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July 12

The Non-Zero Difference Puzzle

Here is an interesting puzzle. I am just curious as to how, using logic, can one solve this puzzle. I have solved it using the brute force method, exactly what the author said NOT to do. But how can it be solved analytically?--A Real Kaiser (talk) 00:10, 12 July 2008 (UTC)

It seems to me like a skewed version of Pascal's triangle. Maybe start there. --Prestidigitator (talk) 05:55, 12 July 2008 (UTC)

Here's a start: Each entry in the top row is required to be between 1 and 9 (inclusive). The maximum possible difference between any pair of those entries is 8, therefore the maximum possible entry in the second row is 8. Each entry in the second row is between 1 and 8. The maximum possible difference between any pair of them is 7, therefore the maximum possible entry in the third row is 7. Keep applying that logic, and you'll find that the maximum allowable number in the bottom row is 1. So fill the bottom 2 boxes in with 1's.

The row above "1 1" must be either "1 2 1" or "2 1 2". From here, trial-and-error is the best method I can come up with. Try the "2 1 2" and work your way up, and you'll find it doesn't work. A useful rule for reducing the search space is: when you have an N in the Nth row from the bottom, its 2 neighbors above must be an "N+1" and a "1", since N+1 is the maximum allowable number in the (N+1)st row from the bottom. So if you use "2 1 2", both 2's must have "3 1" or "1 3" over them. That means the next row up is "1 3 1 3", "1 3 3 1", "3 1 1 3", or "3 1 3 1", none of which match the "1" in between those 2's. So the second row from the bottom must be "1 2 1". --tcsetattr (talk / contribs) 07:01, 12 July 2008 (UTC)

Using that method actually doesn't lead to that big of a tree; many possibilities cancel out quickly. Until the last few rows, you can ignore reverses. Also, because of the planted threes, the highest number in the row will not be on the either edge after the bottom row. If you want to see the answer, the spoiler is in the source.

Integration question

Can anyone help with integrating this?

$\int {\frac {u^{a}}{1+bu}}\ du$

Here, a is a complex number (with real part -1, if that helps; the imaginary part is a parameter), and b is a real number. Mathematica's online integrator was not much help (http://integrals.wolfram.com/index.jsp?expr=(x^a)%2F(1+%2B+b+*+x)&random=false). I suppose it probably doesn't have a closed form integral? Eric. 85.178.19.137 (talk) 18:39, 12 July 2008 (UTC)

Is the Hypergeometric series form no use at all? just out of curiosity was it for anyhting specific? would the bounds be anything?87.102.86.73 (talk) 21:13, 12 July 2008 (UTC)

Thanks, I took another look at that and see it wasn't quite as bad as I feared. This integral arose in calculating the location of a vehicle traveling at constant acceleration and turning rate, undergoing drag (which occurs for this year's ICFP contest, www.icfpcontest.org). I was hoping that a sufficiently nice result for the integral would give some insight in optimizing vehicle navigation, but unfortunately not. (The lower bound is zero, and the upper bound is another parameter, so I just left it is an indefinite integral.) Eric. 85.178.58.188 (talk) 09:48, 13 July 2008 (UTC)

If the integration limits are zero and infinity then put bu= x and use that:

$\int _{0}^{\infty }{\frac {x^{-p}}{1+x}}dx={\frac {\pi }{\sin \left(\pi p\right)}}$ Count Iblis (talk) 23:07, 12 July 2008 (UTC)

How do I compute the odds to see if any National Football League franchises are going to the Super Bowl this year?

When's the draft? That should help. Otherwise, how do Las Vegas oddsmakers compute sports odds? Are there people who watch and take into account everything?. ("The quarterback just drank a Gatorade and put on high heels, -there goes another point against his team..." etc.)--Baseball and and and Popcorn Fanatic (talk) 19:30, 12 July 2008 (UTC)

I think the main factor in working out the odds are how people are betting. If lots of people are betting for a particular team, that suggests they are more likely to win, so you lower the odds, and vice versa. I'm not sure to what extent the bookmakers use knowledge of the sport to try and work out for themselves who's likely to win - quite possibly not at all. --Tango (talk) 21:07, 12 July 2008 (UTC)

Are you talking about something like ranking of sports team? Because if we rank them, then the top few (the best teams) will go to super bowl. If this is what you talking about, then I did a presentation on it (different ranking schemes) some time ago. I can explain more in detail if you want.--A Real Kaiser (talk) 01:39, 13 July 2008 (UTC)

What I'm getting at is whether it's horses, football, baseball, etc. sports oddmakers are remarkably accurate.I want to impress my friends and bocome RICH! How do I achieve this skill? —Preceding unsigned comment added by Baseball and and and Popcorn Fanatic (talk • contribs) 06:34, 13 July 2008 (UTC)

Unfortunately, probabilities alone won't solve this problem. A typical probabilities approach might be to figure out which portion of the time a given team has won in the past, and use that to predict the future expected probability of the event. However, as the make-up of each team changes each year (and even within a year due to injuries, etc.), such a method is rather unreliable here. Instead, you need to basically use knowledge of the sport to figure out who is likely to win games. Things like "pitcher A has a good south-paw pitch, which should be effective against hitter B, but not hitters C or D; therefore, I predict that pitcher A's team will lose". There's a lot more to it than just that, of course, but that's a simple example. The first step, I would imagine, is to get to know the game or sport inside and out. StuRat (talk) 15:19, 13 July 2008 (UTC)

The reason bookmakers make so much money is because they offer worse odds than are realistic - add together the probabilities of each horse in a race winning and you'll find it adds up to more than 1. There's quite a large margin of error in which they still average a profit. --Tango (talk) 16:58, 13 July 2008 (UTC)

July 13

Powers

Hello. If 60^a = 3 and 60^b = 5, then why does $12^{\frac {1-a-b}{2(1-b)}}=2$? Thanks in advance. --Mayfare (talk) 02:23, 13 July 2008 (UTC)

The reason for that is because

$60^{a}=3\Rightarrow a={\frac {\ln 3}{\ln 60}}$

$60^{b}=5\Rightarrow b={\frac {\ln 5}{\ln 60}}$

by taking the natural log of both sides, using their properties and then dividing. After that you just plug in these values of a and b into $12^{\frac {1-a-b}{2(1-b)}}$ and simplification (making extensive use of properties of logs) will give you 2. Is there a particular step that is giving you trouble?--A Real Kaiser (talk) 02:48, 13 July 2008 (UTC)

That'll get you to the answer, but it looks neater if you work with $a={\frac {\log _{12}3}{\log _{12}60}}$ and $b={\frac {\log _{12}5}{\log _{12}60}}$. Also, simplify the rational expression ${\frac {1-a-b}{2(1-b)}}$ first so you don't have to substitute b in 2 places. --tcsetattr (talk / contribs) 06:05, 13 July 2008 (UTC)

Question about baseballs and wind resistance

If x were to represent velocity, angle, and wind resistance, would it travel in a standard trajectory? My understanding is no.--Baseball and and and Popcorn Fanatic (talk) 06:28, 13 July 2008 (UTC)

Are you asking whether a baseball that travels with a non-zero velocity at a non-zero angle in air would follow a symmetrical path? If so, the answer is indeed "no". Here's an analysis of the ball's trajectory:

(1) At the moment the ball is hit, it is both moving vertically and horizontally. Both the horizontal and vertical speeds decrease due to air resistance.

(2) The vertical speed eventually becomes 0 and the ball begins to accelerate in its fall until its terminal velocity is reached, if it ever is reached. It will stop when it hits the ground. The acceleration while the ball was climbing was -9.8 m/s^2 plus the acceleration applied by air resistance; the acceleration during the fall is 9.8 m/s^2 minus the acceleration applied by air resistance. As you can see, the ball takes longer to reach the ground from its maximum point, and therefore end its movement, then it took to reach the maximum point.

(3) The horizontal speed of the ball continues to decrease until the ball hits the ground. Since both the horizontal speed and the vertical speed decrease, the path might be parabolic under certain circumstances, but I'm unsure about whether this is true. --Bowlhover (talk) 07:25, 13 July 2008 (UTC)

I think Baseball and Popcorn fanatic was not asking about symmetry, but about consistency. That is to say, if you through again with the same velocity and angle, would you get the same results? There are several factors effecting the ball's motion: initial velocity and angle, spin, wind, altitude. All these effect the ball's trajectory. The more of these that are accounted for, the more close would be the theoretical trajectory to the actual trajectory. If we try to be very precise, then wind itself is not constant over time or space, hence rather complicated. So I think that the answer is that the motion is not really standard. Oded (talk) 08:30, 13 July 2008 (UTC)

There are some dynamical systems that exhibit sensitivity to initial data. This is sometimes called the butterfly effect. The meaning of this is that event if you now all the parameters (in this case, the wind, initial velocity, spin, etc) but you knew them only up to some finite accuracy (which is practically always the case), you could not predict the resulting motion. I would guess that the motion of a baseball through the air is to a high degree of accuracy not sensitive to initial data, and does not suffer from this effect. But I might be wrong. Is there an expert out there who might know if that is correct? Oded (talk) 17:39, 13 July 2008 (UTC)

Further problems with exponential growth model

I've tried modelling Wikipedia's growth with two different exponential functions, but both functions are inappropriate for my use. I've tried an exponential of a quadratic function: $e^{ax^{2}+bx+c}=y$ and a quadratic of an exponential function: $ae^{2x}+be^{x}+c=y$ but both functions have the quadratic coefficient as negative, meaning the curve turns downwards, and according to the models, Wikipedia will never reach five million articles.

Is there any other way I can have a function either in the form $e^{f(x,a,b,c)}=y$ or in the form $f(e^{x},a,b,c)=y$ so I could express x in terms of a, b, c and $\ln y$ and all parameters a, b and c would be independent of each other? JIP | Talk 09:20, 13 July 2008 (UTC)

I would question the assumption that Wikipedia's growth is (or will permanently remain) exponential. I'd expect the rate of growth to slow once these events occur:

1) We already have articles for most topics in which people have an interest.

2) Most people interested in participating in Wikipedia have already done so, and perhaps they've even got bored and moved on.

So, trying to model the growth with the wrong type of equations would then lead to poor results. StuRat (talk) 14:25, 13 July 2008 (UTC)

For e^f(x) you want df(x)/dx to tend to zero as x tends to infinity? there are so many functions I wouldn't know where to start (maybe f(x) = ax/(bx+c) or variations of that)

Once again I suggest looking at the logistic function which is derived from dA/dt = kA(A_max-A) where A=articles..87.102.86.73 (talk) 14:31, 13 July 2008 (UTC)

I tend to agree that Wikipedia growth will probably not be exponential. It was an idea worth considering anyway. I abandoned the idea of exponential growth and tried a simple quadratic model: $ax^{2}+bx+c=y$ but it suffered from the same problem: the curve turns downwards, and the apex of the curve is below five million, so according to the model, Wikipedia will never reach five million articles. I tried formulas ${\frac {ax}{bx+c}}=y$ and ${\frac {ax+b}{cx}}=y$, but they are equivalent to $ax-ybx=cy$ and $ax-cyx=-b$ respectively, and the only possible solution to either of them is a=0, b=0, c=0. I think there has to be at least one term which has at least one of x and y but none of a, b, and c. I shall have to look into logistic functions but I do not understand enough of them yet. JIP | Talk 16:49, 13 July 2008 (UTC)

Why not try y=e^ax/(bx+c)? I think that could give a good fit..

By the way the logistic function is quite simple - simpler than the article makes it seem.. Can you solve equations of the form dy/dx=kx to show that y=e^kx? If so you should have no problem with the logistic function.

The idea behind the logistic function (in this case) is that

a. more articles mean more readers

b. more readers means more editors

c. more editors means more articles

d. But. it's assumed that there is an absolute limit on how many articles there can be - this is the origin of the (A_maxM-A) term - meaning that the chances of an editor being able to write a new article is proportional to (A_maxM-A)/A_maxM .87.102.86.73 (talk) 16:58, 13 July 2008 (UTC)

$y=e^{\frac {ax}{bx+c}}$ won't work. If $y=f(x,a,b,c)$ already gives a degenerate solution, then $y=e^{f(x,a,b,c)}$ isn't going to make it any better, because it's equivalent to $\ln y=f(x,a,b,c)$. But thanks for the information about logistic functions, I'll have to see what I can come up with. JIP | Talk 17:18, 13 July 2008 (UTC)

I'm not sure you're quite right, or maybe I haven't understood properly?

you say "...they are equivalent to ax − ybx = cy and ax − cyx = − b respectively, and the only possible solution to either of them is a=0, b=0, c=0" but if y=1 when x=0 then ln(y)=0 when x=0 so the equation becomes ax - bln(y) = cln(y) or a.0-b.0=c.0 which means that a,b,c can be anything from the initial conditions? Also the limit is y=e^a/b at infinite x (time). Is this not good?. I would ignore the behaviour when x<0 because then wikipedia does not exist??87.102.86.73 (talk) 17:41, 13 July 2008 (UTC)

As the size of wikipedia, y, is nonnegative, choose a growth model that gives nonnegative values only. So look at log y. At time x=0 when wp was created the size was zero, and in the far future the size of wp may be limited by some value. So try the function log(y)=a−b/x. Plot log(y) as a function of 1/x. Fit a and b. Then y=e^a−b/x is your growth model. Bo Jacoby (talk) 10:28, 14 July 2008 (UTC).

Thank you, this is definitely a model worth considering. However, I would want to have three independent other parameters instead of just two. But no matter how I try to disguise a quadratic equation ($e^{ax+b+{\frac {c}{x}}}=y$ or $e^{a+{\frac {b}{x}}+{\frac {c}{x^{2}}}}=y$), the discriminant always becomes negative before five million. I don't think functions involving a quadratic function of x are going to work. But I don't know how else to write a function of x with three independent other parameters. My first thought was $e^{a+{\frac {b}{x+c}}}=y$ but then I realised that the equivalent non-exponential formula, $a+{\frac {b}{x+c}}=\ln y$, isn't linear in terms of a, b, and c any more, and thus the linear equation system solver can't solve it any more, and it would be far too much trouble to solve by hand. The linear system was enough trouble as it was. JIP | Talk 18:40, 14 July 2008 (UTC)

In your formula, $a+{\frac {b}{x+c}}=\ln y$, the parameter c has the meaning that x₀=−c is the point of time when y=0. Note that b<0 in order to obtain a steady growth of y from y₀=0 to y_max=e^a when x goes from x₀ to infinity. Setting k=−b the formula is $y={y_{\max }}\cdot e^{-{\frac {k}{x-x_{0}}}}.$ Bo Jacoby (talk) 07:29, 15 July 2008 (UTC).

How to solve the nonlinear equations? Look. Suppose you have three observations, (x₁,y₁), (x₂,y₂), and (x₃,y₃). Then the three unknown parameters (x₀,y_max,k) satisfy the three equations

$y_{i}={y_{\max }}\cdot e^{-{\frac {k}{x_{i}-x_{0}}}}$ for i=1,2,3.

Conventionally, unknowns are called x and y and z and constants are called a and b. So rename the variables: x_i→a_i, ln y_i→b_i, x₀→x, ln(y_max)→y, k→z. The equations look like this:

$b_{i}=y-{\frac {z}{a_{i}-x}}.$

Getting rid of the fraction produces three polynomial equations

$a_{1}b_{1}=-xy+b_{1}x+a_{1}y-z$

$a_{2}b_{2}=-xy+b_{2}x+a_{2}y-z$

$a_{3}b_{3}=-xy+b_{3}x+a_{3}y-z$

Subtracting the first equation from the two other equations gives two linear equations

$(b_{2}-b_{1})x+(a_{2}-a_{1})y=a_{2}b_{2}-a_{1}b_{1}$

$(b_{3}-b_{1})x+(a_{3}-a_{1})y=a_{3}b_{3}-a_{1}b_{1}$

which are solved by the standard method. Substituting the solutions x and y into one of the nonlinear equations gives the last unknown z:

$z=-xy+b_{1}x+a_{1}y-a_{1}b_{1}$

Bo Jacoby (talk) 11:42, 18 July 2008 (UTC).

optimum algorithm for choosing from sequential lineup

I recall once hearing a math puzzle. I think it was framed in terms of the prince of a kingdom needing to choose a bride for his princess. He lines up all the eligible women in his kingdom, and starts going down the line, one at a time. Each woman he looks at, he may choose her for his bride. But if he does not choose her, he must move down the line, and may not come back to her, even if she turns out to be the best choice. So when discarding a choice, he must not only weigh how suitable she is, but also how likely he thinks it is that a better choice will come along. Assume the suitability of the women form a total order.

I don't recall where I heard this puzzle, nor whether I've got the framing correct; neither whether I've described the prince's quandary correctly, nor whether the puzzle actually uses a line of brides for illustration. I also don't recall what the solution was. I seem to think there is some optimum algorithm for the prince. Sounds something like this: he should look at 1/e percent of the choices, then choose the first one that's better than the 1/e'th one. Or something like that.

I can't come up with a google query that finds it for me, but that may be because I'm misremembering the specifics. So does it sound familiar to anyone here? Can you set me straight? -lethe ^{talk +} 16:12, 13 July 2008 (UTC)

Just a guess but if the attractiveness forms a linear order eg one princess has attractiveness 1, another 2 etc, and the prince knows how many princesses there are it should be easy to pick one in the top 3...87.102.86.73 (talk) 16:50, 13 July 2008 (UTC)

This is the secretary problem. Michael Slone (talk) 17:52, 13 July 2008 (UTC)

Yep, that's what I was looking for. Thanks, Michael! 97.113.58.212 (talk) 19:36, 13 July 2008 (UTC)

I've encountered this problem is real life when getting gasoline while driving an unfamiliar route. I have to look at several gas stations to get an idea for what the range of prices are, but must choose one before I run out of gas. I could technically go back to a previously passed station to get gas, but that would waste both time and gas, so usually isn't an option. StuRat (talk) 18:02, 13 July 2008 (UTC)

You might have read it in the excellent "Fifty Challenging Problems in Probability With Solutions" by Frederick Mosteller, dirt cheap as it's in Dover. There the problem is stated pretty much as you described it, and the description of the solution is very good (better than the one in the wiki article for the secretary problem). 86.15.141.111 (talk) 10:49, 14 July 2008 (UTC)

Growing Numbers

Someone posed this question to me and I have solved it by writing a program. The problem is this. To "grow" a number is to take a two digit number and add its digits to itself until a three digit number is obtained. For example, 90 and 75 both take two steps to grow because
90+9+0=99 -> 99+9+9=117
75+7+5=87 -> 87+8+7=102.
The question is what is the difference between the largest and the smallest numbers which take exactly three steps to grow? I am just looking for hints as to how to do it analytically. I already have the answer. Thanks people!--A Real Kaiser...NOT! (talk) 04:55, 14 July 2008 (UTC)

I got 65 and 81 for my answers by writing an integer program (where each variable is a digit of some growth):

max: 10*x1 + x2
s.t. 11*x1 + 2*x2 - 10*x3 - x4 = 0
     11*x3 + 2*x4 - 10*x5 - x6 = 0
     10*x5 + x6 <= 99
     11*x5 + 2*x6 >= 100
     0 <= xi <= 9
     xi integer

Maple and many spreadsheet programs will solve such programs.

maple code
maple code
with(Optimization); LPSolve(10*x1 + x2, {11*x1 + 2*x2 - 10*x3 - x4 = 0, 11*x3 + 2*x4 - 10*x5 - x6 = 0, 10*x5 + x6 <= 99, 11*x5 + 2*x6 >= 100, x1 <= 9, x2 <= 9, x3 <= 9, x4 <= 9, x5 <= 9, x6 <= 9}, assume={nonnegative, integer}); [65, [x1 = 6, x2 = 5, x3 = 7, x4 = 6, x5 = 8, x6 = 9]] LPSolve(-10*x1 - x2, {11*x1 + 2*x2 - 10*x3 - x4 = 0, 11*x3 + 2*x4 - 10*x5 - x6 = 0, 10*x5 + x6 <= 99, 11*x5 + 2*x6 >= 100, x1 <= 9, x2 <= 9, x3 <= 9, x4 <= 9, x5 <= 9, x6 <= 9}, assume={nonnegative,integer}); [-81, [x1 = 8, x2 = 1, x3 = 9, x4 = 0, x5 = 9, x6 = 9]]

Often solving analytically just means reducing the problem to a form with some known method of solution, such as an integer program. Compare to using the quadratic formula or writing x = A⁻¹b. -- KathrynLybarger (talk) 05:50, 14 July 2008 (UTC)

Well, this is how I solved the problem except that I used MATLAB. But I was just wondering if I was sitting on my desk without a calculator or a computer with just a pencil and a sheet of paper, how would I figure it out in a few minutes. I think I got it now. I just needed to spend some more time thinking on this problem.--A Real Kaiser...NOT! (talk) 22:04, 14 July 2008 (UTC)

Wright Omega function

Hello,

I have recently created the article about the Wright Omega function, which appears for example in the resolution of x = ln(x).

A definition of the function is :

$\omega (z)=W_{{\big \lceil }{\frac {\mathrm {Im} (z)-\pi }{2\pi }}{\big \rceil }}(e^{z})$

where W_k is the k-th branch of the Lambert W function.

If you look at a few graphs that are on that article, you may notice it looks like there is only one discontinuity, at x < 0, y = ±π, which corresponds to changing the branch. But then most other branch-changes are seamless, as illustrated in this image (computed with the Lambert W definition) : http://s4.tinypic.com/r0bdzo.jpg

(Red is W₀, Blue is W₁, etc...) (Also see http://i34.tinypic.com/2lu8lcw.png which shows how the functions only fit precisely in those intervals of width 2π, by showing them functions in slightly expanded intervals)

I would like to know how come all these branch-changes are seamless, how did the person who first discovered this function realise it was possible to stitch together the different branches of the Lambert W function to give this function ?

Also, does anyone have any references concerning this function (preferably freely available on internet), as I was not able to find much except from a few books.

Thanks. --Xedi^Talk 22:30, 13 July 2008 (UTC)

July 14

Vector fields on the torus

Let's consider vector fields on surfaces with unitary vectors. Is it possible that there is only one such vector field without critical points on the torus up to diffeomorphic coordinate changes? (That means there are not two topologically distinct vector fields.)--Pokipsy76 (talk) 10:54, 14 July 2008 (UTC)

There are three possible interpretations of the phrase "up to diffeomorphic coordinate change" that I can think of. In order to explain this, it is perhaps better to first introduce some terminology (which will also give you some pointers for further reading). Your vector field is a section of the unit tangent bundle. The group of diffeomorphisms of the surface acts naturally on the unit tangent bundle. So the first interpretation of your question would be that you mean up to such diffeomorphisms. More generally, there are diffeomorphisms of the unit tangent bundle that preserve the structure of the fibre bundle, in the sense that they take fibers to fibers. In other words, to each such diffeomorphism there is an associated diffeomorphism of the surface such that there is a commutative diagram involving these two diffeomorphisms and the projection maps of the fiber bundles. Finally, there are just the self-diffeomorphisms of the fiber bundle thought of as a differential manifold without additional structure. I believe the answers are no for the first interpretation, and yes to the other two. But I will not elaborate until I know which interpretation you are actually interested in. I should say that I'm not an expert on topology in general and fiber bundles in particular, but I have some background. I hope I have not discouraged you with all this terminology. I'll be happy to try to help if it causes too much grief and the links I have provided are not very helpful. Oded (talk) 15:12, 14 July 2008 (UTC)

What I had in mind is this: a diffeomorphism from the torus to itself induces a transformation of the vector field acting by multiplication for the differential

$(x,v)\mapsto (\phi (x),D\phi (x)v)$

this is what happens if the tangent vector of a curve is mapped to the tangent vector of the image curve.--Pokipsy76 (talk) 21:26, 14 July 2008 (UTC)

In that case, I think the answer is no. The diffeomorphism would take the flow curves of one vector field to the flow lines of the other. If you look at the vector field ${\bigl (}\cos(2\pi x),\sin(2\pi x){\bigr )}$ on the torus $\mathbb {R} ^{2}/\mathbb {Z} ^{2}$, then you can see that most of its flow lines will not be closed. On the other hand, the flow lines of the vector field $(1,0)$ are all closed. Oded (talk) 21:57, 14 July 2008 (UTC)

Manhattan distance

File:Manhattan.png

In the picture at right, as the number of steps N increases, the distance along that route from A to B stays the same, per Manhattan distance. As N approaches infinity, the distance stays the same, but the route is indistinguishable from a straight line from A to B, which is obviously shorter than the Manhattan distance (${\sqrt {2}}$ vs. 2 for a unit square, say). How does one explain this unintuitive result? Thanks. --Sean 14:51, 14 July 2008 (UTC)

I think that's covered in the linked article, which redirects to Taxicab geometry - it's not Euclidean geometry. In Taxicab geometry the straight line also has a length 2. --LarryMac | Talk 15:09, 14 July 2008 (UTC)

More generally, "length" is not a continuous function on curves. Suppose that you need to walk 10 miles on a road. If every few steps forward you decide to cross the road, you will need to walk much more than 10 miles, but if you measure the road itself it will appear as if you have traveled just 10 miles. Does this make more sense now? Oded (talk) 15:27, 14 July 2008 (UTC)

There was at least one previous discussion of this "paradox" on this desk.. the "user:froth" asked and I seem to remember it getting a bit argumentative (probably my fault)..

However should you wish to read it then Wikipedia:Reference_desk/Archives/Mathematics/2008_March_5#Equilateral right triangle / shortest distance between points - as you will see however the answers consists of mostly me arguing with "user:tango"...

Still it might help...87.102.86.73 (talk) 16:50, 14 July 2008 (UTC)

That did get rather heated, didn't it? No hard feelings? :-) --Tango (talk) 17:34, 14 July 2008 (UTC)

Of course not - this used to happen to me a lot - but I'm trying to cut down... Something about the internet being impersonal - I've also made contributions to Wikipedia:Lamest edit wars.. (not saying which) do I get a barnstar?87.102.86.73 (talk) 17:53, 14 July 2008 (UTC)

There's also this previous discussion Wikipedia:Reference_desk/Archives/Mathematics/2008_February_3#Distance —Preceding unsigned comment added by 87.102.86.73 (talk) 16:52, 14 July 2008 (UTC) which should be more helpful.87.102.86.73 (talk) 16:54, 14 July 2008 (UTC)

I think the bottom line is that the convergence of the stepped line to the straight line isn't uniform which allows strange things to happen. I don't fully understand it myself, so I won't try and explain further. --Tango (talk) 17:34, 14 July 2008 (UTC)

Sorry, Tango, but the convergence is uniform when the curves are parameterized properly. The point is that the convergence of the paths is not really relevant here, since length is not a continuous function on curves. Think, for example, of the function f that is 1 on the irrationals and 0 on the rationals. You have a sequence of irrationals x_i that converges to $1$. Then f(x_i)=1, x_i converges to zero, but f(0)=0. It is just a fact of life, not all functions are continuous. (That's not completely the end of the story, there are stronger notions of convergence of paths for which length is continuous, but that's probably not going to be helpful to discuss them presently.) Oded (talk) 17:48, 14 July 2008 (UTC)

I can recall three separate occasions on which this same question has been asked. Isn't it only logical that someone (who knows a lot more than me) create a separate article for this counterintuitive result? And from then on, just refer people to that page instead of linking, typing, and arguing everytime.--A Real Kaiser...NOT! (talk) 21:59, 14 July 2008 (UTC)

Yes please! --Tango (talk) 23:03, 14 July 2008 (UTC)

So I understand it even less than I thought I did... great! --Tango (talk) 23:03, 14 July 2008 (UTC)

This seems somewhat related to the issue of not being able to precisely define the length of a coastline, since the more detail you include, the longer the coastline gets. [[1]] (talk) 22:15, 14 July 2008 (UTC)

We've got an article on that, but this case has the opposite outcome; it stays the same! --Sean 23:31, 14 July 2008 (UTC)

Thanks for the responses so far, folks, but I forgot to mention that I'm only 8 years old, and while I can understand the ideas of driving taxi-cab style from A to B and driving straight from A to B, I don't understand how just declaring that these real-world operations are in a different geometry or that length is not continuous on the curves (in my rectilinear diagram!) resolve the paradox. Would it be possible for someone to put it in terms as simple as the problem statement? Thanks. --Sean 23:31, 14 July 2008 (UTC)

In simplest possible terms, then: Why should the length of the 'taxicab route' have anything to do with the length of the straight line? Less simply: there's really no connection. I could give you a route as close as you wanted to the straight line that was infinitely long. Algebraist 23:41, 14 July 2008 (UTC)

And while I'm here: mathematicians' use of the word 'curve' is rather perverse. It certainly doesn't need to be curved. Algebraist 23:44, 14 July 2008 (UTC)

You can boil it down to simple arithmatic

say to have N steps - so each step goes L/N right and L/N up (the box is L by L)

So the distance for eachs step is 2L/N

It takes N steps, so the distance is (2L/N) x N =2L

ie it's independent of N - the number of steps.

