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February 23

Econometrics question: Autocorrelation

Hi everyone... am having trouble with a specific aspect of first-order autocorrelation.

The setup is y_t=B*y_(t-1)+u_t ; u_t=p*u_(t-1)+e_t ... y_t and u_t are covariance-stationary processes.

I'm supposed to eventually find cov(y_(t-1), u_t) and I've started off by breaking the covariance into cov(y_t/B , u_t) and cov (-u_t/B , u_t). I'm not sure this is the right approach, am becoming convinced it probably isn't. Anyways, continuing from there I've got the second covariance (with the u's) solved, but the first one, no matter how I break it down, still comes back to something in the form of cov(y_t , u_t).

Any ideas of how to get around this? Thanks very much for any thoughts. rabbit84.92.181.246 00:44, 23 February 2006 (UTC)

repeating decimals

is .999... repeating to infinity equal to 1?

• Yep! See Recurring decimal#The case of 0.99999.... --jpgordon∇∆∇∆ 07:44, 23 February 2006 (UTC)

And also Proof that 0.999... equals 1, and for some additional background real number. -- Meni Rosenfeld (talk) 12:16, 23 February 2006 (UTC)

And if you want your head to explode, see Talk:Proof that 0.999... equals 1, where people identified only by IP addresses try to bring that fact into dispute, and manage to use "phd" as an insult. Confusing Manifestation 12:24, 23 February 2006 (UTC)

You're right, that argument is hilarious. Sad to see that the anon posting there, clearly has some mathematical knowledge, and yet can mix up facts so badly. All the more reason to read real number - The more one knows about where real numbers come from, the less he is likely to make ridiculous claims. -- Meni Rosenfeld (talk) 12:46, 23 February 2006 (UTC)

It's one of the funniest I've ever seen -- all six or so archives as well. Is there a special place for mathematical WP:BJAODN? --jpgordon∇∆∇∆ 03:33, 24 February 2006 (UTC)

What's sadder still is the endless stream of misguided "helpers" who can't see that they're being trolled. --KSmrq^T 18:34, 25 February 2006 (UTC)

Quadratic formula

In the quadratic equation article, it says an alternate form of the quadratic formula is:

$x_{1,2}={\frac {2c}{-b\pm {\sqrt {b^{2}-4ac\ }}}}$

Am I correct in saying that this would not work where c is equal to zero, e.g. in the case "x² - 5x = 0"? That formula would return zero, but the actual answer would be 0 or 5. --210.246.30.87 07:35, 23 February 2006 (UTC)

Congratulations! You have uncovered a longstanding instance of error, delusion, or vandalism. The formula is totally bogus. It went in via this edit (later merged en bloc into quadratic equation). The amazing thing is that this has survived unchallenged for almost two years. -R. S. Shaw 09:49, 23 February 2006 (UTC)

It's not vandalism at all - the formula is perfectly valid. It is used whenever you need to worry about numerical errors from accumulating if you use the so-called "normal" or "canonical" expression for the quadratic formula. --HappyCamper 09:59, 23 February 2006 (UTC)

The expression arises from resolving the radical (is this the correct term for the procedure, as it seems to be an awful mouthful). Here, for example, we have

$x_{+,-}={\frac {-b\pm {\sqrt {b^{2}-4ac\ }}}{2a}}={\frac {\left(-b\pm {\sqrt {b^{2}-4ac\ }}\right)\left(-b\mp {\sqrt {b^{2}-4ac\ }}\right)}{2a\left(-b\mp {\sqrt {b^{2}-4ac\ }}\right)}}={\frac {4ac}{2a\left(-b\mp {\sqrt {b^{2}-4ac}}\right)}}$

Finally, note how some terms magically cancel out on the top and bottom...we get the celebrated (at least in numerical methods and analysis circles)

$x_{+,-}={\frac {2c}{-b\pm {\sqrt {b^{2}-4ac\ }}}}$

of course, valid when a is not zero, and when c is not zero. So, to answer the original question, yes, you are correct to say that the formula fails when c is zero. When c is zero, the conjugate term vanishes:

$\left(-b\mp {\sqrt {b^{2}-4ac\ }}\right)$

Sort of subtle, but do you see how? This should give you a hint to figure out when the alterative formula would be advantageous over the other one...Let me update the article :-) --HappyCamper 10:12, 23 February 2006 (UTC)

But you're not entirely correct to say that the formula incorrectly gives zero when $c=0$. When you take the minus sign it correctly gives 0, because 0 is one of the roots as it always is when $c=0$. When you take the plus sign to find the other root, the denominator becomes zero so the formula gives 0/0, which you recognize as meaning "Now would be a great time to use an alternative formula". So yes, the formula fails, but it doesn't lie to you and tell you an incorrect 0. —Blotwell 12:09, 23 February 2006 (UTC)

In numerical analysis the question of stability of quadratic root-finding is an early cautionary example, illustrating the dangers of naively pulling a formula out of a mathematics text. Assuming a is nonzero and d =b²−4ac is positive, typical code might be

if (b > 0)

 then q := -0.5 * (b + sqrt(d))

 else q := -0.5 * (b - sqrt(d))

r1 := q / a

r2 := c / q

A copysign function could be used instead of the test, for languages that adequately support IEEE floating point. --KSmrq^T 18:52, 25 February 2006 (UTC)

3-torus

This is not a factual question, it is a request. Can anyone generate an animation of the 3-torus? I was thinking something similar to the pentatope one you can see at the right, where each frame is a 3D slice of it. ☢ Ҡi∊ff⌇↯ 10:20, 23 February 2006 (UTC)

Grr...edit conflict....um...this request sort of begs the question: What does a 5d torus look like? Does it even exist? Can 5-D objects have holes in them? Are holes only found in 3-D objects? I wonder if any topologists are here on Wikipedia? Hm... :-) :-) :-) --HappyCamper 10:27, 23 February 2006 (UTC)

Can't say that I know much topology, but...

• I see no reason why 5-D objects shouldn't have holes.
• The torus article, I believe, supplies enough information to decide how does this 3-torus look like, for those able to understand it (myself not included).
• The pentatope is 4-D, making it reasonable to represent it with a 3D animation. But how would you represent a 5-D object? Looks tricky to me.
• Um... whom did you have an edit conflict with? Yourself? I understand, it happens to me all the time.

-- Meni Rosenfeld (talk) 12:29, 23 February 2006 (UTC)

Uh... The 3-Torus is a 4-D object, so an animation would be adequate. I can imagine a 4-D sphere, but a 4-D torus clogs my imagination. ☢ Ҡi∊ff⌇↯ 13:06, 23 February 2006 (UTC)

Ah, I see. So, an animation of the 3-torus would start as a sphere, that would split into two and then join again? ☢ Ҡi∊ff⌇↯ 14:22, 23 February 2006 (UTC)

In general the n-Torus is just S^n-1 X Sⁿ, where Sⁿ is the n-dimensional sphere. (Basically, you take a (n-1)-sphere and rotate is around an axis... think of 3D case, where you rotate a circle around).

All the more reason for me to believe it is a 5-D object... But like I said, I don't really know what I am talking about. -- Meni Rosenfeld (talk) 13:43, 23 February 2006 (UTC)

Ummm..... I think the statement "n-Torus is just S^n-1 X Sⁿ, where Sⁿ is the n-dimensional sphere" is incorrect. In the case n = 3, this would mean that the 3-torus is $S^{2}\times S^{3}$, which is incorrect because the latter space is simply connected whereas the 3-torus is not simply connected (in fact its fundamental group is isomorphic to the sum of three copies of Z). Also it's not entirely clear to me that "the 3-torus is a 4-D object". I can see easily how to embed it in five dimensions: in fact since it's just $S^{1}\times S^{1}\times S^{1}$, we already know that the first $S^{1}\times S^{1}$ can be embedded in three dimensions (that's just an ordinary 2-torus), and then the last $S^{1}$ can be embedded in two dimensions (since it's just an ordinary circle), giving five dimensions altogether. How do you squeeze it into four dimensions? I'm not saying it can't be done; I'm just not a geometer and can't see immediately how it would work. Dmharvey 18:11, 23 February 2006 (UTC)

