eigenvalues belong to the usual Mandelbrot set, and so you don't really get anything new.

On the other hand, Julia sets are a different matter altogether, since the iterates of a general matrix A under a transform like $M\mapsto M^{2}+C$ do not usually commute with one another, and thus cannot be reduced to a common upper-triangular form in general. I remember that there was a picture of a (3D slice of a) quaternion Julia set on the cover of one my math textbooks way back in high school, so people do draw that sort of things, and that's really a particular case of a "matrix Julia set". Bikasuishin (talk) 00:41, 29 February 2008 (UTC)

COMPUTE

COMPUTE: JUNE 18, 2008 + 31 WEEKS + 2 DAYS

THANK YOU.--Goon Noot (talk) 23:23, 28 February 2008 (UTC)

Here is a calendar, it shouldn't be hard to do. In the future, please avoid posting in all caps. It is the online equivalent of shouting. 134.173.92.241 (talk) 23:58, 28 February 2008 (UTC)

Here is a calculator. --hydnjo talk 01:48, 29 February 2008 (UTC)

February 29

Pretty

What are some of the most visually appealing bits of mathematics you can think of? Geometric, algebraic, anything. So far, I've been able to find a number of well-known fractals (Mandelbrot, Julia, Lorenz, Apollonius, Sierpinski, Menger, Koch, Weierstrass, Newton, Peano, Hilbert, Klein) and general methods for generating others, irregular tilings of the plane with colored shapes, the Poincare disk and variations on it, colored graphs of algebraic varieties and renderings of 2-manifolds, impossible shapes in the style of Escher, a few things evocative of the infinite (the Hilbert Hotel, power series), a number of pretty symbols and formulas (always looking to expand), the Platonic and Archimedean solids, and a few things that would only work with animation and several more dimensions. If anyone could add detail to one of those things, or even better give me something totally new to include, I'd be forever greatful. Black Carrot (talk) 04:59, 29 February 2008 (UTC)

This may not be the right thing, but I've found some beautiful youtube videos such as

http://www.youtube.com/watch?v=JX3VmDgiFnY -- Mobius transformations on the plane and the Riemann sphere

http://www.youtube.com/watch?v=J-fcRzvRBqk -- 3D Fractal - Menger sponge

http://www.youtube.com/watch?v=SBxBQYmPIM4 -- The E8 Lie Group

http://www.youtube.com/watch?v=d1Vjsm9pQlc -- The Alexander Sphere

Visualising mathematics is so important, and I seriously seriously beleive it could get a lot more people interested in it. I'm by no means any mathematician but my understanding of most concepts I have learnt so far have been because I have seen them visualised in some way. Damien Karras (talk) 07:01, 29 February 2008 (UTC)

fractal stellated dodecahedrons http://www.clowder.net/hop/Keplrfrct/Keplrfrct.html scroll down - some more stuff on this persons (David Hop) page.87.102.38.45 (talk) 12:12, 29 February 2008 (UTC)

There are also proceedural formulars that give a very good appearance of wood grain in a plane - look nice (provided you choose a nice palette)87.102.38.45 (talk) 12:12, 29 February 2008 (UTC)

I remember some guy/girl left their doctorate on the web - about filling hemispheres with smaller spheres - that to me was 'pretty' can't find it as yet but in the meantime look at Apollonian sphere packing87.102.38.45 (talk) 12:19, 29 February 2008 (UTC)

Cellular automaton eg Conway's Game of Life can be visually entertaining87.102.38.45 (talk) 12:32, 29 February 2008 (UTC)

How about the Koch Sphereflake??? (3D version of the Koch snowflake) A math-wiki (talk) 12:37, 1 March 2008 (UTC)

Learning maths from scratch

I always want to learn maths all over again, but I don’t know how to start with. Now, I’m undergraduate, I “parted” with everything science since high school, and I did almost nothing more than very simple statistics (like poisson distribution, but I don’t really remember it) and calculus (I don’t remember how to differentiate). So, if I want to know “advanced maths”, like what is being taught at university, what sorts of books (maybe in English) may I use (as an adult learner)?

