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Re: Re: )
On Wed, 10 May 2006 11:00:26 +0000 (UTC), Andrzej Kozlowski
<akoz at mimuw.edu.pl> wrote:
>
> On 9 May 2006, at 15:35, Maxim wrote:
>
>> On Sat, 6 May 2006 06:20:52 +0000 (UTC), Andrzej Kozlowski
>> <akoz at mimuw.edu.pl> wrote:
>>
>>>
>>> laws of arithmetic do not hold). Some of them are hard to explain: I
>>> can't see any good reason at all why Infinity^Infinity is
>>> ComplexInfinity, and it seems to contradict the the most basic rule
>>> that x^y is always real when x and y are positive reals. Besides, as
>>> I mentioned earlier, Infinity and ComplexInfinity do not belong
>>> together in any topological model known to me (you need a
>>> "topological model" to be able to consider the issue of continuity)
>>> and should never appear in the same formula. I can only consider this
>>> as a bug, and a rather silly one.
>>>
>>
>> I think this is as it should be: we need to consider all the sequences
>> converging to Infinity, and, for example, Limit[(x + I*Pi)^x, x ->
>> Infinity] == -Infinity. So in that sense Infinity^Infinity ==
>> ComplexInfinity: when z = z1^z2 and z1, z2 go to Infinity the absolute
>> value of z always tends to Infinity but the argument can be arbitrary.
>>
>> Maxim Rytin
>> m.r at inbox.ru
>>
>
>
> One can perhaps make some sense of this even topologically if one
> assumes that map {z,w}->z^w has as its target space the Riemann
> sphere and as its domain the Cartesian product of 2-discs. In other
> words it is not a map of the form X x X -> X, as one usually
> supposes. But then it is quite a different map from the one that gives
>
> 2^Infinity
>
> Infinity
>
> and so on. I certainly cannot conceive of any sensible topology on
> the set that is the union of the complex plane, the set of
> DirectedInfinities and the point ComplexInfinity, can you? Any such
> topology would have pretty shocking properties.... And if there is
> no topology involved than what is Limit supposed to mean? Still, I
> admit, that even if things that don't make any mathematical sense may
> sometimes be acceptable in a symbolic algebra program, for purely
> practical reasons, but then they are pretty likely to lead to
> confusion and contradictions. This is in fact the current state of
> affairs in this regard and it makes me feel that the whole
> Mathematica approach to this business of Infinities maybe misguided.
>
> Of course, as we are not talking about mathematics or any kind of
> empirical science this is all ultimately a matter of taste and as we
> well know there is no accounting for tastes... or sense of humour.
>
> Andrzej Kozlowski
>
>
I think what Limit could do here in principle is taking the limits of
Abs[f[z]] and Arg[f[z]] separately. If the limit of the absolute value is
infinite and the limit of the argument exists, return DirectedInfinity. If
the limit of the argument doesn't exist, return ComplexInfinity.
Currently Mathematica just isn't being very consistent here:
In[1]:= Limit[2^(x + I*Pi/(2*Log[2])), x -> Infinity]
Out[1]= DirectedInfinity[I]
In[2]:= Limit[2^(x + I*Pi/(4*Log[2])), x -> Infinity]
Out[2]= Infinity
The explanation for this can be as follows: if we simplify the limit
expressions first, the first one yields I*2^x and the second remains
unchanged. Then substituting x = Infinity yields I*2^Infinity ==
DirectedInfinity[I] and 2^(Infinity + I*Pi/(4*Log[2])) == 2^Infinity ==
Infinity respectively. But in that case it just means that Mathematica
used some unallowed operations (cannot just substitute x = Infinity as we
don't have continuity), not that there really is some contradiction.
If we accept the convention of taking the limits of Abs and Arg, then
Out[1] is correct and Out[2] isn't (should be DirectedInfinity[1 + I]).
Also then you're quite right that the results for Infinity^Infinity and
2^Infinity are inconsistent with each other -- they should be the same
(either both ComplexInfinity or both Infinity).
Another thing that is needed for completeness is specifying that
Mathematica always assumes that the limit variable goes along a straight
line from the origin:
In[3]:= Limit[Re[z], z -> I*Infinity]
Out[3]= 0
This result wouldn't be correct for an arbitrary path: for instance, if z
= 1 + I*t and t is real, then z also goes to I*Infinity and Re[z] == 1.
Maxim Rytin
m.r at inbox.ru
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