Wikipedia:Reference desk archive/Mathematics/February 2006

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1 February 1
- 1.1 Radian Use in Rates of Change problems
- 1.2 *Plotting good looking fractals
- 1.3 Number Theory Problem
- 1.4 Pythogoras theorem
- 1.5 Odd differential operator
- 1.6 Harmonographs
2 February 2
- 2.1 Cube rotation
- 2.2 Analytic number theory
- 2.3 Solutions
3 February 3
4 February 4
- 4.1 Drawing phase portraits on the surface of a cylinder
- 4.2 Laplace transform calculation
5 February 5
- 5.1 Combinatorics problem
6 February 6
- 6.1 what is an interger?
7 February 7
- 7.1 A Calculating
- 7.2 Change heading in 3 dimensions
8 February 8
- 8.1 I don't like maths
- 8.2 Math help
9 February 9
- 9.1 Semiprimes
- 9.2 Integral question
10 February 10
- 10.1 Toroid/Torus
  - 10.1.1 Surface or Solid?
  - 10.1.2 Non-circular toroids?
- 10.2 Suggestion given to BP guy
- 10.3 Homotopy lambda-calculus
11 February 11
- 11.1 Area of a regular polygon?
- 11.2 Infinite Sums
12 February 12
- 12.1 How is a Lorenz curve calculated?
- 12.2 Covariant derivative
13 February 13
- 13.1 Area of a rectangle?
- 13.2 number with 98 zeroes
14 February 14
- 14.1 Ohnoes
- 14.2 Reverse digits squaring
- 14.3 Trigonometry/Golden Time on a clock face
- 14.4 Pi in base 6???
15 February 15
- 15.1 Sine Curve
- 15.2 Estimating the area under the curve
16 February 16
- 16.1 Miniature Golf- Obstacle at an Angle
- 16.2 multiplication of negative integers
- 16.3 A partial differential equation
- 16.4 Big Bang
17 February 17
- 17.1 "amount in millions"
- 17.2 length of months of indian solar calendars
- 17.3 triple integral
- 17.4 semiprimes
18 February 18
19 February 19
- 19.1 The length of a curve when the metric is not positive semidefinite
- 19.2 find the inverse function
- 19.3 Pythagorean-like problem
- 19.4 Statistics: probability
20 February 20
- 20.1 Powerball
- 20.2 Trigonometry question
- 20.3 Non-integer numeral-system bases
- 20.4 Trajectory of a projectile with air drag
21 February 21
- 21.1 ovaloid : definition (finite projective geometry)
- 21.2 one billion seconds
- 21.3 How to calculate confidence intervals?
  - 21.3.1 Wrong solution:
  - 21.3.2 Correct solution
22 February 22
- 22.1 pseudo-BCH formulas
- 22.2 Gödel's theorem of incompleteness.
23 February 23
- 23.1 Econometrics question: Autocorrelation
- 23.2 repeating decimals
- 23.3 Quadratic formula
- 23.4 3-torus
- 23.5 Sub-groups of general linear group
- 23.6 Estimation methods
- 23.7 Statical Mechanics
24 February 24
- 24.1 Lagrangian dual of a logarithmic barrier problem
- 24.2 Mathematics for the masses
- 24.3 Sample size
25 February 25
- 25.1 Pythagorian Triples
- 25.2 Golden Times on a clock face
- 25.3 What sort of a function is X^X?
- 25.4 Higher Maths Formula
26 February 26
- 26.1 Mechanical aproximations of functions
- 26.2 x^2 = 2^x
- 26.3 A new form of logic?
- 26.4 What is this beautiful abstract nonsense trying to say to me?
- 26.5 Converting trigonometric expressions into algebraic form
27 February 27
- 27.1 Parabolas and straight lines
28 February 28
- 28.1 G. Cantor
- 28.2 Integration
- 28.3 "square root" of exponential function

[edit] February 1

[edit] Radian Use in Rates of Change problems

Problem: A line of light emanating from a lighthouse makes one revolution every 10 seconds. The lighthouse is located 4km off a straight shoreline. How fast does the light move along the shoreline when it forms a 45 degree angle to a line from the lighthouse perpendicular to the shoreline?

It basically boils down to this equation, and we are trying to find dx/dt when θ = 45 degrees

tan(θ) = x/4km

(differentiate)

sec²(θ) dθ/dt = (1/4km) dx/dt

(plug 45 degrees in)

2 dθ/dt = 1/4km dx/dt

(isolate dx/dt)

dx/dt = 8km dθ/dt

Now is where I am confused. The resultant answer is supposed to be in km, so we can't put in 45 degrees, since that has a unit of degrees. We are supposed to put in π/5s (the radian equivalent of 1rev/10 sec) and get

8πkm/5s

because radians are "unitless," being the ratio of the arc swept over the radius. However, why radians? I could easily invent another measure that is "unitless" but would have a different numerical value. For example, I could I could use "diams" (named after diameter), that would equal the ratio of the arc over the diameter. It would effectively equal half of the radian value, so the answer given would be half of the correct answer that putting in radians would give me (and would therefore be incorrect). --JianLi 01:21, 1 February 2006 (UTC)

You have keep in mind that the sinus function which accepts radians is a different function than the sinus function which accepts degrees. Let's call them sinrad and sindeg. Now it is true that d/dx sinrad = cosrad. However it is not true that d/dx sindeg = cosdeg. Note that sindeg(x) = sinrad(πx/180). Therefore d/dx sindeg(x) = π/180 cosdeg(x). Hope that this clarifies things. —Ruud 01:47, 1 February 2006 (UTC)

Ah...I see now. Thanks --JianLi 03:22, 1 February 2006 (UTC)

[edit] *Plotting good looking fractals

Is there a good free software that can make good looking fractals, phase portraits of chaotic attractors, etc at a good resolution. The only tool I have is matlab, but using it I can't make good looking high resolution plots. Thanks! deeptrivia (talk) 01:27, 1 February 2006 (UTC)