Obviously by eye it looks like as N becomes big that the line approximates a diagonal with length sqrt(2) x L

But the structure is still there; because we originally said it was- it just becomes so small that it can't be seen or accurately described (when N tends to infinity)...

In other words no matter how small they are if we say the path is composed of up/right rather than diagonal then it always remains up/right..87.102.86.73 (talk) 23:50, 14 July 2008 (UTC)

Well... as you saw, for any finite number of steps, the distance remains the same as 2 (for the unit square). However, as mentioned, this uses a different metric of measurement - that of distance = change in x + change in y.

Now, suppose that we took a 2-step path. We can change this into a one step path by "pushing out" the indentation. This preserves the distance, as the overall change in x and change in y is unaffected. Thus, we can transform an n-step path to a n-1 step path and vice versa. Thus, repeating this, even if we had an infinite number of steps, we could push out each one, and transform it into a one-step path. Thus, the distance with infinite steps is the same as a single step.

But, this manhattan metric is not what we use normally to consider length. We use the Pythagorean Theorem - that square of the direct distance is equal to the sum of the squares of the perpendicular sides. -mattbuck (Talk) 23:59, 14 July 2008 (UTC)

Let me attempt to fuel the fire a little bit more. I am going to assume that I am working in a Euclidean plane using ONLY the Euclidean metric to measure distances. We are not using the taxi-cab metric. My goal is to get from point a to point b where a and b are the diagonally opposite vertices of a rectangle with dimensions A and B. I will only be walking in straight lines starting from a. Let n be the number of right-angle turns I make in my journey from point a to point b and define f(n) to be the total distanace I have traveled from a to b making n turns. f(1)=A+B because starting from a, I walk along one edge, make a right turn, and walk along the other edge to get to b. Since the lengths of the two edges (independent of the order) are A and B, I have traveled a total distance of A+B. Remember that this distance is measured in the Euclidean norm. The values of f(2)=f(3)=f(4)=f(5)=...=f(k)=A+B for any integer k because the total horizontal distance I traveled and the total vertical I traveled will be the same so the total distance from a to b will always be A+B. Now the question is why does $f(\infty )={\sqrt {A^{2}+B^{2}}}$ while $\lim _{n\rightarrow \infty }f(n)=A+B$? These two quantities are obviously not the same (unless A or B is zero). The $\lim _{n\rightarrow \infty }f(n)=A+B$ because f(n)=A+B for all n. But why does $\lim _{n\rightarrow \infty }f(n)\neq f(\infty )$. Can someone please explain that? We have a constant sequence which is converging to some other number than the values of the terms inthe sequence. Remember, to measure all distances, I am using Euclidean metric, NOT the taxi-cab metric. I feel your pain Tango. I thought I had it down too but I don't think so.--A Real Kaiser...NOT! (talk) 01:07, 15 July 2008 (UTC)

But $f(\infty )\neq {\sqrt {2}}$, it is equal to 2. Unless you move up and along simultaneously, which means the function no longer applies, you move in discrete (but infinitessimal) distances up and along an infinite number of times. The root 2 is the displacement, NOT the distance. -mattbuck (Talk) 01:20, 15 July 2008 (UTC)

In the limit, you have a straight line. The length of a straight line is what it is, regardless of how you constructed the line. A curve is just a set of points, nothing else matters (you probably need to parametrise it to calculate the length, but I'm pretty sure the length is independent of the choice of parametrisation - if it's not, then this is even weirder that I thought!). This example shows that f is not a continuous function (on the extended real line - it's continuous at any finite point, since it's constant at all finite points, it's only at infinity that things go wrong). Why that happens, I really don't know, but it's clear that it does. --Tango (talk) 01:50, 15 July 2008 (UTC)

Surely in the n=infty case, you should be taking infinitely many right-angle turns? It's easy to do this without infinitesimal movements (go up and right 2⁻ⁿ on the nth leg or whatever) and the total distance travelled is 2. More ontopic: I think this is the kind of thing that would have worried 17th (and possibly 18th) century mathematicians, with their naive expectation that behaviour at the limit would nicely reflect finite behaviour. Now we're used to the idea that lots of things (in this case the length operator) don't commute with limits, we shouldn't be very surprised. Algebraist 10:11, 15 July 2008 (UTC)

Well, yes, I was assuming each turn was the same size, without that assumption there is nothing strange going on at all. --Tango (talk) 17:01, 15 July 2008 (UTC)

For everyday use, the length of a path is measured simply by a number of steps sufficient to walk that path. Taking shorter steps, x, you expect the number of steps, N(x), to be greater, so the total length of the path is computed as the length of the step times the number of steps, L(x) = x·N(x). Still L(x) does depend a little on the length of each step, because long steps cut corners. Mathematicians use tiny steps. The length of a smooth curve, like a circle arc, does not depend much on the length of the step when the step is tiny compared to the radius of curvature.

L = lim_x→0L(x) = lim_x→0(x·N(x)).

In the example of a zig-zag curve, there is a second number to be taken into account: the distances between corners, = y. The number of steps, N(x,y) and the distance L(x,y)=x·N(x,y) depend on both. If you take the limit of tiny steplength first then you get the Manhattan distance,

lim_y→0(lim_x→0L(x,y)) = lim_y→0(L_Manhattan) = L_Manhattan

because tiny steps do not cut corners. If you take the limit of tiny cornerdistance first then you get the Euclidean distance,

lim_x→0(lim_y→0L(x,y)) = L_Euclid,

because finite steps cut sufficiently small corners. So the moral is that you cannot always switch the order of taking limiting values.

lim_x→0(lim_y→0L(x,y)) is not necessarily equal to lim_y→0(lim_x→0L(x,y)).

and the intuitive distance lim_{(x,y)→(0,0)}L(x,y) is not well defined. Bo Jacoby (talk) 13:32, 20 July 2008 (UTC).

Continuous distance

If one defines a sequence of functions f_n: [0,1] → R² by f_n(t) = ( t, sin(n*t*π)/(n*π) ), then | f_n(t) - f_m(t) | ≤ 1/min(n,m) for all t, and indeed, | f_n(t) - (t,0) | ≤ 1/n for all t. Hence f_n approaches the function t -> (t,0) uniformly in t. Every f_n is smooth.

The arclength of f_n is the integral from t=0 to 1 of sqrt( 1 + f'_n(t)^2 ), which is the integral of sqrt(1+cos(n*t*π)^2) from t=0 to 1. I believe this integral is constant in n, and strictly larger than 1.

However, the limit function t → ( t, 0 ) clearly has arclength 1.

Hence "arclength" as a function from the space of smooth functions from [0,1]→ R² to the nonnegative reals is not a continuous function when the domain is given the topology of uniform convergence.

However, the derivative of f_n is ( 1, cos(n*t*π) ) which does not converge to the derivative of t → ( t, 0 ), so there is no reason to expect anything defined in terms of derivatives to be continuous. In particular, there is no reason to expect arclength to be continuous. To me at least, this is a typical reason to use sobolev spaces (though with p=∞, if you don't mind a little extra continuity, the higher hölder spaces might be simpler).

Question: Is $\int _{0}^{1}{\sqrt {1+\cos(nt\pi )^{2}}}~dt$ constant in n? Does the constant have a name? Is there a reference for this?

BTW: In case someone dislikes the integral, the same conclusion is reached in a brutish way by using g_n(t) = ( t, sin(n*n*t)/n ). Here the arclength diverges to infinity, rather than remaining constant and inequal to the arclength of the limit function. JackSchmidt (talk) 19:42, 14 July 2008 (UTC)

$\int _{0}^{1}{\sqrt {1+\cos(nt\pi )^{2}}}~dt={\frac {2{\sqrt {2}}{\text{EllipticE}}\left[{\frac {1}{2}}\right]}{\pi }}$

Integrals are not something I do for fun. For this, I use Mathematica. Mathematica can evaluate the integral symbolically if n=1,2,3,100 (and gets the same answer, though the calculation for 100 takes longer). So there is some knowlege out there which can evaluate the integral at least when n is integer. But there is really no need to evaluate the integral to see that this is constant when n is integer. You can look at it geometrically. If you scale the path by a factor of 1/2 and put two of these one after the other, you go from n to 2n and the length does not change. The same argument shows that you can go from n=1 to n=k without changing the length. If you do not believe in geometry (I am sure that you do), you can get the same result by substitution. You make the substitution u=n t, and then use the periodicity of cosine. Oded (talk) 21:15, 14 July 2008 (UTC)

Thanks! The geometric reason is very clear. I tried the simple substitution before, but didn't change the limits of integration... oopsy. The periodicity of cosine just rewords your geometric argument, so I like it: n copies of 1/n'th the length. It also clear both geometrically and basic inequality pushing that the constant is strictly larger than 1, so this solves my problem (self-contained example of arclength not continuous, but not divergent). JackSchmidt (talk) 22:02, 14 July 2008 (UTC)

I played some more and Mathematica can evaluate the indefinite integral:

