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Re: Re: Questions regarding MatrixExp, and its usage

On 27 Dec 2005, at 09:58, Andrzej Kozlowski wrote:

> On 27 Dec 2005, at 09:42, Andrzej Kozlowski wrote:
>> On 27 Dec 2005, at 08:19, Andrzej Kozlowski wrote:
>>>  Now you should  know that in general this is not going to hold  
>>> in all of complex plane (but will hold in most).
>> I wrote the above quite thoughtlessly: obviously there no sense in  
>> which the equality "holds in most of the complex plane". Clearly  
>> in every sense it holds just as often as it does not. Sorry about  
>> that; I replied too quickly.
>> Andrzej Kozlowski
> One more correction is needed: in a certain obvious sense the  
> equation does not hold "more often" than it holds since Exp is a  
> surjective mapping of the compelx plane to itself which covers it  
> infinitely many times (the fibre is Z - the integers).
> Because of the holidays I am now constanlty in a rush and can't  
> find enough free time even to write a proper reply!
> Andrzej Kozlowski

Actually, even the above is not strictly correct: Exp is a surjective  
mapping form the complex plane to the complex plane minus the point  
0. Now that I have a little bit of time I can try to analyse the  
entire problem more carefully. (In fact, I have not taught complex  
analysis for over 15 years and I have become a little bit rusty. So  
when I first saw this post I thought the problem lied in the branch  
discontinuity of Log, which is why I wrote the relations was true in  
"most of the complex plane". Of course I was completely wrong in this  

Let's again define the function

f[x_, y_] := E^(x*y) - E^(y*Log[E^x])

We want to investigate where in the complex plane this is 0. This is  
by definition the same as

f[x, y]

E^(x*y) - (E^x)^y

First, this is going to be zero for any real x and and an arbitrary y:



Secondly, suppose we have any pair of complex numbers a,b where f[a,b] 
==0. That is:

a /: f[a, b] = 0;

Then we have

    f[a + 2*Pi*I, b]]]

E^(a*b + 2*I*b*Pi) - (E^a)^b

This will be zero if an only if b is an integer:



So for every pair (a,b) for which the identity holds and b is not an  
integer we can generate uncountably many pairs for which it does not  
hold by simply adding 2*Pi*I to a. For example:



FullSimplify[f[2 + 2*Pi*I,

(-1 - I)*E^(3/2)

On the other hand, we can get pairs of complex numbers for which the  
identity holds provided the imaginary part of the first complex  
number is not large:



However, for complex numbers with large imaginary part:

Simplify[f[1 + 12*I, 2 + 3*I]]

(-E^(-34 + 27*I))*
   (-1 + E^(12*Pi))

it is easy in this way to give a complete description of the pairs  
(a,b) for which f is 0, but I will skip it and turn to matrices.

In this case, while I am not 100% sure, I tend to believe the  
situation to be quite analogous. We are interested in the equation


I believe this will hold for real matrices B and (probably) all  
complex p but will not hold in general. In fact I believe most what I  
wrote above can be generalised to this case, although the statements  
and proofs would be more complicated.

Let's just illustrate this in the case of a 2 by 2 random  matrix.



Let's take some complex p, e.g. 1+I

N[MatrixExp[B*(1 + I)]]==N[MatrixPower[MatrixExp[B],1+I]]


To produce a case where the relationship does not hold just imitate  
the procedure for complex numbers given above. First we add to   2Pi  
* times the identity matrix to B:

Z = B + 2 Pi*I IdentityMatrix[2];

For p take any non-integer number, real or complex:



Andrzej Kozlowski

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