OK, I think I see now. The statement should be replaced by "the n-torus is just $S^{1}\times T^{n-1}$", where $T^{n}$ is the n-torus. Then you can embed the n-torus in n+1 dimensions as follows. Draw the circle $S^{1}$ in your first two dimensions. We're going to "drag an n-1 torus around the circle". So we have one spare dimension (in the direction of the radius of the circle) plus n-1 dimensions perpendicular to our first two. By induction the n-1 torus that we are dragging requires n dimensions, which is (n-1) + the spare one, so we have enough room. So: a 3-torus needs 4 dimensions, a 4-torus needs five dimensions, etc. That was clear as mud. Sorry. I guess the way to think about it is that in the ordinary 2-torus case, we're dragging a 1-torus (which also happens to be a 1-sphere) around a circle. The correct generalisation is to drag a 2-torus around, not a 2-sphere. Perhaps this helps to visualise the thing, which was the original question :-) Dmharvey 18:28, 23 February 2006 (UTC)

OK, I messed up there... what I meant to say was that the n-torus was $S^{1}\times S^{n-1}$. Mind you, that was still not correct. Tompw 18:49, 23 February 2006 (UTC)

So, what a 3-torus would look like as an animation then? ☢ Ҡi∊ff⌇↯ 08:20, 24 February 2006 (UTC)

I think I can work out what it looks like from one particular "direction", and maybe someone skilled in animation will be able to make an animation for you (I have no idea how to do that). Let's call the dimensions A, B, C, and D. Our main circle (call it M) will be in dimensions A and B. Let R be the radial direction of that circle, so we're going to rotate a 2-torus T which lives in dimensions R+C+D around our circle. Now, I haven't yet specified exactly how to embed T in those three dimensions; there are a few choices, and I think the versions of the 3-torus you get will look kind of different. I'm going to choose the embedding where the axis passing through the "hole" in the middle of the 2-torus T runs along dimension C. Now let's slice the 4-dimensional object we have created. I'm going to take a slice in the A+B+D plane. What this corresponds to is taking a slice of T perpendicular to the C axis and rotating it around the circle M. Such a slice of T looks like two circles with the same centre but different radii.

So here's what the animation looks like: let t (time) run from say 0 to 1. (t is moving along the C axis.) When 0 < t < 1, it looks like an annulus rotated around the circle; or in other words it looks like a "thin 2-torus" contained "inside" a "fat 2-torus". At time t = 1/2 the fat torus has maximal radius and the thin torus has minimal radius. At times t = 0 and t = 1 the two tori coalesce into a single 2-torus, before vanishing completely.

I'm sure other embeddings give interesting results too, I think that's the easiest one to visualise. In the animation you might want to make the outer one transparent and the inner one a different colour so you can see what's going on. Dmharvey 13:27, 24 February 2006 (UTC)

You can represent any n-D object in 3-D or 2-D projections, knowing that you miss a lot ; quasicrystals are a 3-D projection of a 6-D grid. An animation here would be gorgeous. --DLL 21:30, 24 February 2006 (UTC)

A topological 3-torus is trivial to illustrate. Show a point moving inside a solid unit cube. When the point hits a face while moving away from the origin, it teleports to the opposite face, moving toward the origin. For a geometric 3-torus, a specific geometry must be chosen. This can be rather entertaining. Consider the familiar 2-torus, commonly seen as the surface of a donut, depicted in 3D space. It has a geometric form in 4D with the fascinating property that it is "flat", as measured by Riemannian curvature of the surface. (A web search for 'flat torus' finds some nice images.) A comparable option for the 3-torus might map (u,v,w) to the 6D point (cos u,sin u,cos v,sin v,cos w,sin w)/√3; but then a series of projections to 3D or 2D must be chosen. --KSmrq^T 06:07, 26 February 2006 (UTC)

Sub-groups of general linear group

Does there exsist any non-trival subgroup of the (real) General linear group that is not a sub-group of the Orthogonal group? Further, does there exsist such a sub-group that doesn't have the Orthogonal group as a sub-group?

(Alternatively: is there a geometry which is an Affine geometry (which deals with invarients under GL_n), but not a Euclidean geometry (which deals with invarients under O_n)? Further, is there such a geometry which does not "contain" Euclidean geometry?) Tompw 13:40, 23 February 2006 (UTC)

There sure is. For example, take the scalar multiples of the identity matrix by integer powers of 2. This is a group because the product of any two scalar multiples of the identity matrix is another scalar multiple of the identity matrix, and the product of any powers of 2 is another power of 2. None of its elements are orthogonal except the identity matrix itself. —Keenan Pepper 14:16, 23 February 2006 (UTC)

Thanks for that one... any others? (Ignoring those in a similar vein) Tompw 18:43, 23 February 2006 (UTC)

There is plenty to be sure. The symplectic group, the diagonal subgroup, the complex general linear group embedded in the real general linear group, and the group of upper triangular matrices are a few that immediately come to mind. -- Fropuff 20:11, 23 February 2006 (UTC)

Estimation methods

In doing a mock maths paper, one question asked me to estimate the sum of six four-digit numbers by a)the front-end method and b)the "cluster" method. None of us have any idea of what these terms meant, and after some google search(wikipedia search on both terms failed, returning results on topic of statistics), the results shows that the front-end method is to truncate the number to have only two most significant figures, then add them up as usual.(notice in case some number has three digits only as in my example, like 729, it takes 700, not 720). But unfortunately, I still can't find out what the "cluster" method is. Any helps would be appreciated. Thanks. --Lemontea 15:18, 23 February 2006 (UTC)

See if this helps. It's the first Google hit from arithmetic estimation cluster. hydnjo talk 21:17, 23 February 2006 (UTC)

Thanks! It helped a lots. Some follow-up questions: a)From the webpage you provided, does that means both methods are not strictly algorithmic--e.g. it's up to the user to determine which number the group cluster to, how many sig. fig. the front-end method should use? b)If we do front-end method all the way to completion(added all digits), then it is no longer an estimation method, is that right? Finally, it seems wikipedia didn't give any mention of these things, maybe we can add them to the corresponding entries. --Lemontea 02:59, 24 February 2006 (UTC)

It seems that "clustering" is a special case say, "estimate the total weight of 24 containers that nominally weigh 10 pounds each (9.8, 10.1, 10.3, 9.9 ....)", so a reasonable estimate would then be 240 pounds. And yes, if you "front-end" to completion then it is no longer an estimate (neglecting mistakes). The idea behind estimating is to quickly get a ballpark answer to the degree of precision that you want but can also be utilized to get the correct answer via mental arithmetic and amaze your friends.  ;-) This could well be the substance of an article on say, Estimating arithmetic. hydnjo talk 05:29, 24 February 2006 (UTC)

Statical Mechanics

(Transferred from the French Reference Desk)

I know that planck's:

${\mathcal {\,}}E=h\nu (e^{h\nu /kT}-1)^{-1}\,$

mulitiplied by 1/n then do another parts of statics. Then do N*n backing to the original formulas

Same or not between these two cases?

note:for 3D

${\mathcal {\,}}E=3h\nu (e^{h\nu /kT}-1)^{-1}\,$

.

.

.

.

.

.

${\mathcal {\,}}E=(1/3)*3h\nu (e^{h\nu /kT}-1)^{-1}\,$

........then others

Between

${\mathcal {\,}}E=h\nu (e^{h\nu /kT}-1)^{-1}\,$

.......then others in Math.....

.

.

.

.

.

.

The same or Not?

• I see that they are not the same. Because of hv/ (e{hv/ kT}-1) do not need to set somewhat coeffiecients, even in this case.

And setting somewhat numbers does not respect Planck.

The official method, is to be the 2nd I wrote before.--HydrogenSu 23 février 2006 à 16:34 (CET)

February 24

Lagrangian dual of a logarithmic barrier problem

Yes, this is a homework question. Yes, I've put some thought into it myself. I just need some help getting over the last little bit.

Here's the primal problem:

minimize $c^{T}x-43\sum _{i=1}^{m}\ln s_{i}$

subject to $Ax+s=b$, $s>0$

Ok, so first I make the Lagrangian:

$L(x,s,u)=c^{T}x-43\sum _{i=1}^{m}\ln s_{i}+u^{T}(Ax+s-b)$

and I can rearrange this so that x & s are separate:

$L(x,s,u)=-u^{T}b+(c^{T}+u^{T}A)x+u^{T}s-43\sum _{i=1}^{m}\ln s_{i}$

and now I find the dual function

$L^{*}(u)=-u^{T}b+\min _{x}[(c^{T}+u^{T}A)x]+\min _{s>0}(u^{T}s-43\sum _{i=1}^{m}\ln s_{i})$

Now, since x is unrestricted, and the dual function can't go off unbounded below, I know that the dual problem must have the constraint $c^{T}+u^{T}A=0$ which means I can reduce my dual function to

$L^{*}(u)=-u^{T}b+\min _{s>0}(u^{T}s-43\sum _{i=1}^{m}\ln s_{i})$

And... now I'm stuck. What, if anything, do I do with the part that's minimized over s? Is there another constraint in the dual problem, or am I stuck with this, or what? moink 02:08, 24 February 2006 (UTC)

The answer, by the way, is to show that $(u^{T}s-43\sum _{i=1}^{m}\ln s_{i})$ is convex and large as s nears zero and as s increases without bound. Then differentiate that expression w.r.t. s, find the critical point (which must therefore be a global minimum) and substitute that into the Lagrangian. moink 16:49, 28 February 2006 (UTC)

Thanks for coming back with that...I was wondering about it myself. --HappyCamper 15:44, 1 March 2006 (UTC)

Mathematics for the masses

Does anybody know of any puzzles or concepts in mathematics that are:

• Entertaining and educational
• Short
• Easy (People tune out at the sign of a pronumeral)

I work at a Science Centre, and am looking for ideas,

Thanks, --Alexs letterbox 07:56, 24 February 2006 (UTC)

Sure, there are lots. Can't think of any right now, but... Perhaps list of paradoxes is a good place to start looking for all sorts of curiosities. And I think maybe platonic solids match your description. I'll let you know if I think of anything else... -- Meni Rosenfeld (talk) 08:06, 24 February 2006 (UTC)

Take a look at recreational mathematics. :D ☢ Ҡi∊ff⌇↯ 08:18, 24 February 2006 (UTC)

Try cut-the-knot? —Blotwell 10:25, 24 February 2006 (UTC)

I like "guess the number of jelly beans in the jar". It touches on geometry, sampling, and probability (if asked to also estimate the number of each color). And after the contest you all get to eat the jelly beans ! StuRat 11:18, 24 February 2006 (UTC)

Sudoku? Proto||type 12:51, 24 February 2006 (UTC)

We might need an article on pronumeral — does it differ from variable? Gdr 16:28, 24 February 2006 (UTC)

I saw some beautiful graphical proofs of - 1. sum of cubes, 2.pythogorean theorem, etc...which they didnt teach me at school. I found it very nice and inspiring for thinking problems in a very graphical PoV. I could search it up again if u want. -- Rohit 18:07, 24 February 2006 (UTC)

How about fractals? The pictures always look stunning and impressive. --HappyCamper 20:11, 24 February 2006 (UTC)

Sample size

Without having to relearn statistics, how large of a sample would be required to determine, with a normal degree of certainty, valid results about the distribution of topics in Wikipedia (so from a total population of 1 million)? Rmhermen 19:41, 24 February 2006 (UTC)

You'd first have to do a preliminary examination to determine how to determine that, since you don't know what kind of distribution to expect. You'd also have to define the problem in greater detail. --BluePlatypus 20:12, 24 February 2006 (UTC)

How many topics are you thinking of splitting it up into? Is there any overlap? If not, it sounds like the chi-squared test is what you'd use. If so, the only way I know of to find out what sample size you need is to input the data you expect, and try it with progressively larger sample numbers until you get what you want, then take a survey somewhat larger than that in case you guessed wrong on the expected values. Black Carrot 21:30, 24 February 2006 (UTC)

1100 is common number used as a minimum, as it gives about a 3% margin of error over a 90% confidence interval, meaning the results will be within 3% of the actual number 90% of the time. Hopefully a statistician here can show the calculations for this and add details. StuRat 22:45, 25 February 2006 (UTC)

February 25

Pythagorian Triples

Is there a way to prove that there are an infinite number of pythagorian triple families (like 3,4,5 is the family in which 6,8,10 and 9,12,15 belong)? I assume this would involve the proof that there are an infinite number of prime numbers, but you also have to prove that they are in the right proportions. — Ilyanep (Talk) 04:18, 25 February 2006 (UTC)

It's also probably an indirect proof — Ilyanep (Talk) 04:19, 25 February 2006 (UTC)

See Pythagorean triple#Generating Pythagorean triples. It shouldn't be too hard to prove that the triple is primitive if m and n are coprime and one of them is even, and then you're done. —Keenan Pepper 04:31, 25 February 2006 (UTC)

I added a sentence to that article which makes the proof trivial. Since all pythagorean primes are represented by complex numbers of the form $(m+in)^{2}$, one needs only to prove that there are infinite number of complex arguments of such numbers (which is also trivial because their arguments are twice the argument of $m+in$.  Grue  16:36, 25 February 2006 (UTC)

Take any odd integer greater than 1 - call it a. Square it, so you get a². Now express a², which is also odd, as the sum of two integers whose difference is 1 - call these b and c (note that b and c will be coprime). Then

$c^{2}-b^{2}=(c+b)(c-b)=c+b=a^{2}$

So (a,b,c) is a Pythagorean triple - for example, starting with a=13 you get (13,84,85). So each odd integer > 1 gives you a distinct Pythagorean triple. This does not generate all Pythagorean triples, but it does generate an infinite number. Gandalf61 16:44, 25 February 2006 (UTC)

You can also go this more "complicated" method. Here's a sketch. Suppose $x^{2}+y^{2}=z^{2}$. Now divide both sides by $z^{2}$, and set a = x/z, b = y/z. The equation then becomes $a^{2}+b^{2}=1$, which is precisely a unit square. Suppose you have a line that goes from the origin to the circle. If the line has a rational slope, the point of intersection must also be a pair of rational points. So, now, you try and parameterize the original variables in terms of the slope, and this will give you the family of all the possible solutions, and you can show that there are an infinite number of them. --HappyCamper 18:26, 25 February 2006 (UTC)

In my universe, a²+b²=1 is precisely a unit circle. Furthermore, the line with slope 1, which I'm sure is a rational slope, intersects the circle at (¹⁄_√2,¹⁄_√2), which I'm sure is not a rational point. --KSmrq^T 19:04, 25 February 2006 (UTC)

Right. HappyCamper might have been misremembering the following fact. First pick an initial point P on the circle (say (0, 1)), and then consider any line with rational slope passing through P. Except for the tangent line, every such line intersects the circle in exactly one other point whose coordinates are rational; and conversely, if you take a point Q on the circle with rational coordinates, then the line PQ has rational slope. Effectively this is an example of a rational map between the projective line and the circle. Dmharvey 19:22, 25 February 2006 (UTC)

Oh yes, thanks for correcting that critical mistake of mine. A while ago, I used this method to find the all the rational solutions to x^2 + xy + y^2 = 1, which of course, is related to triangles which have angles of 120 degrees instead of 90 degrees. (The 3-5-7 triangle is an example) --HappyCamper 20:50, 25 February 2006 (UTC)

Which nicely brings us to the generation of Pythagorean triples via a rational parameterization of a unit circle. Let the fixed point be (−1,0). Then points on the circle are given by x = (1−t²)/(1+t²), y = 2t/(1+t²). Here t is, indeed, the slope; and clearly a rational slope gives a rational point. More directly, for any rational t we have a Pythagorean triple obtained from 1−t², 2t, 1+t² by clearing denominators. Conversely, a rational point, (x,y), guarantees a rational slope, t = y/(1+x). It suffices to restrict attention to 0 ≤ t ≤ 1, the first quadrant. --KSmrq^T 23:45, 25 February 2006 (UTC)

Golden Times on a clock face

Does anyone have a solution to the following question that would be accessible to an average high school student? "If a rectangle is drawn inside the circle touching where the hands of a clock are pointing, at how many times will the rectangle be 'golden'?" Visit http://en.wikipedia.org/wiki/10:08 for clarification. There is an image at http://en.wikipedia.org/wiki/Talk:10:08 [Don]

Well, the answer is either 22 or 44 times in 12 hours, depending on whether you also allow the hands to point to the short side of the embedded rectangle. Suppose x is the fraction of the circle taken up by the long side of the embedded golden rectangle. Then the times are:

