Let me a bit elaborate on what I wrote yesterday. On Wed, 29 Jun 2005, Henning Thielemann wrote:
> I think matrices and derivatives are very different issues. I have often > seen that the first derivative is considered as vector, and the second > derivative is considered as matrix. In this spirit it is used like > x^T * (D2 f)(x) * x It should be h^T * D2 f x * h (if at all the transposition notation is used) if the curvature of f at x is meant. > but this is only abuse of the common multiplication definitions. A good > interpretation and notation should seamless extend to higher derivatives. > But the interpretation above does not work in higher dimensions. > > I like the following type for derivation. > derive :: ((i -> a) -> b) -> ((i -> a) -> (i -> b)) > Here i is the index type, (i -> a) is the vector type, b is the type the > vector function maps to. Its derivative has the same type of argument, but > the result is a vector with indices of type i. You see that it is easy to > repeat the application of 'derive', just replace b by say i->b. The second > derivative yields vectors of type (i -> i -> b). This can be interpreted > as matrix because it has two indices. But this is certainly not a matrix > which represents a linear mapping as usual, but it is a matrix > representing a bilinear form. The only thing we need is a multiplication > to reduce one level of indices. > mul :: (i -> c) -> (i -> b) -> b > Though, what we still need is a general (overloaded?) definition of the > scaling of b by c and a sum of b. That is b must be a vector space with respect to c. 'derive' in this form is a bundled partial derivative, but it can be identified with the total derivative. That is derive f x i is the partial derivative of f at x with respect to the i-th component derive f x is the total derivative of f at x. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe
