       Re: Differentiation ?

• To: mathgroup at smc.vnet.net
• Subject: [mg14694] Re: [mg14668] Differentiation ?
• From: Jurgen Tischer <jtischer at col2.telecom.com.co>
• Date: Sun, 8 Nov 1998 21:15:51 -0500
• References: <199811070710.CAA01906@smc.vnet.net.>
• Sender: owner-wri-mathgroup at wolfram.com

```Hi Yves,

In:= s = Sum[y[i]^2 - 2 mu y[i] + mu^2, {i, 1, n}]

In:= Evaluate /@ D[s, mu]

Out= Sum[2*mu - 2*y[i], {i, 1, n}]

You need the Evaluate/@ because Sum has Attribute HoldAll. Now if you
insist in the formula you ended up with (or better with the correct
version) you would have to write some simplification rules to apply to
the result, something like:

In:= simplifySum={
Sum[x_+y_.,{n_,na_,ne_}]/;FreeQ[x,n]:>(ne-na+1)x+ Sum[y,{n,na,ne}],
Sum[x_ y_,{n_,na_,ne_}]/;FreeQ[x,n]:>x Sum[y,{n,ne,na}]}

In:= Evaluate/@D[s,mu]/.simplifySum

Out= 2*mu*n + Sum[-2*y[i], {i, 1, n}]

Now to the second problem:
First of all, I think your formula (apart from the trivial error of a
factor of 2 of the first term) is false. In any case it's a problem of
interpretation. Let me show what I mean with an example, using matrices
in M(2,2). So my interpretation of your formula  E'E = Y'Y - B'X'Y +
B'X'XB is that there you have a function f[B_]:=Y'Y-B'X'Y+B'X'X B where
Y, X are constants (in M(2,2)), and you are searching the derivative.
Now by linearity of the derivative we can interchange derivation and
addition so we have: the first term is constant, so the derivative is
0, the second term is linear (in b) so the derivative is THE SAME
linear function and the third term is quadratic, so by the rules for
derivatives of bilinear forms (see for example Dieudonne', A treatise
in analysis) we get all together D[f,B][U]==U'X'Y+U'X'X B+B'X'X U. So
far the treatment without Mathematica, now lets do it with Mathematica.
We identify M(2,2) with R^4 (just use Flatten on the matrices), so the

In:=
x={{x11,x12},{x21,x22}};
y={{y11,y12},{y21,y22}};
b={{b11,b12},{b21,b22}};
u={{u11,u12},{u21,u22}};

In:= f[{b11_,b12_,b21_,b22_}]=
Flatten[Transpose[y].y-Transpose[b].Transpose[x].y+
Transpose[b].Transpose[x].x.b];

In:= df=Outer[D,f[{b11,b12,b21,b22}],{b11,b12,b21,b22}];

(* This df is in M(4,4) and represents the derivative in the usual
matrix form. Lets check if the two results coincide. *)

In:= df.Flatten[u]==
Flatten[-Transpose[u].Transpose[x].y+Transpose[u].Transpose[x].x.b+
Transpose[b].Transpose[x].x.u]//Simplify

Out= True

This brings us back to your original question: Is it possible to
implement this type of derivation in Mathematica. As I see it, if you
need a concrete derivative of that type (in finite, low order
dimension) the method of identifying M(n,n) with R^(n^2) is viable. If
you want derivation in Banach spaces in a theoretic fashion, you would
have to implement it, and I think that would be quite a challenge
(think only how to implement the rule for multilinear functions and how
to identify a multilinear function automatically).

Jurgen

Yves Gauvreau wrote:
>
> Hi,
>
> I saw this equation in a book (Sum => greek SIGMA)
>
>         Sum ei^2 = Sum (yi^2 - 2 mu yi + mu^2)
>
> That easy to implement but there's this one
>
>         D[ Sum ei^2, mu] = 2 Sum yi - 2 n mu
>
> and this one to
>
>         E'E = Y'Y - B'X'Y + B'X'XB
>
>         D[E'E,B] = -2 X'Y + 2X'XB
>
> the ' mean Transpose[]
>
> How can I implement these type of Differenciation ?  Is it possible to
> do that in Mathematica ?
>
> Thanks
> Yves

```

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