Re: Derivative of Dot[]
- To: mathgroup at smc.vnet.net
- Subject: [mg91102] Re: [mg91055] Derivative of Dot[]
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Wed, 6 Aug 2008 05:05:33 -0400 (EDT)
- References: <200808050759.DAA09628@smc.vnet.net>
On 5 Aug 2008, at 09:59, Eitan Grinspun wrote:
> I would like to compute the gradient F' of a scalar-valued function F
> of a vector-valued argument, without knowing a-priori the dimensions
> of the vector.
>
> I am having some trouble with a very simple case.
>
> Consider the following function:
>
> F[x_] := Dot[x,x]
>
> Evaluating this function works as expected: F[{1,2}] evaluates to 5.
>
> I differentiate this function w.r.t. its sole argument, F' evaluates
> to 1.#1+#1.1&
>
> This is reasonable, and as expected. I would think that, since the
> argument of Dot must be a list, the derivative of Dot would have been
> designed to return something that is useful. While I presume that this
> is the case, I have been unable to move ahead.
>
> I evaluate the derivative at {1,2}: F'[{1, 2}] returns 1.{1,2}+{1,2}.1
>
> The Dot is not defined for scalar arguments, and therefore Simplify
> does not reduce this further. I could of course program a rule so that
> Dot[1,x_]->x, but my intent here is to understand why the derivative
> of Dot was designed the way it was---presumably there is a reason, and
> there is a proper way to make use of the derivative.
>
> Once I have the derivative, I should be able to contract it with a
> (tangent) vector to obtain the (linearized) change the (scalar)
> function value:
>
> F'[{1, 2}].{3,4}
>
> Alas, this returns (1.{1,2}+{1,2}.1).{3,4} which does not simplify
> (even after Distribute) because Dot does not operate on scalar
> arguments.
>
> I'd like some help in understanding how to use Derivative with Dot (it
> was evidently designed to be used, or there would not be a rule built
> in).
>
> Sincerely,
>
> Eitan
>
The problem is that you are deriving wrong conclusion (that the
formula you get must have some meaning) from a correct premise (that
Mathematica has some meaningful buil-in rules for differentiating
Dot). The built in rule is:
D[X[t] . Y[t], t]
X[t] . Derivative[1][Y][t] + Derivative[1][X][t] . Y[t]
where of course X[t] and Y[t] should be vectors of the same dimension
dependent on t. It is not valid for scalars. But what you are doing in
effect is defining:
X[t] := t; Y[t_] = t;
and then getting
D[X[t] . Y[t], t]
1 . t + t . 1
which is meaningless since X[t] and Y[t] cannot be scalars.
Andrzej Kozlowski
- References:
- Derivative of Dot[]
- From: "Eitan Grinspun" <eitan@grinspun.com>
- Derivative of Dot[]