Wikipedia:Reference desk/Mathematics

June 21

Write naturally in math, automatically translate to TeX

Is there a product out there where I can write mathematical equations naturally on a surface, and it would automatically convert it to TeX for me? --HappyCamper 01:32, 21 June 2007 (UTC)

No. Learn to type TeX quickly; that should be more useful. (This unsigned comment added by 129.78.64.102.)

That sounds about right. State-of-the-art PDAs have enough trouble with simple letters and numbers, the mind boggles at what it'd take to get them to understand complex formluas. Black Carrot 04:06, 21 June 2007 (UTC)

Hmm...nevermind, I just found what I was looking for today. Thanks! --HappyCamper 11:41, 21 June 2007 (UTC)

Ah, wanna share? —Bromskloss 13:28, 21 June 2007 (UTC)

"Wanted: math grad student for typing job, $10/hr or ramen+shower." [edited] Tesseran 01:00, 22 June 2007 (UTC)

dodecagons and hexagons

How do I calculate the area of a dodecagon and the area of the hexagon if I only know the length of one side, let's say 7cm?Invisiblebug590 06:32, 21 June 2007 (UTC)

Dodecagon—12 sided figure

Hexgon—6 sided figure

You can't—unless you assume/guarantee something more about the shapes (e.g. that they are regular convex polygons). --CiaPan 07:18, 21 June 2007 (UTC)

Have you read our dodecagon and hexagon articles? they would seem to answer your question (for regular versions). 213.48.15.234 07:15, 21 June 2007 (UTC)

P.S. the dodecagon and hexagon are regular polygons.

Invisiblebug590 06:11, 23 June 2007 (UTC)

See Regular_polygon#Area 67.182.185.143 22:37, 23 June 2007 (UTC)

Maybe i've turned stupid

How would you solve $2^{x}+3^{x}=13$ for x?

I can see that the answer's x=2 but i have no idea how to go about getting the solution through manipulation of the equation. 213.48.15.234 07:40, 21 June 2007 (UTC)

I had enough trouble sleeping before you asked. :) The only thing I know is that you need logarithms to remove the x variables from the exponent. I just don't see how to do it because the "simple" rules of logarithms don't give a simplification for log of the left side of your equation. YechielMan 07:29, 21 June 2007 (UTC)

X=2 is certainly one answer, but I doubt it's the only correct answer. -- JackofOz 07:32, 21 June 2007 (UTC)

I've tried both logarithms and exponent manipulation, but the old not being able to do much with added exponents of different bases is throwing me off. Jack, I can't see another real answer, maybe there are some imaginary ones. 213.48.15.234 07:36, 21 June 2007 (UTC)

Since $2^{x}+3^{x}$ is a strictly increasing function of $x$, there can be only one solution. —Bromskloss 08:01, 21 June 2007 (UTC)

$\log(a+c)=\log a+\log(1+e^{\log c-\log a})$ Assuming Natural Log.

Put a = 2^x and put c=3^x

$\log(2^{x}+3^{x})=x\log 2+\log(1+e^{x\log 3-x\log 2})$

$\log(2^{x}+3^{x})=x\log 2+\log(1+e^{x(\log 3-\log 2)})$

Now solve

$x\log 2+\log(1+e^{x(\log 3-\log 2)})=\log 13$

$xF+\log(1+e^{xG})=H$

where F = Log 2

where G = Log 3 - Log 2

where H = Log 13

So we have

$x+{\frac {\log(1+e^{xG})}{F}}={\frac {H}{F}}$

$x+{\frac {\log(1+e^{xG})}{\log K}}=L$

where e^K=F

where L=H/F

$x+\log(1+e^{xG}-K)=L\,$

Raise to the e

$e^{x}\times (1+e^{xG}-K)=e^{L}\,$

$e^{x}\times (M+e^{xG})=N\,$

where M=1-K

where N=e^L

Now expand

$Me^{x}+e^{x}e^{xG}=N\,$

$e^{x+P}+e^{x(1+G)}=N\,$

$e^{x+P}+e^{xQ}=N\,$

211.28.131.152 07:47, 21 June 2007 (UTC)

Well, i could do all of that from your first formula, that's exactly the point I got stuck, though :p 213.48.15.234 07:53, 21 June 2007 (UTC)

shoot he's still going 213.48.15.234 08:11, 21 June 2007 (UTC)

This transformation seems incorrect to me:

$x+{\frac {\log(1+e^{xG})}{\log K}}=L$

$x+\log(1+e^{xG}-K)=L\,$

Logarithmic identity is ${\tfrac {\log A}{\log B}}=\log _{B}A\neq \log {(A-B)}.$

--CiaPan 08:34, 21 June 2007 (UTC)

I believe you're correct there CiaPan, i'm still only a little closer to a solution. :/ 213.48.15.234 09:00, 21 June 2007 (UTC)

If simple manipulations could solve these kinds of equations, I'm sure Fermat's last theorem would have been proven much earlier. Some related (but much tougher obviously) problems are described in Mihăilescu's theorem. nadav (talk) 09:39, 21 June 2007 (UTC)

Let me just emphasize these statements:

• X=2 is certainly one answer, but I doubt it's the only correct answer. -- JackofOz 07:32, 21 June 2007 (UTC)
• Since $2^{x}+3^{x}$ is a strictly increasing function of $x$, there can be only one solution. —Bromskloss 08:01, 21 June 2007 (UTC)

Now, for the complex solutions, $n^{r\cdot \operatorname {cis} (\varphi )}=n^{r\cdot \cos(\varphi )}\cdot \operatorname {cis} (r\cdot \ln(n)\cdot \sin(\varphi ))$, so we get (for "cis", see this section):

$2^{r\cdot \cos(\varphi )}\cdot \cos(r\cdot \ln(2)\cdot \sin(\varphi ))+3^{r\cdot \cos(\varphi )}\cdot \cos(r\cdot \ln(3)\cdot \sin(\varphi ))=13$

$2^{r\cdot \cos(\varphi )}\cdot \sin(r\cdot \ln(2)\cdot \sin(\varphi ))+3^{r\cdot \cos(\varphi )}\cdot \sin(r\cdot \ln(3)\cdot \sin(\varphi ))=0$

Try solving this system, and let me know what you get. -- Jokes Free4Me 09:44, 21 June 2007 (UTC)

Oh, are we looking for non-real solutions as well? —Bromskloss 09:51, 21 June 2007 (UTC)

Not really, I was just speculating, however I would like to see if it's possible to find x=2 from the equation. From the answers i've gotten so far, it doesnt look like it is easy to do. 213.48.15.234 09:55, 21 June 2007 (UTC)

One needs to use a root-finding algorithm.--Patrick 11:31, 21 June 2007 (UTC)

I haven't worked out the details, but I strongly suspect there are no complex solutions beyond x = 2. Putting x = a+bi, |2^x + 3^x|² is at least (3^a – 2^a)² and at most (3^a + 2^a)²; if 13² is not in this range, we can't have a solution. This implies that 2 ≤ a < 2.61. Examining this range in small increments, with 0 ≤ b < 2π, does not suggest any second solution lurking there.  --Lambiam^Talk 15:54, 21 June 2007 (UTC)

Your intuition is wrong. I did a quick graph of $2^{x+yi}+3^{x+yi}=13$ in the OS X grapher application and got what looked like a periodic curve (but probably isn't), which shouldn't be terribly surprising. Consider the simpler case of $e^{z}=c$ for some real $c>1$. Remember that $e^{z}$ is a transcendental function and periodic. The functions $2^{z}$ and $3^{z}$ are similarly shaped and adding them gives you something which is vaguely periodic (I think that the periods will interact in an irrational fashion so there probably does not exist non-zero $w\in \mathbb {C}$ such that $2^{z+w}+3^{z+w}=2^{z}+3^{z}$. There are in fact an infinite number of solutions to $2^{z}+3^{z}=13$ but only one real solution. Donald Hosek 16:57, 21 June 2007 (UTC)

My intuition here is mostly useless, but empirically looking at the imaginary part of the sum via a contour plot in Mathematica, with x between -3 and 4, and y between -3 and 3, and one contour line at 0 while sampling on a grid of 3000 points on each axis, shows one contour which strongly seems to lie on the real axis. Zooming the y axis in to between +/- 0.001 supports this hypothesis. This doesn't prove anything, but does suggest only the aforementioned root 2. Baccyak4H (Yak!) 17:28, 21 June 2007 (UTC)

Have mathematica give you a graph of solutions to $2^{z}+3^{z}=13$. I'm not sure what you're trying to prove with looking at the imaginary part of the graph. I'd spend some time also just looking at graphs of the exponential functions. Look at the real part, imaginary part and modulus (absolute value) to get some sense of what's going on, then you should be able to get some sense of what happens with the function $f(z)=2^{z}+3^{z}$ and what the set of solutions to $f(z)=13$ should look like. Mathematica should be able to graph those for you as well (you may need to use the $(x,y)$ form of the equation I cited above to get it to graph the solutions in a real plane... I'm too old and cheap to know mathematica well). Donald Hosek 17:57, 21 June 2007 (UTC)

Sorry if I wasn't clear. I did just what you suggested: made a contour plot of Im( 2^(x+i*y) + 3^(x+i*y) ) with x and y being the two axes of the plot. The idea is for the sum to be 13, the imaginary part must be 0. Allowing one contour value at 0 shows where this happens (indeed, the intersection of those contours with those of the real part at value 13 are in principle all of the roots). The only contour which took value zero was apparently coincident with the x axis, meaning y=0, and thus implying all roots indeed are real. Baccyak4H (Yak!) 18:17, 21 June 2007 (UTC)

Update: This suggests a way to prove it. Ignore the 13 for a moment and try to show that 2^x + 3^x is real implies x is real. Baccyak4H (Yak!) 18:19, 21 June 2007 (UTC)

Not sure how accurate it is, but Maple turned up no complex solutions. Dlong 18:57, 21 June 2007 (UTC)