$\int {\sqrt {1+\cos ^{2}t}}~dt={\sqrt {2}}{\text{EllipticE}}\left[t,{\frac {1}{2}}\right].$

Therefore, all that remains for the human to do in order to verify the above equality (if we really care to) is to look up the definition of the ellipticE function and verify by differentiation. Oded (talk) 21:21, 14 July 2008 (UTC)

I leave it to the interested reader :) Your simple proof of constancy is enough for me. JackSchmidt (talk) 22:02, 14 July 2008 (UTC)

I cannot help taking this opportunity to mention some related current research mathematics that I am excited about. Suppose that you use a (say smooth) curve $\gamma :[0,t]\to \mathbb {R} ^{2}$ do drive some differential equation. What does that mean? It means that you are looking at the solution $u:[0,1]\to \mathbb {R} ^{n}$ of a differential equation of the form $u'(t)=F{\bigl (}t,\gamma (t),\gamma '(t){\bigr )}$ for some smooth F. A nice example is that you have a sphere that you roll around on the plane while keeping the contact point on the curve, and your u consists of the position and rotation of the sphere. The map thus induced from u to $\gamma$ is not continuous with respect to the uniform convergence of paths. A British mathematician Terry Lyons had the idea to look at some extra structure on the path which I think is "less" than having the derivative but enough in order to solve all such equations. In this way, the enhanced path carries some structure with it (which I think of as something like a coordinate frame). One point is that you can take the limit of such enhanced curves where the resulting curve (after you forget the extra structure) is highly non-smooth, but still the differential equation makes sense. I don't know much beyond this. Oded (talk) 21:45, 14 July 2008 (UTC)

It sounds very interesting. I know very little about ODE, mostly only parabolic and elliptic 2nd order PDE (virtually no smoothness assumptions at least), and that was a while back. This sounds a lot like the typical gist of a PDE proof: solutions to nice equations carry some extra structure, complete the space they live in, and prove there is a solution to the not nice equation, then prove the resulting solution is a little better than you assumed. I like the idea that you can do this without even assuming the existence of a whole derivative. I guess for a path, you basically want the derivative to point in the direction of travel, so if you just had a consistent coordinate frame, you'd at least have some consistent notion of which way you were pointing. If you know of a survey article, I'd be interested to look through it. I doubt I could read a research article in ODE, but if that is all there is, it might be a good excuse to learn the basics of ODE. JackSchmidt (talk) 22:02, 14 July 2008 (UTC)

Terry's homepage has a link to his list of publications. I assume that you can find the best of what is available there. Here is a nice illustrative example to contemplate. Take your sequence of curves that converge to the straight line segment, take the straight line segment itself, and take another sequence of curves converging to it where the n'th cruve is periodic with period 1/n in the same way that your curves are (periodic is not the precise term here, but you know what I mean) but in each period makes a little clockwise loop of diameter of order 1/n. Then you roll the sphere on the curve in each of these examples, but the curve is so close to the line segment that you cannot visually tell the difference. The ending result of the rotation of the sphere at the end might be different. Thus, on the line segment itself one can put more than one type of these enhanced structures, though only one of these is compatible with its derivative. In that sense, I think it is a bit different than the standard thing that happens in the ODE PDE framework. No? Oded (talk) 22:18, 14 July 2008 (UTC)

It seems standard in very (very) broad strokes, but the actual structure being placed on the functions strikes me as quite new, and very natural. I'll let you know when I've had a chance to read through a few of the articles, but it might not be until the weekend. JackSchmidt (talk) 18:17, 15 July 2008 (UTC)

July 15

Shell/disk volume integrals different?

Something is confusing me. I was trying to remember the volume integrals I learned ages ago, and came across the shell and disk methods at Solid of revolution (hazy memories now). So I thought ah!, I'll do a really easy example: f(x) = x, a=0, b = 1 (a=lower x limit, b=upper x limit). So:

• Disk method: $V=\pi \int _{0}^{1}x^{2}dx=\pi [x^{3}/3]_{0}^{1}=\pi /3$ (good, the volume of a cone is correct!).
• Shell method: $V=2\pi \int _{0}^{2}x\times xdx=2\pi [x^{3}/3]_{0}^{1}=2\pi /3$ (uh oh).

Clearly I'm making a fully elementary mistake. But surely if f(x) = x, then f(x)^2 === xf(x), and everything is symmetric with this straight line and these limits? Have I even so got the axis of rotation wrong in the shell method?79.76.150.219 (talk) 01:43, 15 July 2008 (UTC)

I think your problem might be that you aren't actually drawing a cone - that formula is for the volume between f(x) and the x-axis, which is a cylinder with a cone removed (point down), not a cone (draw the graph and shade in the relevant part to check). The volume of that *is* 2pi/3. The question, then, is why did you get the wrong answer with the disk method? That, I'm baffled by... --Tango (talk) 01:59, 15 July 2008 (UTC)

Oh, and I'm assuming that 2 in the upper limit on the second integral is just a typo, you've switched back to the correct 1 in the next expression. --Tango (talk) 02:01, 15 July 2008 (UTC)

The correct integral for the shell method is:

$\int _{0}^{1}(1-x)\times 2\pi x\,dx.$

Drawing a picture might be helpful. siℓℓy rabbit (talk) 02:09, 15 July 2008 (UTC)

You mean f(x) should be 1-x for a cone, not x? I agree with that, but why does the disk method get the right answer with the wrong function? --Tango (talk) 16:54, 15 July 2008 (UTC)

Revolution is about the x-axis (at least given the form of integral used for discs), so it is a cone for f(x) = x, and the discs give the right value. The catch is that for shells you still want revolution about the x-axis (while the formula on the Solid of revolution page assumed revolution about the y-axis, giving a cylinder with a cone removed as you suggest) so to construct your shell you need to be careful, since your cylinders lie horizontally, and thus have radius f(x)=x but have height 1-x (distance from function to our upper bound b=1 running horizontally) so we have cylinders of area $2\pi x(1-x)$ giving the integral form supplied by Silly rabbit -- Leland McInnes (talk) 20:06, 15 July 2008 (UTC)

D'oh. Yes, of course, one formula was revolution about the x-axis, the other about the y-axis, leading to different integrands to compute the same volume. Thanks. 79.76.150.219 (talk) 20:49, 15 July 2008 (UTC)

Well spotted! Thank you! --Tango (talk) 21:17, 15 July 2008 (UTC)

Happy Numbers

We have an article on happy number and in the article, it describes the "proof" that all integers either end up in a cycle of 1,1,1,1,... or a cycle containing the number 4. Everything is fine until the last line which says that every number above 99 drops below 99 and then an exhaustive search shows us that only these sequences are possible. Without referring to this computer proof, how can we prove that all number between 1 and 99 eventually get into one of the two cycles?--A Real Kaiser...NOT! (talk) 05:39, 15 July 2008 (UTC)

I wouldn't call that a computer proof. While 99 cases is a bit of a stretch, one could verify it by hand. This isn't the four color theorem. 69.106.57.217 (talk) 06:46, 15 July 2008 (UTC)

It's less than fifteen minutes' work. I've spent much longer than that trying to understand a proof, let alone produce one. Algebraist 09:59, 15 July 2008 (UTC)

I suspect the OP actually meant a proof not involving proof by cases, not specifically a computer proof. GromXXVII (talk) 10:55, 15 July 2008 (UTC)

In that case, I doubt anything satisfactory is possible. You can streamline the case analysis with a little effort, but there doesn't seem to be much point. Algebraist 11:00, 15 July 2008 (UTC)

Drilling Holes in a Cube

We have a perfect cube with sides 10 units each. A hole has been drilled along its main diagonal (all the way across) with a drill whose radius was one unit. What is the new surface area? I got the answer as $600-2\pi +2\pi {\sqrt {300}}$. 600 was the original surface area before the drilling. After the drilling, $2\pi$ was lost and then $2\pi {\sqrt {300}}$ (the surface area of the hollow cylinder) was gained. Do you guys agree?--A Real Kaiser...NOT! (talk) 05:49, 15 July 2008 (UTC)

I think your last two terms are incorrect. The drill was through a diagonal, so it would not have met the faces at a right angle. The hollow shape is also not quite a cylidner as its end shapes are not circles. Maelin (Talk | Contribs) 09:25, 15 July 2008 (UTC)

Almost agree - you equation is very nearly right - but not perfect as pointed out above.87.102.86.73 (talk) 12:42, 15 July 2008 (UTC)

To get it absolutely right you'd need to calculate the surface area of the ellipses projected onto the cube corners by the cylinder, as well as the surface area of the trigonal pyramid caps removed from the corners by making the tube..87.102.86.73 (talk) 13:16, 15 July 2008 (UTC)

My answers: The total "lost" area is $2\pi {\sqrt {3}}$ (each face loses ${\frac {\pi }{\sqrt {3}}}$). The area of the inner cylinder-ish surface is $20\pi {\sqrt {3}}+2\pi {\sqrt {2}}-6{\sqrt {6}}$ making the total surface area of the object: $600+18\pi {\sqrt {3}}+2\pi {\sqrt {2}}-6{\sqrt {6}}\approx 692.133993088$. How about that, "do you guys agree" now? --tcsetattr (talk / contribs) 22:12, 15 July 2008 (UTC)

Hemispheres

If I mark three distinct points on a perfect sphere randomly, what are the chances that two of the points are on the same hemisphere?68.126.250.171 (talk) 06:32, 15 July 2008 (UTC)

There are only two hemispheres in a sphere. Therefore possible distributions for your three points are 3-0, 2-1, 1-2 or 0-3. In all cases one hemisphere has at least 2 points. -- SGBailey (talk) 08:06, 15 July 2008 (UTC)