My “junior high school” maths is about factorizing, polynomials, simple geometry (calculating angles) and so on; which areas do high school maths (I mean, for learning science subjects) and university maths cover? I’m looking for books for learning more advanced things (compared to my present level) and I don’t just concentrate on calculus or statistics (or something like that). Any suggestions? --Fitzwilliam (talk) 14:10, 29 February 2008 (UTC)

From my experience, university maths tends to recap all you learnt at school in the first year - try just going along to the first year maths lectures. -mattbuck (Talk) 14:39, 29 February 2008 (UTC)

Especially maths lectures targeted at science students, rather than maths students. My Maths department (at a UK uni, I know it's a little different elsewhere) offers a module called "Mathematics for Scientists and Engineers" which covers a wide range of mathematical topics at a fairly basic level (by Uni standards) without assuming much (if any) prior knowledge. If there are similar modules are your uni, those would be the ones to go to. --Tango (talk) 16:11, 29 February 2008 (UTC)

I'd go to a Maths Department. They love to find new students. Imagine Reason (talk) 23:15, 29 February 2008 (UTC)

What is your motivation? Do you want to learn advanced maths primarily because you feel it will be useful, or is it more for fun? If you have forgotten how to differentiate, maybe you should take a course in calculus (actually analysis) anyway, using a text that does not just give the rules and formulas but also precise definitions of concepts like the real numbers, limit, and continuity, and rigorous proofs. Other topics you may study that don't immediately require much prior knowledge are linear algebra and projective geometry. Also consider elementary number theory and combinatorics. A nice book is Concrete Mathematics; although aiming at hopeful computer scientists, it is also quite valuable for mathematicians. I'd also advice you not to go immediately very deep into one field of maths, but to first build up a fairly broad basic knowledge of various fields. Much of the more advanced stuff in maths requires some knowledge of other fields.  --Lambiam 00:26, 1 March 2008 (UTC)

Regression Question

I have a dataset and I've just run a regression on it using R. I have a model which is predicting well, but I don't know how to tell from the results I have whether the relationship between the two variables ("part" is the DV, "conf" is the IV) is inverse or not. I've been doing a lot of research and everything I've come across as only shown me regression equations where it's quite easy to see the direction of the gradient, but it's not so clear from the output of R. Here's the model:

Generalized linear mixed model fit using Laplace Formula: part ~ 1 + conf + (1 | id) + (1 | word)

  Data: align
Family: binomial(logit link)
  AIC   BIC logLik deviance
490.1 507.7 -241.1    482.1

Random effects:

Groups Name        Variance Std.Dev.
word   (Intercept) 2.977549 1.72556
id     (Intercept) 0.097771 0.31268

number of obs: 601, groups: word, 32; id, 20

Estimated scale (compare to 1 ) 0.9142627

Fixed effects:

           Estimate Std. Error z value Pr(>|z|)    

(Intercept) -3.1001 0.4014 -7.724 1.13e-14 *** conf 1.7941 0.2734 6.563 5.29e-11 *** --- Signif. codes: 0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Correlation of Fixed Effects:

    (Intr)

conf -0.505

So can anybody tell me where I should be looking to find out the direction of the correlation (The data is too cloudy to do it simply by studying a graph)? --Kiltman67 (talk) 14:52, 29 February 2008 (UTC)

Edit: The format of R's outputs hasn't fitted in very nicely, but to my eye it's still readable. If anyone has any problems I'm willing to alter it. --Kiltman67 (talk) 14:54, 29 February 2008 (UTC)

I don't understand the question. By "inverse relationship", do you mean a negative correlation (as one goes up, the other goes down)? Does the program not output correlation coefficients? What do you mean by "Laplace Formula"? Apparently not Laplace's formula. The use of the term "gradient" suggests that you may be referring to Laplace's method, but it is not clear how that relates to regression analysis (which I assume you are attempting to do). I cannot interpret the formula "part ~ 1 + conf + (1 | id) + (1 | word)"; what are the meanings of "~" and "|" here, and what are "id" and "word"?  --Lambiam 00:40, 1 March 2008 (UTC)

