[Date Index]
[Thread Index]
[Author Index]
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
Prev by Date:
**Re: Re: Simplify**
Next by Date:
**Re: Re: Simplify**
Previous by thread:
**Re: Derivative of Dot[]**
Next by thread:
**Re: Derivative of Dot[]**
| |