${\frac {12}{11}}(n+x),{\frac {12}{11}}(n+1-x)$ and optionally ${\frac {12}{11}}(n+{\frac {1}{2}}+x),{\frac {12}{11}}(n+{\frac {1}{2}}-x)$ hours since midnight or noon, where n is an integer in the range from 0 to 10. Arthur Rubin | (talk) 17:27, 25 February 2006 (UTC)

Using "SohCahToa" trigonometry, I found that the central angle that intercepts the long side of the embedded golden rectangle is approximately 116.565 degrees. That makes your "x" = 116.565/360 = 0.32379. If we let n=9, the (12/11)(9+x) formula gives 10:10:17.07, one instance of "golden time" that I found earlier by a different means. However, I confess that I do not understand why your formulas work. Please explain how you derived them. Thanks! [Don]

OK, the position of the hour hand at time y hours (measured as a fraction of the circle) is

${\frac {y}{12}}\mod 1$

The position of the minute hand is

$y\mod 1$

so the angle between them is

$\theta (y)={\frac {11}{12}}y\mod 1$

Solving θ(y) = x gives the desired result with the minute hand to the "right" (clockwise) of the hour hand with the two hands forming the long side of a golden rectangle. The other formulas cover the other orientations. Arthur Rubin | (talk) 18:19, 26 February 2006 (UTC)

Arthur, I appreciate your interest in my "Golden Time" inquiry. Unfortunately, I don't understand your latest explanation. If I look at 10:10:17.07 = 10.171409246, as an example, what would y/12 mod 1 and y mod 1 produce? (It is my hope that with your patient tutorage, I will someday be able to use this problem with my high school trigonometry class.) [Don]

I would have thought the puzzle of determining the time from the angle between the hour and minute hands would have been somewhere on Wikipedia, but I can't find it. For the meaning of "mod", see the article on modular arithmetic, ignoring anything relating to multiplication and division. In the instant case (10:10:17.07 above):

y mod 1 = 0.171409246, i.e. the minute hand is 17.1% of a revolution clockwise from noon.

y/12 mod 1 = 0.8476174372, i.e. the hour hand is 84.8% of a revolution clockwise from noon.

(y - y/12) mod 1 = 0.323791809, i.e. the angle between the hands is 116.565 degrees.

Arthur Rubin | (talk) 15:43, 27 February 2006 (UTC)

It was good for me to review "clock" or modular arithmetic. I now understand that mod 1 returns the decimal part of a number. In our problem, it is the fraction of a revolution from noon. Please respond to a couple more questions.

1. How does one "Solve θ(y) = x" or "x = 11/12y mod 1"? Is it appropriate to add an integer "n" to x when dropping the "mod 1" since mod 1 only returns the remainder when dividing a number by 1?

2. How did you know that n should be an integer from 0 to 10 rather than from 0 to some other number? [Don}

What sort of a function is X^X?

A variable to a constant is a parabola, a constant to a variable is exponential, but what is a variable to a variable?

It is neither of those. X^X is equivalent to e^{X ln(X)}, but I don't know a name for a class of functions that contains that. —Keenan Pepper 17:22, 25 February 2006 (UTC)

I've never really thought of it before, but perhaps Transcendental function? Beautiful beautiful topic: Differential Galois theory. --HappyCamper 18:36, 25 February 2006 (UTC)

Also, note that the function y = x^c is only a parabola if c=2. In general, if c is a constant integer, the function is called a polynomial. ×Meegs 03:11, 26 February 2006 (UTC)

Since we're on this topic, how do we find the lowest point of $f(x)=x^{x}$? Derivating it we get $f'(x)=x^{x}\ln x+x^{x}=e^{x\ln x}\ln x+e^{x\ln x}=e^{x\ln x}(\ln x+1)$, but I can't go any further. ☢ Ҡi∊ff⌇↯ 05:12, 26 February 2006 (UTC)

Psh, I guess I was too tired when I tried this yesterday.

$e^{x\ln x}(\ln x+1)=0$

This will happen when either $e^{x\ln x}$ or $\ln x+1$ are zero, but this only happens for $\ln x+1$. So we have $\ln x=-1$, which gives us $x=e^{-1}$. Bah! ☢ Ҡi∊ff⌇↯ 05:22, 26 February 2006 (UTC)

Define g(x) = ln (f(x)), where f(x) = x^x. Then, since the logarithm is a monotonically increasing, you can just find the minimum of g(x), which in turn will let you find the minimum of f(x). The answer should be "reasonably nice number". Sort of a nifty trick which is very useful in statistics and detection theory. Not to mention statistical thermodynamics. --HappyCamper 05:23, 26 February 2006 (UTC)

Nice trick. Thanks for that! ☢ Ҡi∊ff⌇↯ 06:28, 26 February 2006 (UTC)

Yes, that trick comes in very handy. Let's say you have a nasty exponentiated multivariable beast. What happens, is that very often, when you take the derivative with respect to a certain variable, a lot of the other irrelevant variables just disappear! --HappyCamper 15:02, 26 February 2006 (UTC)

$e^{x\ln x}(\ln x+1)=x^{x}(\ln x+1)$. Which is strictly increasing and isn't real for x < 0 and undefined at x = 0. —Ruud 05:24, 26 February 2006 (UTC)

I just got to $e^{x\ln x}$ because I thought that'd be important for the solution, but in the end leaving it as x^x would be the same. Bleh. ☢ Ҡi∊ff⌇↯ 06:27, 26 February 2006 (UTC)

On second thought, it seems to be quite an interesting function when x < 0. —Ruud 05:45, 26 February 2006 (UTC)

Agreed. I've been trying to make a 3D plot of it for a while but I don't have any program to do so. ☢ Ҡi∊ff⌇↯ 06:27, 26 February 2006 (UTC)

Looks like an exponentially damped corkscrew for x<0.

-lethe ^{talk +} 07:08, 26 February 2006 (UTC)

Higher Maths Formula

Hi there, I searched for one of the numerous parabola formulas: $y=k(x+a)(x+b)$ formula, but got no hits; I also searched the parabola article. Can anyone give me the specific article for this formula? Thanks, ^KILO-_LIMA 22:47, 25 February 2006 (UTC)

Assuming k,a,b are constants and x is a variable, then it's a form of Quadratic equation. --BluePlatypus 22:59, 25 February 2006 (UTC)

It's called a merkwaardig product in Dutch but I wouldn't know how to translate that into English. Maybe irreducible polynomial is of any help? Cheers, —Ruud 23:01, 25 February 2006 (UTC)

I've heard it being called "Factored form", but I don't think there is a Wikipedia article on it, and no section on it in the quadratic equation article. --Borbrav 01:51, 26 February 2006 (UTC)

The article on polynomials calls it 'factored form'. It's not specific to quadratics. --BluePlatypus 03:17, 26 February 2006 (UTC)

Yes, I have heard it referred-to as the "factored form" as well. Among other things, this form makes it easy to find the roots (the "x-intercepts") of a function. The first section of the article Factoring, entitled "Factoring a Quadratic Equation" should help you in dealing with parabolas. ×Meegs 03:23, 26 February 2006 (UTC)

February 26

Mechanical aproximations of functions

I've been interested on mechanisms to aproximate certain functions in the physical world. For example, you can draw a circle by fixing a point somewhere and rotating a fixed length thing around. An ellipse can be drawn with a string with ends fixed on the foci. Sure, easy, because they can be defined like that anyway. But, I'm looking for methods for a few particular functions, not sure if they are even possible, but it's worth asking...

1. sin x
2. x² (Keenan Pepper's idea will do)
3. 1/x
4. e^x
5. ln x
6. (e^θ,θ) (logarithmic spiral)
7. the catenary

Note that I'm not looking for straight geometric algorithms, like using a ruler\straightedge and compass, or careful measures (except on the case of certain, necessary proportions), but mechanical devices that actually "plot" these curves in their natural movement. Also, they must be table-top gadget things, so saying I could throw a sphere on a plane to plot x² isn't really what I want. These devices can include strings, pulleys, rods, wheels, gears, trails, etc.

Additionally, I'd like to know if there's any interesting, physical method to approximate e, something akin to Buffon's needle?