Have you tried XaoS? —Keenan Pepper 03:53, 1 February 2006 (UTC)

Shouldn't we care more about what's on the inside of the fractals? Sure, they might be bad-looking, but does that mean we should ignore and reject them? Joe 21:02, 5 February 2006 (UTC)

Seems like I'll need to install compilers, libraries et al. before I can get this to work. Does this allow me to plot things based on my equations or just lets me change parameters? Can I make jpg files using this?? Do you know any programs that I can just install and start using? Thanks! deeptrivia (talk) 04:09, 1 February 2006 (UTC)

Not really. If you're using Windows, just follow this link, which you'll also find on the sidebar of their old (non-wiki) documentation page. —Ilmari Karonen (talk) 00:04, 2 February 2006 (UTC)

Am trying Chaospro. It's looking good. deeptrivia (talk) 04:24, 1 February 2006 (UTC)

Unemployment rate vs Inflation rate. Time taken for economy to stabilize.

Thanks, XaOS is working! But actually I wanted to make fractals based on my own equations. For example, see the fractal on the right. This is the best resolution I can get with matlab, but I want more! I hope someone will have something for me.deeptrivia (talk) 02:45, 4 February 2006 (UTC)

Many years ago, John Walker, Rudy Rucker and I wrote some software for the soon-to-be-cancelled Autodesk Science Series. You can download James Gleick's Chaos the Software and Cellab, among other things; I know that Chaos still works under Windows (fifteen years after being written!), though it was a DOS program with tons of direct screen I/O. It won't solve your needs, but it is fun to play with. --jpgordon ∇∆∇∆ 03:03, 4 February 2006 (UTC)

Thanks! I'll surely try it out. deeptrivia (talk) 03:11, 4 February 2006 (UTC)

How did you make that Unemployment v Inflation graph? Black Carrot 04:32, 5 February 2006 (UTC)

As far as I remember, using the "image" function in matlab. By the way, I'm still looking for a useful software, so the question is still open. deeptrivia (talk) 05:00, 5 February 2006 (UTC)

I mean, what formula did you use? Black Carrot 01:57, 7 February 2006 (UTC)

Based on the following equations:

deeptrivia (talk) 02:07, 7 February 2006 (UTC)

[edit] Number Theory Problem

Hi,

Is there anyone can show me the method on how to proof the below number theory's problem:

For any interger n there are always n consecutive integers that all of which are not primes.

Thank you very much

Hint: Consider the integers (n+1)!+2, (n+1)!+3, (n+1)!+4... —Keenan Pepper 03:51, 1 February 2006 (UTC)

[edit] Pythogoras theorem

Many people say the pythogoras theorem was not given by Phytogoras. So my question is who gave the pythogoras theorem?(was it phytogoras or someone else)

See the section entitled "History" in Pythagorean theorem. Says it better than I can - but do come back if you're confused by it. enochlau (talk) 08:09, 1 February 2006 (UTC)

See Sulba Sutras. deeptrivia (talk) 14:07, 1 February 2006 (UTC)

[edit] Odd differential operator

I wonder if there is a reference in the literature for this operator?

$D(f(x))_p = \frac{f(px) - f(x)}{px - x}$

And of course, is there a witty proof that lim p -> 1 is rigorously the normal differential operator? --HappyCamper 07:09, 1 February 2006 (UTC)

That's called the q-derivative (usually that parameter is denoted q instead of p. For "quantum", I believe). It has a red link on derivative (generalizations). It's used in quantum groups sometimes. -lethe ^{talk +} 11:12, 1 February 2006 (UTC)

See Category:Q-analogs although it seems we don't yet have an article on this one. q-series are older than quantum mechanics; the earliest q-series I know of is the basic hypergeometric function, which is 19th century. The relationship between the q-derivative and q-series can be seen by taking the q-derivative of x^n. (left as a homework exercise). linas 14:57, 1 February 2006 (UTC)

I created a very stubby q-derivative article, and split out a distinct article q-analog, away from q-series. linas 16:03, 1 February 2006 (UTC)

I also expanded on the "why its quantum-like" in the q-analog article. linas 16:18, 1 February 2006 (UTC)

Is this a function that is heavily used in number theory? How I encountered this was actually through statistical mechanics and a very cutting edge research compilation... --HappyCamper 06:58, 2 February 2006 (UTC)

Maybe not "heavy", but it does show up in number theory and string theory and combinatorics, and so perhaps not surprisingly in stat mech. I've seen it in the context of the Ising model, the theory of which does invoke Riemann surfaces of high genus, etc; the Ising model is stat mech; there's been some recent major progress in that field. Although I would also be curious to look at your reference. linas 06:28, 5 February 2006 (UTC)

As for the limit as p -> 1, it depends on what you mean by limit. But it is true that this is equal to the derivative on a compact set, but the convergence is non-uniform on non-compact sets. As an example, let xεR¹. Writing p = 1+ε and considering Taylor series gives

$\frac{f(1+\epsilon)x) - f(x)}{\epsilon x} = f'(x) + O(\epsilon x),$

and it's easy to see that this limit is uniform only if x is constrained to be in some bounded set. --Deville (Talk) 07:01, 5 February 2006 (UTC)

[edit] Harmonographs

After much searching, I have found the parametric equations that describe harmonographs such as the ones at Questacon and the Wollongong Science Centres in Australia. They are:

$\begin{matrix} x\left(t\right) & = & A_x(t) \sin \left(w_xt+p_x\right) + A_s(t)\sin(w_st+p_s) \\ y(t) & = & A_y(t) \sin \left(w_yt+p_y\right) \\ A_x(t+1) & = & d_x \times A_x(t) , |d_x| < 1 \end{matrix}$

Where A_x and A_y are the amplitudes, p_x and p_y are the phase shifts, w_x and w_y are the period setters and d_x is the decay factor. The site I visited here says that A_s, p_s and w_s are all to do with the rotation.