This is an instance of the Pigeonhole principle. Maelin (Talk | Contribs) 09:26, 15 July 2008 (UTC)

Is it possible that the questioner meant 'all three points are on the same hemisphere' which is slightly less trivial?? (are you still there?)87.102.86.73 (talk) 12:40, 15 July 2008 (UTC)

They may also have meant that the hemispheres are defined before the points are added, such as the Earth's northern and southern hemispheres. In that case, you could have all three points on the border (equator) between the two hemispheres. So, would those be considered to be on the same hemisphere or not ? StuRat (talk) 05:09, 16 July 2008 (UTC)

The whole sphere/hemispheres part is irrelevant. We can recast this as follows : if a set S is partitioned into two disjoint sub-sets, then for any three members of S, at least two of the three will be members of the same sub-set. As Maelin says, it's a simple application of the pigeonhole principle. As StuRat says, the original formulation unfortunately introduces some ambiguity around the status of points on the boundary between the hemispheres. Gandalf61 (talk) 13:43, 16 July 2008 (UTC)

Another interpretation is that the hemisphere may not be fixed in advance. In other words, take three points at random on the sphere, and then ask: is there a hemisphere that contains all three points? The answer to this question is yes, almost surely. (The exceptional cases all occur when the three points lie on the same great circle.) siℓℓy rabbit (talk) 14:12, 16 July 2008 (UTC)

Oh, wait. I was answering 87.102.86.73's modification of the problem (all three points rather than two out of three). siℓℓy rabbit (talk) 14:14, 16 July 2008 (UTC)

What is this?

Can anyone identify this? I think it might be a cipher or code of some kind. Or, I could be completely wrong :P. VIVID (talk) 13:56, 15 July 2008 (UTC)

53 6f 6d 65 74 69 6d 65 73 20 61 20 67 69 72 6c 20 63 61 6e 20 67 69 76 65 20 79 6f 75 20 61 20 6c 6f 6f 6b 20 61 6e 64 20 61 20 73 6d 69 6c 65 20 74 68 61 74 20 63 61 6e 20 6d 61 6b 65 20 79 6f 75 20 66 65 65 6c 20 6c 69 6b 65 20 61 20 6d 69 6c 6c 69 6f 6e 20 62 75 63 6b 73 2e 20 20 4f 72 20 69 6e 20 74 68 69 73 20 63 61 73 65 2c 20 61 20 62 69 6c 6c 69 6f 6e 2e

Is is hexadecimal? some context would help -where's it from etc?87.102.86.73 (talk) 14:14, 15 July 2008 (UTC)

It is simple ascii in hex format. It says "Sometimes a girl can give you a look and a smile that can make you feel like a million bucks. Or in this case, a billion.". -- Q Chris (talk) 14:19, 15 July 2008 (UTC)

So it was a (simple) cipher of sorts .. see the hex column ofAscii#ASCII_printable_characters which one of you two is the girl?87.102.86.73 (talk) 14:22, 15 July 2008 (UTC)

Natural oracles giving P≠NP

In Computational Complexity: A Modern Approach, Arora and Barak construct a set B such that P^B≠NP^B. This B is explicitly constructed to ensure that no B-Turing machine decides the set of lengths of strings in B in polynomial time. But is there a more natural set (or one that was not specifically constructed to prove this)? --Taejo|대조 16:01, 15 July 2008 (UTC)

Sure -- it's the empty set. No one knows how to prove it, of course. --Trovatore (talk) 22:36, 15 July 2008 (UTC)

Lol, of course. Naturally, I meant a known natural set. --Taejo|대조 21:09, 17 July 2008 (UTC)

Converting between PPPs of different years

Sigh, OK, help me out on this. Math ain't my strong suit, much less international economics. (And no, this isn't homework. It's work work.)

I'm trying to take a number of different datasets relating to expenditures in terms of Purchasing power parity of a given "current year" and trying to combine them to be in terms of PPP of a fixed year.

So let's say I have spending figures, in "million current PPP$" for the year 2008, that look like this:

	2002	2003	2004
China	1162.4	1362.7	1630.1

And let's say I have spending figures "in millions current PPP$" for the year 2005 that look like this:

	2002	2003	2004
China	71358.6	84646.7	102622.9

OK. Obviously what "current PPP" means has changed a lot in those years. Let's say I have data from the 2005 report that is not in the 2008 report, and I'd like to convert all of the data to "millions of 2005 PPP$". Can I do that with just the information above? Would it be a straightforward conversion factor? I mean, it seems like the 2005 data is around 62 times the 2008 data. Would just using a factor like that produce junk data on the whole, though?

(My problem, if isn't clear, is I have lots of reports, each of which give values I need for different time scales in "current PPP$", which of course changes from year to year. I'm trying to aggregate the data. I'm not having a lot of luck. Or put another way, I am trying to get PPP values that are both currency AND time independent.) --140.247.248.35 (talk) 21:38, 15 July 2008 (UTC)

With only one datapoint, it could be anything really. It could be a straight conversion factor, but you'd need a 2nd data point to falsify that hypothesis. -mattbuck (Talk) 22:10, 15 July 2008 (UTC)

Well I'm only converting between these two types of data sets. I've given you three datapoints above—the total datapoints are only about 10 per country (and I think it's pretty clear that these calculations would have be done from country to country separately, as they are based on different national currency changes). So...? --98.217.8.46 (talk) 23:08, 15 July 2008 (UTC)

Are those genuine figures? A factor of 1/62 in 3 years doesn't sound right. Unless I'm completely misunderstanding this (I'm not an economist), the difference should be purely due to inflation in the US dollar (I'm assuming the dollars in question are US, they usually are for this kind of thing), and that should give a much smaller difference (in the order of 10% rather than 6200%) and it should be in the other direction... --Tango (talk) 23:30, 15 July 2008 (UTC)

Do I understand the data correctly? Was Chinese expenditure in 2002 equal to 1162 million "2008 US dollars" and 71359 million "2005 US dollars" (where 1162 million "2008 US dollars" represents the value of goods and services that can be purchased with Chinese Yuan equivalent to 1162 million US dollars in 2008 at 2008 purchasing power implied exchange rates)? From this graph and my calc, the Chinese Yuan appreciated by only 13.7% from 2005 to 2008 (in US dollar terms) so something ridiculous must have happened to purchasing power for my interpretation to be correct. Zain Ebrahim (talk) 13:00, 16 July 2008 (UTC)

I just editted my post above (diff). Zain Ebrahim (talk) 13:39, 16 July 2008 (UTC)

Oh, haha. I made a gaffe. The 2008 figures are Argentina, whereas the 2005 figures are China. Yeah, OK. Here's the correct 2008 figures:

	2002	2003	2004
China	39444.7	46944.6	57669.6

Which still gives us a change of -55%, which is pretty weird.. --140.247.240.177 (talk) 20:13, 16 July 2008 (UTC)

That still doesn't make sense... it's closer to the right size change, but it's still in the wrong direction. Due to inflation, a 2008 US dollar is worth less than a 2005 US dollar, so the same quantity measured in 2008 dollars should be larger than when measured in 2005 dollars... --Tango (talk) 22:10, 16 July 2008 (UTC)

If my definition above is correct then it would be possible (though unlikely) for the number to decrease over time if the purchasing power of 1 yuan increased relative to the purchasing power of one dollar over 2005 to 2008 or if PP exchange rates stayed the same but China had higher inflation. Zain Ebrahim (talk) 11:34, 17 July 2008 (UTC)

Say China's expenture in year Y was X Yuan (not adjusted for anything). To get this in "2008 US dollars", first you'd convert it into 2008 Yaun (i.e. increase to adjust for inflation (time)) and then convert that into dollars using purchasing power implied exchange rates (to adjust for currency) as of 2008. If I understand you correctly, you have data for ten years at "2005 US dollars" but for fewer years at "2008 US dollars" and you want to convert the data points you don't have in the 2008 report into "2008 US dollars", right? Or is it the other way around?

So the X Yuan spent in year Y is converted into "2005 US dollars". To convert this into "2008 US dollars", all you need are three peices of information:

• The 2005 PP implied exchance rate,
• Chinese inflation over 2005 to 2008 and
• The 2008 PP implied exchange rate

It doesn't depend on Y but it does depend on the country so (if all my assumptions are correct) using the factor seems fine as long as you do it separately for each country. Zain Ebrahim (talk) 11:34, 17 July 2008 (UTC)

Can't you just use US inflation? I thought the point of PPP was that you didn't need to worry about where it was being spent, that's all taken care of in the exchange rate. --Tango (talk) 16:28, 17 July 2008 (UTC)

That would work if the expenditures were first adjusted for currency and then for time - I assumed the opposite. I think adjusting for time first is better because it captures the time-value of money in China for China's data. Zain Ebrahim (talk) 19:09, 17 July 2008 (UTC)

Factorial differentiation

I know that the factorial function can't be differentiated because it is only defined on the (positive?) integers but I don't see why this prevents it from being differentiable. Can someone explain please? Thanks 92.2.122.213 (talk) 22:13, 15 July 2008 (UTC)

Well, the definition of the derivative involves evaluating the given function arbitrarily close to a given point. For example, to get an approximation of the derivative of the factorial function at 3, you might form the fraction

${\frac {3.000001!-3!}{0.000001}}$

But if there's no such thing as 3.000001!, how are you going to do that?

