Wikipedia:Reference desk/Archives/Mathematics/2009 July 6

= July 6 =

Del
Why is $$\nabla (\vec c \cdot \vec X) = 2 \vec c$$ where c is constant and X is the identity funtion (ie X(Xn) = Xn)? 76.67.76.234 (talk) 04:23, 6 July 2009 (UTC)


 * I find myself trying to guess what the question means. Your X with an arrow seems to indicate a vector in Rn, and the nabla would usually mean the gradient operator, so I'm thinking
 * $$\nabla f(\vec X) \, $$
 * is the gradient&mdash;a vector field&mdash;of the scalar-valued function &fnof;. If g is the identity function then you'd write
 * $$ g(\vec X) = \vec X \, $$

and then
 * $$ \vec c \cdot \vec X \, $$
 * would be a scalar-valued function, so you could take its gradient, which would be a vector-valued function. But what you'd get is
 * $$\vec c,\text{ not }2\vec c. \,$$
 * And if that's what you meant then your functional equation X(Xn) = X would not make sense. So I'm wondering what you have in mind. Michael Hardy (talk) 05:31, 6 July 2009 (UTC)


 * also note $$\nabla (\vec X \cdot \vec X) = 2 \vec X$$, if you want the "2" --pma (talk) 05:56, 6 July 2009 (UTC)

Isomorphism
I have a small question related to group isomorphism which has been bothering me for sometime. My intution says that isomorphic (normal) subgroups of a group G have the same structure, and so when we take the factor groups we should have the two factor groups isomorphic. Yet this is not true. In $$(\mathbb{Z},+)$$ we have two isomorphic subgroups $$n\mathbb{Z}$$ and $$m\mathbb{Z}$$ where $$m\ne n;m,n\in \mathbb{Z}^+$$. But clearly the two factor groups $$\mathbb{Z}_n$$ and $$\mathbb{Z}_m$$ are not isomorphic. Shouldn't isomorphic structures behave in the same way? Can someone please explain. Thanks--Shahab (talk) 16:23, 6 July 2009 (UTC)
 * Your intuition is miscalibrated. nZ and mZ are isomorphic as groups, so their behaviour as groups is the same. But you aren't considering them as just groups: you're considering them as subgroups of Z. They are not isomorphic as subgroups of Z (which would mean there was an automorphism of Z which restricted to an isomorphism between them). If they were, then the quotients would indeed be isomorphic, as one would expect. For an example where this works, consider Z×Z and the subgroups Z×{0} and {0}×Z. These are isomorphic not just as abstract groups but as subgroups, and indeed the quotient is isomorphic to Z in both cases. Algebraist 16:34, 6 July 2009 (UTC)


 * In other words, what counts is not only the subgroups being isomorphic objects, but also, the way they are put into the ambient group. You can find plenty of analogous situations in all mathematics. One for all: any simple closed curve is homeomorphic to a circle, but it may be embedded in R3 in many ways. --pma (talk) 17:02, 6 July 2009 (UTC)


 * Thank you both--Shahab (talk) 17:30, 6 July 2009 (UTC)
 * Another example: $$\mathbb{Z}$$ and $$2\mathbb{Z}$$ are isomorphic as sets (they have the same cardinality), but not as subsets of $$\mathbb{Z}$$. $$\mathbb{Z} \backslash \mathbb{Z}$$ is very different from $$\mathbb{Z} \backslash 2\mathbb{Z}$$ (\ is set-theoretic difference). -- Meni Rosenfeld (talk) 13:04, 7 July 2009 (UTC)

question about computer library programs for solving partial differential equations
On someone elses behalf, please see Reference_desk/Science - question 2. —Preceding unsigned comment added by 83.100.250.79 (talk) 21:53, 6 July 2009 (UTC)