Does anybody have or know of graphing software that can graph a parametric equation with a decaying function incorporated?

Replace the last equation with its analytic solution $A_x(t) = a_0\exp(t \log d_x)\!$ , where

| d x | < 1

as before but

a 0

is an extra constant you need to choose. Then you can simply plug this formula for

A x (t)

into the first ~~two~~ equations. Any ordinary package which graphs parametric equations will be able to handle this. —Blotwell 03:20, 2 February 2006 (UTC)

Also, does anybody know the fractal formed on the complex plane, which is the set of points z such that z² is a real number?

I think this description is probably missing something: the set $\{z \in \mathbb{C} : z^2 \in \reals\}$ is simply $\reals \cup \reals{}i$ , the set of numbers that are purely real or purely imaginary. It is not a fractal. —Ilmari Karonen (talk) 23:54, 1 February 2006 (UTC)

--Alexs letterbox 10:46, 1 February 2006 (UTC)

[edit] February 2

[edit] Cube rotation

Eh... This looked so simple, so trivial, but it's being terrible. I created a cube on POV-Ray, <1,1,1>/2 to -<1,1,1>/2 (side = 1). Well, I want to rotate it and then use the difference command to remove everything below 0y, creating then a tetrahedron with sides = $\sqrt{2}$ and height $\frac{\sqrt{3}}{2}$ . Cutting is easy, but the rotation has been an issue. I made an animation to help you guys understand what I'm trying to do. Here, x = red, y = green, z = blue.

First thing, I rotate 45° on the z axis. Then, I have to find the angle of rotation on the x axis to make the three vertices lie on the xz plane. Well, I first tried 45°. It didn't work, and I figured why. Well, I then did some math and came with the angle $\arctan \sqrt{2}$ . This almost works, but something is wrong and I can't figure what. Can't be that I need to rotate the other axis, because rotating on the y axis won't change anything. As you can see, the top vertice matches the z axis perfecly, but the other three just don't.

Worth noting, POV-Ray treats rotate <1,1,0> (two axis in one command) differently than rotate <1,0,0> and then rotate <0,1,0> (one axis per command). Not a clue why, though. Anyone can clear this out?

So, anyway, how do I do this? ☢ Ҡieff⌇↯ 01:38, 2 February 2006 (UTC)

You already have the cube oriented corretly, you just need to translate it downwards a bit. Anyway, wouldn't it be easier to use a prism? —Ilmari Karonen (talk) 02:06, 2 February 2006 (UTC)

Eh, doesn't make sense :o The cube is centered on the origin, the cut plane should be the xz plane, right? I think it's a matter of rotation only, not translation. And the issue is not the prisma itself, but I'm doing a certain 3D structure here that's based on 3 perpedincular cylinders, and I want to make it look like a tripod lying on the floor, but I keep getting a foot right and the other two off. The cube is just an easy way to deal with the problem. ☢ Ҡieff⌇↯ 02:27, 2 February 2006 (UTC)

No, Ilmari Karonen's right. When you orient the cube with a vertex uppermost, as you correctly have, there's a top vertex, a bottom vertex and 6 others. Three of them are adjacent along an edge from the top vertex, and they're all on the same horizontal plane which is above the origin. The other three are adjacent to the bottom vertex and they're all on a different horizontal plane, symmetrically located below the origin. If you were thinking they were all at the same horizontal level you had something in mind more like an octahedron.

By my quick calculations (could be wrong), the vertical distance from the top vertex to the upper plane is sqrt(3), so that's the height of the tripod and you should cut off everything below the plane y = 1 - sqrt(3) (and lower accordingly). —Blotwell 03:02, 2 February 2006 (UTC)

(edit conflict) The cut plane should be parallel to the xz plane, but you need to move either the plane or the cube vertically to get a tetrahedron. In the animation on the right, look closely at the spot where the x axis meets the cube. After the first rotation, this point is in the middle of one of the edges of the cube, and it will stay there during the second rotation. But the cut plane should meet the cube at the upper end of this edge. —Ilmari Karonen (talk) 03:19, 2 February 2006 (UTC)

I've rendered it semi-transparent and yes, I see it now. I was, indeed, having the properties of the octahedron in head... :/ I've now found that the cut plane lies on $\frac{\sqrt{3}}{6}$ , by the way. (my Rubik's Cube was a lot of help on figuring this out, though.) Anyway, thanks everyone :) ☢ Ҡieff⌇↯ 03:33, 2 February 2006 (UTC)

[edit] Analytic number theory

In a very amateur way I have enjoyed working through books on elementary, classical number theory. I am of course aware that there is a huge field of analytic number theory, but I have no insight into this area at all (and only first year undergraduate knowledge of analysis). Is anyone aware of a textbook that might introduce me to some of the main techniques and results of analytic number theory without assuming that I am a mathematical genius or throwing me right in at the deep end? Thanks for any help you can give. Maid Marion 13:50, 2 February 2006 (UTC)

Well, "first year undergraduate knowledge of analysis" might not be enough, depending on what that covers. It really is essential to know the most basic parts of complex analysis, for example the Cauchy integral formula, and what a complex analytic function is. Then again, it's possible to get a simple version of Dirichlet's theorem on arithmetic progressions working without complex analysis. Dmharvey 02:20, 4 February 2006 (UTC)

I recommend A Primer of Analytic Number Theory : From Pythagoras to Riemann by Jeffrey Stopple (published by Cambridge, ISBN 0521012538). It is aimed at undergraduates and specifically does not assume a prior knowledge of complex analysis. Gandalf61 09:30, 4 February 2006 (UTC)

Thanks guys - very interesting. I'll try to get hold of the Stopple book. Maid Marion 15:24, 4 February 2006 (UTC)

