Re: Assumptions question (1/m^x,x>1,m=Infinity)

*To*: mathgroup at smc.vnet.net*Subject*: [mg31077] Re: Assumptions question (1/m^x,x>1,m=Infinity)*From*: hugo at doemaarwat.nl (BlackShift)*Date*: Sun, 7 Oct 2001 03:11:42 -0400 (EDT)*Organization*: Rijksuniversiteit Groningen*References*: <9pmdb3$62v$1@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On Sat, 6 Oct 2001 07:54:11 +0000 (UTC), Andrzej Kozlowski <andrzej at bekkoame.ne.jp> wrotf: >Still, there are several possible ways to deal with some problems of >this kind. However, the example you give is just too trivial to serve as >a good illustration and actually ought to be done by hand. Still, here Can do it by hand of course, but since the formula's got quite large at the moment I wanted to make the substitution it wasn't that easy to do by hand... >is one way one could try to use Mathematica to solve this sort of >problem. Unfortunately it is not guaranteed to work in other cases, >since Integrate with Assumptions on which it depends on is very erratic. > >Observe that: >In[2]:= >Integrate[(-m^(-1 - x))*x, {m, 1, Infinity}, > Assumptions -> {Re[x] > 1}] + Integrate[(-m^(-1 - x))*x, > m] /. m -> 1 > >Out[2]= >0 > >must be the result you wanted. Hmm, that's actually quite true, I get that formula by integration, so I can use the assumption there already (why didn't I think of that). Nonetheless, isn't it possible to do it afterwards? That would be a 'nicer' thing to do, since it is part of a model, where m doesn't have to be Infity in all cases >There are other ways, but you would have to present your real problem >first. I think I can make mathematica do what I want now, but if you know a method to make the assumption that m is infinity later on in the calculations that would be very great, so here is the real problem (x=x, MU=m) Background: It is a model of starforming in galaxies, Phi[M_]dM is the ratio of stars formed with a mass between M and M+dM x is just a parameter for the model In[1]:= Phi[M_]=Cp*M^(-1-x) To normalize Cp (which is just a normalization factor) I Integrate over al possible M, from ML (lower mass, about .1 solar masses) to MU (upper mass, about 32 solar masses) In[2]:= subC=Solve[Integrate[Phi[M],{M,ML,MU}]==1,Cp][[1]] 1 Out[2]= {Cp -> -------------} 1 1 ----- - ----- x x ML x MU x with which I calculate further, with Cp in the expresions, in the final result /.subC them. In some cases it is better to choose for ML and MU the numbers 0.1 and 32, but sometimes it is algebraically easyer to assume ML->0 or MU->Infinity, so sometimes I want to do /.subC/.{ML->0.1,MU->32}, and sometimes /.subC/.{ML->0.1,MU->Infinity}, depending on the equations, but the latter isn't possible: In[4]:= subC/.{ML->0.1,MU->Infinity} Out[4]= {Cp -> Indeterminate} Because it is indeterminate when x=0, which is not the case (x~1.3). Is there anyway I can do this? without explicitly giving a value for x, since that is the last valuable to enter (so I can test what value for x is likely) BTW, why doesn't mathematica not result with an If statement which says that it is only indeterminate when x=0 and give Cp->0 otherwise. That would be more logical behavoir I think. groetjes, hugo

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**Re: Assumptions question (1/m^x,x>1,m=Infinity)**

**Re: Assumptions question (1/m^x,x>1,m=Infinity)**