Teaching the adjoint method of calculating the gradient

I'm lecturing next week, and one of the topics is the adjoint method of gradient calculation. The audience is graduate students from a wide array of backgrounds, the most common of which is engineering. The level of mathematical knowledge varies greatly. They've been introduced to the Lagrangian but haven't explored it in any depth. I would like some help in figuring out how to present the adjoint method. I would like to be mathematically accurate without being too complicated, and if possible, would like to give the students some kind of intuition. I can't do anything horribly mathematically incorrect or the more mathematically-educated will be confused, and I can't do too much math without leaving the less mathematically-educated behind. This class is more aimed at people using the methods, rather than writing them, so they need to understand advantages/disadvantages and a basic idea of how it works. I also only have a few minutes for this topic.

Here's one way to present it:

The ultimate goal is to solve an optimization problem:

$\min _{\mathbf {x} }J(\mathbf {x,u} )$

subject to $\mathbf {R} (\mathbf {x,u} )=\mathbf {0}$

where J is a scalar objective function (for example, drag), $\mathbf {x} \in \mathbb {R} ^{n}$ is a vector of design variables (for example, defining the shape of a wing), $\mathbf {u} \in \mathbb {R} ^{k}$ is a vector of state variables (for example, flow variables), and $\mathbf {R} \in \mathbb {R} ^{m}$ is a vector of constraints, often defining a system of PDEs (for example, the Navier-Stokes equations). The method we're using to solve this problem is a gradient based optimization algorithm, so we need the gradient of the objective function with respect to the design variables ${\frac {dJ}{d\mathbf {x} }}$.

Define a vector $\mathbf {\psi } \in \mathbb {R} ^{m}$ that we will call the adjoint vector. Form the quantity

$I(\mathbf {x,u} )=J(\mathbf {x,u} )-\mathbf {\psi } ^{T}\mathbf {R} (\mathbf {x,u} )$.

Now take a differential of each side of this equation:

$\delta I={\frac {\partial J}{\partial \mathbf {x} }}^{T}\delta \mathbf {x} +{\frac {\partial J}{\partial \mathbf {u} }}^{T}\delta \mathbf {u} -\mathbf {\psi } ^{T}\left({\frac {\partial \mathbf {R} }{\partial \mathbf {x} }}^{T}\delta \mathbf {x} +{\frac {\partial \mathbf {R} }{\partial \mathbf {u} }}^{T}\delta \mathbf {u} \right)$.

This must vanish. (um, why?) Rearranging

$\delta I=\left({\frac {\partial J}{\partial \mathbf {x} }}^{T}-\psi ^{T}{\frac {\partial \mathbf {R} }{\partial \mathbf {x} }}^{T}\right)\delta \mathbf {x} +\left({\frac {\partial J}{\partial \mathbf {u} }}^{T}-\psi ^{T}{\frac {\partial \mathbf {R} }{\partial \mathbf {u} }}^{T}\right)\delta \mathbf {u} =0$.

We have specified nothing about $\psi$ so far, and are free to choose it as we like. We choose it such that:

${\frac {\partial J}{\partial \mathbf {u} }}^{T}-\psi ^{T}{\frac {\partial \mathbf {R} }{\partial \mathbf {u} }}^{T}=0$

or

${\frac {\partial \mathbf {R} }{\partial \mathbf {u} }}\psi ={\frac {\partial J}{\partial \mathbf {u} }}$

which we are going to call the adjoint equation.

We can then get the gradient of J with respect to x by first solving the adjoint equation and then using

${\frac {\partial J}{\partial \mathbf {x} }}={\frac {\partial \mathbf {R} }{\partial \mathbf {x} }}\psi$

You can see I'm confused here because I ended up with a partial derivative when I thought we were looking for the total derivative.