There are 2 complex solutions...somewhere around r = 6.524959 and theta = +- 1.1651783. --HappyCamper 19:51, 21 June 2007 (UTC) <--- update -- actually, I take that back. I think there are an infinite number of solutions. It's easier to see them by solving with r and theta. --HappyCamper 19:55, 21 June 2007 (UTC)

Good stuff... my efforts didn't look far out enough. Baccyak4H (Yak!) 20:06, 21 June 2007 (UTC)

If someone could double check these, it would be great. I don't have Mathematica at my disposal. Looks like the next furthest solution is r = 11.3275949 and theta = 1.362440594. --HappyCamper 20:37, 21 June 2007 (UTC)

Check. Also r = 17.4819... and θ = 1.4525....Baccyak4H (Yak!) 20:48, 21 June 2007 (UTC)

• Check. I get r = 17.48192845163..., θ = 1.452539552949... --HappyCamper 21:24, 21 June 2007 (UTC)

Come on guys, take a look at the graph of the exponential function. Note that the real values for $e^{z}=c$ for $c\in \mathbb {R} ^{+}$ are not isolated points, but are actually continuous sets. It shouldn't take a genius to realize that $a^{z}=c$ for $a$ positive and not 1 will be similar other than scaling (think back to high school algebra for change of base). Now with a bit more effort, you should be able to come up with the idea that the solutions for $2^{z}+3^{z}=c$ will also be continuous sets. Donald Hosek 20:55, 21 June 2007 (UTC)

Whether you meant it to or not, your tone seems like it could offend some people, especially since they're trying their best to help with a problem that resists immediate solution. That aside, though, what you've written is false. If c > 0 is real, for example, the solution set of z such that e^z = c consists of exactly those points of the form Log c + 2kπi for k some integer. This set, as you probably know, is discrete (made up of isolated points), rather than continuous. In fact, the solution set is discrete for any complex c; this is captured by the observation that the exponential map is a covering map (onto the plane minus the origin).

Your second observation is correct: the base e here is not special. Your last claim is also false; the best way to see this in general is a dimension argument. The function that sends z to 2^z + 3^z is an entire function, and in particular a smooth map from the complex plane (a 2-dimensional real manifold) to itself. Then any regular value (here, any regular value of c) has discrete, or 0-dimensional, preimage (solution set). By Sard's theorem (PM), the set of regular values has full measure, which means that if there are any values of c such that the solution set is continuous, those values form a set of measure 0. Thus your assertion holds extremely rarely, if at all. Tesseran 00:59, 22 June 2007 (UTC)

The dimension argument needs a little more care -- zero-dimensional is not in general synonymous with "discrete". For example the Cantor space is zero-dimensional. But I seem to recall that, if the zeroes of an entire function accumulate, then the function is constantly zero. I think you just figure out what the Taylor series has to be, centered at the accumulation point. That should show that the solution set is discrete, without invoking Sard's theorem. --Trovatore 19:39, 23 June 2007 (UTC)

In this case, we know that the solution set is a manifold, and a 0-dimensional manifold is discrete (has the discrete topology). As you say, any entire function which vanishes on a set containing a limit point is identically 0, and this implies discreteness immediately. (A much nicer argument.) Tesseran 19:40, 24 June 2007 (UTC)

Thanks! --Trovatore 22:49, 24 June 2007 (UTC)

It looks like the solutions a+bi with 2 ≤ a < 2.61 occur with an average frequency for b of once in every interval of length $\pi /((1/ln(2))-(1/ln(3)))=5.9001941$, that is, twice for every time that the argument of $3^{z}$ overtakes that of $2^{z}$.--Patrick 08:28, 22 June 2007 (UTC)

Actually, the 2 ≤ a < 2.61 I quoted from above should be 2 ≤ a < 2.7047.--Patrick 08:39, 22 June 2007 (UTC)

$2^{x}+3^{x}=2^{2}+3^{2}$

That's all there is to it. The above demonstrates quite clearly how 2 can be the only real solution.

As for imaginary solutions, put x = a + bi and you equate real and imaginary parts and get our original equation. I don't think there are any imaginary solutions.

Fan pressure

I have a room which in the room a person doing painting and sanding. There are a lot of dust. I want to suct all the dirt particles. Is there any formula to calculate the static pressure that we need to able the fan suct all the dirt particles?

I don't think we can help you practically, other than to advise that you use a vacuum cleaner, but try to clarify the situation. Are the particles floating in the air or sitting on the floor? And how do you know when they have been removed? YechielMan 15:12, 21 June 2007 (UTC)

Any changes in static pressure wont help you, as they will cause no net force on your dust particles. The only way your fan will help is if it causes an air flow strong enough to remove your particles. I agree that a hoover is the best option, though a fan may help ventilate your room to get rid of the smell of paint, as well as the smallest floating dust particles. Good luck. 195.137.96.79 06:27, 22 June 2007 (UTC)

Be sure to use one fan blowing in and another blowing out on the opposite side of the room, for maximum airflow. Close any other windows, as they would interfere with proper airflow. Also, beware that air saturated with saw dust can explode if exposed to an ignition source. So, avoid smoking or any other activity with an open flame in that room. StuRat 06:45, 22 June 2007 (UTC)

You may find health and safety regulations could be of help. If you are running a real buisness then there is surly regulations as to the amount of particles in the air, and there may well be standards for type of ventilation you need. Ignore advice about domestic vacuum cleaners which will not be upto the job. --Salix alba (talk) 19:08, 22 June 2007 (UTC)

Unless it's a domestic setting, and you're just doing a bit of DIY. A vacuum cleaner would be just fine for this. 195.137.96.79 03:40, 24 June 2007 (UTC)

June 22

Coupled differential equation

Just for kicks, I'd like to solve this coupled set of differential equations

${\frac {df}{dx}}=\alpha g(x)-xf(x)$

${\frac {dg}{dx}}=\alpha f(x)+xg(x)$

Is there a closed form power series for the answer? --HappyCamper 02:08, 22 June 2007 (UTC)

I think so, and it doesn't look too bad. If we let $f(x)=a_{0}+a_{1}x+a_{2}x^{2}+\ldots$ and $g(x)=b_{0}+b_{1}x+b_{2}x^{2}+\ldots$, then you can substitute that all into the DE to get recurrence relations for $a_{n}$ and $b_{n}$, which are coupled in two groups - the $a_{2m}$ and $b_{2m+1}$, and the $a_{2m+1}$ and $b_{2m}$. Working them out for the first few terms in each series suggests that you could probably find a closed formula, since each one comes in a form of either $p(\alpha )a_{0}$ or $p(\alpha )b_{0}$, where p is a polynomial of degree n (when looking at the nth coefficient). Confusing Manifestation 03:34, 22 June 2007 (UTC)

Mathematica produces a direct closed form solution:

${\begin{cases}f(x)=C_{2}\,e^{{\frac {x^{2}}{2}}-ax+\ln(e^{2ax}-e^{C_{1}})}\\g(x)=f(x){\frac {e^{2ax}+e^{C_{1}}}{e^{2ax}-e^{C_{1}}}}\end{cases}}$

which you could calculate the power series for if you want. --mglg_(talk) 21:43, 22 June 2007 (UTC)

Wow - even better! Thanks! --HappyCamper 23:19, 22 June 2007 (UTC)

Obviously, I could have re-written that answer more simply as:

${\begin{cases}f(x)=C_{2}\,e^{{\frac {x^{2}}{2}}-ax}(e^{2ax}-C_{3})\\g(x)=C_{2}\,e^{{\frac {x^{2}}{2}}-ax}(e^{2ax}+C_{3})\end{cases}}$

--mglg_(talk) 00:13, 23 June 2007 (UTC)

Both were very useful to me. Thanks again!! --HappyCamper 15:05, 23 June 2007 (UTC)

Oops, I was full of it (as usual?): my expressions above are not solutions to your problem but to the much simpler problem without the minus sign. With the correct sign, I fail to find anything very useful (Mathematica spits out integrals of complicated sums of hypergeometric functions). --mglg_(talk) 02:08, 24 June 2007 (UTC)

Why does the extra minus sign complicate things so much? Is it because the symmetry is sort of removed from the system? --HappyCamper 17:59, 24 June 2007 (UTC)

Do maximal ideals of the homogeneous coordinate ring correspond to the points on it?

Hello,

let $Y$ be a projective variety, in $\mathbb {P} ^{n}$ . Let$S$ be the ring of polynomials in the variables $x_{0},\ldots ,x_{n}$. Let $I(Y)$ be the homogeneous ideal, generated by the homogeneous polynomials vanishing on $Y$. Let $S(Y)=S/I(Y)$ be the homogeneous coordinate ring.

Now, how much is this "like the affine case"? I mean, does every (homogeneous?) maximal ideal of $S(Y)$ correspond to a specific point on the projective variety $Y$ ?

Note that I am cautious when I write homogeneous, since I'm not even sure whether or not$S(Y)$ is a graded ring.