Well, that'll be all. :P ☢ Ҡi∊ff⌇↯ 13:25, 26 February 2006 (UTC)

Okay, at least for sinx, if you agree to some level of sophistication, you can create a mechanical device which when you rotate it, it moves a long sheet of paper and simultaneously moves a pen in an essentially harmonic motion, thus plotting a sine wave on the paper (I hope you understood what I meant, couldn't find the best words). As for some other curves, and without using such gizmos, I have an idea which I haven't really thought out through, but it just might be crazy enough to work: Suppose you wrap 2 strings around a wheel in opposite directions - so that when the wheel rolls, it releases one string and pulls the other. If the strings are kept in tension, this should keep the difference of lengths of the strings constant. So, by wrapping them around properly placed pulleys\nails, and attaching them to a pen, you can create a curve where the difference of distances between 2 points is constant - Which I believe is a hyperbola, essentially plotting 1/x. With the proper setting, I think this idea can allow plotting a curve where any given linear combination of distances between any number of arbitary points is constant - Opening many possiblities, and with some tweaking perhaps including a parabola. What do you think? -- Meni Rosenfeld (talk) 14:23, 26 February 2006 (UTC)

For the sine, the problem with the rotating wheel is to extract the movement in one axis without distortion. I've been thinking of a few devices, but they end up with either a cycloid or some weird shape I can't name. ☢ Ҡi∊ff⌇↯ 14:53, 26 February 2006 (UTC)

The catenary's an easy one: put a pen on the rim of a circle and roll the circle I'll have to think about the others. Confusing Manifestation 14:31, 26 February 2006 (UTC)

You're mixing it up with the cycloid. Funnily I, too, mistook it for the cycloid at first, but followed the link to make sure. Besides, I'm not sure your idea is easy to implement in practice. -- Meni Rosenfeld (talk) 14:34, 26 February 2006 (UTC)

Yes, sorry, I just checked the article to confirm what I thought. I suppose to make a catenary you could try to find some way to make a "string" that can be frozen in place somehow ... Confusing Manifestation 14:41, 26 February 2006 (UTC)

Nah, not what I'm looking for :P ☢ Ҡi∊ff⌇↯ 14:53, 26 February 2006 (UTC)

I'm not sure about the transcendental functions, but the conic sections should be possible. For the parabola, fix one end of a piece of string to the focus, and let the other end move along a line parallel to the directrix but on the opposite side of the focus. Draw the parabola by moving the pen so that the string remains taut.

The hyperbola is much trickier. I need to think about it longer. —Keenan Pepper 16:16, 26 February 2006 (UTC)

This one will do. Thanks! ☢ Ҡi∊ff⌇↯ 18:16, 26 February 2006 (UTC)

For x², would it work to choose the focus and the directrix and attach a string that is fixed to the focus and can move along the directrix, attach your pencil to the midpoint of the string, pull it taut, and trace? I guess this would work for any conic section, so it would do 1/x as well (simply adjust the eccentricity). -lethe ^{talk +} 16:18, 26 February 2006 (UTC)

Not the directrix itself, but a line parallel to the directrix. The same thing does not work for a hyperbola. —Keenan Pepper 16:51, 26 February 2006 (UTC)

I don't understand. Which parallel line will you choose? A conic section is defined as the locus of points with fixed ratio of distance between the focus and directrix. -lethe ^{talk +} 16:55, 26 February 2006 (UTC)

Any parallel line; it doesn't matter. If the string can slide easily along the line, and the string is taut, then it's perpendicular to the line. How far away the line is only affects how much of the parabola you can draw. —Keenan Pepper 17:01, 26 February 2006 (UTC)

For the parabola, you're right, it doesn't matter how far the parallel line is. But for the other conics it does, you have to use the directrix. As far as not being able to draw very much, I guess we'll have to use an elastic string but make sure that the pencil stays at the right part (hmm... that would be difficult). -lethe ^{talk +} 17:06, 26 February 2006 (UTC)

Wait, elastic? What are you talking about? In my construction, the string must have a fixed length, and the pen can move along the string as long as the string remains taut... —Keenan Pepper 17:13, 26 February 2006 (UTC)

In my construction, the pen remains fixed exactly 1/(1+e) percent of the way from the directrix. This insures that the ratio between the distances is always e. The ratio remains fixed, but the total length will grow to infinity in a hyperbola or parabola, so we need an elastic string. -lethe ^{talk +} 17:19, 26 February 2006 (UTC)

I think my method is more practical. How are you supposed to keep the elastic straight? —Keenan Pepper 17:22, 26 February 2006 (UTC)

Keep the elastic straight by pulling taut. The more serious problem I guess is how you keep the ratio fixed in an elastic string. -lethe ^{talk +} 17:27, 26 February 2006 (UTC)

Maybe some sensitivity to tension: if the two parts of the elastic have equal tension, then the ratio can be kept fixed. -lethe ^{talk +} 17:45, 26 February 2006 (UTC)

Agreed. Also, the fact there'll be a limit to the drawing is irrelevant. I don't intend to use this to infinity, you know. ☢ Ҡi∊ff⌇↯ 17:25, 26 February 2006 (UTC)

I don't need elastic because it goes to infinity, only because the length changes. I didn't mean to suggest that an elastic string will actually allow you to draw an infinite graph in finite time. -lethe ^{talk +} 17:27, 26 February 2006 (UTC)

Sure, but still, the elastic wouldn't be a good idea. I think there must be some more ingenius way without using strings. ☢ Ҡi∊ff⌇↯ 17:36, 26 February 2006 (UTC)

Well if you don't like my elastic, see Keenan's idea. -lethe ^{talk +} 17:45, 26 February 2006 (UTC)

Yeah, his works great. ☢ Ҡi∊ff⌇↯ 18:14, 26 February 2006 (UTC)

For the hyperbola, I think you can do it by fixing one end of the string to one focus and letting the other end move around a circle centered at the other focus. A practical problem is getting the string to move smoothly around the circle without sticking and slipping. —Keenan Pepper 16:56, 26 February 2006 (UTC)

I don't understand why focus/directrix won't work for the hyperbola. It's basically an exact mechanical implementation of the definition. -lethe ^{talk +} 17:11, 26 February 2006 (UTC)

Well, I managed to make a approximation of sine. The method needs some polishing but it works. The others seem to be a lot harder to derive mechanically, so I'm dropping the idea. Still, I wonder if there's any method to approximate e, like Buffon's needle for pi. Any thoughts on that? ☢ Ҡi∊ff⌇↯ 15:29, 2 March 2006 (UTC)

x^2 = 2^x

There are three solutions for $x^{2}=2^{x}$, x = 2, x = 4 and the other one is negative. I have absolutely no idea how to find this, and I've tried everything I know. How do we solve things like these? ☢ Ҡi∊ff⌇↯ 14:47, 26 February 2006 (UTC)

Did you mean "how to find that these are the only roots" or "how to find the negative root"? If the former, that's easy by analysing the functions. If the latter, this is an essentially non-algebraic equation so the root can't be given explicitly in terms of elementary functions, but it should be possible to represent it with non-elementary functions (I can try to work that out if you'd like), and you can approximate it with any numerical root-finding algorithm (it is roughly -0.76666469596212309311). Hope that's what you meant... -- Meni Rosenfeld (talk) 14:58, 26 February 2006 (UTC)

Oh, so I was trying in vain... Nice to know. ☢ Ҡi∊ff⌇↯ 15:39, 26 February 2006 (UTC)

You may be looking for Lambert's W function. —Ruud 15:21, 26 February 2006 (UTC)

Yeah, that does it. Thanks! ☢ Ҡi∊ff⌇↯ 15:39, 26 February 2006 (UTC)

This is (sequence A073084 in the OEIS) - Fredrik Johansson 15:24, 26 February 2006 (UTC)

I wonder if anything can be said by letting x = a + bi ... when you do this, you get two sets of equations by equating real and imaginary parts

${\frac {a^{2}-b^{2}}{2^{a}}}=\cos(b\ln 2)$

${\frac {2ab}{2^{a}}}=\sin(b\ln 2)$

Dividing the first by the second, we get

${\frac {2ab}{a^{2}-b^{2}}}=\tan(b\ln 2)$ - and ah! This can be written as a quadratic equation

$a^{2}\tan(b\ln 2)-2ab-b^{2}\tan(b\ln 2)=0\;\,$ and I'm sure something can be said about the admissibility of these roots, the conditions on a and b, et cetera... --HappyCamper 15:45, 26 February 2006 (UTC)

Since b = 0, this doesn't help. Fredrik Johansson 16:18, 26 February 2006 (UTC)

Why is b = 0 ? --HappyCamper 18:54, 26 February 2006 (UTC)

We're considering the function only for real values of x. ☢ Ҡi∊ff⌇↯ 19:00, 26 February 2006 (UTC)

Oh. Hm...a pity. :-) --HappyCamper 19:04, 26 February 2006 (UTC)

A new form of logic?

