Re: Re: Derivative of Dot[]

*To*: mathgroup at smc.vnet.net*Subject*: [mg91155] Re: [mg91119] Re: Derivative of Dot[]*From*: "Eitan Grinspun" <eitan at grinspun.com>*Date*: Thu, 7 Aug 2008 04:43:05 -0400 (EDT)*References*: <200808060908.FAA22542@smc.vnet.net>

>>I differentiate this function w.r.t. its sole argument, F' evaluates >>to 1.#1+#1.1& > >>This is reasonable, and as expected. > > Expected yes given the way the form F' works, but reasonable? A > dot product of vector and any other vector is a scalar. If you > give specific values for the vector components, I would think it > would be reasonable to return a value zero. Further, the > gradient of a vector isn't given by the derivative of the dot > product. Instead, it is the sum of partial derivatives of each > component with respect to the corresponding basis function. That > is the gradient of vector {x, y w, z^2} is given by > > In[27]:= Tr@D[{x, y w, z^2}, {{x, y, z}}] > > Out[27]= w+2 z+1 I would say that the derivative of x.x with respect to some other vector y is given by 2x.(dx/dy), where dx/dy is a tensor. When mathematica returns 1.#1+#1.1&, it is reasonable if the 1 is interpreted as the identity map, so that, e.g., (#1.1&)[x].(dx/dy)=x.1.(dx/dy)=x.(dx/dy). Unfortunately, x.1 does not simplify to x, as it would if 1 were interpreted as the identity map. Eitan

**References**:**Re: Derivative of Dot[]***From:*Bill Rowe <readnews@sbcglobal.net>