[edit] Solutions

Does anyone here know where I can get the solutions to the International Mathematical Talent Search. I found the questions at http://www.math.ca/Competitions/IMTS/. I've searched a while on google, and didnt get any, so if you dont get the solutions, I might have to post the questions which I find difficult here again ;-) ! -- Rohit 17:37, 2 February 2006 (UTC)

[edit] February 3

(no questions today)

Oh no, not again... ☢ Ҡieff⌇↯

So just for your entertainment: in a mirror, why is left and right exchanged but not up and down? (I know the answer, but just as you seem so disappointed ;-)
(-; Safety warning: trying to perform a headstand in front of a mirror may damage your health, the mirror and anything that can't get away quick enough ;-) The Infidel 21:59, 4 February 2006 (UTC)

They are exchanged, you just don't realize it because you see horizontal reflection and vertical reflection of a human in different ways, thanks to our vertical line of near-symmetry and our curious insistence on remaining upright while talking to each other. Were you freed from such restraints, you would see clearly that a mirror not only reflects horizontally and vertically, but diagonally, northeasterly, substandardly, and many other ways besides. Black Carrot 04:44, 5 February 2006 (UTC)

No: write top rigth botom left on a circle, apply your transformation and see that it's just a rotation of a half circle The Infidel 20:38, 6 February 2006 (UTC)

In fact, I think it doesn't reflect wither up and down or left and right but it does reflect back and front: when I try to comb my hair or shave with the aid of a mirror, I often move my hand forwards when I should reach to the back or in the other way, but I never confuse the left or right side of my head or the top with the bottom. – b_jonas 10:20, 5 February 2006 (UTC)

You're right, but why did you have to tell it right away? I'd loved to hear theories about polarized light and the psychology of starfish staring at a mirror, or even, if I'm really mean, about some strange accidents in the bathroom involving a mirror ... The Infidel 20:38, 6 February 2006 (UTC)

I'm always feet up in a mirror and right is always right. hydnjo talk 01:25, 7 February 2006 (UTC)

Do you get a lot of footprints on your mirror? —Ilmari Karonen (talk) 14:24, 7 February 2006 (UTC)

[edit] February 4

[edit] Drawing phase portraits on the surface of a cylinder

Potential energy and phase portrait of a simple pendulum. Note that the x-axis being angle, wraps onto itself after every 2π radians.

I have to draw a phase portrait of a dynamical system given by the following equation.

$\frac {1}{2} a \dot {\theta^{2}} = g \left(1-\frac {1}{2} \lambda \cos \theta \right) \cos \theta + C$

Since the x-axis is angle, the manifold is $\Bbb{R} \times \Bbb{S}$ . I want to draw the phase portrait on the surface of a cylinder so that $2π$ and 0 coincide. How can I do this in Matlab/Maple ? On the right is an example of a similar situation. deeptrivia (talk) 02:16, 4 February 2006 (UTC)

I don't understand; just graph that equation above, won't that give you the phase diagram of your system? What more do you want? -lethe ^{talk +} 05:25, 4 February 2006 (UTC)

I want to map it on the surface of the cylinder, so that 0 and 2π on the theta axis coincide, rather than being on opposite sides. This is somewhat like saying that I want to represent it as a globe (only one dimension is curved though, making it a cylinder instead of a sphere) as opposed to a map. I want to keep this cylinder transparent (probably not 100% transparent) so that things on the far side can be seen. deeptrivia (talk) 05:31, 4 February 2006 (UTC)

Oh, maybe I see what you mean. Like, you want it graphed on the surface of a 3d cylinder. The graph you have is actually on a topological cylinder alreadyl, once you decree that 0 an 2π are identified, but I guess you want it to actually look like the cylinder embedded in R³. I tried graphing it. How's this look? -lethe ^{talk +} 07:37, 4 February 2006 (UTC)

That's exactly what I want. How did you make it? Thanks! deeptrivia (talk) 12:51, 4 February 2006 (UTC)

I plotted something similar to what Confusing Manifestation suggested below, except that I used cylindrical coordinates. He is correct, this method suffers from the defect that it requires you to solve for ω. I had to graph two plots, one for the positive square root and one for the negative. I could have avoided this by plotting the whole relation in cylindrical coordinates and intersecting with the cylinder. -lethe ^{talk +} 13:13, 4 February 2006 (UTC)

In terms of formulas, if you're looking to have something like what lethe drew, you want to plot

(c o s θ, s i n θ, f (θ))

, where

f (θ)

is the function you're plotting. Of course, given that the phase portrait is more of a relation, you'll need to either split it into a number of functions or find another paramater to include so that it draws the full portrait. Confusing Manifestation 09:53, 4 February 2006 (UTC)

Alright. And how do I make the cylinder below it, say, in MATLAB? deeptrivia (talk) 12:51, 4 February 2006 (UTC)

Thanks Lethe! What software did you use, and how did you plot the cylinder? deeptrivia (talk) 13:16, 4 February 2006 (UTC)

I used grapher, though I don't suspect it much matters what program you use, they all do parametric plotting (though I can't give you specific instructions for MATLAB, since I don't use that. I could get you there in Mathematica). The cylinder was plotted with the equation r=1. -lethe ^{talk +} 13:21, 4 February 2006 (UTC)

Thanks a ton, Lethe! I think I'll manage it now. deeptrivia (talk) 13:24, 4 February 2006 (UTC)

[edit] Laplace transform calculation

I have a question on[1] Anyone please help me.

(On which,why do for-all $\mathcal ,isinx=odd,$ ,therefore they all equal to 0?)