The upshot of this method is that if the original set of constraints is a PDE, so is the adjoint equation, and it takes a similar amount of time to solve. So the time for the gradient calculation is independent of n, as opposed to a finite difference calculation. —Preceding unsigned comment added by Moink (talk • contribs) 15:47, 29 February 2008 (UTC)

The logistic map

How would I go about showing that the logistic map is chaotic when r = 4? I've tried to show sensitivity to initial conditions, using the solution:

$x_{n}={\frac {1}{2}}\lbrace 1-\cos[2^{n}\arccos(1-2x_{0})]\rbrace$

(from http://mathworld.wolfram.com/LogisticMap.html)

...but I'm not having much luck. I'm looking for something a bit more analytic than a cobweb diagram. I considered expressing x₀ in binary, so that the multiplication by 2ⁿ would just result in a shunting of the decimal point, but I'm not sure how helpful an avenue that would be to go down, or even how to go about showing it - apart from that, I'm at a loss. Any help would be great, thanks, 81.102.34.92 (talk) 17:36, 29 February 2008 (UTC)

Binary expansion is a good idea, but not applied to x₀. Define

$z_{n}={\frac {1}{2\pi }}\arccos(1-2x_{n})\,,$

and consider how the binary expansion of z_n relates to that of z₀.

 --Lambiam 23:59, 29 February 2008 (UTC)

March 1

Lebesgue Integration Question

My question deals with the relationship between Lebesgue and Riemann Integration on the real line. If one has a continuous function $f:R\to R$ such that the improper Riemann integral over $R$ is infinite, then is the Lebesgue integral over $R$ also infinite?

A concrete example would be a function such as $f(x)=min(1,1/|x|)$. Is it possible for this function's Lebesgue integral to be finite? —Preceding unsigned comment added by 74.15.5.191 (talk) 01:19, 1 March 2008 (UTC)

No. Bottom line is that the only time the Lebesgue and Riemann integrals differ is when the integrand is so discontinuous that the Riemann integral is not defined (because the upper and lower integrals differ). Now watch, someone will come up with some hypothesis I forgot to include, but this is the usual case anyway. --Trovatore (talk) 01:25, 1 March 2008 (UTC)

Oh, since you actually asked two questions, I should specify that "no" is the answer to the second question. The answer to the first question is "yes", at least if the value of f is always nonnegative. --Trovatore (talk) 01:26, 1 March 2008 (UTC)

Ok I think I see why now - I thought the idea about "so discontinuous" would only work on compact intervals, but it can be applied to the whole real line as well, the proof would be something like this: If the Riemann integral is infinite, then for any a > 0, we can find a compact interval such that the integral over that interval is greater than a. But on compact intervals, for continuous functions, Riemann and Lebesgue are equivalent. So we have a series of compact intervals on which the Lebesgue integral is growing without bound... Hence the Lebesgue integral is infinite. 74.15.5.191 (talk) 01:45, 1 March 2008 (UTC)

So you do need the hypothesis that f is nonnegative. Otherwise consider the function that takes the value 2 on the interval [2n,2n+1) for every natural number n, and the value −1 on the interval [2n+1,2n+2). The improper Riemann integral converges to +∞, but only because you're taking the intervals in a fixed order. If you try to take the Lebesgue integral you wind up with ∞−∞. I don't remember for sure how the definition of Lebesgue integral deals with this sort of case; it isn't really very important so it might differ from author to author. --Trovatore (talk) 02:15, 1 March 2008 (UTC)

In the Lebesgue theory f integrable implies $|f|$ integrable, so this function is not Lebesgue integrable on the reals.Silverfish70 (talk) 12:14, 1 March 2008 (UTC)

Large Numbers

This may seem like a simple question to some but, how can a number be larger than all numbers (i.e. $\aleph _{1}$) yet not be infinite? Thanks, Zrs 12 (talk) 01:49, 1 March 2008 (UTC)

In ordinary mathematical usage, $\aleph _{1}$ is infinite. Once in a while you'll come across a context in which "infinite" is used to mean absolutely infinite, and in this case $\aleph _{1}$ may be described as transfinite to disambiguate. --Trovatore (talk) 01:55, 1 March 2008 (UTC)

Cool, thanks. Zrs 12 (talk) 02:13, 1 March 2008 (UTC)

Arc Symbol

Hello. Is there a symbol for circular arcs? If so, what does it look like? I could not find anything on the Internet. I regret the simplicity of this question. Thanks in advance. --Mayfare (talk) 05:44, 1 March 2008 (UTC)