Many thanks Evilbu 12:02, 22 June 2007 (UTC)

There is an analog of the Nullstellensatz for projective varieties [1]. Also, for a graded ring A and homogeneous ideal I, $A/I$ will always be graded. See Graded algebra. nadav (talk) 06:26, 24 June 2007 (UTC)

Thank you very much. That's a very clear explanation, but unfortunately it doesn't really goes into depth about the maximal ideals. Do you know something about it. My instructor tried to convince me that points correspond to maximal ideals, but I don't see it?Evilbu 10:38, 24 June 2007 (UTC)

Applications of Pure Math

I've heard several times that lots of things that were once thought the silly, idealistic, nigh-childish fancy of pure mathematicians have found actual uses. I fact, I've heard there's some famous quote along the lines of "All results in mathematics, no matter how abstract, are in danger of being applied." The only examples I can think of are certain parts of number theory in the form of RSA (and related cryptographic systems), and nonEuclidean geometry in the form of relativity. Does anyone know of others? Black Carrot 17:16, 22 June 2007 (UTC)

Maybe non-Archimedean metrics have a place in quantum physics? (Google p-adic analysis physics or see p-adic and adelic physics) iames 19:34, 22 June 2007 (UTC)

Our article on group theory lists, next to cryptography, applications in the sciences. There is also the use of category theory in physics; see for example http://math.ucr.edu/home/baez/categories.html and http://math.ucr.edu/home/baez/quantum/. The concept of Von Neumann algebras can be considered to belong to abstract algebra, but was of course motivated by the possible application to quantum mechanics. See also Wikipedia:Reference desk/Archives/Mathematics/2007 June 18#Useless mathematics for an almost perfectly complementary question.  --Lambiam^Talk 06:22, 23 June 2007 (UTC)

If uses of category theory in theoretical physics (also computer science) are considered applications, then perhaps uses of foundations of mathematics topics in philosophy also qualify. Boolean logic was famously applied as a model for digital electronics. nadav (talk) 06:44, 23 June 2007 (UTC)

Thanks, that's a good start. Lot of good suggestions. Yeah, I remember that discussion. It looks like the only example he got was hyper-inaccessible cardinals, which someone else said actually do have applications. Real shame. Let me know if you think of any more. Black Carrot 04:23, 24 June 2007 (UTC)

June 23

Euclidean spaces

Hello. Suppose $k\geq 3$, x,y are in R^k, |x-y| = d > 0 and r > 0. Prove that if 2r > d, there are infinitely many z in R^k such that |z-x| = |z-y| =r. If 2r = d, there is exactly one such z and if 2r < d there is no such z.

All I can think of is to take a sphere centered at x with radius r and then somehow taking a cone with vertex y and base containing that circle on the sphere for which |z-y| = r. But I have no idea how to prove anything analytically. Any help will be appreciated. --Shahab 10:18, 23 June 2007 (UTC)

For the "no such" part, think triangle inequality, which for R^k is a consequence of the Cauchy–Schwarz inequality. The case of strict equality can be used to handle "exactly one such", although it is not hard to do that from first principles. For 2r > d, visualize two equally large spheres with radius r centred at, respectively, x and y. One sphere is the locus of ||z−x|| = r, the other of ||z−y|| = r. Where do you have that ||z−x|| = ||z−y|| = r? It should not be hard to come up with a formula or construction for some vector z_c parametrized by a value c such that ||z_c-x|| = ||z_c-y|| = r for all c, where different choices of c give you infinitely many different vectors z_c.  --Lambiam^Talk 10:53, 23 June 2007 (UTC)

Thankx for the reply. However I still have trouble in constructing z_c. In the case k = 3 actual spheres intersect and so I can use the result that their intersection is a circle. But I don't know how to do that in general. I think I can handle the other cases. Can you explain the construction of z_c?--Shahab 16:47, 23 June 2007 (UTC)

A fact that may be helpful is that the set of points equidistant from two given points, the perpendicular bisector, is flat. In 2D it's a line; in 3D, a plane; and so on. The set of points at a fixed distance from one given point is a "sphere", in a generalized sense. In 2D it's a circle; in 3D, a spherical surface; and so on. Now intersect. --KSmrq^T 19:23, 23 June 2007 (UTC)

WLOG we can assume that $\mathbf {x}$ lies at the origin and $\mathbf {y} =d\mathbf {e} _{1}+0\mathbf {e} _{2}+0\mathbf {e} _{3}+\cdots +0\mathbf {e} _{k}$. Let $\mathbf {z} (\varphi )=(d/2)\mathbf {e} _{1}+{\sqrt {r^{2}-(d/2)^{2}}}(\mathbf {e_{2}} \sin \varphi +\mathbf {e_{3}} \cos \varphi )$, $0\leq \varphi <2\pi$. These form a circle around $(d/2)\mathbf {e} _{1}$, with the axis along $\mathbf {e} _{1}$ and with radius ${\sqrt {r^{2}-(d/2)^{2}}}$. We verify

${\begin{aligned}||\mathbf {z} (\varphi )-\mathbf {x} ||&={\sqrt {(d/2)^{2}+({\sqrt {r^{2}-(d/2)^{2}}}\sin \varphi )^{2}+({\sqrt {r^{2}-(d/2)^{2}}}\cos \varphi )^{2}}}\\&={\sqrt {(d/2)^{2}+r^{2}-(d/2)^{2}}}=r\end{aligned}}$

and

${\begin{aligned}||\mathbf {z} (\varphi )-\mathbf {y} ||&={\sqrt {(d-d/2)^{2}+({\sqrt {r^{2}-(d/2)^{2}}}\sin \varphi )^{2}+({\sqrt {r^{2}-(d/2)^{2}}}\cos \varphi )^{2}}}\\&={\sqrt {(d/2)^{2}+r^{2}-(d/2)^{2}}}=r\end{aligned}}$.

Did I just do someone's homework? ;-) You could complete it by proving that the number of $\mathbf {z} (\varphi )$, $0\leq \varphi <2\pi$, are infinite. You could also do better than me and find all $\mathbf {z}$ that satisfies the criterion. I only give one circle of solutions, no matter how high the dimension $k$ is. (And find my mistakes. No intentional ones, I promise!) Bonne chance. —Bromskloss 20:31, 23 June 2007 (UTC)

June 24

Shuffling

I heard something about odds that I don't think is right, but I'm not sure I've worked it out correctly. Here's how you set it up: Get a deck of cards, and have someone shuffle it thoroughly. Ask them to name two numbers Ace through King (like Jack and 5, say). Bet them that a Jack and a 5, somewhere in the deck, have at most one card between them. The claim is that you have more than a 99% chance of winning that bet. I think the odds of winning would be less than 82%, probably a fair bit less. What do you think? Black Carrot 00:06, 24 June 2007 (UTC)

Well, you have 16 different jack+five pairs. The probability of a given five being separated by zero or one cards from a given jack is approximately 4/52. The probability of none of the 16 pairs being that close is approximately (1-4/52)¹⁶. So I would say the probability you are looking for should be approximately 1-(1-4/52)¹⁶~72.2%. Taking into account end-of-deck effects, but still approximating the pairs as independent, I end up with 1-(1-202/52²)¹⁶~71.1%. --mglg_(talk) 01:21, 24 June 2007 (UTC)

While the magnitude there sounds about right, I would point out that those 16 pairs are strongly non-independent, most notably that other Js and 5s can be between any one particular pair of them!. This suggests looking at the pairs in a consecutive fashoin, where there are only seven pairs, and the only non independence is that a card between one such pair cannot be between another.

The 8 Js and 5s are in a certain order, and in between them are 7 gaps, with two gaps on either end. To lose the bet for any ordering of those 8, each of the inner 7 gaps need to have at least 2 other cards in it. One can combinatorially compute how many all the gaps are filled up with two cards. For any assignment of pairs of cards to the gaps, one can combinatorially compute how many ways the rest of the cards can be inserted anywhere (including at the ends). Thus the number of shuffles which lose is a product of three terms: # orderings of the 8 cards, # ways separating all 7 adjacent pairs, # ways to insert the rest. Then divide into the total number of ways to arrange 52 cards and subtract from one. There may be other ways of breaking the problem down into independent nested binnings which are easier to calculate, but I strongly suspect that considering the 8 cards sequentially like this will makes things loads easier. Baccyak4H (Yak!) 02:41, 24 June 2007 (UTC)

See, the thing about that is, getting an accurate answer would take a lot of work (even more than Baccyak suggests, since in many orderings of the 8 cards two of the same could land right next to each other) and most answers I could get including the word "about" seem a bit vague. It's hard to tell how close they'd land. So, I tried to get a strict upper bound on it. The way I saw it, we could assume the Jacks had been placed first, then place the 5s. There are four Jacks, and at most 16 cards within two of a Jack, meaning at least 32 remaining. Placing the first 5, the odds are 32/(16+32)=2/3 that it would land away from the Jacks. Assuming all of them landed away from Jacks, the odds for each would be the same but with the 32 (the count of cards still unclaimed that aren't near Jacks) going down once each time. Their product, the lower bound on the odds of losing the bet, would be 1798/9729 = 0.1848, putting an easy upper bound on the odds of winning at 82%. Does that sound right? Black Carrot 03:17, 24 June 2007 (UTC)

I agree with your upper bound, Black Carrot – I just came back here to enter the same thing! I was going to express it as $1-{\binom {32}{4}}/{\binom {48}{4}}={\frac {7931}{9729}}\approx 81.52\%$. As you say, the exact answer is likely to be considerably lower than this, because there are large probabilities of having less than 16 "winning slots" free after placing the jacks. But the upper bound suffices to prove that the "99%" claim is false. --mglg_(talk) 04:12, 24 June 2007 (UTC)

Thanks. Great minds think alike, I guess. :) Black Carrot 04:25, 24 June 2007 (UTC)

That, of course, being equally true about mediocre minds... --mglg_(talk) 04:38, 24 June 2007 (UTC)

I've run a simulation and, modulo mistakes in my implementation, the result is around 73.5%. -- Meni Rosenfeld (talk) 11:06, 24 June 2007 (UTC)

Wow, he hit it about on the head. I've been trying to work it out Baccyak's way, but the result I've gotten is a bit hard to believe. I should have gotten a much tighter upper bound, but instead I got that the odds of winning are strictly less than 94%. I'd like someone to check my work. I get the number of ways of placing eight cards in eight spots as 8!, the number of ways of choosing 14 cards from what remains as (44 C 14), the number of ways of arranging those between the original 8 as 14!, the number of ways of placing the remaining 30 cards as f(30) to be defined in a moment, and the number of ways of ordering those 30 as 30!. The total number of orderings that would lose is greater than that, because it counts nothing that does not lose but leaves out situations where cards of the same type (5 and 5, say) can be closer than two from each other. However, it seems like it should provide a tighter bound than the previous one, because it counts all the same cases and more. f(30) is the number of ways of dropping 30 identical pegs into 9 baskets, which would be a Figurate number of 8 dimensions and length 31. That makes it equal to (38!/30!)/8!. The total product, divided by 52!, simplifies to 44!38!/(52!30!), about 6.5%. Black Carrot 13:29, 24 June 2007 (UTC)

