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MathGroup Archive 2005

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Re: Bypassing built-in functions in differentiation

  • To: mathgroup at smc.vnet.net
  • Subject: [mg62647] Re: [mg62577] Bypassing built-in functions in differentiation
  • From: <bsyehuda at gmail.com>
  • Date: Wed, 30 Nov 2005 00:06:47 -0500 (EST)
  • References: <200511290944.EAA08566@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hi Ofek,
If have an access to the Mathematica Journal, look at Volume 8, Issue 4.
Chapter "In and Out" by Paul Abbott (page 529) gives a full solution for
matrix derivatives
yehuda
On 11/29/05, Ofek Shilon <ofek at simbionix.com> wrote:
>
> Dear MathGroup.
>
> consider the following statement:
>
> Dt[Transpose[a]]
>
> which evaluates to
>
> Dt[a] Transpose`[a]
>
> that is, mathematica treats Transpose as a function and uses the chain
> rule. i can try and bypass this behaviour manually:
>
> Unprotect[Dt];
> Dt[Transpose[x_]] := Transpose[Dt[x]]
>
> but now consider expressions like -
>
> Dt[Transpose[a].b]
>
> which still produces:
>
> Transpose[a].Dt[b] + Dt[a] Transpose`[a] b
>
> which is a bit surprising. i can of course bypass this behaviour
> manually as well:
>
> Dt[Transpose[x_].y_] := Transpose[Dt[x]].y + Transpose[x].Dt[y]
>
> which gives the desired result, but then check the following -
> Dt[(Transpose[a].b)^2]
>
> etc. etc.
>
> i tried also to define -
> Dt[Transpose[x_]] =1
>
> which produces readable results, but discards the (correct, and needed)
> 'Transpose' head over a factor in the differentiation.
>
> There has to be a general solution. is there a 'hook' where i can
> interfere with the derivative computation? (i thought user definitions
> would suffice, but apparently not)
>
>
> thanks,
>
> Ofek
>
>


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