--HydrogenSu 12:16, 4 February 2006 (UTC)

your question seems to be unclear. I think perhaps you want to ask why

$\int^\infty_{-\infty} e^{-2at}i\sin \omega t\,dt$

equals 0. I think the explanation is supposed to go like this: a function f is odd if for all x, f(x)=f(–x). Odd functions satisfy

$\int^a_{-a}f(x)\,dx=0,$

since the positive area on one side of the y-axis cancels out with the negative area on the other side. With some care, this result can be extended to improper integrals. For even functions, you can't cancel out the area, but you can say that the total area from –∞ to +∞ is twice the area from 0 to +∞. This is used in the subsequent step, and accounts for the 2 in front of the cosine.

The sine function is indeed an odd function, and the cosine function is indeed even. However, the integrand above is not the sine function, but rather the product of a sine and an exponential. As far as I can see, the integrand is not odd, and therefore this step of the calculation is an error. Nor is the 2 in front of the cosine correct.

(It's

d ω

, not

d t

, so it actually is odd. The Infidel 21:51, 4 February 2006 (UTC))

Oh, you're right! I totally mucked that up. I'm sorry, yes, it i odd, and that integral is 0. -lethe ^{talk +} 02:30, 5 February 2006 (UTC)

This calculation seems to be proceding with a Fourier transform, which isn't really appropriate for doing a Laplace transform.

What I would suggest instead would be to start with the identity

$\sin at = \frac{e^{iat}-e^{-iat}}{2i}$

then square it, and integrate the resulting exponentials. -lethe ^{talk +} 13:10, 4 February 2006 (UTC)

After trying it myself, I think that the above suggestion isn't so great, and you'd be better served with the identity

$\sin^2 x = \frac{1-\cos 2x}{2}$

-lethe ^{talk +} 14:10, 4 February 2006 (UTC)

Thanks. I've seen your reply. If I have any more qut. ,ask here again.--HydrogenSu 12:04, 5 February 2006 (UTC)

[edit] February 5

[edit] Combinatorics problem

Dear all:

Does anyone know what the following combinatorial expression sums up to?

${n \choose 1}^{3} + {n \choose 2}^{3} + {n \choose 3}^{3} + \ldots + {n \choose n}^{3}$

Thanks

206.172.66.136 01:59, 5 February 2006 (UTC)

I remember thinking about this once. There definitely is a simple identity involving squares, but I couldn't find one involving cubes. Not that that means anything. Dmharvey 02:07, 5 February 2006 (UTC)

With a question like this, you compute the first half-dozen terms then type them into the Online Encyclopedia of Integer Sequences. The result lists an asymptotic formula, a recursive formula, an integral formula and a convolution relation, but no alternative combinatorial expression. Do any of those help? —Blotwell 08:28, 5 February 2006 (UTC)

Thank you very much for your help Blotwell. Yes it was very useful. I was able to find the 'Frenel numbers' entry in MathWorld, which solved everything. thank you again!!! 129.97.252.63 03:18, 6 February 2006 (UTC)

Careful with the spelling; it's "Franel". --KSmrq^T 03:44, 6 February 2006 (UTC)

LOL, sorry I screwed up. :-))))))))))))))))))))) 129.97.252.63 23:41, 6 February 2006 (UTC)

You can use imaginary numbers to help out here...the cube root of unity plays a role... --HappyCamper 02:08, 7 February 2006 (UTC)

[edit] February 6

[edit] what is an interger?

—The preceding unsigned header was added by 152.163.100.11 (talk • contribs) . probably meaning integer

If you were paying attention to our pages, you'd have seen the little search box at the left. From it, you'd easily find our article on integers. ☢ Ҡieff⌇↯ 05:05, 6 February 2006 (UTC)

Note that searching for "interger" doesn't turn up anything relevant. Fredrik Johansson - talk - contribs 07:30, 6 February 2006 (UTC)

Unless, of course, you try "interger" on a search engine like Google or Yahoo! that suggests a correct spelling ("Did you mean: integer"). A spelling suggestion feature here would be useful. --jh51681 19:24, 6 February 2006 (UTC)

I just use a redirect whenever I notice a common misspelling. Redirects are cheap. —Keenan Pepper 23:26, 6 February 2006 (UTC)

Perhaps this WP article would be helpful. hydnjo talk 23:33, 6 February 2006 (UTC)

"Interger vitae, scelerisque purus". (Pure in life, and free from crime). linas 23:42, 6 February 2006 (UTC)

Random etymology: The Latin word integer literally means "untouched", hence "pure". Its current English meaning comes from "untouched" -> "unbroken" -> "whole" -> "whole number". —Keenan Pepper 23:54, 6 February 2006 (UTC)

Once again proving that there are no stupid questions ... actually, there are very stupid questions, or at least ones that are asked in the wrong place, but even you still have a chance of learning something very interesting from them. Confusing Manifestation 09:05, 7 February 2006 (UTC)

[edit] February 7

[edit] A Calculating

$\, \int_{0}^{\infty}\, X^2\frac{1}{e^{X}-1}dX= ..................$

Is 2.54? How to do? Thanks.--HydrogenSu 13:17, 7 February 2006 (UTC)

I don't know how to do it, but Mathematica tells me the answer is 2.40411 though. enochlau (talk) 13:32, 7 February 2006 (UTC)

The Integrator gives the indefinite integral in terms of polylogarithms, so it's over my head. —Keenan Pepper 14:25, 7 February 2006 (UTC)

If using $\mathcal\, \Gamma(z)\,$ functions,how to do (it)? (Extra:-Remembered in street-talk something like in movie ID4 the "it" can be ignored? I'm an Asian.)--HydrogenSu 19:16, 7 February 2006 (UTC)

I heard from my physics teachre he said: $\, \int_{0}^{\infty}\, X^3\frac{1}{e^{X}-1}dX=\frac{\pi^4}{15}$ can be calculated by that kind of special functions.--HydrogenSu 19:21, 7 February 2006 (UTC)

I just now guess that $\, \int_{0}^{\infty}\, X^2\frac{1}{e^{X}-1}dX=\frac{\pi^3}{ab}$