As far as I'm aware no such symbol is in generally understood use in mathematics. If you need a symbol, you might use ◝, unicode U+25DD, ("upper right quadrant circular arc"), but you would have to explain its meaning. Or if no confusion can arise, appropriate ∡, unicode U+2221 ("measured angle"), which is not in common use at least in mathematical texts. Again, you would have to explain the meaning to which you put this symbol.  --Lambiam 12:33, 1 March 2008 (UTC)

You could use something like ${\widehat {AB}}$ (preferably a curve over the top, I can't find one in LaTeX) for the arc from A to B and ${\overrightarrow {AB}}$ for the chord from A to B. It's not universal, but it's easy to understand. You should always explain your notation the first time you use it if it isn't 100% standard. --Tango (talk) 13:21, 1 March 2008 (UTC)

Is there any discovery that could destroy mathematics?

Could a discovery be made that would destroy mathematics, in the same way that we don't have Humorism anymore because of modern medicine? What would such a discovery look like? —Preceding unsigned comment added by 79.122.77.105 (talk) 07:52, 1 March 2008 (UTC)

New notation constantly destroys (old) mathematics, replacing it with a new mathematics. The use of the positional system slowly replaced the roman numerals. The use of computers has replaced the use of mathematical tables and slide rules. The use of complex exponentials is slowly making trigonometry to vanish from the litterature. The next step might be that programming languages makes the traditional mathematical notation obsolete, such that the square root sign and the integral sign and the bar used for fractions will be used no more. Bo Jacoby (talk) 09:36, 1 March 2008 (UTC).

Most unlikely - mathematics (can be/) is founded in logic, and it's rock solid in that respect. That which is already known cannot be disproved.. In this respect it differs from science theories.

I suppose if we discover some sort of edenic utopia, or it's found that thinking or using a pen causes cancer people might give up on it..(joke)87.102.79.228 (talk) 09:48, 1 March 2008 (UTC)

Or maybe they'll find a vaccine against mathematics...

Throughout the history of mathematics you find preoccupations or misconceptions that are hard to understand now (Is it meaningful to add squares to cubes/represent quantities by letters/imagine non-Euclidean geometries? Do imaginary numbers/infinitesimals/actual infinities exist?). Presumably, a few centuries from now future mathematicians will look at what is considered normal now and wonder: what were they (i.e. we) thinking? Perhaps mathematics can be founded in logic, but which logic? How can we know that that logic offers the "solid rockbottom" you want as foundation? And even if mathematics can be founded in such logic, as it is practiced it isn't.

I can imagine the standards of proof going up, as attempted in Project FlysPecK for Kepler's conjecture, with the typical present-day proof becoming to be considered unacceptably non-rigorous, just like we do not accept most pre-Cauchy/Weierstrass arguments purporting to establish limits anymore as proofs. A proof transported from a far future back to today might, conversely, be utterly incomprehensible to us. Even if the field is still called "mathematics", present-day mathematicians may be considered then like today's chemists look at alchemists.  --Lambiam 12:00, 1 March 2008 (UTC)

On a side note, I have often been occasionally bothered by the idea of sufficient rigor for a proof, it would seem to me, that if one can understand all of the notation in a give problem (and in it's proof) then the question of sufficient rigor is one that can be answered through simple logic. Take for example a final result like $y={\frac {x+3}{x}}$ I would content that one should not have to write $x\neq 0$ as it is directly implied by the notation, if one understands what that a fraction is another notation division. (I should also say that somewhere in the proof one had to know the preceding fact.) A math-wiki (talk) 12:54, 1 March 2008 (UTC)

If you're given the fraction a priori, then you may not need to specify that x in non-zero, but usually you would start with something like xy=x+3 and manipulate it into the fraction, and that manipulation requires you to *assume* x is non-zero, and you should always specify your assumptions. --Tango (talk) 13:09, 1 March 2008 (UTC)

The difference of x to the power of n upto nth position defined

I have not found this formula anywhere.

xⁿ - (x-1)ⁿ- (x-2)ⁿ - .... (x-n)ⁿ = n!

I have verified this through MS Excel.

Anupamdutta (talk) 07:36, 1 March 2008 (UTC) Anupam Dutta