Note that the gamma function is differentiable. --Trovatore (talk) 22:34, 15 July 2008 (UTC)

(edit conflict) The derivative of a function f at x is (loosely speaking) the limit of the difference quotient ${\frac {f(a)-f(b)}{a-b}}$ as the two points a and b both get arbitrarily close to the point x. If a and b can only take integer values, they can't approach each other arbitrarily closely — the smallest possible difference between two different integers is 1. The only way for two integers to get closer than that is for them to be equal — and that won't work, since then the difference quotient would become 0/0, which is undefined. Of course, you can still take finite differences: for example, (x+1)! - x! = x·x!. Also, the factorial function can be extended so that it has values on (almost) all real (and even complex) numbers, and that function can then certainly be differentiated. —Ilmari Karonen (talk) 22:39, 15 July 2008 (UTC)

You might be interested in Stirling's approximation for factorials which is derived thus:

n! = product(from x=1 to x=n) of x

therefor ln(n!)=sum(from x=1 to x=n) of ln(x) remember ln(abcde)=ln(a)+ln(b)+ln(c)+ln(d)..etc

Therefor ln(n!) can be approximated by the integral of ln(x) between 1 and n

ie ln(n!) ~ xlnx - x + 1 I've ignored the stuff that minimises the error -see Stirling's approximation for that..

Therefor n! ~ e^{xlnx - x + 1} = e^xlnx+e^-x+e¹ = x^x+e^-x+e = e(x/e)^x

So as a rough rule of thumb you can say that the slope of the graph formed by joining the points of n! vs. n is given by d/dx(e(x/e)^x) which you might be able to calculate..

For a better version read the rest of the article!87.102.86.73 (talk) 22:55, 15 July 2008 (UTC)

OK I see what you're saying. So if you have a function involving the factorial function, eg $f(x)=x!*x^{n}$, how do you go about differentiating it? Or does this depend on the function? 92.2.122.213 (talk) 18:29, 16 July 2008 (UTC)

You can't differentiate it, any more than you can differentiate the factorial function itself, and for the same reason. By the way, the answer involving Stirling's approximation is a bit misleading -- the derivative of an approximation of a function doesn't have to be a good approximation to the derivative of the function.

Really, if you want a ("natural") differentiable function that interpolates the factorial function, you have to learn about the Gamma function. It's not that hard, so quit looking for shortcuts. --Trovatore (talk) 18:47, 16 July 2008 (UTC)

(Yes I gave the stirling approximation 'concept' as a method of just estimating the slope of a graph that almost fits the points given by n! in fact if you read more of that article it too goes onto mention the gamma function)

Also it's worth noting that the gamma function does give a way of getting n! as a continuous function, but that factorial as most people know it just isn't really a curve or line - it's just a set of points - and as such doesn't really differentiate.

As for your function y=x!*x^n then as usual dy/dx = xⁿd/dx(x!) + nx^n-1*(x!) (see [[chain rule] product rule] How you interpret d/dx(x!) depends on what your equation means - it may be meaningful to use a stirling-derived approximation, or not - but the questions seems to be how do I differentiate Gamma(n) with respect to n ?? 87.102.86.73 (talk) 19:05, 16 July 2008 (UTC)

July 16

Solitary numbers

Why is 18 a solitary number even though (18, σ(18)) ≠ 1 ? Same for 45, 48, etc. Thanks, --Think Fast (talk) 01:51, 16 July 2008 (UTC)

According to Friendly number, a number that is coprime to the sum of its divisors is solitary, but the converse is not true, 18 being a counterexample. --Tango (talk) 01:57, 16 July 2008 (UTC)

I read the article. Can you explain why 18 is a counterexample? --Think Fast (talk) 02:31, 16 July 2008 (UTC)

The formulation in Friendly number seemed easy to misunderstand so I tweaked it.[2] PrimeHunter (talk) 02:39, 16 July 2008 (UTC)

The converse (which is false) would be: If a number is solitary then it's coprime to the sum of its divisors. It is this for which 18 is a counterexample: 18 is solitary (not proved here), but 18 is not coprime to its sum of divisors 39. PrimeHunter (talk) 02:46, 16 July 2008 (UTC)

This is all good, but why is 18 a solitary number? It is not prime, it is not a prime power, and (18, σ(18)) ≠ 1 (i.e., it is not coprime to the sum of its divisors). But nonetheless, it is a solitary number. Can anyone explain what makes it a solitary number even though it does not satisfy any of these criteria? --Think Fast (talk) 03:04, 16 July 2008 (UTC)

Presumably it is a solitary number because it is the only positive integer n such that σ(n)/n = 13/6. In other words, your question is "Why are there no positive numbers n (other than 18) such that the sum of their divisors is 13n/6?" or so. JackSchmidt (talk) 03:42, 16 July 2008 (UTC)

I don't know if it helps, but this was discussed in 2004 at sci.math:

We know that 2*3=6 divides x, so we know that x=2^a * 3^b * n where n is 
relatively prime to 6.  Now it is easy to see that if sigma(x)/x = 13/6 
and x != 18 then a=b=1. (Can you see why?)  Now if x=6*n with n 
relatively prime to 6 then the numerator of sigma(x)/x reduced to lowest 
terms will be even, and the denominator will be relatively prime to 6, 
in particular it will not equal 13/6.

The only other reference appears to be some sort of closed mailing list. JackSchmidt (talk) 04:01, 16 July 2008 (UTC)

Thank you for finding this. It answers my question. --Think Fast (talk) 05:30, 16 July 2008 (UTC)

Divisor function

This isn't really a reference desk question, but I hoped one of you maths guys could check the table of values I just added to divisor function. Is it ok? (There - I managed to ask a question...) -- SGBailey (talk) 08:24, 16 July 2008 (UTC)

It doesn't contain any errors, if that's what you're asking. Algebraist 10:09, 16 July 2008 (UTC)

I added Table of divisors to Divisor function#See also. Your table lists the first 15 of 1000 entries. PrimeHunter (talk) 12:32, 16 July 2008 (UTC)

For future reference, if you want help with writing/editing/proofreading mathematics articles, the best place to go is Wikipedia talk:WikiProject Mathematics. --Trovatore (talk) 00:58, 17 July 2008 (UTC)

Numbers

Hello, I do not speak english well and I do not understand the next number: How much is thirty hundred thousand in number? Thank you --Humberto (talk) 14:22, 16 July 2008 (UTC)

"Thirty hundred thousand" is not how one would phrase that number correctly, so I can't blame you for not understanding it. I think that it is supposed to be 3,000,000; which is three million. Paragon12321 (talk) 14:27, 16 July 2008 (UTC)

I do not speak English well, either, however... Simple 'googling' returns lots of results:

http://www.google.com/search?q=%22thirty+hundred+thousand%22

which may indicate it is not an ordinary number, but rather a synonym for "lots of", "a large number". --CiaPan (talk) 14:42, 16 July 2008 (UTC)

It actually sounds like english as spoken by a non-perfect foreign language speaker of english, or possibly a translation from another language written in psuedo-archaic terms.. If that makes sense.

Also is 'thirty-hundred' a normal term in sanskrit or something, as many of the pages turned up by google seem to have a vedic connection?87.102.86.73 (talk) 17:01, 16 July 2008 (UTC)

Numbers in the teens are often used before "hundred" to indicated thousands, so "thirteen hundred" is the same as "one thousand three hundred". I've never heard numbers greater than 19 used in this sense, though, nor have I heard this technique used on longer numbers. « Aaron Rotenberg « Talk « 16:46, 17 July 2008 (UTC)

Dedekind cuts and the proff that 0.999.... = 1

I have a question about these two statements in the proof:

"Every element of 0.999… is less than 1, so it is an element of the real number 1."

I feel that while this states that every cut belonging to 0.999... includes values < 1 -(1/10)^n, it does not necessary exclude the value 1-(1/10)^n on each set. If that is the case, how does the proof jump to the conclusion that "Every element of 0.999… is less than 1, so it is an element of the real number 1" so immediately. Are there any missing steps?

There are some missing steps, but they are arithmetic in nature. If we want to fill in the missing steps we would say something like: What are the elements of 0.999...? Each is some rational smaller than some number of the form 0.999..99. Since the latter number is smaller than one, also the rational is smaller than one. Oded (talk) 19:01, 16 July 2008 (UTC)

The next question I have regarding this quote: "Conversely, an element of 1 is a rational number \begin{align}\tfrac{a}{b}<1\end{align}, which implies \begin{align}\tfrac{a}{b}<1-(\tfrac{1}{10})^b\end{align}"

How does the first imply the 2nd? If I test out that statement with values for a and b, it could work. But won't I have to test it infinitely to prove to myself that it does work, therefore defeating the purpose of this proof. And then the would "implies" would have a really loose meaning.

You are right: the proof should be improved at this point, and I will improve it. (It was not incorrect, but assumed that the reader could fill in some missing steps, which is quite reasonable. But it is better if these steps are more elementary.) Oded (talk) 19:01, 16 July 2008 (UTC)

Sorry. I tried to improve, but if I try to put in all the details, I would have to prove that $1/b>10^{-b}$. While this can certainly be done, I think it is better to accept on faith that the reader knows these things. Otherwise, this would be just too involved. Oded (talk) 19:19, 16 July 2008 (UTC)

I am only a student so maybe the proof is rigorous and I just can't understand the jumps. However, these are serious questions and the proof seems to take leaps instead of steps, leaving gaps in the proof. If it is doing so, I hope the author can continue his proof. If it isn't skipping steps, I hope my question would be answered in a way I can understand.

98.210.254.18 (talk) 16:33, 16 July 2008 (UTC)Quang Pham

Where is this proof you are referring to? I looked at the article Dedekind cut and found no such proof. Oded (talk) 18:08, 16 July 2008 (UTC)

It's in the 0.999... article - [3]. --LarryMac | Talk 18:11, 16 July 2008 (UTC)

The infinite set that corresponds to real number $1-(1/10)^{n}$ is

$U_{n}=\{x\in \mathbb {Q} |x<1-(1/10)^{n}\}$

Real number 0.999... is the union of all sets $U_{n}$

$U_{\infty }=\bigcup _{n=1}^{\infty }U_{n}$

and real number 1 is the set

$V=\{x\in \mathbb {Q} |x<1\}$

What needs to be proven is that the two sets are equal $U_{\infty }=V$.

Clearly every $U_{n}$ is a subset of V, thus so is their union $U_{\infty }\subset V$.

For the other inclusion ${\tfrac {a}{b}}<1$ implies ${\tfrac {a}{b}}\leq {\tfrac {b-1}{b}}=1-{\tfrac {1}{b}}<1-({\tfrac {1}{10}})^{b}$. Thus a/b is an element of the set $U_{b}$.

${\tfrac {a}{b}}\in U_{b}\subset U_{\infty }$

This is true for every element a/b in V, thus V is a subset of $U_{\infty }$. Tlepp (talk) 19:04, 16 July 2008 (UTC)

Intersecting Spheres

Imagine an (n+1)-simplex in n-dimensional space, with each edge of unit length. I'm pretty sure that the intersection of the unit (n-1)-spheres around all but three of its vertices would form a 2-sphere containing the remaining three. Is this true? And if so, how could I estimate the radius of this sphere? Black Carrot (talk) 20:06, 16 July 2008 (UTC)

Yes, it is true. The intersection of a sphere with a sphere is a sphere (or a point or empty) you known that each sphere in your intersection reduces the dimension by 1. By symmetry, the center of the 2-sphere will be the average of the vertices which are the centers of the spheres in the intersection. That should facilitate the calculation of the radius. Oded (talk) 20:19, 16 July 2008 (UTC) PS: in calculations involving the m-simplex, it is usually easiest to think of it as the convex hull of the m unit vectors in m-space, rather than embed it in (m-1)-space. Oded (talk) 20:23, 16 July 2008 (UTC)

Fractional derivative

Are there any 'real world' applications, meaningful quantities/measurables, etc of fractional derivatives? (that you know of)? thanks.87.102.86.73 (talk) 00:35, 17 July 2008 (UTC)

[4] might be of interest. --Tango (talk) 00:56, 17 July 2008 (UTC)

Wow. Where to begin. Let's see. Fractional derivatives are found in the integrodifferential equations corresponding to Levy processes (see Feynman-Kac formula for the duality), which are often used to model a variety of different physical and financial processes (see the article). Existence in a fractional Sobolev space might mean that you can derive higher regularity (and thus better bounds for numerical convergence, etc.) for a function. Fractional powers of the Laplacian can manifest as Dirichlet-to-Neumann operators (see this paper) for other partial differential equations, which means that you can estimate, say, the heat flow through a wall without solving the Laplace's equation inside the wall. There's lots more fun to be had. G'luck! RayAYang (talk) 01:54, 17 July 2008 (UTC)

Do we have an article on the p-laplacian, Δ^p? We have biharmonic equation, but I didn't see any p other than 1 and 2. JackSchmidt (talk) 20:19, 17 July 2008 (UTC)

July 17

Is timescales mathematics

Is timescales mathematics? Would it be acceptable as a mathematical document? 122.107.219.245 (talk) 14:10, 17 July 2008 (UTC)

That article is discussing an area of mathematics, yes. I don't know what you mean by "acceptable as a mathematical document". --Tango (talk) 16:24, 17 July 2008 (UTC)

If you're asking whether you should use it as a reference or citation in a paper, then the answer is probably no. However, it has links at the bottom that may be acceptable, such as the one to New Scientist. Black Carrot (talk) 23:13, 17 July 2008 (UTC)

Naive Bayes classifier for non-mutually-exclusive classes?

How does a naive Bayes classifier change if some of the classes are not mutually exclusive? NeonMerlin 17:27, 17 July 2008 (UTC)

choice of weights for linear regression

A question on choice of weights for weighted linear regression. I need to solve for x a problem of type $Ax=B$ where $A$ is size m by n, $x$ is size n by 1, and $B$ is size m by 1, m >> n > 1, given the weights vector $W$ of length m. In other words, I would like to minimize $\sum _{i=1}^{m}W_{i}\left(B_{i}-\sum _{j=1}^{n}A_{ij}x_{j}\right)^{2}$. My question is: what is the appropriate choice of weights when different points i, i = 1...m, have different variances of $B$ and different numbers of counts $C$ that went into determination of the variances of $B$ ? Let me explain my question in a bit more detail. I measure $B$ for any given $A$, many times. $A$ is a discrete variable. $A$ values may be assumed exact for all practical purposes. Measured value of $B$ varies from measurement to measurement even for exact same $A$. Each possible value $A_{i}$ of $A$ is sampled $C_{i}$ times, $2<C_{i}<10^{5}$. Note that $C_{i}$ varies by 4 orders of magnitude. What is the best choice of $W$ in that situation? I have tried three choices so far, none of which is really good. Choice 1. If I use $W_{i}=1/var(B_{i})$ I bias my regression towards points that may have low variance beacuse thay have been poorly sampled. Indeed, the lower is $C_{i}$ the higher is the variance of variance of B, obviously; so for some points with low $C_{i}$ the value of $var(B_{i})$ will come out very low by chance. That is bad. Choice 2. If I use $W_{i}=C_{i}$ , I bias my regression towards points with high $C_{i}$, but these points not necessarily have lower $var(B_{i})$; also, any two points with same large $C_{i}$ will have the same weight even though I can reliably claim that $var(B)$ is larger in one than in the other. That is also bad. Choice 3. I can use Bayesian average instead of mean when evaluating $var(B_{i})$, but, first, I do not know if that is legitimate, and, second, the choice of constant for the Bayesian average is quite arbitrary. So, to repeat my question, what should I use for $W$ ??? Please help! Thank you in advance, --OcheburashkaO (talk) 22:31, 17 July 2008 (UTC)

Maybe I'll be back later, but weights proportional to reciprocals of variances is the usual thing. 75.72.179.139 (talk) 02:59, 18 July 2008 (UTC)

You wrote:

Each possible value $A_{i}$ of $A$ is sampled $C_{i}$ times, $2<C_{i}<10^{5}$.

Did you mean each value of B_i? You said observations of A could be taken to be exact. If individual observations of B can be taken to be equally uncertain, then using weights proportional to the number of observations makes sense. But apparently this equality of uncertainty does not hold, according to what you're saying. More later........ Michael Hardy (talk) 03:20, 18 July 2008 (UTC)

I'm curious; (without giving too much away :-) what is the nature of the measurement that sampling counts vary so much?

And a little confused; in the cases I'm familiar with, A is the variable to be solved for. That is, finding an $A$ that minimizes $\sum _{i=1}^{m}W_{i}\left(B_{i}-\sum _{j=1}^{n}A_{ij}x_{j}\right)^{2}$. It's not a constant until you're done. Saintrain (talk) 16:48, 18 July 2008 (UTC)

Rational Angle

Is the angle between two faces of a regular tetrahedron, arccos(1/3), a rational multiple of pi? If not, how would you prove it? Black Carrot (talk) 23:21, 17 July 2008 (UTC)

Typing "arccos(1/3)" into my calculator, dividing by pi and pressing the fraction button fails to give a fraction, which would suggest it's not rational (it could be rational with a denominator too large for my calculator to handle, I suppose). I wouldn't know where to start with proving it analytically. --Tango (talk) 00:01, 18 July 2008 (UTC)

One thing to be careful about with many calculators is that they have much more precision when dealing with fractions. That is for instance, the CAS in the TI line will do fractions with hundreds of digits, but if you start out with a decimal they only store around 9-15 digits, depending on the model – so getting a fraction from a decimal can be tricky.

I don’t remember what model, but I used a calculator once that I actually got to round “0.3333333333333” or so to 1/3, so all of them might not be entirely accurate either. GromXXVII (talk) 11:14, 18 July 2008 (UTC)

I don't think that arccos(1/3) is a rational multiple of pi. And one way to prove it would be by contradiction. Assume that this arccos(1/3)/pi is rational and get a contradiction like pi being a rational number also.--A Real Kaiser...NOT! (talk) 04:26, 18 July 2008 (UTC)

It isn't. Let θ=arccos(1/3). Then the complex number

$\alpha =e^{i\theta }={\frac {1}{3}}+i{\frac {2{\sqrt {2}}}{3}}$

has degree 2 over the rationals, and so cannot be a root of any cyclotomic polynomial since the minimal polynomial of α is not cyclotomic. siℓℓy rabbit (talk) 06:00, 18 July 2008 (UTC)

How does one tell that the polynomial is not cyclotomic? 24.227.163.238 (talk) 17:03, 18 July 2008 (UTC)

Is arccos(1/3) an irrational algebraic [i.e. not transcendental] multiple of pi (or pi^2 sqrt(pi) 1/pi etc)? I.e. could you describe it using a finite number of operations on integers and pi? --Random832 (contribs) 18:40, 18 July 2008 (UTC)

July 18

Exponential Exponents

Exponents or powers or the superscripts that in math mean to multiply the base the exponent's number of times. There are also numbers with exponents that have exponents. If you have an infinite number of exponents, does that make the number infinity or another infinite number?

i.e.

$2^{2^{2^{2^{2^{2^{2^{2}...}}}}}}=\infty$

or

$2^{2^{2^{2^{2^{2^{2^{2}...}}}}}}\neq \infty$