Wikipedia:Reference desk/Archives/Mathematics/2011 September 8

= September 8 =

Did I use neighborhood right?
this has been bugging me. I am a monitor (kinda like a high-school version of an AP, I am an upperclassman and do this in an underclassman calculus class during my apparently mandatory study hall). I got to teach today :) since the teacher is out sick and the substitute was not a math teacher. There was a question based on a problem I presented from the regular teacher's notes that if there were a function defined say, over (-∞,0] and (3,∞), what would be the limit at 2 (a modification of the problem in the notes, which had it defined over R but with a jump at x=2). I said it would not exist because f(x) is not defined within a neighborhood of x=2. Did I use the term right? (It's been a while since I took calc and I went the discrete route in my math studies instead, so I'm not sure; I mused as much out loud but if anyone in the class thought I used it wrong they were afraid to correct me because they're all freshies and I'm the scary senior ;)) thx — Preceding unsigned comment added by 24.92.85.35 (talk) 00:06, 8 September 2011 (UTC)
 * Yes, that's correct.--Antendren (talk) 10:58, 8 September 2011 (UTC)

Maximising winnings on Euromillions
Back when I was in a syndicate for the National Lottery (the game they named Lotto), we used to buy tickets in a way that maximised the winnings. So if we chose the numbers 1-7 we'd buy 123456, 123457, 123467, 123567, 124567, 134567 & 234567. (We actually did this for 8 numbers, because we were a large syndicate) This would mean that if we got 3 numbers, we would win 4 times, and if we got 4 numbers we'd win on every line and so on. Basically, we didn't really expect to win the jackpot, but wanted to increase the actual winnings we made.

I was thinking of setting up another syndicate, but for the Euromillions. I can see how we'd do a similar method if we held the lucky stars still, but is there a better way to do it? At the moment, I'm seeing we could buy 6 tickets if we want 6 numbers with 2 fixed lucky stars, or 18 tickets if we want 6 numbers with 3 lucky stars. I'm sure I could sit down and work out the levels of probability, but I thought I'd see if some illustrious reference deskers had any thoughts. WormTT  &middot; &#32;(talk) 08:34, 8 September 2011 (UTC)
 * For information, Lotto takes 6 random numbers between 1-49, with prizes if you get 3 or more matching. Euromillions has takes 5 random numbers from 1-50 and 2 lucky stars from 1-11, with prizes for 2 main numbers upwards.  WormTT   &middot; &#32;(talk) 08:38, 8 September 2011 (UTC)

(-1/3)^(-1/3)
So google gives this answer as 0.721124785 - 1.24902477 i. But -(3^(1/3)) is also a solution? Why do they give the complex solution? --163.202.48.109 (talk) 11:10, 8 September 2011 (UTC)
 * Since it was a negative number that you were getting the root of they treated it as a complex number and gave the principal value of the cube root. Getting a real cube root of a negative real number only works for rational exponents and is a bit of a mess, even if they did do it they would have to recognize 1/3 as a rational number rather than just some random real number. Dmcq (talk) 13:01, 8 September 2011 (UTC)
 * What he said. Google doesn't "know" that 1/3 is rational, and so doesn't extract the real root.  Here is how to get Google to report the real root: .  Sławomir Biały  (talk) 13:16, 8 September 2011 (UTC)

I'd also recommend that you read the root of unity article and, depending on your level. From that you'll see that
 * $$ (-1/3)^{-1/3} = -\sqrt[3]{3}, \sqrt[3]{3}(1/2 - i \sqrt{3}/2), \ \text{or} \ \sqrt[3]{3}(1/2 + i \sqrt{3}/2).$$

Maybe you already knew how to do that, but I thought I'd mention it anyway. — Fly by Night  ( talk )  15:10, 8 September 2011 (UTC)

Toric Configuration Space
Say I've got a robotic arm that operates in a two dimensional plane. The has a "shoulder" and an "elbow", both joints can rotate about 360°. I'm interested in the final position of the "hand". The configuration space is a torus. But clearly, different shoulder and elbow configurations can lead to the hand being in the same place. Taking the quotient space $T/~$, where $p ~ q$ if and only if the hand position for joint configuration $p$ is the same as that of $q$. Topologically, the space of final positions of the hand is an annulus, but what's the topology of $T/~$? There's obviously an injective mapping from $T/~$ to the annulus; but I'm not sure about continuity. — Fly by Night  ( talk )  17:55, 8 September 2011 (UTC)
 * I'm not really an expert on this but the closed map lemma (see Open and closed maps) implies that the map T→A is closed and our article on quotient maps says this implies that the map is a quotient map (see the last paragraph in the Properties section). That proves that T/~ is homeomorphic to the annulus unless there's something I'm missing.--RDBury (talk) 21:08, 8 September 2011 (UTC)


 * If I understand the configuration of your arm correctly, then you have two segments A and B, with A attached to a fixed pivot at the "shoulder" and B attached to A at the "elbow", and each joint free to rotate independently through 360 degrees in the same plane. So if A and B have lengths a and b respectively, and we assume a > b, then the "hand" at the far end of arm B can reach anywhere in the annulus a-b &le; r &le; a+b.
 * Let's call the shoulder O, the elbow joint P and the hand Q. Then the two parameters of your configuration space are the angle that the arm A makes with some fixed direction - call this &alpha; - and the angle that arm B makes with arm A - call this &beta;. So OPQ is a triangle with |OP| = a, |PQ| = b, angle OPQ = 180 - &beta;, and you want to find the polar co-ordinates (r, &theta;) of Q in terms of a, b, &alpha; and &beta;.
 * The cosine rule tells us that
 * $$r = |OQ| = \sqrt{|OP|^2 + |PQ|^2 - 2|OP||PQ| \cos( \angle OPQ)} = \sqrt{a^2+b^2+2ab\cos \beta}$$
 * and the sine rule tells us that
 * $$\sin( \angle POQ) = \frac{|PQ| \sin ( \angle OPQ )}{|OQ|} = \frac{b \sin \beta}{r}$$
 * so we have
 * $$\theta = \alpha + \angle POQ = \alpha + \sin^{-1}(\frac{b \sin \beta}{r})$$
 * Now, if we are given r and &theta; and want to find &alpha; and &beta;, then we have
 * $$\cos \beta = \frac{r^2 - a^2 - b^2}{2ab}$$
 * which gives two possible values for &beta; (except when &beta; is 0 or 180 degrees). Once we fix &beta; then we have
 * $$\alpha = \theta - \sin^{-1}(\frac{b \sin \beta}{r})$$
 * So the projection from configuration space to the annulus is, in general, 2-to-1, except when &beta; is 0 or 180 degrees, when it is 1-to-1. Gandalf61 (talk) 11:33, 10 September 2011 (UTC)


 * Topologically the map T → A is just the obvious flattening. If you have the torus defined in cylindrical coordinates (r, θ, z) = (a+bcosβ, α, bsinβ) for shoulder and elbow angles α and β, the map onto the annulus isn't a straight projection, but it's close.  The radial distance is preserved.  The points in the torus that map down to a spoke in the annulus form a sort of slanted circle in the torus, since height in the torus translates into radial movement in the annulus.  Think of squashing a slinky that's lying on its side. Rckrone (talk) 05:09, 11 September 2011 (UTC)


 * Just an observation: Suppose the elbow is a socket rather than a gear (so it permits motion through a sector of a sphere). Then I think motion planning becomes a more interesting question.  Sławomir Biały  (talk) 22:31, 11 September 2011 (UTC)