Re: Re: Re: )
- To: mathgroup at smc.vnet.net
- Subject: [mg66457] Re: [mg66442] Re: Re: )
- From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
- Date: Sun, 14 May 2006 02:57:32 -0400 (EDT)
- References: <200605050902.FAA28575@smc.vnet.net> <e3hf84$m1h$1@smc.vnet.net> <200605090635.CAA18518@smc.vnet.net> <e3sh4a$luu$1@smc.vnet.net> <200605120603.CAA17394@smc.vnet.net> <A97105AC-D52E-4163-A9FA-5EDFDBE620FF@mimuw.edu.pl>
- Sender: owner-wri-mathgroup at wolfram.com
*This message was transferred with a trial version of CommuniGate(tm) Pro* On 13 May 2006, at 21:04, Andrzej Kozlowski wrote: > We have already seen that: > > > Infinity^Infinity > > ComplexInfinity > > but note that actually: > > Limit[z^z, z -> Infinity] > > Infinity > > The choice of direction does not make any difference here: > > > Limit[z^z, z -> Infinity, Direction -> I] > > Infinity > > Moreover: > > > Limit[z^z, z -> ComplexInfinity] > > Infinity > > (I don't get this one) > Actually, I do get this one. It seems reasonable that Direction has no effect on Limit[z^z, z -> Infinity] since there is only straight line direction towards DirectedInfinity [1]. But, the situation is different in the case of ComplexInfinity. The seemingly strange answer Limit[z^z, z ->ComplexInfinity] Infinity derives from the default direction of Limit (towards 1). So now I do get it now, although I still do not like it since I would much prefer the Riemann sphere model to be used consistently here and the answer ComplexInfinity to be returned. There is a certain duplication involved in the above answer, the two inputs (with one with Infinity and one with ComplexInfinity are interpreted by Mathematica to mean the same thing - because of the default direction). But I can live with this. This example is even more interesting: Limit[z^z, z -> ComplexInfinity, Direction->I] 0 This is completely reasonable, since ComplexExpand[ Abs[(a*I)^(a*I)]] E^((-a)*Arg[I*a]) so the modulus tends to 0. This, indeed agrees with Limit[z^z,z->I Infinity] 0 So on the whole, I think limit already works in a fairly reasonable way although I would still prefer and Assumptions based approach or maybe one based on an option Model, with values such as RiemannSphere and DirectedInfinities. In particular Limit[1/z, z->0, Model->RiemannSphere] should return ComplexInfinity while Limit[1/z,z->0,Model->DirectedInfinities] should return Infinity (which is what happens by default). On the other hand, direct arithmetical operations on infinite quantities in mathematica, still seem to me by and large meaningless. Andrzej Kozlowski Tokyo, Japan
- References:
- When is x^y = != E^(y*Log[x])
- From: ted.ersek@tqci.net
- Re: )
- From: Maxim <m.r@inbox.ru>
- Re: Re: )
- From: Maxim <m.r@inbox.ru>
- When is x^y = != E^(y*Log[x])