Where $\mathcal\,b-a=3\,$ . I'm going to guess in my room what they are. Such interesting.--HydrogenSu 19:26, 7 February 2006 (UTC)

Heh, that doesn't tell much. Any positive number can be written in the form

π 3 / (a b)

where

a - b = 3

, and that integral is positive. – b_jonas 21:39, 7 February 2006 (UTC)

It's just a tech for memory,not need to really operate in math. Too trouble. I'm learning Physics,not math.--HydrogenSu 07:57, 10 February 2006 (UTC)

It's $2ζ(3)$ , according to the second formula given in Riemann_zeta_function#The_Riemann_zeta_function_as_a_Mellin_transform.--gwaihir 21:31, 9 February 2006 (UTC)

[edit] Change heading in 3 dimensions

We're at point (x1, y1, z1) looking at angle $θ$ . We want to face point (x2, y2, z2). I know that to face a new angle in two dimmensions, we could rotate by the arctan((y2-y1)/(x2-x1)). But how do I do this in three dimensions? 216.158.57.50 18:07, 7 February 2006 (UTC)

In 3D there are two angles that describe a direction, not just one. For example, there is the angle as projected into the XY plane and the angle above the XY plane. See spherical coordinates and cylindrical coordinates. StuRat 21:17, 7 February 2006 (UTC)

The section title uses the word "heading", by which I'm guessing you mean compass heading. But suppose I'm at an air show watching a helicopter fly straight toward me and over my head. As the plane approaches and my gaze follows it, the heading of my gaze does not change, yet I must move my head. When the plane is directly overhead and my gaze is perfectly vertical, every possible heading still leaves my gaze fixed on the plane. Now suppose that I am videotaping the flight, and that the plane pauses to hover. Still leaving the plane centered in the viewfinder, I can rotate the camera around the line of sight, changing neither the azimuth (heading) nor the altitude, yet changing the camera orientation. So your problem is underspecified, and based on a false premise. The problem in 3D has subtleties and complexities for which 2D does not prepare us. --KSmrq^T 06:03, 8 February 2006 (UTC)

See also flight dynamics. —Keenan Pepper 15:21, 8 February 2006 (UTC)

[edit] February 8

[edit] I don't like maths

can you do all my math homework for me, it's hardMaths are for kids 18:48, 8 February 2006 (UTC)

Yeah, sure. The answer is 42. Have a nice day! —Ilmari Karonen (talk) 19:28, 8 February 2006 (UTC)

Only if you say the magic word and emphazi it suitly. – b_jonas 19:29, 8 February 2006 (UTC)

Why do Brits call it "maths" instead of "math" ? Do you also study "Englishes", "histories", and "governments" ? StuRat 20:02, 8 February 2006 (UTC)

Do you study mathematic? Fredrik Johansson - talk - contribs 20:06, 8 February 2006 (UTC)

We generally only study one mathematic subject at a time, yes. Do you study them all at once, and hence study "maths" ? StuRat 21:53, 9 February 2006 (UTC)

I study several of them; not necessarily at the same time, but that's not a requirement. Fredrik Johansson - talk - contribs 21:58, 9 February 2006 (UTC)

Well, in Australia when it comes to school subjects we have Maths and Sport, whereas, I am told, in the US it's Math and Sports. And yet there are those who would say Australia is one of the most Sports-crazy countries in the world (noting the plural in particular given we play at least 4 codes of football, cricket, netball, and countless others). Confusing Manifestation 02:18, 9 February 2006 (UTC)

It's such a shame that people nowdays are so anti-mathematics...

"I am accustomed, as a professional mathematician, to living in a sort of vacuum, surrounded by people who declare with an odd sort of pride that they are mathematically illiterate." — David Mumford

Amen to that. ☢ Ҡi∊ff⌇↯ 04:10, 9 February 2006 (UTC)

Hear Hear. — Ilyan e p (Talk) 04:20, 9 February 2006 (UTC)

Ditto. I recommend Innumeracy by John Allen Paulos, if you haven't read it yet. --JianLi 01:02, 10 February 2006 (UTC)

I'm British English, why do I say Maths, no idea! Who said language so logical. I suppose I would say 'Maths' was better because mathematics ends in an 's'--βjweþþ (talk) 20:01, 9 February 2006 (UTC)

In my neverending quest to be politically correct, I always write "mathematics" in full, never "math" or "maths". This is every bit as important as men leaving the toilet seat down. One compromise might be to take a note from the non-sexist pronoun s/he, and write math/s. Note that British and American English differ also in examples like "Ford Motor Company pay their workers" versus "Ford Motor Company pays their workers". --KSmrq^T 01:11, 10 February 2006 (UTC)

Thus We might take a guess of French men would say it as "MATHE" 'cause end of Mathématique in it.

And Chi. say:ㄕㄨˋ ........Chineses in TW used to be use tongues to be lazy in writting.

"I love you=ㄨㄛˇ ㄞˋ ㄋ一ˇ"

[edit] Math help

where was the first math instrument found?

In the woodwinds section--64.12.116.11 22:50, 8 February 2006 (UTC)

Hah...that made me Laugh out loud :\ I'm such a band geek. — Ilyan e p (Talk) 04:20, 9 February 2006 (UTC)

Depends what qualifies as an "instrument". The earliest form of counting probably involved keeping one pebble for every animal in your heard, and pairing them off one by one. The Latin word calculus literally means "pebble". So do pebbles count as "math instruments"? —Keenan Pepper 23:13, 8 February 2006 (UTC)

The abacus goes way back, too, although not as far as the pebble. StuRat 05:46, 9 February 2006 (UTC)

The earliest mathematical tools with any degree of permanence were probably tally sticks such as the Ishango Bone. Gandalf61 11:10, 9 February 2006 (UTC)

[edit] February 9

[edit] Semiprimes

The density of the primes are 1/ln(P), is there a similar result for semiprimes? Can this result be edited to only include square free numbers?