Goedel's incompleteness theorem only applies to first order logic, right? If so, would that mean that it might be possible to create a different form of logic that is sound, complete, etc. AND is capable of doing what Goedel's incompleteness theorem says that first order logic is incapable of (i.e. is also capable of making a complete and consistent system of math)? —The preceding unsigned comment was added by 86.138.233.25 (talk • contribs) 15:49, 26 February 2006 .

In a sense, yes. Let's specialize to the case of the natural numbers (too hard to talk about all of "math" because it's not clear where math ends). You could have a generalized "logic" that would, by fiat, "conclude" that a statement about the natural numbers is true, just in case it is true. But it wouldn't be very helfpful to you when trying to make deductions, because you wouldn't know whether something was a correct deduction, until you knew whether the conclusion was true.

Bottom line is, if your "logic" is one such that the question of what constitutes a correct deduction is checkable by a fixed computer program, it can't get you out from under the Gödel theorems. (Rant: will everyone please notice there are two Gödel incompleteness theorems?) --Trovatore 16:03, 26 February 2006 (UTC)

I think no. Gödel's incompleteness theorem is very general, and it would probably apply to any extended logic as well, becuase that extended logic would be able to do anything first-order loic can do. – b_jonas 12:22, 27 February 2006 (UTC)

No, it's simply not true that the first incompleteness theorem applies to "any extended logic"; see the example I give above. For a less contrived example, it doesn't apply to second-order logic. A few simple axioms completely determine the whole theory of the natural numbers, in second-order logic.

But while that fact about second-order logic is important for theoretical purposes, it doesn't make much practical difference in terms of trying to figure out what's true about the natural numbers. That's because no one knows how to make deductions, in general, in second-order logic (and checking whether such deductions are correct provably cannot be reduced to a computer program. --Trovatore 14:51, 27 February 2006 (UTC)

What is this beautiful abstract nonsense trying to say to me?

Could someone please explain what the Nine lemma is, and why it is important? Where is this lemma used, and what is it really trying to say in the article? --HappyCamper 15:58, 26 February 2006 (UTC)

Well, I guess there is some theorem that uses the nine lemma, that's why it's called a lemma. I'm not sure which, but I can tell you that the closely related snake lemma is the key for proving the existence of the derived functors, which are studied in homological algebra. I've always viewed the 3x3 lemma (its name as I learned it) as an application of the snake lemma, but I suppose it must have been used in the proof of a theorem as well. -lethe ^{talk +} 16:35, 26 February 2006 (UTC)

Converting trigonometric expressions into algebraic form

How do I go from $\csc(\arctan \left({\frac {\left(x\right)}{\sqrt {2}}}\right))$ to ${\frac {\sqrt {x^{2}+2}}{x}}$? I've been staring at problems of this sort for a couple days, and I'm at a complete loss. --Theshibboleth 20:19, 26 February 2006 (UTC)

In other words, you're trying to find $csc(\theta )$ where $tan(\theta )={x \over {\sqrt {2}}}$. Picture a right triangle in which one of the angles is $\theta$, the opposite leg is x, and the adjacent leg is ${\sqrt {2}}$. Now it should be easy. —Keenan Pepper 21:39, 26 February 2006 (UTC)

February 27

Parabolas and straight lines

Hi, I am aware of the no-homweork questions, but this question is really beginning to annoy me; and I hope you won't be bothered by helping me complete it! It concerns parabolas and straight lines:

File:Parabolas34.JPG

Question 9(a) and (b) I have done correctly. However I am having problems with (c). It begins:

$-2x^{2}+4x+6=2x+2$ - The $2x+2$ is from the line. Then, becuase this is a quadratic question, it must equal zero, so: $-2x^{2}+2x+4=0$. Then change to a positive $x^{2}$ term: $2x^{2}-2x-4=0$. This then factorises to $(2x-4)(x+1)=0$. So $2x-4=0$ so $x=2$ and $x+1=0$ so $x=-1$. Now this is where the problem occurs. Becuase we have the x-coordinate, put it back into the formula to get the y-coordinate. So using $x=-1$: $y=2x^{2}-2x-4$, $y=2+2-4$, and $y=0$. So the coordinates are $(-1,0)$—but looking at the graph this is already there.

So, using $x=2$: $y=2x^{2}-2x-4$, $y=8-4-4$, and therefore $y=0$. So the coordinates for this one are $x=(2,0)$—but this is clearly not correct by looking at the graph. Does anybody know where I am going wrong? I appreciate to the highest on this question. If you cannot answer, then no problems caused. Thank you very much again. ^KILO-_LIMA 20:33, 27 February 2006 (UTC)

You need to go back to the original equation $-2x^{2}+4x+6=2x+2$ and work from that. In other words, the parabola shown on your graph is $y=-2x^{2}+4x+6$, and the straight line is $y=2x+2$. Chuck 21:00, 27 February 2006 (UTC)

(Chuck Carrol was faster than me in an edit conflict, but I'll post this anyway.)

You're susbtituting the polynomials back to the wrong formula. $x=2$ is indeed the solution, but then $-2x^{2}+4x+6=-2\cdot 2^{2}+4\cdot 2+6={}$$6=2\cdot 2+2=2x+2$.

Anyway, I think this doesn't count a homework question, because you show that you have done efforts to solve it, not just ask us to solve it, and because you ask a concrete question. So, I think there's no problem posting this. – b_jonas 21:06, 27 February 2006 (UTC)

And you were nice about it, which never hurts! — QuantumEleven | (talk) 08:52, 1 March 2006 (UTC)

February 28

G. Cantor

From what I've read, I know that the general consensus among mathematicians is that 'Cantor was right'...and they agree even more because of Quantum Mechanics (I don't know why)... I'm also aware that Cantor was religious and sort of 'prooved' the existence of God with his theory about transfinite numbers and said that God was the 'actual infinite' or 'actus purisimus' or something like that... My question is...If mathematicians accept as truth Cantor's theory, and supposedly it's 'proved' (I don't know how)...then doesn't that mean that all mathematicians are theists?...that would be the logical consecuence...but I'm sure this is not the case...so why...I mean, how do non-theist mathematicians 'escape' Cantor's reasoning?.--Cosmic girl 02:34, 28 February 2006 (UTC)

Cantor's Absolute Infinite doesn't follow logically from his formal mathematical works; in fact, it's inconsistent with them. It's just a beautiful way of integrating his faith with his mathematics. —Keenan Pepper 03:39, 28 February 2006 (UTC)

Hi Pepper!...why is his Absolute Infinite inconsistent with his mathematical works? I guess I know why, but maybe I'm wrong, can you explain me?.--Cosmic girl 04:27, 28 February 2006 (UTC)

Cantor was certainly a believer, but I'm not aware that he ever published a proof of God's existence. Are you sure you aren't confusing him with Gödel? He did have such a purported proof; see Gödel's ontological proof. --Trovatore 03:42, 28 February 2006 (UTC)

Nope...I'm not :) lol.thanks for your answers to the Gödel question by the way! XD --Cosmic girl 04:34, 28 February 2006 (UTC)

The premise of the question is nonsense. Mathematics can only prove mathematical theorems; it can prove nothing about physical reality, much less about metaphysical reality. Furthermore, the reasoning in this question is horrible: It rambles from a handwave "consensus among mathematicians" (who?) that "Cantor was right" (about what?) and speculation that Cantor "sort of proved" (is that vague enough?) the existence of "God" (defined how?) to a question of "escaping" (eh?) Cantor's reasoning. Atheists are a diverse group, but I would expect most to treat such arguments with even less interest and credibility than an alleged sighting of the Loch Ness Monster drawing crop circles. --KSmrq^T 03:50, 28 February 2006 (UTC)