The way I looked at it (to find the upper bound) is to first assume the jacks are evenly spaced, so there are 4 different cards near each jack. This gives us a total of 16 cards of the remaining 48 non-jacks which we're concerned with. The chances that each five will be in those 16 cards is thus 16/48 = 1/3. The chance that each five is not one of those 16 cards is therefore 2/3. This gives us the probability that none of the 4 fives is near a jack as (2/3)⁴ or 16/81 = 19.75%, and the P that one or more five is near a jack is 80.25%. However, if we allow for the possibility that the jacks are near each other, or near the top or bottom of the deck, both would lower the probability of a nearby five to below 80%. StuRat 01:27, 25 June 2007 (UTC)

StuRat, your result is an approximation of the upper bound that has already been reported twice above. Your approximation is to ignore the fact that no more than one five can occupy the same position. This is why you get 65/81 instead of the exact answer 7931/9729 for the upper bound. --mglg_(talk) 01:50, 25 June 2007 (UTC)

Since the question was "is 99% correct" or possibly "is less than 82% correct", an exact answer is not needed, we only need to prove that 99% (or 82%) is above the upper bound to answer the question, which I have done. I believe my method was the simplest approximation which gave an answer accurate enough to answer this question (so simple, in fact, that you can do the math in your head). It is also helpful to confirm earlier answers, especially when using a slightly different method, as a check. StuRat 15:44, 25 June 2007 (UTC)

I can report an exact value for the probability of 3230755238/4389805875 = 0.735967678297..., which is consistent with Meni Rosenfeld's reported experimentally determined 73.5%. What I've done is count, for various lengths, the number of sequences (shuffle prefixes) for each of the patterns *···*@@, *···**J, *···**5, *···*J@, and *···*5@, in which * is any card, J is a Jack, 5 is a 5, and @ is any other card than Jack or 5; additionally, in the sequence no J and 5 must occur as neighbours or almost neighbours. Given all counts for length n for each of these patterns of the sequences with exactly u Js and v 5s, the same counts can be cheaply computed for length n+1. From the resulting data, with a little effort the probability can be determined.  --Lambiam^Talk 16:27, 25 June 2007 (UTC)

Thanks Lambiam! I knew you would come through and enlighten us in the end. --mglg_(talk) 17:55, 25 June 2007 (UTC)

I don't understand. Could you explain that a little more? Black Carrot 04:06, 26 June 2007 (UTC)

A rethink

Carrot, up above you came up with an approximation of about 6.5%, trying my line of reasoning. In hindsight, I realize the exact formulation I suggested was needlessly complicated, but nonetheless (IMO) still on the right track. I tried a rethink and want to share it here.

Forget which cards are which, except for a) eight target cards; b) the rest. This can be seen by noting your problem is isomorphic to selecting eight integers from 1 to 52 without replacement and seeing whether the smallest difference between any two integers is three or greater (three because there are two integers between integers that are three apart). I will tackle this formulation.

How many total ways can this be done, win or lose? This is 52 choose 8. Only the positions of the key cards matter here. Which particular key card is, e.g., J♥ or which non-key card is 2♣ is immaterial.

How can we enumerate the winning shuffles? Here, piggybacking on Lambiam's notation, let "*" be a key card (J or 5, in your example), let "x" be any single non-key card, and let "@" be any sequence (perhaps of length zero) of non-key cards. These are placeholders for the cards' positions, i.e., 1 to 52. Then a winning hand (does not have an interval of at most one card) can be enumerated as such (spacing for clarity):

@ *xx @ *xx @ *xx @ *xx @ *xx @ *xx @ *xx @ * @

Note the last "*" is not followed by "xx". (We could have written "@ xx* @ xx* ...", but the results will be the same.) The "*"s are exchangeable, as are the "x"s, so we can ignore their exact identity. Now each such winning shuffle (or sample of 8 numbers) is completely determined by the "@"s. These form a composition of (52 - 3*(8-1) - 1) = 30 into 9 parts, including the possibility of zeros (if two "*xx" are adjacent). The formula for the number of such compositions of n into r parts is (n + r - 1) choose (r - 1). In this case, it is 38 choose 8.

Thus the number of winning hands is

${\frac {\dbinom {38}{8}}{\dbinom {52}{8}}}$

which reduces to 479446 / 7377825 or about 6.49847%. The losing chance is thus about 93.50153%. I have done a simulation in R, and with 4000000 runs, am getting 6.5090% with SEM of 0.0123%. This is very close to your estimate above. (Code available on request—I am working to generalize this to any number of key cards from any size desk with any size gaps.)

An aside. The change of losing the bet if asked that there would be no cards adjacent to each other (that is, there was at least one card between each of the Js and 5s), then I get ~71.2% I wonder if this is what was really calculated elsewhere in several places. Baccyak4H (Yak!) 19:45, 26 June 2007 (UTC)

"This can be seen by noting your problem is isomorphic to selecting eight integers from 1 to 52 without replacement and seeing whether the smallest difference between any two integers is three or greater (three because there are two integers between integers that are three apart). " If I understand you correctly, this isn't correct, since you're counting, e.g., Jack-something-Jack as a successful hit. Donald Hosek 20:55, 26 June 2007 (UTC)

I think there's been a misunderstanding. If I challenged you to this, I would win if a Jack and a 5 were in close proximity (with about 73.5% likelyhood), and lose if no Jack was near a 5 (with about 26.5% likelyhood.) Whether a Jack is near a Jack, or a 5 near a 5, has no influence on the bet. That's why all the other analyses, except for my use of yours (which arrived at the same fraction you did :) ), specifically mentioned Jacks and 5s, in two groups of four. Black Carrot 02:46, 27 June 2007 (UTC)

J-something-J is a win for the poser (you lose!). There are 8 Js and 5s (in this case) total. Do not be confused between Js and 5s. Here, the important distinction is between 8 key cards (both Js and 5s in your example) and the rest. If (say) 2 Js were only one card apart, you would lose. But you would as well if 2 5s were only one card apart, or any J and 5. Thus you can look at the 4 Js and 4 5s as being from a group of 8 "key" cards. I am reading Carrot's reply to be questioning of Donald's conclusions; if so, I agree.

Can anyone point out a flaw in my reasoning? If not, I can provide a general solution (as alluded to above) to the probabilities of winning such a generalized bet. Baccyak4H (Yak!) 03:24, 27 June 2007 (UTC)

Um, no. Black Carrot's reply (as well as his original problem statement) confirms Donald's observations, and questions yours. You win if at any place there are a J and a 5 (not J and J or 5 and 5) close to each other (separated by at most one card). So the Js and the 5s are different, you can't treat them as just "8 key cards". -- Meni Rosenfeld (talk) 07:33, 27 June 2007 (UTC)

Mystery Series

Something came up on a Facebook forum, and I can't remember the answer to it. Well, I can, but it seems I'm remembering it wrong. Imagine a tower of identical bricks, each shifted to the right as far as it will go without falling off. In other words, the center of balace of each is right on the edge of whatever it's sitting on. How far over is each one, and how far does the tower extend to the side? I remember it being an example of the harmonic series in action, and I remember the result being that, given an impractically tall tower, you can extend the right edge arbitrarily far. I'm working it out now, though, and I'm getting the displacement of each to be 1/2^n from the one below it, with the result that you can never get even a block's length away from the base. As this is patently false, experiment being my guide, I have to assume I made a mistake somewhere. Any ideas? Black Carrot 23:58, 24 June 2007 (UTC)

For the tower to have the maximum stable extension, the edge of each brick should coincide with the combined center of mass of all the bricks above it. If you work this out, brick n (from the top) will extend by 1/(2 n) beyond the brick below it, for unit length bricks. This means that the total overhang of an n-brick tower will be ${\frac {L}{2}}\sum _{k=1}^{n-1}{\frac {1}{k}}$, which does go to infinity with n, albeit very slowly. Not all sequences that start with 1/2 and 1/4 are 1/2ⁿ... --mglg_(talk) 00:36, 25 June 2007 (UTC)

The first Google hit on "harmonic series" bricks is http://www.antiquark.com/2005/04/harmonic-series-and-bricks.html which includes 2 nice pictures. I have seen the problem before and the Google search has many relevant hits. Maybe we should mention this nice problem somewhere (I searched shortly and couldn't find it in Wikipedia). PrimeHunter 00:45, 25 June 2007 (UTC)

On page 3 of the Google search I get:

"In response to a legal request submitted to Google, we have removed 1 result(s) from this page. If you wish, you may read more about the request at ChillingEffects.org."

I haven't seen such a message before. The link doesn't work for me. PrimeHunter 00:49, 25 June 2007 (UTC)

I found a name. Does somebody (not me) want to write Leaning tower of Lire? [2] Or maybe it should just be a section somewhere. PrimeHunter 01:02, 25 June 2007 (UTC)

In reality, I assume the weight of the bricks would creat an enormous amout of stress that would eventually break any material. nadav (talk) 01:31, 25 June 2007 (UTC)

That's what it was! Thanks. Black Carrot 03:35, 26 June 2007 (UTC)

June 25

Additional Mathematics Project Work 2007 (Form 5)

Question: Ang, Bakar and Chandran are friends and they have just graduated from a local university. Ang works in a company with a starting pay of RM2000 per month. Bakar is a sales executive whose income depends solely on the commission he receives. He earns a commission of RM1000 for the 1st month increases by RM100 for each subsequent month. On the other hand, Chandran decides to go into business. He opens a cafe and makes a profit of RM100 in the first month. For the first year, his profit in each subsequent month is 50% more than that of the previous month.

In the second year, Ang receives a 10% increment in his monthly pay. On the other hand, the commission received by Bakar is reduced by RM50 for each subsequent month. In addition, the profit made by Chandran is reduced by 10% for each subsequent month.

