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August 24

A term

What is a word for "non-arbitrarily close" as in "We can find the equation of a function f if we know its degree to be n and n+1 "non-arbitrarily close" points on f"? 68.76.159.51 (talk) 00:12, 24 August 2010 (UTC)[reply]

Discrete, maybe? As an aside, finitely many points are always discrete. Invrnc (talk) 03:01, 24 August 2010 (UTC)[reply]

Whoops, I mean finitely many distinct points. Invrnc (talk) 03:02, 24 August 2010 (UTC)[reply]

I'm not sure what you are describing is meaningful, or at least I don't really understand what it could mean. Any two points in R² have some positive distance between them unless they're equal. You could say the points are distinct as Invrnc mentioned to describe that no two of them are equal. If you want some at least some minimum distance d between any pair of points you could say they are separated by at least a distance d. Rckrone (talk) 05:22, 24 August 2010 (UTC)[reply]

I, too, don't understand exactly what you want. The suggestions above are good if you meant what the respondents think you may have. If OTOH you meant "not necessarily close, just any old points" then arbitrary is the usual term.—msh210℠ 16:20, 24 August 2010 (UTC)[reply]

Certainly there is a unique nth degree polynomial that fits prescribed values at n + 1 distinct points in the domain. I suspect that is what is meant. I don't know what "non-arbitrarily close" means, unless maybe it's a clumsy way of saying "distinct". Michael Hardy (talk) 01:52, 25 August 2010 (UTC)[reply]

Set formula conversion

I was looking at Dice's coefficient and Jaccard index. In some papers, it is claimed that Dice's coefficient is twice the Jaccard index. In others, it is claimed that Dice's coefficient cannot be translated to Jaccard index. In the articles here, it claims that the relationship between Dice's coefficient (D) and Jaccard index (J) is D=2J/(1+J). It appears that it is converting |X|+|Y| = |X∪Y|+|X∩Y|. Is any of this correct? If I could translate Dice's coefficient directly to Jaccard index, it would be helpful. -- kainaw ™ 12:08, 24 August 2010 (UTC)[reply]

I am not familiar about this particular topic, but |X| + |Y| = |X ∪ Y| + |X ∩ Y| is correct, which implies that the conversions D = 2J/(1 + J) and J = D/(2 − D) are also correct, given the definitions in the two articles. Therefore the other papers you mention either use different definitions or are in error.—Emil J. 12:48, 24 August 2010 (UTC)[reply]

Actually, now I noticed that the Dice's coefficient article contradicts itself. It says that he coefficient is defined as twice the shared information (intersection) over the combined set (union), and then it gives the expression 2|X ∩ Y|/(|X| + |Y|). However, twice the intersection over the union is actually 2|X ∩ Y|/|X ∪ Y|, which is exactly twice the Jaccard index. Go figure.—Emil J. 12:53, 24 August 2010 (UTC)[reply]

Because these come from botany, it is very possible that there are multiple formulas running around with the same name. I will try to hunt down Dice's original paper and see what his formula was. -- kainaw ™ 13:14, 24 August 2010 (UTC)[reply]

Quaternion question

If I have a quaternion $\Omega$ , which I know is equal to $2{\dot {q}}q^{\dagger }$ , where the overdot denoted differentiation with respect to time, how is it possible to find $q$ ? I know $q$ is a unit quaternion if that helps.--Leon (talk) 18:13, 24 August 2010 (UTC)[reply]

Could you explain your notation a little, please. What does the

\scriptstyle \dagger

represent? Is that a footnote marker that you have copied verbatim from a text, or does it have some other notation meaning? — Fly by Night (talk) 19:32, 24 August 2010 (UTC)[reply]

The dagger refers to quaternion conjugation. ALSO, I forgot to explicitly mention that the quaternion varies as a (differentiable) function of time but is always a unit quaternion.--Leon (talk) 19:57, 24 August 2010 (UTC)[reply]