Hahaha...okok! you don't have to yell at me! I'll explain myself... first of all, by consensus among mathematicians I meant that Cantor's theory is regarded as 'true' and 'proved' now, and in his time he was criticized...and I made it clear that I didn't know how is it 'proved', since I don't know the math of it in the first place and I admit it. second...by God I mean 'actual infinity' like Cantor did,and by 'escapinc Cantor's reasoning' I mean that since he said he 'prooved the existence of an actual infinite which he equated with God'...I believe there must be detractors...what I asked was if the detractors had consistent arguments and theorems like Cantor's to 'disprove' 'actual infinity' or 'god'.

ps. take it easy KSmrq...I'm dumb.--Cosmic girl 04:32, 28 February 2006 (UTC)

Cantor did not equate what's ordinarily called the "actual infinite" with God. In fact he went out of his way not to. That's where the word "transfinite" came from; as I understand it: He wanted a word for something that wasn't finite, but wasn't the Infinite either. --Trovatore 04:36, 28 February 2006 (UTC)

He did equate the actual infinite with God...and I believe transfinite numbers stand for numbers that are composed of an infinite ammount of numbers that are infinite as well...not something that isn't finite nor infinite...but correct me if I'm wrong.--Cosmic girl 04:42, 28 February 2006 (UTC)

OK, you're wrong :-). No, he simply did not equate what most people call "the actual infinite" with God. He didn't call it "actual infinite", but rather "transfinite". Your facts are simply wrong here. --Trovatore 04:46, 28 February 2006 (UTC)

Are you sure? ok, I bet you are right...but my understanding was that the actual infinite was a 'real infinity' and transfinite numbers where infinite numbers but only potentially... :S. --Cosmic girl 05:02, 28 February 2006 (UTC)

You're mixing up two (quite different) distinctions: potential-vs-actual, and transfinite-vs-absolute-infinite. The set of all natural numbers, for example, is an "actual" infinity, but not an "absolute" infinity, because there are bigger things (such as the set of all real numbers). The article Absolute Infinite claims that he identified "absolute infinity" with God, but if you look at his writings on it, I think that's an oversimplification. I seriously don't think he thought God was any sort of mathematical object. He was much too traditionally Christian for that. --Trovatore 05:18, 28 February 2006 (UTC)

I see...thanx :).--Cosmic girl 16:57, 28 February 2006 (UTC)

Integration

Hello mathematicians, I've been having fun with Euler's equation and have managed to juggle a homework equation up to the point where I'm left with

$\int {\frac {dy}{\sqrt {1-A(y+1)^{2}}}}$, where A is a constant. This looks to me like I should be getting y equal to something with arcsin or arccos, and need a push in the right direction. Sorry this is a homework question guys, but pretend it isn't

(Can someone put this into maths syntax if possible, ;) ): Thanks --131.251.0.8 10:02, 28 February 2006 (UTC)

Check out Trigonometric substitution. You need to work a little on the integral to bring it to the form discussed there, but it isn't difficult. -- Meni Rosenfeld (talk) 11:36, 28 February 2006 (UTC)

Incidentally, Euler's equation already redirects to Euler equations, and when it comes to Euler it helps to specify things a bit more (see List of topics named after Leonhard Euler to see why) Confusing Manifestation 11:45, 28 February 2006 (UTC)

Actually, scratch that, we have both Euler equations which are for fluid dynamics and Euler's equations for rigid rotation. Could we perhaps have some clarity here? Confusing Manifestation 11:47, 28 February 2006 (UTC)

The user probably searched for the article with a space between Euler and 's, just like in the above link, so he\she didn't find the article. ☢ Ҡi∊ff⌇↯ 12:13, 28 February 2006 (UTC)

Cool, I've done that, and have ended with a dodgy equation in the form:

2cos(ci+i) = cos(ci), with c constant and i the sqrt(-1). Surely there's no solution to cos(i) is there? cosine only applies to real numbers surely. --131.251.0.7 12:23, 28 February 2006 (UTC)

Cosine applies to complex numbers (see also Cosine#Relationship to exponential function), but you shouldn't have gotten there - I guess you used a wrong substitution. Follow the example in Trigonometric substitution which is identical to your case after some simplification. -- Meni Rosenfeld (talk) 13:02, 28 February 2006 (UTC)

Drat, once again my impatience causes my edit to be lost. (Closed window before seeing edit conflict.) Anyway, trigonometric substitution is truly the correct choice. The core trigonometric identity is cos² θ+sin² θ = 1, which we can solve for either trig function. So, letting B²=A, set sin θ = B(y+1); then the denominator neatly simplifies to cos θ. Next, dy must be converted; but since y = −1+(sin θ)/B, clearly dy = (cos θ) dθ/B. Thus the form in the integrand becomes trivial to integrate as a function of θ, and all that remains is to convert θ back to a function of y in the result. We are playing fast and loose here, and ignoring issues that would arise with a definite integral. (What if A is zero? Does the sign of B matter? Do we lose a needed branch by taking the positive square root, cos θ = +√(1−sin² θ)? Should there be limits on the range of y?) Depending on your instructor's tastes, this is probably acceptable for such a homework question.

"square root" of exponential function

Richard Feynman once asked if there was a "natural" "square root" of the exponential function - that is, a function f defined on the reals such that

$f\left(f(x)\right)=e^{x}$

It's "obvious" that there are continuous solutions, and less obviously $C^{\infty }$ solutions, and I've been told there are real analytic solutions, but is there a "natural" solution?

One can find all continuous solutions by letting a be an arbitrary real strictly between 0 and 1, and f an arbitrary monotone increasing continuous function from [0,a] onto [a,1]. One can then extend f uniquely to a solution of the specified equation.

It's not too difficult to do the same with $C^{\infty }$ solutions -- all you need to do is for the initial f to be $C^{\infty }$ and the formal derivatives of f(f(x)) must match the derivatives of e^x at x=0.

Also "obviously", if there is a "natural" solution g of

$g\left(e^{x}\right)=g(x)+1,$

then we could take

$f(x)=g^{-1}\left(g(x)+{\frac {1}{2}}\right)$

Any ideas? Arthur Rubin | (talk) 16:00, 28 February 2006 (UTC)

How many real analytic functions are there? If there's only one, then you could consider that a "natural" solution.

However, I think it's quite possible that there is no single "natural" solution. Iirc, Concrete Mathematics says that there's no single best way to extend Stirling numbers to arbitary real arguments. – b_jonas 21:02, 28 February 2006 (UTC)

I've looked this up in the Art of Computer Programming chapter 4.7 (which is in volume 2), and if I understand it correctly, it says that for an analytic function $f$, there are exaclty two analytic functions $g$ for which $g(g(x))=f(x)$, with some conditions of course. – b_jonas 21:10, 28 February 2006 (UTC)

Interesting question. Confusing. I got the idea to work with polynomial approximations first (so as to give hint of what an analytic solution might look like). So take f(x) to be a polynomial of degree n, and try to match up terms with the polynomial $e_{N}(x)=\sum _{k=0}^{N}x^{k}/k!$ where $N=n^{2}$. Works fine for n=1, breaks for anything larger, since the problem is over-determined: there are $n^{2}+1$ equations for $n+1$ unknowns. One might hope for some "happy accident" that would make most of these equations be degenerate, but I don't see how. I tried this by pencil and paper, and, unless I made a mistake, there is no solution for n=2. So I don't see how there can be an analytic solution, unless I'm overlooking something. linas 01:17, 1 March 2006 (UTC)

That doesn't in itself show the non-existence of an analytic solution. Consider $f(f(x))=x^{2}$; the solution $f(x)=x^{\sqrt {2}}$ is analytic but $a(ax^{2}+bx+c)^{2}+b(ax^{2}+bx+c)+bx+c=x^{2}$ does not have a solution. Here is a list of references to similar problems: [1]. Kusma (討論) 02:07, 1 March 2006 (UTC)

Err, well, I phrased that badly, in several ways. The very first sentance posing this problem shows that there are zillions of analytic functions that "solve" this (pick a real a, s.t. etc.) However, most of these choices will not be differentiable at x=1. And if I force $C^{\infty }$ at x=1, then they won't be differentiable at x=0. Also, all I've managed to demonstrate is that working with monomials is a trap (as it often is). So I'm trying to dream up other sequences of functions that approximate/converge to an answer ... linas 02:01, 1 March 2006 (UTC)

Further follow up. The proof that there exist $C^{\infty }$ solutions is as follows, with some additional effort. Suppose f is $C^{\infty }$ on [0,a].