1. a) How much does each of them receive at the end of the 1st year? (2 or more method are required for this ques.) [max: besides progressions method, what can i use?? show me..]

b) What is the percentage change in their total income for the 2nd year compared to the 1st year? Comment on the answers.

c) Ang, Bakar and Chandran, each decided to open a fixed deposit account of RM10000 for 3years without any withdrawal. - Ang keeps the amount at an interest rate of 2.5% per annum for a duration of 1month renewable at the end of each month. - Bakar keeps the amount at an interest rate of 3% per annum for a duration of 3months renewable at the end of every 3months. - Chandran keeps the amount at an interest rate of 3.5% per annum for a duration of 6months renewable at the end of every 6months.

(i) Find the total amount each of them will receive after 3years. (ii) Compare & comment on the difference in the interests received. If you were to invest RM10000 for the same period of time, which fixed deposit account would you prefer? Give your reasons.

Further Exploration

2. a) When Chandran’s 1st child, Johan is born, Chandran invested RM300 for him at 8% compound interest per annum. He continues to invest RM300 on each of Johan’s birthday, up to and including his 18th birthday. What will be the total value of the investment on Johan’s 18th birthday?

b) If Chandran starts his investment with RM500 instead of RM300 at the same interest rate, calculate on which birthday will the total investment be more than RM25000 for the 1st time.

PLEASE GIVE ME THE FULL ANSWERS A.S.A.P..THANK YOU..

Go away and do your homework. Come back when you at least have started some of the problems.—Preceding unsigned comment added by 203.49.208.227 (talk • contribs)

Non zero y so that x.y = 0

Hello.

My question is as follows:

If $k\geq 2$ and x is in R^k, prove that there exists y in R^k such that y is non zero but x.y = 0. Is this true if k = 1?

The way I have thought is as follows: R is an integral domain so the result is not true for non zero x if k is 1. For $k\geq 2$ we want that for non zero x,

$\sum _{i=1}^{k}x_{i}y_{i}=0$ there exists a non zero $(y_{1},\ y_{2}\cdots y_{k})$. But this is obvious as the set $\{x_{1},\ \cdots x_{k}\}$ is linearly dependent in R. Hence the result. What I want to know is there some other simpler way of constructing y. Thanks.--Shahab 09:41, 25 June 2007 (UTC)

Without giving it away, consider what it means for a sum to be zero. Some of the summands must be negative. This tells you the sign of some of the components of the vector y. Make another manipulation to easily force the sum to be zero. That will handle the case of k even. The case where k is odd is handled similarly, but with a minor twist.

Seems to me the simplest and clearest approach for k>1 is to construct a specific example. Set $y_{1}=y_{2}=...=y_{k-1}=1$. What value does $y_{k}$ have to take to make x.y equal 0 ? Is the vector y in R^k? If so, you are done. Gandalf61 11:30, 25 June 2007 (UTC)

This approach is pretty much equivalent also.

Gandalf's solution, as presented, fails in the case $x_{k}=0$. Of course, for a nonzero x, at least one coordinate is nonzero, and you can adapt the argument accordingly. -- Meni Rosenfeld (talk) 13:36, 25 June 2007 (UTC)

Ah, yes. Let's just say "details are left to the reader" ! Gandalf61 14:22, 25 June 2007 (UTC)

I'd just like to point out that it's not necessarily true that if a sum is zero, then some term must be negative. Tesseran 06:44, 26 June 2007 (UTC)

Yes, excluding the case where every summand are zero, however some summand(s) must be negative, if we are talking about a field of characteristic zero (which, in this case, I believe we are).

The vectors (1,0) and (0,1) are orthogonal over any field, of any characteristic. The sum to compute their dot product is 1x0 + 0x1 = 0, in which no summand is negative. Tesseran 17:12, 26 June 2007 (UTC)

Every summand is zero! That is the exclusion case I provided.

help me in my add maths project work

how to solve the questions.pls help me.i am lack of time.

Wikipedia's reference desk is not a service to do people's homework, so you should not waste what little time you have asking us to do so. However, if there is anything you do not understand and wish us to clarify, we will be more than happy to help. -- Meni Rosenfeld (talk) 15:08, 25 June 2007 (UTC)

Ellipse

I'm trying to create a drawing program much like paint and the circle drawing tool can obviously draw ellipses from knowing where the mouse's last click position was and where it is now. So how to find out an ellipe's two focus points from it's height and width? I saw nothing about it in the ellipse article, or it was too obscure to decipher. 62.20.156.137 16:51, 25 June 2007 (UTC)

Why do you want to find the foci? The easiest way to draw an axis-parallel ellipse is with the parametric equations $x=a\cos {t}$ and $y=b\sin {t}$, where a is the horizontal radius (half the width), b is the vertical radius (half the height), and t varies from 0 to 2π. However, if you insist on finding the foci, the relevant information is in the first section ("Eccentricity") of the ellipse article - the distance of each from the center, assuming $a>b$, is $a\epsilon$, where $\epsilon ={\sqrt {1-{\frac {b^{2}}{a^{2}}}}}$ is the eccentricity. If $a<b$ you need to reverse their roles. -- Meni Rosenfeld (talk) 17:21, 25 June 2007 (UTC)

counter intuitive probability

A friend of mine blew my mind with a bit of probability the other day, but she did a lousey job explaining it. The scenario is of a game show where one is offered three curtains, one of which is a prize and the others are something else. After one is picked, one of the false curtains is revealed and the contestant is offered the chance to change thier curtain of choice. She said that one should always change curtains, but couldn't articulate why. Is there a name or an article for this scenario?

142.33.70.60 17:45, 25 June 2007 (UTC)

Yes, the article is called Monty Hall problem. --mglg_(talk) 17:52, 25 June 2007 (UTC)

And you are welcome back if you want more persuasion that it's actually true. :-) —Bromskloss 18:48, 25 June 2007 (UTC)

Thanks :) Actually, it turns out that the arguement we were having revolved around a glossed over detail, covered in the article--that it all depends on Monty's rules(If he -always- has to show a goat curtain or not). I was positive that it was 50/50, but now that I know Monty always has to show the curtain, I understand what's going on.142.33.70.60 20:12, 25 June 2007 (UTC)

solve the equation

find the value of x and y where x^y+y^x=17 and x^x+y^y=31

Well I notice that in both equations, we're looking at examples of symmetric functions, meaning that we can exchange $x$ and $y$ without changing the equations. So if there are any solutions (and there are), they'll come in pairs. I'm not sure if that helps at all, though. Donald Hosek 17:59, 25 June 2007 (UTC)

Guess some simple values of x and y and try them. Then keep in mind Donald's tip. --mglg_(talk) 19:01, 25 June 2007 (UTC)

Finding a solution to this problem by guessing is pretty straightforward (even easier if you have a grapher which can graph the equations, the domains of the functions are kind of interesting, actually). I'm starting to wonder if there's any way of leveraging the symmetry to be able to generalize this to solving the systems $x^{y}+y^{x}=k$, $x^{x}+y^{y}=m$. Is there anything interesting to the values of $(k,m)$ which have integral solutions in $(x,y)$? Donald Hosek 19:50, 25 June 2007 (UTC)

That they have integral solutions? Black Carrot 08:09, 26 June 2007 (UTC)

Finding patterns graphically I find interesting. Why don't you plot the $(k,m)$ that satisfy the criterion (and show us too)? —Bromskloss 14:23, 26 June 2007 (UTC)

Solution: x = 2, y = 3; y = 2, x = 3. Probably the only trivial solution. Graphing, numerical approximations or "guessing" will obtain the other solutions, if there are any. Keep in mind that x and y are both nonzero, and that x and y can't both be negative. Also, y ≠ x under any circumstances.

Here's my advice on graphing: sketching the first curve is difficult. It does not touch either of the coordinate axes (in fact they're both asymptotes). The curve exists only in the 1st, 2nd and 4th quadrants. Also plot some symmetric points (reflected in the line y = x, which is also a diagonal asymptote), including (2, 3) and (3, 2). As someone said above, it's an interesting curve.

Now attack the next curve. The y-intercept of the next curve is somewhere between 3 and 4 (as the function y^y is monotone increasing). The x-intercept is the exact same value, again, somewhere between 3 and 4. Note that the points are also reflected in the line y = x, which is an asymptote. Unlike the other curve, this curve touches the coordinate axes. This curve, also, exists only in the 1st, 2nd or 3rd quadrants.

The local extrema of both curves can not be determined by elementary means, so don't bother with them. You should use this information to roughly plot both curves, and from there you can find approximate solutions. You can then use Newton's method to produce more accurate and more approximate solutions. The graph-sketching will be very hard as the curve is discontinuous at so many points and oscilliates. The values at the extreme points of the axes of the curve explode between large positive and negative values.

Actually unsigned, your arithmetic is a bit off. Neither curve can touch the axes. For the first curve, set $x=0$. Then you have $0^{y}+y^{0}=1$ for all non-zero values of $y$. The behavior of exponentiation is not well-defined on the reals when the base is less than zero, so unless we make a definition of exponentiation which will depend on a branch of the complex log function and will by necessity exclude a ray from 0 in its domain (usually the negative reals), we can't talk intelligently about the graph of this function outside of the domain of positive reals, so our graphs will only be in the first quadrant if we assume $x,y\in \mathbb {R}$. There may be complex solutions to this system, but I'd hate to even consider them. Donald Hosek 18:28, 27 June 2007 (UTC)

Good point there, Donald, I had a lapse in concetration and gave $0^{0}$ a numerical definition, when it doesn't have one. I am still not convinced that the two curves are exclusively in the first quadrant though.