[edit] Integral question

$I_{4}=\, \int_{0}^{\infty}\,x^{4}e^{-{\alpha}x^2} dx=\frac{3}{8}\sqrt{\frac{\pi}{\alpha^{5}}}$

How to derive this? --HydrogenSu 14:37, 9 February 2006 (UTC)

See Integration by parts, and the Error function. --BluePlatypus 14:53, 9 February 2006 (UTC)

My Soluntuon by Using $\mathcal\, L(t^a)\,$ :--HydrogenSu 18:03, 16 February 2006 (UTC)

$\mathcal\, \, \int_{0}^{\infty}\,x^4e^{-{\alpha}x^2}dx =\frac{1}{2}\, \int_{0}^{\infty}\,x^{2((5/2)-1)}e^{-{\alpha}x^2}d(x^2)\,$

Where $\mathcal\, \frac{x^4}{x}=x^3=x^{2{\cdot}\frac{5}{2}-2}\,$

The $\mathcal\, \, \int_{0}^{\infty} \,e^{-{\alpha}x^2}x^{2{\cdot}((5/2)-1)}d(x^2) \,$ which just satisfies with

$\mathcal\, \, \int_{0}^{\infty} \,e^{-{\alpha}t}t^{(x-1)}dt=\frac{1}{\alpha^{x}}\Gamma(x) \,$

So $\mathcal\, \frac{1}{2}\, \int_{0}^{\infty}\, e^{-{\alpha}x^2}x^{2((5/2)-1)}d(x^2)\,$

$\mathcal\, =\frac{1}{2\alpha^{5/2}}\Gamma(\frac{5}{2})\,$

$\mathcal\, =\frac{1}{2\alpha^{5/2}}\frac{3}{2}\Gamma(\frac{3}{2}) \,$

$=\frac{3}{8}\sqrt{\frac{\pi}{\alpha^5}}$

Where $\mathcal\, \Gamma(\frac{1}{2})=\sqrt{\pi} \,$ ,and $\mathcal\, \Gamma(x)=(x-1)! \,$

Typed the above by JAVA,also contribute them to other users!

--HydrogenSu 19:01, 9 February 2006 (UTC)

The Error function certainly has do with it. The antiderivative of (x^4)*exp(-a * x^2) contains two exponential terms and one Erf term. --BluePlatypus 08:45, 10 February 2006 (UTC)

HydrogenSu - based on your recent contribution history, it is clear that you are using the services of the reference desk pages to help with your homework only - an account should not be used for this purpose. There is a reason why homework questions are frowned on - the volunteers here not to do your homework! It has nothing to do with your nationality or freedom or whatnot - those sorts of politics can simply stay off of the reference desk - Wikipedia simply isn't a forum for that. Respectfully, in the future please be mindful of that. --HappyCamper 13:13, 10 February 2006 (UTC)

What is this "USA time" anyway? I live in the USA now, and ever since coming here, I don't seem to have any time at all! Dmharvey 14:42, 10 February 2006 (UTC)

I responded your question above yesterday and moved it on Language Section.--HydrogenSu 11:31, 11 February 2006 (UTC)

[edit] February 10

[edit] Toroid/Torus

[edit] Surface or Solid?

According to the Wikipedia articles, the Toroid is a solid while the Torus is its surface. According to the American Heritage Dictionary,

a toroid is

a A surface generated by a closed curve rotating about,
but not intersecting or containing, an axis in its own plane.
b. A solid having such a surface.

and a torus is

A toroid generated by a circle; a surface having the shape
of a doughnut. In this sense, also called tore2

Which definitions are right? I'm confused --JianLi 03:10, 10 February 2006 (UTC)

I would say "toroid" or "torus" may refer to either the surface or the solid. To be specific, refer to a "toroid surface" / "torus surface" or a "solid toroid" / "solid torus". StuRat 04:38, 10 February 2006 (UTC)

What the rest of the world chooses to do is up to them, but in mathematics a torus is a surface, as is a sphere. --KSmrq^T 13:55, 10 February 2006 (UTC)

The dictionary definition you give of torus says it is a surface, which is consistent with Wikipedia's definition. In any case, one should note that the page on torus makes it clear the contexts of usage given are in mathematics, e.g. "in geometry..." and "in topology...". The dictionary definition of toroid indicates there is ambiguous usage as a solid or surface; this is reflected also in the toroid article, where the toroid is described as a solid, but the example of coils given actually are "hollow". --C S (Talk) 14:12, 10 February 2006 (UTC)

Exactly. One should always be a bit careful about looking up specialized terms in a dictionary (unless it's a specialty dictionary, of course), since the dictionary intends to give the common usage. So what is commonly called a torus need not be what topographers call a torus. Most people just say "doughnut shape", though. :)--BluePlatypus 18:19, 10 February 2006 (UTC)

I thought a toroid was a psuedo-torus, where the revolving circle can intersect itself...is that not true? --HappyCamper 18:52, 10 February 2006 (UTC)

Actually, I did make an effort to use only the mathematical definitions. For the full definitions, see torus and toroid and look under the definitions labeled "Mathematics" --JianLi 21:05, 10 February 2006 (UTC) In any case, if the terms are used interchangeably (as StuRat said), or Torus is a surface while Toroid may be either (as Chan-Ho said), why do we have separate articles? Does anybody think a merge is in order? --JianLi 21:05, 10 February 2006 (UTC)

[edit] Non-circular toroids?

