> On this subject, I have never quite understood the domain of d. 
> 1 .  For example:
> 
> f=:***:
>    f i:5
> _25 _16 _9 _4 _1 0 1 4 9 16 25
>    (f d.1) 2
> |domain error
> |       (f d.1)2
> 
> f seems a perfectly reasonable function to differentiate: f' is
> continuous.  While I would not expect this to be differentiated
> symbolically, I would have thought that either polynomial approximation or
> secant slope (the methods indicated in the Dictionary) would 
> work.  Or maybe there are stability issues?

f d. 1 is a symbolic differentiator and is table driven plus a few rules.  
In this case I believe it tried to use the product rule and then
foundered on the monad * which is not differentiable.   In contrast:

   (1&o. * 2&o.) d. 1
(2&o. * 2&o.) + 1&o. * -@(1&o.)

For * * *: to be differentiable the function itself would have to
be in the table.



----- Original Message -----
From: John Randall <[EMAIL PROTECTED]>
Date: Sunday, October 14, 2007 4:29
Subject: Re: [Jgeneral] Brief example
To: General forum <[email protected]>

> Roger Hui wrote:
> > 0. The following is a shorter equivalent for "newton":
> >    newton=:1 : '- u % u d. 1'
> >
> Thanks for this: I should have seen it myself.
> 
> > 1. Showing convergents:
> >
> > 2. Rational numbers:
> >
> 
> These are good examples, and I should do them.  I would 
> like to include
> ^:_ , since it also shows tolerant equality.
> 
> Thanks, Raul, for the symbolic derivatives.
> 
> On this subject, I have never quite understood the domain of d. 
> 1 .  For
> example:
> 
> f=:***:
>    f i:5
> _25 _16 _9 _4 _1 0 1 4 9 16 25
>    (f d.1) 2
> |domain error
> |       (f d.1)2
> 
> f seems a perfectly reasonable function to differentiate: f' is
> continuous.  While I would not expect this to be differentiated
> symbolically, I would have thought that either polynomial 
> approximation or
> secant slope (the methods indicated in the Dictionary) would 
> work.  Or
> maybe there are stability issues?
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