The problem is that you need to expand yourself to complex numbers to get out of positive reals, and once you do that, you're faced with the difficulty of defining just what you mean by, say, $-1^{1/2}$ (two possibilities) or worse still $-1^{\sqrt {2}}$ (infinite possibilities). Defining $c^{z}$ requires choosing a branch of the complex log function, since log is a multi-valued function over $\mathbb {C}$. Most of the time, the "standard" domain chosen for log over $\mathbb {C}$ specifically excludes the ray $(-\infty ,0]\subset \mathbb {R}$. We could potentially try and talk about values of the curves over $\mathbb {Z} \setminus \{0\}$, I suppose, since we do have a well-defined approach to integer exponents. One we expand to non-integral exponents, then we start dealing with the fact that we get multi-valued functions which don't have the obvious "right" choice that comes out of staying in the positive reals. It's possible to do, but not practical. Donald Hosek 17:19, 4 July 2007 (UTC)

June 26

Simplex

Given a simlex of n+1 vertices, slice it with hyperplanes along its axes of symmetry - that is, for each pair of vertices, construct a hyperplane that connects the point halfway between those two with all other vertices. How many chunks will this slice the simplex into, and will they all have the same volume? Black Carrot 04:12, 26 June 2007 (UTC)

Well, the number of edges of a simplex is a triangular number. So you'll have $E={{n+1} \choose {2}}$ edges, and that should give you $2^{E}$ chunks. I don't know about the equal volume idea though. - Rainwarrior 06:07, 26 June 2007 (UTC)

Er, that was an overestimate. I think all of the planes will be intersecting at one point. So each new division would really only be passing through two of the existing spaces (adjacent to the edge, and adjacent to the other (n-1) vertices. My guess would be 2E instead. - Rainwarrior 06:44, 26 June 2007 (UTC)

Actually, thinking about it in terms of Barycentric coordinates, I think you're right that the volumes should be equal. Planes connecting (n-1) points to the midpoint of each edge should be at the barycentre, which means the simplex created by (n-1) points and this centre point should have the same volume as every other simplex from every other group of (n-1) points and this centre, and finally each of these simplexes is divided exactly in two by the remaining plane, which should make your hypothesis true. - Rainwarrior 06:32, 26 June 2007 (UTC)

Yeah, I was pretty sure about the symmetry. The way I figured it, any set of (n-1) planes is mapped to itself by reflection across the remaining plane, it having been chosen as an axis of symmetry. Since the planes are the only thing defining the boundaries of each section other than the surface of the simplex, which is also symmetric across each axis, chunks map to chunks. So, given any starting chunk, it's symmetric with each chunk it's adjascent to by reflection, and etc. The harder part is the actual number of chunks. In 3 and fewer dimensions, I can just draw it, which gives 1, 2, 6, 12, but I'm not sure how to generalize that. Black Carrot 07:48, 26 June 2007 (UTC)

Okay, sorry, I miscounted. The last one should be 24, which suggests factorials or something similar. In general, it seems that the chunks of each simplex can be made by chunking each of its one-less-dimensional faces, and extruding those faces to the center, which would prove that factorial idea nicely. Does anyone support that notion? Black Carrot 07:57, 26 June 2007 (UTC)

Is it really 24? Mentally it comes out to 12 for me, but maybe I'm wrong. - Rainwarrior 05:11, 27 June 2007 (UTC)

For each pair of vertices (A,B), the bisectrix hyperplane divides hyperspace into the points closer to A and closer to B. Therefore there is a bijection between the permutations of the vertices and the chunks into which the hyperplanes partition space. The interior points of each chunk share the closeness ranking of the vertices.  --Lambiam^Talk 09:36, 26 June 2007 (UTC)

I'm afraid I don't follow you. Black Carrot 02:31, 27 June 2007 (UTC)

I think I get it now. Given each permutation of n+1 vertices (A,B,C...), a set of n divisions (A,B) (B,C) (C,D)... can be defined. Each bunch of divisions can be put into correspondence with a chunk by picking the one on the first side of each - The one on the A side of (A,B), the B side of (B,C), etc. Is that about right? Black Carrot 03:01, 27 June 2007 (UTC)

If there are n vertices, there are n(n–1)/2 pairs, each giving a division; for example, for n = 3, the vertices {A,B,C} give rise to the divisions (A,B), (A,C) and (B,C). Except on the bisectrix for (B,C), each point is closer to B or closer to C. In the ASCII art below, the chunk labelled ABC corresponds to the points for which A is closest, then B, and finally C. And for chunk CAB, C is closest, followed by A, with B coming in last. The orderings in which B precedes C (ABC, BAC and BCA) are on one side of the (B,C) bisectrix; the others are on the other side.

                       A
                       .
                      /:\
                     / : \
                    /  :  \
                   /   :   \
                  /    :    \
                 /     :     \
                /  ABC : ACB  \
               /..     :     ..\
              /   ...  :  ...   \
             /       ..:..       \
            /  BAC  ...:...  CAB  \
           /     ...   :   ...     \
          /   ...      :      ...   \
         / ...         :         ... \
        /..     BCA    :    CBA     ..\
       /_______________:_______________\
      B                                 C

In general, there are n! orderings, each of which gives a different chunk.  --Lambiam^Talk 13:29, 27 June 2007 (UTC)

Thanks. That's helped a lot. Black Carrot 08:14, 29 June 2007 (UTC)

What do Tesseracts have to do with Time Travel?

I was just curious, but what do Tesseracts have to do with time travel?68.105.139.161 05:29, 26 June 2007 (UTC)

I'm not sure they do. Tesseracts are four-dimensional analogs of cubes. Since most people are only familiar with physical space, which appears to be 3-dimensional, a convenient way to visualize the fourth dimension is to think of it as a time dimension. Other than that, I don't think they are in any way related to time or time travel. -- Meni Rosenfeld (talk) 07:30, 26 June 2007 (UTC)

(after edit confict) In our article on Tesseracts, the secrtion Tesseracts in art and literature mentions several references to the notion or term in science fiction. Presumably, the connection is that tesseracts are 4D objects, and in some fictional-scientific way help to achieve time travel – after all, is time not the mysterious fourth dimension?  --Lambiam^Talk 07:34, 26 June 2007 (UTC)

For some reason the words "Time Cube" came to mind. Confusing Manifestation 13:14, 26 June 2007 (UTC)

If you consider space-time as a four dimensional vector space, then the shape a 3-D cube forms moving forward through time is a Tesseract (the fourth-dimensional length depends on the length of time that elapses). Dugwiki 15:30, 26 June 2007 (UTC)

Nothing, except in the book A Wrinkle in Time. --71.146.129.86 17:20, 26 June 2007 (UTC)

Question about Xfig

I do not know whether this is the right place to ask, or whether it should be the computer desk.

With Xfig, I used ARC drawing by specifying three points. I drew four arcs to form a simple closed curve. Then I used Edit, fill color with blue, fill style with filled, clicked apply and clicked done. I managed to paint the area of EACH ARC with blue color. I did RTFM but probably overlooked something.

Question: How do I paint the interior of this simple closed curve with blue color? Thank you in advance. Twma 05:56, 26 June 2007 (UTC)

Rather than ARC you'll need to define your curve as a "closed spline", this will define a closed region. While the four arcs appear to define a closed region to the program they are seperate objects, so fillings not possible. I've only got a windows port so I don't know if theres a way to join the arcs together so they create a closed spline. --Salix alba (talk) 12:06, 26 June 2007 (UTC)

Thank you. In four color theorem, the arcs are shared by neighboring regions although I over simplify the problem in the above setting. Twma 02:05, 27 June 2007 (UTC)

I still need help. Thank you in advance. Twma 02:36, 28 June 2007 (UTC)

Digitizing volumes

Is there any open code available for digitizing arbitrary 3d shapes (using a uniform grid)? Or else, any algorithms that I could implement? The volume (input) is known in terms of several randomly located points on its surface. The output will be a set of points located on a uniform grid. Thanks! deeptrivia (talk) 16:40, 26 June 2007 (UTC)

Possibly the convex hull of the points would give a polgonization of the surface, you could then intersect to polygons with the grid lines to give a regular set. The Qhull algorithm, is very efficient. While solving the problem this would be less than ideal as it would square off the surface. An alternative would be to consider algorithms for isosurface and Level set and meshing may help solve the more genaral problem. There is a very an extensive litrature on all these topics and source forge seems to have a good few project devoted to meshing. --Salix alba (talk) 23:17, 26 June 2007 (UTC)

Your problem statement confuses me. Are you trying to begin with sparsely sample surface points, infer a surface (maybe smooth, maybe not), and generate a "uniform" grid on that inferred surface (for some meaning of uniform)? Depending on the initial sampling and what you know about the surface, this could be difficult. I'd suggest visiting the comp.graphics.algorithms newsgroup, browsing, then posting a more detailed request there. --KSmrq^T 01:30, 27 June 2007 (UTC)

June 27

Number of possibilities

I have 8 balls and 3 gates; all the balls must pass through a gate exactly once before a possibility is created. How do I calculate the number of possibilities? (The order in which the balls pass through the gates is irrelevant, let's assume the balls are all exactly the same). --The Dark Side 01:11, 27 June 2007 (UTC)

For a simple example like this, one approach is to list all the possible separate cases, of which I believe there are 10. Working out the possibilities in each case and adding them gives the answer, which I won't provide in case this is homework. But if it were about 162 balls and 57 gates, I'm not sure how I'd go about it. -- JackofOz 01:27, 27 June 2007 (UTC)

I don't understand the question. To me, it looks like it's about the number of compositions of 8 into exactly 3 parts, in which case the answer is 21. What are these "possibilities" we are counting? nadav (talk) 04:00, 27 June 2007 (UTC)

I'm going to take this opportunity to point out that the article on compositions could use some work. I'll try to make some time, but if one of you Wikipedia all-stars wants to take a crack at it... 209.189.245.114

Depends what you mean, the question is a little too vague. If we number the balls 1 thru 8 and the gates A,B,C then I imagine that one configuration is A=12345678, B=empty, C=empty. Another is A=158, B=27, C=346 etc. There are thousands of combinations like this. On the otherhand, as you say the balls are all the same, maybe it is only the number of balls rather than which balls, so my previous two examples would change to A=8, B=0, C=0 and A=3, B=2, C=3. If you can define what you seek better, it will be answerable. Please give a couple of examples - probably a "typical" example and an "extreme" example. -- SGBailey 11:44, 27 June 2007 (UTC)

Assuming gates are distinguishable but balls are not, then I make it 45 possibilities. It's like allocating 8 indistinguishable balls to 3 distinguishable bins (we must have an article on that, but I can't find it) so it is ${10 \choose 2}$. (Nadav - I don't think you included cases where some gates have 0 balls). Gandalf61 14:56, 27 June 2007 (UTC)

