Wikipedia:Reference desk/Archives/Mathematics/2014 March 14

= March 14 =

Jacobians and vector field differentials
Am I correct in saying that the Jacobian matrix is essentially taking the gradient of each component of a vector field?

In what sense does $$d\mathbf{w}=J_{\mathbf{w}} \cdot d\mathbf{x}$$? There, w is a vector field, x is position (in column vector form, for matrix multiplication purposes), and the dot product not a real dot product but an "inner product" (I'm not too sure of this) in the sense that each entry of the Jacobian appears represent the (scalar) product of a partial derivative and a dyadic unit tensor for each unit vector pair (according to an expert I asked). Further, am I correct in asserting that taking that "inner product" of the Jacobian with a vector produces a directional derivative of the vector field?

The dyadic tensor article is of little help for me because "juxtaposition" (of vectors) is not something that makes sense to me.--Jasper Deng (talk) 05:28, 14 March 2014 (UTC)


 * If you take "vector field" to mean function on Rn with values in Rm, where each of the domain and codomain are represented as column vectors, then the Jacobian matrix of such a function $$\mathbf w:\mathbb R^n\to\mathbb R^m$$ is the m&times;n matrix whose rows are the gradients of the components of w. In the formula $$d\mathbf{w}=J_{\mathbf{w}} \cdot d\mathbf{x}$$, the "dot" is ordinary matrix multiplication of an m&times;n matrix by an n&times;1 column vector.  (This is actually a form of the chain rule.)  Finally, if u is a unit vector in Rn, then the directional derivative of w along u is $$J_{\mathbf w}\mathbf u$$.
 * All of this assumes that you are working in Euclidean space with a Cartesian coordinate system. It is no longer true in a general coordinate system (or on a manifold), since strictly speaking the gradient doesn't transform correctly under coordinate transitions.  It is contravariant, which means that it transforms as a column vector rather than a row vector.  Instead of the gradient, you need to use the differential, which transforms as a row vector.   Sławomir Biały  (talk) 11:52, 16 March 2014 (UTC)

how to replace 1.66^-18 to scientific form?
I know that it equals to 1.66*10^-17 (because someone tell me that...), but how can I understand and do that by myself? May I do that by my calculator? (I have fx82 of Casio) 5.28.179.184 (talk) 23:40, 14 March 2014 (UTC)
 * First of all, if you mean $$1.66^{-18}$$, then that is not even correct - that equals about 0.00010915771. --Jasper Deng (talk) 04:52, 15 March 2014 (UTC)
 * Please double-check the problem; if it actually states $$1.66 \times 10^{-18}$$, then that is already in proper scientific notation. -- Kinu  t/c 04:59, 15 March 2014 (UTC)
 * Sorry, I don't understand what the problem is. The origin question is "How many moles - million molecules water they are". So, the formula is: $$n=\frac{N}{Avogadro}$$, then, according this formula we get: $$1.66^{-18}=\frac{10^6}{6.02*10^{23}}$$. Someone tell me (and I saw that he's right) that $$1.66^{-18}=1/6*10^{-17}$$. My question was (and is) how can I reach to the scientific form ($$1/6*10^{-17}$$) by the calculator, not by someone else...5.28.179.184 (talk) 08:38, 15 March 2014 (UTC)
 * The answer is $$1.66 \times 10^{-18}$$ (to 3 significant figures). This is already in scientific notation form. The expression $$1.66 ^{-18}$$, although it looks similar, has a very different value. What may be confusing you is that your calculator probably displays the answer as $$1.66 \text{E}{-18}$$, which is $$1.66 \times 10^{-18}$$ in E notation. Gandalf61 (talk) 10:56, 15 March 2014 (UTC)
 * Actually, the calculator likely does display it as $$1.66^{-18}$$ (that's how it looks on my Casio calculator). That's pretty harmless if you know what you're doing and how the calculator displays information, but it can certainly be a source of confusion. -- Meni Rosenfeld (talk) 17:30, 16 March 2014 (UTC)
 * Are you sure there isn't a tiny ×10 between 1.66 and -18? The online Casio calculator documentation I found shows that, including for Casio fx82MS. PrimeHunter (talk) 18:43, 16 March 2014 (UTC)
 * My Casio fx-7400G Plus displays scientific numbers without the ×10 or an E.--Salix alba (talk): 18:56, 16 March 2014 (UTC)
 * OK, it appears to vary. See for a tiny ×10 in a Casio fx-300MS. PrimeHunter (talk) 19:12, 16 March 2014 (UTC)


 * Some calculators do not display the x10 see these pictures.
 * https://lh5.ggpht.com/QP2gGoYmn9hOjFqc1Hng4hI7mNjQoURc6YrVbgTXpluKBTnzhoFiS25I-YJJHW5Afw=h900
 * https://lh4.ggpht.com/RNUDeFEdrZhzQ3L-EJvULhx2Y2mO7iuDdNJHjfuNrpcsblic6mP5IU3kWbPz0LVYuU8=h900

202.177.218.59 (talk) 22:52, 16 March 2014 (UTC)