Write, formally,

$f(x)=\sum _{k=0}^{\infty }{a_{k}x^{k}}$ for x near, but greater than 0, (with a₀ = the a specified above in the continuous solution).

$f(x)=\sum _{k=0}^{\infty }{b_{k}(x-a)^{k}}$ for x near, but less than a.

(These 'equalities' mean that all derivatives are equal, not necessarily that the series converge.)

If, working as formal power series, $f(f(x))=\sum _{k=0}^{\infty }{\frac {x^{k}}{k!}}$, then the resulting extension will be $C^{\infty }$.

Now, given an arbitrary sequence a, (satisfying the constraints that $0<a_{0}<1$ and $0<a_{1}$) we can solve for the corresponding sequence b, and construct a $C^{\infty }$ bridge f, such that $f'(x)\,>0$ for $0\leq x\leq a$, which will then satisfy all the conditions, and can be extended to a full $C^{\infty }$ solution.

Arthur Rubin | (talk) 02:49, 1 March 2006 (UTC)

Furthermore, $f(x)=x^{\sqrt {2}}$ is not analytic in any sense near x=0, as $f''\,$ does not exist there.

Yes, well, silly me. Where I was trying to go with this was to look at sequences of functions, since "naturalness" often manifests itself in terms of some related sequence which has some particularly nice properties. But I'm taking a naive aproach here, clearly there's a raft of literature, which Kusma references. linas 03:07, 1 March 2006 (UTC)

March 1

Third order covariance matrix

Can anyone give me a clear definition of a third order covariance matrix? The only thing I can find on the net is at mathworld but it is not clear to me what is meant by the superscript notation mn --Will

The superscript on the letter V is just for notational convenience. The m and n on the right side are exponents though. Whenever m + n = 3 (and presumably m and n are positive) you have your "third order" covariance matrix. From this definition, there are 4 different "third order" matricies found by setting (m,n) = (0,3),(1,2),(2,1),(3,0)...in some books, only (1,2) and (2,1) are considered as "third order covariance matricies". --HappyCamper 19:08, 1 March 2006 (UTC)

March 2

how di i make a combination of eighteen numbers in groups of six

I'm afraid you'll have to suitly emphazi your question. —Keenan Pepper 03:24, 2 March 2006 (UTC)

I'll make a wild guess and assume you meant "How many possibilities are there to select 6 elements of a set of 18 elements, without repetition and without order being important". In that case, the answer is

${18 \choose 6}={\frac {18!}{6!12!}}=18564$

-- Meni Rosenfeld (talk) 07:29, 2 March 2006 (UTC)

convert 2'-9" in inches

Can you really not do this in your head? If not, there's always Google Calculator. —Keenan Pepper 05:48, 2 March 2006 (UTC)

In case it's the notation that's giving you problems, see our article prime (symbol). In the context of length, ′ denotes a foot, ′′ an inch. ×Meegs 06:01, 2 March 2006 (UTC)

The notation is certainly giving me problems. Does the questioner want two feet plus nine inches, or two feet minus nine inches? —Blotwell 13:02, 2 March 2006 (UTC)

Wikipedia growth prediction question (third-degree polynomial equation)

If I know that Wikipedia reached 500,000 articles on March 17 2005, 666,666 articles on August 4 2005, and 1,000,000 articles on March 2, 2006, and I know that Wikipedia "currently" has x articles, then how can I make this into a third-degree polynomial approximating Wikipedia's growth and solve it to get the time Wikipedia will reach a given milestone? JIP | Talk 08:17, 2 March 2006 (UTC)

Okay, first you need to convert the dates to explicit amount of days. I'll leave it to you to work that out, but let's assume you get March 2005 = 90 days, August 2005 = 270 days, March 2006 = 450 days. Now you write:

$f(x)={\frac {500000(x-270)(x-450)}{(90-270)(90-450)}}+{\frac {666666(x-90)(x-450)}{(270-90)(270-450)}}+{\frac {1000000(x-90)(x-270)}{(450-90)(450-270)}}$

Simplify, and solve using Newton's method. -- Meni Rosenfeld (talk) 08:31, 2 March 2006 (UTC)

Wait, you got me mixed up. With 3 points of data you only get a quadratic polynomial. So you can solve it directly. -- Meni Rosenfeld (talk) 08:34, 2 March 2006 (UTC)

Or perhaps you meant there's an additional point, the current time x₀ and the current articles y₀? Stick it to the formula and you get

$f(x)={\frac {500000(x-270)(x-450)(x-x_{0})}{(90-270)(90-450)(90-x_{0})}}+{\frac {666666(x-90)(x-450)(x-x_{0})}{(270-90)(270-450)(270-x_{0})}}+{\frac {1000000(x-90)(x-270)(x-x_{0})}{(450-90)(450-270)(450-x_{0})}}+{\frac {y_{0}(x-90)(x-270)(x-450)}{(x_{0}-90)(x_{0}-270)(x_{0}-450)}}$

And see also Lagrange polynomial. -- Meni Rosenfeld (talk) 08:40, 2 March 2006 (UTC)

This isn't my best day. You can also solve a third-degree polynomial directly, but I guess Newton's method is simpler. -- Meni Rosenfeld (talk) 08:42, 2 March 2006 (UTC)

Beware, though, that an exactly-fit polynomial will probably not make such an accurate a model of WP's growth, and could actually be astonishingly bad. Depending on what that fourth data point is, the cubic could easily predict negative growth in the future (as well absurd things back between your data points). There is some discussion about modeling the growth exponentially at Wikipedia:Modelling Wikipedia's growth that might interest you. ×Meegs 09:11, 2 March 2006 (UTC)

Fourier transform of a positive semidefinite function

Under which additional conditions (on the function f(x)) is the Fourier transform of a positive semidefinite function f(x) again positive semidefinite? — Preceding unsigned comment added by 193.144.64.159 (talk • contribs) 11:06, 2006 March 2 (UTC)

(Please sign your questions.) The Fourier transform is a change of basis. Contemplate the definition of positive semidefinite. --KSmrq^T 13:06, 2 March 2006 (UTC)

Area bounded by a hyperbola

I need to prove that the area bounded by a unit hyperbola and a line from the origin to the hyperbola is equal to twice the angle of that line.

I know that the point where the line and the hyperbola meet is (cosht, sinht). Using that, I was able to find the equation of the line: y = xtanht. Im running into trouble with the integral. I think I did it right. I wish I could show it better, but I can't get the syntax of the math language to work: I figured I have to integrate along the y axis, from sinht to 0, because sinht is the y value of the intercection. Assuming I did it right, I got an integral of sqrt(y^2+1) - y/tanht. Please let me know if I am on the right track, and if you can help me out with the math formatting, that would help too. Thanks a lot. --Chris 17:07, 2 March 2006 (UTC)

Looks okay to me. Now all you need is to look at List of integrals of irrational functions to solve the integral

$\int _{0}^{sinh(t)}\left({\sqrt {y^{2}+1}}-{\frac {y}{tanh(t)}}\right)dy$

Btw, an efficient way to do math formatting is to seek an equation that uses the symbols you need, copying the code and altering it to match your expression. -- Meni Rosenfeld (talk) 18:05, 2 March 2006 (UTC)

Hey, thanks for helping out. Just one question though: is it supposed to be the difference of the two integrals, not the sum?

You're right, I fixed it. -- Meni Rosenfeld (talk) 19:10, 2 March 2006 (UTC)

Riemann hypothesis, who cares?

I've been told that the Riemann hypothesis has many broad implications, if only it were proven, but what are those implications? And why should I, a person who has an interest in math but doesn't spend all day looking at numbers (though I read quite a bit about philosophy, science and computers) care about it?

(Also, couldn't we construct a statement along the lines of "X is true iff the Riemann hypothesis is true." And then see if X is true or not, and use that information to either prove or disprove the Riemann hypothesis?) -86.138.233.25 20:09, 2 March 2006 (UTC)