What you have is a nonlinear differential equation. You can solve it numerically, but it is non-trivial. You have a relation that defines q multiplied by dq/dt - in otherwords, not a linear combination of q and its derivatives. Your constraint to be unit-quaternion is not a proper boundary-condition, but it can serve as a regularization condition for a numerical solver. Nimur (talk) 20:42, 24 August 2010 (UTC)[reply]

Could the expression $2{\dot {q}}q^{\dagger }$ have emerged from a chain rule? Did you mean that q varies or that Ω varies? Are you saying Ω is a known unit-quaternion-valued function of time, or maybe that Ω is a fixed (constant) and known unit quaternion? Michael Hardy (talk) 01:29, 25 August 2010 (UTC)[reply]

\Omega

is a function of time, as is q. I know

\Omega

and I know that q is a unit quaternion. I want to find q. I also know that

\Omega

is a pure (complex) quaternion, which can be demonstrated from the above expression, knowing q is a unit quaternion for all time.--Leon (talk) 07:23, 25 August 2010 (UTC)[reply]

If

q=a+bi+cj+dk\,

where a, b, c, d are real, then q is a unit quaternion precisely if

1=qq^{\dagger }=a^{2}+b^{2}+c^{2}+d^{2},\,

and the derivative of that with respect to time is

0={d \over dt}1={d \over dt}(a^{2}+b^{2}+c^{2}+d^{2})=2a{\dot {a}}+2b{\dot {b}}+2c{\dot {c}}+2d{\dot {d}}.

If I look at $2{\dot {q}}q^{\dagger }\,$ and expand it, I get

2a{\dot {a}}+2b{\dot {b}}+2c{\dot {c}}+2d{\dot {d}}+(a{\dot {b}}-b{\dot {a}}+c{\dot {d}}-d{\dot {c}})i+\cdots \,

and the sum of the first four terms vanishes as above. Now $a{\dot {b}}-b{\dot {a}}\,$ is the numerator in an application of the quotient rule, as is $c{\dot {d}}-d{\dot {c}}\,$ , etc. I don't know where this will take us. So just some preliminary scratchwork. Michael Hardy (talk) 02:02, 25 August 2010 (UTC)[reply]

The equation is equivalent to $\Omega =2{\dot {q}}q^{-1}$ on account of |q| = 1. If it were complex numbers rather than quaternions, then the solutions to this equation would be $\textstyle q(t)=\exp \left({\frac {1}{2}}\int _{t_{0}}^{t}\Omega (u)\,du+c\right)$ , where c is any constant (with real part 0 to ensure |q| = 1). I don't know whether one can do such thing with quaternions, but since the complex numbers are a special case (where the j and k components of q and Ω are 0), the solution will have to involve some quaternion generalization of the exponential.—Emil J. 11:43, 25 August 2010 (UTC)[reply]

Well, actually, would it work to just take this expression as is, and interpret it in quaternions using the exponential here? The only problem is the chain rule. The definition ensures that

\exp(z)\in \mathbb {R} (z)\simeq \mathbb {C}

, hence exp(z) commutes with z for any z; is this enough to make (exp(f(t)))' = f'(t)exp(f(t))?—Emil J. 12:10, 25 August 2010 (UTC)[reply]

I don't think it is, though I might be wrong. I know that doesn't hold for general matrices, but it might hold in some sense for quaternions. —Preceding unsigned comment added by Star trooper man (talk • contribs) 12:15, 25 August 2010 (UTC)[reply]

I guess you are right, I'd need f(t) to commute with f'(t). The solution above is valid if q(t) commutes with its derivative, or Ω(t) commutes with

\textstyle \int _{t_{0}}^{t}\Omega (u)\,du+c

. IIUIC this condition is equivalent to the values of q(t) being confined to a fixed subfield isomorphic to C, so it is rather a degenerate case.—Emil J. 13:15, 25 August 2010 (UTC)[reply]