In addition to the surface/solid confusion, note that the dictionary allows a toroid to be made by any closed curve, not just a circle. How do you reconcile that? Are dictionary definitions just unreliable for mathematics, even if it was listed as a "mathematics" definition? --JianLi 21:08, 10 February 2006 (UTC)

Well, if you get a square and rotate it in a circular orbit, the resulting shape will be a toroid with a square cross-section. The torus is just the case where this toroid's closed curve happens to be circle. From what I know, the word torus is usually used for the surface alone, where toroid is the solid with such surface. ☢ Ҡi∊ff⌇↯ 21:39, 10 February 2006 (UTC)

I call any section rotated about an axis to be a "surface of revolution" or a "solid of revolution". StuRat 02:03, 11 February 2006 (UTC)

Well, sure. Take a cone for example. Those are rotations when the inner radius is zero. If the radius is > 0, you'll have a toroid. ☢ Ҡi∊ff⌇↯ 03:06, 11 February 2006 (UTC)

I call them all "surfaces of revolution" or "solids of revolution". I only call them a "torus" or "toroid" if the section which is rotated is a full circle in the plane of the axis of revolution, but not touching that axis. I therefore consider "toroids" to be a proper subset of all "surfaces/solids of revolution". StuRat 16:50, 11 February 2006 (UTC)

Oh, that. Well, aren't they? I always thought so. ☢ Ҡi∊ff⌇↯ 23:27, 11 February 2006 (UTC)

[edit] Suggestion given to BP guy

<post removed> - Relegated to talk pages. See BluePlatypus and HydrogenSu. Thank you. --HappyCamper 13:26, 10 February 2006 (UTC)

[edit] Homotopy lambda-calculus

Since too many questions are being answered too quickly, I thought I'd ask if anybody has heard of the homotopy lambda-calculus talked of by Vladimir Voevodsky (2002 Fields medal) in this lecture, or can guess what it's about? Is it something to do with motivic cohomology? --- Charles Stewart^(talk) 19:14, 10 February 2006 (UTC)

I went to this lecture. He didn't get to the "homotopy" part of it... it's a four lecture series. He pretty much outlined plain vanilla lambda calculus. That's it.

[edit] February 11

[edit] Area of a regular polygon?

The Regular polygon article says the following gives the area of a regular polygon:

$A=\frac{nt^2}{4\tan(180^\circ/n)}$

What does t stand for?

t is the apothem, I think. ☢ Ҡi∊ff⌇↯ 02:11, 12 February 2006 (UTC)

If t were the apothem the formula would be $\frac{1}{2}pt$ (where p is the perimeter and t is the apothem) wouldn't it? — Ilyan e p (Talk) 02:19, 12 February 2006 (UTC)

Yeah I know, but you don't see p in there, do you? I'd think both are equivalent, the only difference is that this one calculates the perimeter inherently based on number of sides or something. I'd have to check out by hand, but I'm too lazy. ☢ Ҡi∊ff⌇↯ 06:54, 12 February 2006 (UTC)

My friend and I deduced an alternate formula, is it correct?

$A=\frac{s^2n(\tan(i/2))}{4}$

Where s is the side length, n is the number of sides, and i is the angle measure of an interior angle. — Ilyan e p (Talk) 18:46, 11 February 2006 (UTC)

On looking at it a second time, they look really similar. Assuming that t = s, then :

$\frac{nt^2}{4\tan(180^\circ/n)} = \frac{s^2n(\tan(i/2))}{4} / \frac{\tan(i/2)}{\tan(i/2)}$

(aside from the fact than tan(180/n) is not equal to tan(i/2) I think, but that seems to be a minor difference. — Ilyan e p (Talk) 18:50, 11 February 2006 (UTC)

$i/2 = \frac{180-(360/n)}{2} = 90-(180/n)$

hmm...somehow I doubt that tan(90-(180/n)) = tan(180/n) ... getting somewhere here though. These math tags are fun though. — Ilyan e p (Talk)

Perhaps if i was the angle measure of a central angle rather than interior angle, then it would work, because then:

$i/2 = \frac{360/n}{2} = 180/n$

— Ilyan e p (Talk) 19:03, 11 February 2006 (UTC)

Yeah, I'd say t is side length and 1/tan(180/n) is (length of half a side)/(length of apothem), making it equivalent to your (1/2)(perimeter)(length of apothem) formula. Black Carrot 02:46, 12 February 2006 (UTC)

I deduced this formula on my own and t is equal to the length of the side.

[edit] Infinite Sums

These will solve Topic:Infinite Sums.

您的問題的確需要"技巧(tech.)"來說明.請不妨把兩邊都平方,做此後,你將發現(9-6)恆等於3:

$(\sqrt{6+\sqrt{6+\sqrt{6+...}}})^2 = 3^2$

第 1輪: $6+\sqrt{6+\sqrt{6+...}} = 9$

$\Longrightarrow \sqrt{6+\sqrt{6+\sqrt{6+...}}} = 9-6=3$

第 2輪: $6+\sqrt{6+\sqrt{6+...}} = 9$

$\Longrightarrow \sqrt{6+\sqrt{6+\sqrt{6+...}}} = 9-6=3$

第 3輪: $6+\sqrt{6+\sqrt{6+...}} = 9$

$\Longrightarrow \sqrt{6+\sqrt{6+\sqrt{6+...}}} = 9-6=3$

第10輪: $6+\sqrt{6+\sqrt{6+...}} = 9$

$\Longrightarrow \sqrt{6+\sqrt{6+\sqrt{6+...}}} = 9-6=3$

第 $\mathcal\, \infty\,$ 輪: $6+\sqrt{6+\sqrt{6+...}} = 9$

$\Longrightarrow \sqrt{6+\sqrt{6+\sqrt{6+...}}} = 9-6=3$

That's why $\sqrt{6+\sqrt{6+\sqrt{6+...}}} = 3$ always stands.

That's what my high school teachers taught. I learned.--HydrogenSu 23:01, 12 February 2006 (UTC)

Hm, an inspiring way to prove things. So I've tried is myself:

$(\sqrt{6+\sqrt{6+\sqrt{6+...}}})^2 = (-2)^2$