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Re: Re: Re: Re: Questions regarding MatrixExp, and its usage

• To: mathgroup at smc.vnet.net
• Subject: [mg63597] Re: [mg63564] Re: [mg63390] Re: [mg63355] Re: [mg63335] Questions regarding MatrixExp, and its usage
• From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
• Date: Fri, 6 Jan 2006 05:24:48 -0500 (EST)
• References: <200601050812.DAA19881@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

On 5 Jan 2006, at 07:01, Michael Chang wrote:

>
> Hmm ...
>
>
>> Table[KroneckerDelta[i ² j, True] *Derivative[j - i][f][r]/(j -  
>> i)!,  { i, 1, k}, {j, 1, k}]
>>
>
> One symbol doesn't appear to be quite right on my screen (or  
> perhaps my understanding fails me!), but in the above line, should  
> this be:
>
> KroneckerDelta[i<=j,True] (a 'funny' symbol seems to appear in my  
> message between the i and j)
>
> and should the Derivative be with respect to 'x', evaluated at  
> 'r' (maybe that's what's implied, but I just wanted to make sure  
> that I'm not mis-interpretating this line as the Derivative with  
> respect to 'r')?
>

I was in a hurry and when I am in a hurry I think like a  
mathematician ;-). There is no point using KroneckerDelta here; it is  
much better to use sim-le If:

p[f_, r_, k_] := Table[If[i â?¤ j,
    Derivative[j - i][f][r]/(j - i)!, 0], {i, 1, k}, {j, 1, k}]p[f_,  
r_, k_] := Table[If[i â?¤ j,
    Derivative[j - i][f][r]/(j - i)!, 0], {i, 1, k}, {j, 1, k}]


For example



p[f, r, 5]


{{f[r], Derivative[1][f][r], Derivative[2][f][r]/2, (1/6)*Derivative 
[3][f][r],
    (1/24)*Derivative[4][f][r]}, {0, f[r], Derivative[1][f][r],  
Derivative[2][f][r]/2,
    (1/6)*Derivative[3][f][r]}, {0, 0, f[r], Derivative[1][f][r],  
Derivative[2][f][r]/2},
   {0, 0, 0, f[r], Derivative[1][f][r]}, {0, 0, 0, 0, f[r]}}

(By the way, I think your e-mail program should be able to render the  
"funny symbol" correctly if it interprets the encoding of the message  
as Unicode.)

>
>
>> The important thing is that all the derivatives required to  
>> define  the above matrices should exist!
>> If they do f[A] will be well defined.  So, if I am right, taking f  
>> [x_]:= x^z ought to give the definition of power of a matrix for   
>> provided all the above derivatives are well defined. If z is  
>> integer  certainly here will be no problems, and the definition  
>> will agree  with the usual one.
>>
>
> Perhaps I do *NOT* understand, but here's an example that I  
> considered with regards to the above definition -- assume that  
> there is simply one Jordan block, and that A=B (S=I), with
>
> A = B = {{0,1,0},{0,0,1},{0,0,0}};
>
> Suppose that we desire to find Exp[8 A] (MatrixExp[8 A]); the  
> minimal polynomial in x is x^3, and for i==j, I can see that the  
> diagonal entries will be Exp[0]==1, and that Exp[A] should be upper  
> triangular -- but the upper off-diagonal terms don't seem to be  
> quite right (using the above formula ... for instance, I don't see  
> how I can get the anticipated '8' and '8^2/2' terms) ...
>

I think your mistake is that although A is a Jordan block 8 A is not.  
So the right thing to do is this:


A = {{0, 1, 0}, {0, 0, 1}, {0, 0, 0}};

In[4]:=
{S, B} = JordanDecomposition[8*A];


f[x_] := Exp[x]


S . p[f, 0, 3] . MatrixPower[S, -1]


{{1, 8, 32}, {0, 1, 8}, {0, 0, 1}}

This seems to be right.


Andrzej Kozlowski

Tokyo, Japan

• References:
  • Re: Re: Re: Questions regarding MatrixExp, and its usage
    • From: "Michael Chang" <michael_chang86@hotmail.com>