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Finding the diversity order of Gamma function

John G
Posted 11 years ago
I would like to simplify the following expression of Gamma function to obtain the diversity order (d) using Mathematica.
We assume that (\[Alpha]/s) \[LessLess] 1
d = Limit[-(Log[(2 - ((2 Gamma[ m, ((m*(\[Alpha]/s)^(1/n))/\[CapitalOmega])])/Gamma[m]))^ M])/Log[s], s -> \[Infinity]]

Thanks for your support!!
POSTED BY: John G
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The expression "d" isn't being evaluated. This means that the limit , as it is written now, must be hard to compute or doesn't have an easy solution.

If you haven't already tried it, please try using the Assumptions option for Limit. For example, you can try assuming that the parameters other than "s" are Real numbers or that they are positive. It is possible that these assumptions are needed to make the problem solvable:
reference.wolfram.com/mathematica/ref/Assumptions.html

Other than solving the limit, I am not sure what you are doing. If you can explain further, this might help us.
POSTED BY: Sean Clarke
John G
John G
Posted 11 years ago
Thanks for your reply. However, the original function is (2 - ((2 Gamma[ m, ((m*(?/s)^(1/n))/?)])/Gamma))^ M , and I want first to simplify it by assuming (?/s) ? 1 , which means that (s) is very high positive value, thus how can we approximate this function when the condition (?/s) ? 1 is met?. By doing so, we can solve the limit.
POSTED BY: John G
There really isn't a way to clearly convey (a/2)<<1 to the Limit function since it isn't reall a rigorous statement. Are you trying to approximate the value of the limit? If this limit has a closed solution, I haven't found it yet.

The best way to convey what you would like to do with this limit might be to come up with a simpler example of the same kind of problem that you know how to solve and see how you would solve it in Mathematica.
POSTED BY: Sean Clarke
Could the Limit be preformed in pieces?  (Mathematicians might correct this.)

Here I take the innermost section, take its limit as s-> Infinity, then stick in in
in place of the original counterpart.
 In[2]:= Limit[-(2 Gamma[m, ((m*(\[Alpha]/s)^(1/n))/\[CapitalOmega])]),             
           s -> Infinity,                                                            
              Assumptions -> {n > 0, m > 0 , 0 < \[Alpha] < s,                           
             M > 0, \[CapitalOmega] > 0}]                                                
 
 Out[2]= -2 Gamma[m]
 
 In[3]:= d=Limit[-(Log[(2-((2 Gamma[m])/Gamma[m]))^M])/Log[s],  s->Infinity,            
            Assumptions->{n>0, m>0 , 0<\[Alpha] <s, \[CapitalOmega]>0}]
Out[3]= Indeterminate 
It appears that one ends up with
Log[0] / Log[Infinity]
POSTED BY: Bruce Miller
John G
John G
Posted 11 years ago
Finding the diversity order of Gamma function
POSTED BY: John G
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