I didn't include them on purpose because I (mistakenly, I guess) interpreted the wording as saying that at least one ball should pass through a gate. From now on, I propose all questions be posed in mathematical symbols only ;) Also, the compositions article is what you are looking for (the article says if the zero case is included then it's called a "weak composition") nadav (talk) 15:00, 27 June 2007 (UTC)

I assume this is homework, so hints only for you! :) Try to figure out the answer for just 1 ball and 1 gate. Simple, right? Then figure it out for 1 ball and two gates. How did it change? Then 2 balls, then 3, then 4. How is the number of possibilities growing? Can you generalize this? Now do the same for increasing gates. Good luck! --TotoBaggins 15:23, 27 June 2007 (UTC)

1 ball and 1 gate obviously means just 1 possibility since each ball must pass through a gate. When I have 2 gates, that works out to be 2 possibilities for 1 ball, 3 for 2, 2 for 3 (4 for 3), 5 for 4, 6 for 5. If I have 2 balls and 3 gates there are 6 possibilities. If I have 3 balls and 3 gates there are 10 possibilities. If I have 4 balls and 3 gates there are 12 (15) possibilities. If I have 3 balls and 4 gates I have 19 (20) possibilities. Unfortunately, I do not see a pattern yet and I cannot give any extreme examples, as I do not even know how to find the answer to this one. SGBailey's idea about "the number of balls rather than which balls" is what I am looking for. As most people are assuming (incorrectly, as school is out anyways) that this is homework, I ask only for someone to help me find the answer (i.e. process). A solution would help me this time, but not the next time I encounter something like this. --The Dark Side 15:55, 27 June 2007 (UTC)

After going through that composition article I found an article on binomial coefficients. This lead me to the formula

${\frac {n!}{k!(n-k)!}}\quad {\mbox{if}}\ n\geq k\geq 0\qquad$

Using that, I got 56 as the answer. Can anyone verify that?--The Dark Side 16:16, 27 June 2007 (UTC)

I got 45. Look at it this way. Suppose there are b balls and g gates, and what you are looking for is f(b,g), or, in your specific case, f(8,3). If there is one gate, there is only one possibility, so f(b,1)=1. If there aren't any balls there's only one possibility, so f(0,g)=1. If there is more than one gate, the number of possibilities with having no balls in the first gate is f(b,g-1). The number with one ball in the first gate is f(b-1,g-1). Since there can't be more balls in the first gate than there are balls. If you add all the possibilities, you get f(b,g-1)+f(b-1,g-1)+...+f(1,g-1)+f(0,g-1), or $\sum _{k=0}^{b}f(k,g-1)$. f(b,2) is $\sum _{k=0}^{b}f(k,1)$, $\sum _{k=0}^{b}1$, or just b+1. f(b,2) becomes $\sum _{k=0}^{b}k+1$, $\sum _{k=1}^{b+1}k+1$, or the b+1th triangular number. f(b,3) is the b+1th tetrahedral number and so on. These can all be found along the diagonals of Pascal's triangle which are all binomial coefficients. Specifically, it's ${b+g-1 \choose b}$, ${\frac {(b+g-1)!}{b!((b+g-1)-b)!}}$, or ${\frac {(b+g-1)!}{b!(g-1)!}}$. — Daniel 16:51, 27 June 2007 (UTC)

I'm in the dark when patterns of passing-throughs are considered the same possibility, and when different. Let's name the balls A, B, C, ... and the gates 1, 2, 3, .... For 2 balls and 2 gates I see the possibilities (1:AB, 2:-), (1:A, 2:B), (1:B, 2:A), and (1:-, 2:AB). You write that there are 3 possibilities; which of the above do you consider the same possibility? Likewise, for 2 gates and three balls, we have (1:ABC, 2:-), (1:AB, 2:C), (1:AC, 2:B), (1:A, 2:BC), (1:BC, 2:A), (1:B, 2:AC), (1:C, 2:AB), (1:-, 2:ABC). You count 2 possibilities. Which of these eight are one possibility, and which the other?  --Lambiam^Talk 22:44, 27 June 2007 (UTC)

I took the liberty of changing the third term in the 2-ball-2-gate example above from "(1:B, 2:-)" to "(1:B, 2:A)" in the hopes of reducing confusion. I hope this isn't too presumptuous of me. Tesseran 23:31, 27 June 2007 (UTC)

Only the number of balls that pass through a gate matters, not the order of the balls or which ball passes through which gate. This means that (1:A, 2:B) and (1:B, 2:A) are the same since they still end up with the same number of balls that passed through the gates. For 2 gates and 3 balls (1:AB, 2:C), (1:AC, 2:B), (1:BA, 2:C), (1:BC, 2:A), (1:CA, 2:B), and (1:CB, 2:A) are all the same. However, I did make a mistake, for 3 balls and 2 gates there are 4 possibilities. --The Dark Side 00:44, 28 June 2007 (UTC)

Each ball either passes through a particular gate or it doesn't. For ball A, there are 2^3 possibilities.

For eight separate balls there are 2^(8*3) = 16777216 possibilities.

202.168.50.40 02:11, 28 June 2007 (UTC)

I just had to find this out myself, for a question I asked further up the desk. The problem as I understand it is to count how many ways there are to place 8 identical balls in 3 baskets, or equivalently, how many choices of nonnegative integers A, B, and C sum to 8. This would be equal to the number of vertices in a barycentric graph of side length 9. As Daniel said above, this is a triangular number, equal to 9(9+1)/2 = 45. The generalization of this to n-1 balls and r+1 baskets is given at Figurate number, and is equivalent to running down diagonals of Pascal's Triangle. Black Carrot 08:12, 29 June 2007 (UTC)

Constant Dollars formula

Greetings;

I am doing some research. But I cannot find what I am looking for anywhere on the Net or the Library. I need TWO mathematical formulas.

The first one I need is the formula for Constant dollars in economics. The second formula I need is the formula for the AVERAGE of gravity on Earth.

How can I find this information?

Mark Carrillo Los Lunas, New Mexico

You could look it up at Wikipedia, for example! Regarding constant dollars, what is it you want to calculate and what data do you have? For average gravity, the standard gravity (9.80665 m·s⁻²) might be appropriate to use. —Bromskloss 14:44, 27 June 2007 (UTC)

Jensen's inequality extravaganza

Hi all, another question. This time it's in realm of real analysis.

Let $(X,\Sigma ,\mu )$ be a measure space with $\mu X=1$, and let f be a function which is positive a.e. on X. Show that

$\lim _{h\to 0}\left(\int _{X}f^{h}\ d\mu \right)^{1/h}=\exp \left(\int _{X}\ln f\ d\mu \right).$

I can use Jensen's inequality to show that for all h > 0, $\left(\int _{X}f^{h}\ d\mu \right)^{1/h}\geq \exp \left(\int _{X}\ln f\ d\mu \right)$, so at least the lim inf of the expression on the left hand side is bound below by the expression on the right hand side, but I'm having issues getting the lim sup to be bound above by the right hand side. Maybe my technique of using lim sup and lim inf is not the way to go on this problem, but I don't know what else to do.

I've also tried writing $\exp \left(\int _{X}\ln f\ d\mu \right)=\lim _{h\to 0}\left(1+h\int _{X}\ln f\ d\mu \right)^{1/h}$ to try to get something to work out. For instance, I can write 1 as $\int _{X}1\ d\mu$ since μX = 1, but I did not achieve any success.

Any help would be appreciated! Thanks in advance.

(For those unfamiliar with my wheelings and dealings, this is not a homework problem; I'm studying for PhD qualifying examination, and this problem appeared on one of the past exams.) –King Bee ^{(τ • γ)} 18:49, 27 June 2007 (UTC)

OK, since no real mathematicians are stepping up, here is a humble physicist's pedestrian approach, which is probably invalid in your more refined spheres: $\partial {f^{h}}/\partial h=\ln {f}$ at h=0 for positive f, so $f^{h}=1+h\ln {f}+O(h^{2})\,$. Thus $\int {f^{h}}=\int 1+h\int {\ln {f}}+O(h^{2})$, and $\int 1=1$ since $\mu X=1$. We also know that $\lim _{h\to 0}\left(1+ah\right)^{1/h}=e^{a}$ as this is one definition of e. From these we have $\lim _{h\to 0}\left(\int f^{h}\right)^{1/h}=\lim _{h\to 0}\left(1+h\int {\ln {f}}+O(h^{2})\right)^{1/h}$ $=\lim _{h\to 0}\left(1+h\int {\ln {f}}\right)^{1/h}=\exp \left(\int \ln f\right)$. Is any of that salvageable to mathematical standards? --mglg_(talk) 23:30, 27 June 2007 (UTC)

Thanks for the help. I might be able to use your idea of taking the derivative of f^h with respect to h, but I don't think this "big oh" notation is going to fly. Especially since when you take that integral, you are not integrating with respect to h; I feel like there might be an issue with passing the limit inside the parentheses.

Thanks for the ideas though! I appreciate the help. –King Bee ^{(τ • γ)} 11:08, 29 June 2007 (UTC)

So what is the dimension of the symplectic lie algebra?

