       Re: Bypassing built-in functions in differentiation

• To: mathgroup at smc.vnet.net
• Subject: [mg62666] Re: [mg62577] Bypassing built-in functions in differentiation
• From: Ofek Shilon <ofek at simbionix.com>
• Date: Wed, 30 Nov 2005 05:40:39 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```Hi Yehuda

I don't have access to the mathematica journal at the moment, but i suspect
their solution
for matrix derivatives relates to matrices with explicit entries, that is -
explicit 2 dimensional arrays.
is this the case?     (i'll try and get my hands on this journal issue
either way, of course)

thanks,

Ofek

_____

From: bsyehuda at gmail.com [mailto:bsyehuda at gmail.com]
To: mathgroup at smc.vnet.net
Subject: [mg62666] Re: [mg62577] Bypassing built-in functions in differentiation

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

'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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