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December 22[edit]

Parametric equations[edit]

what is a good, detailed, and comprehensive text for understanding parametric equations? I tried your article but it wasn't much help and I tried Spivak but his Appendix on the subject is much to short and undetailed. 24.92.70.160 (talk) 02:34, 22 December 2010 (UTC)[reply]

Basic probability question[edit]

So I'm very, very rusty on basic probability, and have been unable to find an answer to this exact question online. So I'd appreciate the help: let's say I have 10 six-sided die. If I roll them all at once, how do I estimate the probability that a given number of dice - let's say, 5 or more - will come up to be 4 or higher? Obviously, each die individually has a 50% chance of coming up 4 or higher. So how do I estimate/calculate the cumulative probability in this case?

Thanks. --Brasswatchman (talk) 04:32, 22 December 2010 (UTC)[reply]

You need to use combinatorics to count the possibilities. Ginger Conspiracy (talk) 05:19, 22 December 2010 (UTC)[reply]

To answer this question, you will need to be familiar with choose notation and the binomial coefficients. Firstly, note that P(≥5 dice show ≥4) = P(5 show ≥4) + P(6 show ≥4) + P(7 show ≥4) + P(8 show ≥4) + P(9 show ≥4) + P(10 show ≥4). Speaking generally, the probability that k of 10 show ≥4 is

{\binom {10}{k}}\left({\frac {1}{2}}\right)^{k}\left({\frac {1}{2}}\right)^{10-k}={\binom {10}{k}}\left({\frac {1}{2}}\right)^{10},

due to your insight that there is a probability of 1/2 that a particular die shows ≥4 (and therefore a 1/2 chance that it doesn't). Hence the overall probability is

\left({\frac {1}{2}}\right)^{10}\left[{\binom {10}{5}}+{\binom {10}{6}}+{\binom {10}{7}}+{\binom {10}{8}}+{\binom {10}{9}}+{\binom {10}{10}}\right]\approx 0.623.

Let me know if you found any of that unclear. See also binomial distribution. —Anonymous Dissident^Talk 05:20, 22 December 2010 (UTC)[reply]

The probability distribution involved is that of the number of successes in a fixed (non-random) number of independent Bernoulli trials with a fixed probability (in this case 1/2) of success on each trial. It's called the binomial distribution, and I'm surprised that some people have answered without mentioning that. Michael Hardy (talk) 05:29, 22 December 2010 (UTC)[reply]

Michael, you have a few bad habits with respect to your Reference Desk contributions. Replies should be one indentation level higher than what they're replying to; replies to the original question should be one level of indentation. Also, your comment suggests that you've only skimmed Anonymous Dissident's answer and haven't noticed that he did mention the binomial distribution. -- Meni Rosenfeld (talk) 09:04, 22 December 2010 (UTC)[reply]

To return to the problem, it's always worth looking for the shortest way of getting an answer. In this case it would be to calculate the probability of getting fewer than 5 dice showing 4 or higher, by summing P(0), P(1), P(2), P(3) and P(4), then subtracting from 1. There's not much difference in effort, but there would be if the problem was to do with, say, more than 1 die out of the 10.→81.135.47.163 (talk) 12:29, 22 December 2010 (UTC)[reply]

Well, not quite as basic a probability problem as I thought, I'll admit. :) Did some reading, let me see if I have this straight... As per binomial distribution - where p is the probability of a given die roll (in the given example 3/6 or 1/2) and n is the total number of die rolled, the specific probability of given number k of die being rolls defined as successful is:

f(k)=({\frac {n!}{k!(n-k)!}})(p^{k})(1-p)^{n-k}

Which means that the probability of k OR MORE successes can be defined in terms of a sum of the given function f(k):

total=f(k)+f(k+1)+f(k+2)...f(n)

Do I have this right? Thank you all for your help. I really appreciate it. --Brasswatchman (talk) 03:59, 23 December 2010 (UTC)[reply]

Yes, that's right. —Anonymous Dissident^Talk 04:33, 23 December 2010 (UTC)[reply]

The question seems fully answered. "Brasswatchman", could I suggest some improvement in your TeX style?

{\text{total}}=f(k)+f(k+1)+f(k+2)+\cdots +f(n)\,

You should never write "..." within TeX; instead use \dots or \ldots or \cdots. And notice also the "+"s before and after the dots and the use of \text at the beginning. Michael Hardy (talk) 20:42, 23 December 2010 (UTC)[reply]

...also, you can allow parentheses to have the right sizes, thus:

\left({\frac {n!}{k!(n-k)!}}\right)

(although you're using more parentheses than there's really an occasion for). Michael Hardy (talk) 20:43, 23 December 2010 (UTC)[reply]

Adjustment for symbolic regression[edit]

How does one adjust a statistical regression's r² goodness of fit statistic for the number of degrees of freedom involved in symbolic regression? Ginger Conspiracy (talk) 05:19, 22 December 2010 (UTC)[reply]

I don't know what you mean by "symbolic regression". I've know of relationships between r² and things like mean squares and F-statistics, and you need to think about degrees of freedom for that, but I haven't heard of adjusting r² for degrees of freedom. I think you need to be more explicit about what you have in mind. Michael Hardy (talk) 20:45, 23 December 2010 (UTC)[reply]

I'm asking about improving Coefficient of determination#Adjusted R2 which only talks about adjustment for the number of coefficients and other variables fitted in a regression model. How do you adjust for trying multiple models, each with potentially different numbers of such variables? Ginger Conspiracy (talk) 04:18, 24 December 2010 (UTC)[reply]

Two translation articles[edit]

Is there a difference between Translation (geometry) and Translation (physics)? Am I right in thinking they should they be merged, or is there a detail I missed? Ariel. (talk) 05:48, 22 December 2010 (UTC)[reply]

They do seem to be talking about the same thing, but the physics version only defines it, while the geometry article goes into detail about the math needed to perform the operation. A case could be made for merging them (you add a tag and put your reasons for requesting the merge on the talk page), or at least linking them to each other. One thing somewhat surprising is that both articles seem to restrict the discussion to 3D. I'd have expected the geometry article to talk about 1D, 2D and perhaps higher dimensions, too. StuRat (talk) 07:08, 22 December 2010 (UTC)[reply]

I'm surprised Translation (geometry) doesn't mention Translation (group theory), seeing as it is just a special case.... Eric. 82.139.80.114 (talk) 14:11, 22 December 2010 (UTC)[reply]

I've marked them with merge tags, I may do it myself later. Ariel. (talk) 01:31, 24 December 2010 (UTC)[reply]

How to define that a set is inside another set[edit]

This puzzled me today. For simplicity, assume a metric two-dimensional space. Assume $A$ is a connected set and $B$ is another connected set, $A\cap B=\emptyset$ . How can we mathematically define that $B$ is inside $A$ , which is rather easy to intuitively visualise? The closest I've come to think is that $\forall x\in B:d(x,A)<d(x,{\overline {A\cup B}})$ , where $d$ is the metric distance function. But that doesn't even work if $B$ is a subset of a bigger "hole" in $A$ , only if $B$ fills the "hole" completely. Is there any simple way to define this? JIP | Talk 18:23, 22 December 2010 (UTC)[reply]

Ok, how about this. Take the boundary of A take the union of the interiors of the closed curves (Jordan Curve theorem). Intuitively this is all that is "inside of A" whether or not "in" A. Then it's clear when B is part of the inside of A. 198.161.238.19 (talk) 21:00, 22 December 2010 (UTC)[reply]

Do you literally mean a connected set in the sense of general topology? I'm afraid that allows for some pretty messy stuff that may be well outside of the intuitions you're starting with. Not that I've quite understood what those intuitions are, or what you mean by "inside" even at an informal level. But maybe you'd like to start with simpler sets; say, connected open sets. --Trovatore (talk) 21:09, 22 December 2010 (UTC)[reply]

Hmm, if you have a (real) vector space, does

\forall a\in A:\exists a'\in A:\exists \lambda \in (0,1):\lambda a+(1-\lambda )a'\in B

say what you mean? —Bkell (talk) 21:12, 22 December 2010 (UTC)[reply]

… er, together with

\forall b\in B:\exists a,a'\in A:\exists \lambda \in (0,1):\lambda a+(1-\lambda )a'=b

, to make sure all of B is included? —Bkell (talk) 21:14, 22 December 2010 (UTC)[reply]

On second thought, this isn't quite enough, I suppose. Or maybe it is, I don't know. I'm not sure I understand the "intuitive" notion you're trying to describe. Suppose the set A is, say, an Archimedean spiral in the Euclidean plane, and B is a connected set disjoint from A. Is B "inside" A? What if A isn't the whole spiral, but just several revolutions of it, and B is a connected set "inside" the spiral? —Bkell (talk) 21:22, 22 December 2010 (UTC)[reply]

Okay, how about just saying that the connected component of

{\overline {A}}

that contains

B

is bounded? —Bkell (talk) 21:35, 22 December 2010 (UTC)[reply]

That might work, but I'll have to think about it a bit more to be sure. JIP | Talk 21:39, 22 December 2010 (UTC)[reply]

I have thought about it a bit, and think that it does indeed work. It even generalises to non-connected sets: a non-connected set is "inside" another non-connected set if each one of its connected components is "inside" one of the other set's connected components. JIP | Talk 18:28, 23 December 2010 (UTC)[reply]

I should perhaps try to clarify my intuitive definition. Both $A$ and $B$ have nonzero area, in other words they have nonzero interiors and are not equal to just their boundaries. I am assuming a metric topology throughout this question, so there is no need to consider non-metric topology. This question generalises to non-metric topology in at least three dimensions, if not more, but does not generalise to non-metric topology. Perhaps one way to help understand the definition would be to say that it is not possible to construct a third set $C$ , $C\cap A=\emptyset ,C\cap B=\emptyset$ , so that the closures of $B$ and $C$ intersect but the closures of $C$ and $A$ do not. (This is possible if $B$ does not completely fill up the "hole" in $A$ but I think it's not if it does.) JIP | Talk 21:38, 22 December 2010 (UTC)[reply]

Having nonempty interiors doesn't help much; there could be all sorts of mess that's not in the interior. What about a set such as, you start with the closed disk centered at the origin with radius 1, and then for points at distance between 1 and 2 from the origin, you add them if they have at least one rational coordinate? There's a connected set with nonempty interior, not equal to its boundary, so it satisfies all the stipulations you've given thus far, but I'm unable to decide from what you've written whether it's "inside", say, the set constructed in the same way, except the inner radius (full disk) is 5/4 and the outer radius (one rational coordinate) is 7/4. --Trovatore (talk) 22:55, 22 December 2010 (UTC)[reply]