Hello,

I'm having an argument here with someone else. We are trying to determine the dimension of the real symplectic lie algebra $sp(2n,\mathbb {R} )$ . That is the Lie algebra of$2n\times 2nA$ satisfying $A^{T}J+JA$ where $J$ stands for the matrix $J={\begin{bmatrix}0&I_{n}\\-I_{n}&0\end{bmatrix}}$

I keep finding that it's $3n^{2}$ Some source I have says it's $n(2n+1)$ . All hints or sources of remarks are welcome. Thank,Evilbu 18:53, 27 June 2007 (UTC)

Well back when I taught algebra, some of my students thought that $2n^{2}+n=3n^{2}$. Donald Hosek 00:23, 28 June 2007 (UTC)

Hey I'm not a noob anymore:). In fact, if all goes well, the exam I'm asking this for should be my very last. But I found it, I forgot to take the transpose in those blockmatrices. That's what you get when beoming "too comfortable" when block matrices are in town....Evilbu 07:05, 28 June 2007 (UTC)

I'm assuming you want the dimension over the reals of the real symplectic Lie algebra

${\mathfrak {sp}}_{2n}\mathbb {R} ,\,\!$

the Lie algebra of the Lie group Sp_2n R of real (2n)×(2n) matrices M satisfying M^TJM = J. (Differentiating the constraint at the identity gives the stated condition, X^TJ+JX = 0.) Do you not trust the answer in our table of Lie groups? We agree with your sources that say n(2n+1). In the case of 4×4 matrices (that is, n = 2), the prediction is dimension 10; yours is dimension 12. You should be able to check this by hand, and also consult other sources to confirm. Our symplectic matrix article agrees that the dimension is n(2n+1), as does symplectic group, but can we trust Wikipedia? You might try proving that any X belonging to the Lie algebra has the form JS, where S is symmetric (sufficiency is easy); then note that S has ∑_k=1…2n k independent entries, implying dimension n(2n+1). --KSmrq^T 07:33, 28 June 2007 (UTC)

June 28

Commutators and "ad"

On the page commutator, what does "ad" stand for? It says that ad(x)(y) = [x,y]. --HappyCamper 04:44, 28 June 2007 (UTC)

The article says that's a useful alternative definition, with the particular bonus that ad(x) is then something that can act like an operator - ad(x)(y) = [x,y], (ad(x))^3(y) = [x, [x, [x, y]]] etc. Confusing Manifestation 05:49, 28 June 2007 (UTC)

Here "ad" is short for "adjoint"; for any Lie algebra g, you get a map from g to the Lie algebra of endomorphism gl(g) = End(g) by mapping the element X of g to the endomorphism "send Y to [X,Y]". This is called the adjoint representation of the Lie algebra g. Tesseran 07:37, 28 June 2007 (UTC)

I've linked Adjoint representation in the article not sure if this is the most appropriate link as Adjoint has several related terms. --Salix alba (talk) 07:46, 28 June 2007 (UTC)

(edit conflict) We need to be careful with capitalization, because we have both

$\operatorname {Ad} \colon G\to \operatorname {Aut} (T_{e}G)\,\!$

and

$\operatorname {ad} \colon T_{e}G\to \operatorname {End} (T_{e}G).\,\!$

Fulton & Harris (Representation Theory, ISBN 978-0-387-97495-8, pp. 106–107) define "ad" using the differential of "Ad", then define the commutator using "ad".

Caution: The term "adjoint" is used in several contexts, and the meaning varies (though there is usually a family resemblance). For example, we have adjoint matrix, a different meaning for adjoint matrix, adjoint operator, and adjoint functors; none of which are the same as the adjoint representation idea under discussion.

Does this help, or merely pile on confusion? :-) --KSmrq^T 08:05, 28 June 2007 (UTC)

I don't know that F&H's choices of notation are relevant -- I'm familiar with Chapter 8 of F&H, and have always found their approach extremely opaque. Regardless of how they choose to endow the tangent space to a Lie group with the structure of a Lie algebra, the adjoint representation of a Lie algebra is inherent, and a priori has nothing to do with Lie groups at all. (There are certainly other ways to give the tangent space a bracket; my personal favorite is to identify the tangent space with the space of left-invariant vector fields, and then the bracket of vector fields gives you your Lie algebra structure. [Not that this is news to you, KSmrq.]) I believe that "ad" for the adjoint representation of a Lie algebra is standard; I haven't seen enough notation introduced for the Lie group concept to say what's accepted there. Tesseran 09:29, 28 June 2007 (UTC)

Our own adjoint representation article uses the Ad/ad notation, and cites Fulton & Harris; therefore I feel it helpful to point out the distinction. It would be horribly confusing to overlook it, especially since I link to that article. Likewise, I feel that it would be unkind to force HappyCamper to stumble through all the different meanings of "adjoint". In other words, my post is intended to provide orientation and an explanatory link.

While we're disambiguating, perhaps we should caution that brackets and commutators come in different flavors as well. I know, "A foolish consistency is the hobgoblin of little minds, …" (Emerson, my emphasis); but sometimes I wish mathematicians could try for a bit more. All the variations make it tough on us encyclopedia editors (not to mention the generations of mathematics students)! --KSmrq^T 11:49, 28 June 2007 (UTC)

OK, so I guess it was a bit of wishful thinking that "adjoint" here has any analogies with any of the other adjoints. --HappyCamper 16:37, 28 June 2007 (UTC)

Sorry, KSmrq; I thought the link was to Adjoint representation of a Lie algebra (which redirects to Adjoint endomorphism). That page treats things from an inherent perspective (until the end). Thanks for your clarifying remark, and sorry if I seemed antagonistic. Tesseran 01:45, 29 June 2007 (UTC)

impact of wind speed hitting a door at full force

An OHS query in context but a mathematical one. If the wind speed hitting a door rose from 40kph to 65kph how much pressure would it add to a door?

Leedale1 06:42, 28 June 2007 (UTC)

Let's start from first principles. The likely factors in determining the pressure are the density of the air and the wind speed. Use dimensional analysis to find a function of density and velocity that has the same dimensions as pressure. Assuming constant density, you now know that presure is proportional to some power of the wind speed. So if the wind speed has increased by a factor of 65/40 = 1.625, you can now find the relative effect on pressure. You can't find the absolute pressure without knowing (a) the density of the air and (b) any numerical constants in the pressure/density/wind speed relation. Gandalf61 12:28, 28 June 2007 (UTC)

Oh, are you sure setting up a formula with dimensional analysis like that will yield a correct result? Mabye it will, often it does, but I don't know why that has to be the case here. —Bromskloss 12:53, 28 June 2007 (UTC)

Yes, dimensional analysis gives the correct result, modulo a constant factor. The actual formula is in a Wikpedia article with initials d.... p... but I thought that Leedale1 might learn more from doing a bit on dimensional analysis than by just blindly putting numbers into a formula. Gandalf61 13:41, 28 June 2007 (UTC)

I get your reference. (But aren't we forgetting the s… p…? If at all the question is well defined.) Anyway, in other situations, couldn't dimensional analysis mislead us? Time dilation in special relativity is $t'=t/{\sqrt {1-v^{2}/c^{2}}}$, but that's not apparent from the dimensions of the variables involved. In general, it seems to me like we run into trouble when there are dimensionless quantities involved or when the equation doesn't consist exclusively of products of powers of the quantities. —Bromskloss 18:54, 28 June 2007 (UTC)

Associated Legendre functions

I feel that the article associated Legendre function can be improved somewhat, see here. To that end one needs to prove that $P_{l}^{(|m|)}\propto P_{l}^{(-|m|)},\quad |m|=0,1,\ldots ,l$. I believe that it is sufficient to show that both functions are solutions of the same associated Legendre equation, although this is a second order differential equation and has two solutions. Concrete questions: (i) Is it indeed sufficient to show that positive and negative m functions are solutions for the proportionality to be true? (ii) How do we show that $P_{l}^{(m)}$, given in its form with the l+m-th derivative of $(x^{2}-1)^{l}$, is a solution?--P.wormer 09:04, 28 June 2007 (UTC)

June 29

What do you call this really large number?

What do you call a number that, strictly speaking, is finite, but is so large that it may as well be infinite? For example, if a book has 500 000 letters in it, there are 500 000 ^ 26 possible books that could exist, and the number is technically finite because I can't print (500 000 ^ 26) +1 without repeating myself, but in practice that number is larger than all the atoms in the universe, right?

Thanks --Duomillia 00:34, 29 June 2007 (UTC)

Actually, that's $26^{500000}$, which is a much larger number. And I think you're asking for a name for numbers that "can't be written in decimal notation in this Universe". Never heard any names for those. — Kieff | Talk 00:57, 29 June 2007 (UTC)

I've seen bigger :P In any case, while there may or may not be some kind of specific term, there's a lot of information on large numbers at the article on, well, large numbers. Confusing Manifestation 03:09, 29 June 2007 (UTC)

Pi

Here is a question that relates to the number pi. Every once in a while, you hear about an individual who has pi memorized to x places (say, 700 places). (1) Why on earth do people do that? (2) Is there any type of "trick" that enables them to perform these seemingly impossible feats? I thought that pi did not have repeating patterns in its numbers, so I can't imagine a useful trick or pattern that assists their memorization of so many numbers. Any insight is appreciated. Thanks. (JosephASpadaro 07:49, 29 June 2007 (UTC))

It's memorized for no other reason than a sense of achievement and for showing off. There's a lot of memory techniques, one is to say walk around your house and memorize a short segment for each piece of furniture. If you can remember a few numbers each time you think of a piece of furniture, you can remember the whole length, because you'r intimately familiar with what furniture is next to each other. EverGreg 08:55, 29 June 2007 (UTC)

It is best to avoid terms like "no other reason than...", as every person may have his or her own reasons for memorizing digits of the number. Another reason I can think of is simply appreciation for the importance of the number, combined with the belief that its essence is embodied in its sequence of decimal digits - and thus that knowing more digits is represents more intimate knowledge about the number. Of course, the reasoning behind this motivation is questionable, as the essence of the number is in its exact value and not in any finite sequence of digits in an arbitrary base, but it is a possible motivation nonetheless.

As for your other question - I'll take a wild guess that the total of all the phone numbers you know by heart easily exceeds a hunderd digits. Adding other information you know by heart, not necessarily numeric, can amount to the equivalent of tens of thousand of digits. And if we include all the information that is stored in your brain - pictorial and the like - my guess is that it can be equivalent to billions of digits. So there's no question that the human mind is more than capable to store a large number of digits. Additionally, some people are better at remembering imprecise information like images, while others excel at recalling exact data like digits - and of course, some people just have a signficantly better memory than others. You can guess which kind the people who memorize myriads of digits of pi are. -- Meni Rosenfeld (talk) 11:17, 29 June 2007 (UTC)