Oleg, Between what John Randall wrote and the elaboration by the author in the original, I think your attempt at correction is unsuccessful.
Here is what Conal Elliott wrote to clarify the equation I reproduced: [begin quotation] The composition on the right hand side is on linear maps (derivatives). You may be used to seeing the chain rule in one or more of its specialized forms, using some form of product (scalar/scalar, scalar/vector, vector/vector dot, matrix/vector) instead of composition. Those forms all mean the same as this general case, but are defined on various *representations* of linear maps, instead of linear maps themselves. [end quotation] John's comments have helped me understand how to think of this in terms of linear maps. I'm not entirely "there", yet, but I'm a whole lot closer than I was. Tracy On Tue, Sep 16, 2008 at 9:33 AM, Oleg Kobchenko <[EMAIL PROTECTED]> wrote: > >... >> >> http://conal.net/blog/ >> >> Perhaps most interesting is the one entitled "Higher-dimensional, >> higher-order derivatives, functionally". In that posting the chain >> rule is given by the following expression: >> >> deriv (f . g) x = deriv f (g x) . deriv g x >> > > Incorrect form: the first "." is composition, the second > should be multiplication. The composition on the right > side should be between f' and first g. > ---------------------------------------------------------------------- For information about J forums see http://www.jsoftware.com/forums.htm
