20170724, 12:24  #1 
Sep 2011
3·19 Posts 
Inverse of Smoothness Probability
The probability that a random number below X is Bsmooth is given by u^{u}, where u=ln(X)/ln(B). However, I would like the do the inverse, that is, given the smoothness probability and B, how do I solve for X?
I have a solution via Newton method. Is there a closed form? 
20170724, 19:53  #2  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}·11·227 Posts 
Quote:


20170725, 09:10  #3  
Sep 2011
3×19 Posts 
Quote:


20170725, 10:07  #4  
Basketry That Evening!
"Bunslow the Bold"
Jun 2011
40<A<43 89<O<88
3×29×83 Posts 
Quote:
The links he provides are sources for p ≈ some_better_function_of(X, B). Find those better functions before trying to invert them. 

20170725, 10:21  #5  
Sep 2011
3×19 Posts 
Quote:


20170725, 16:33  #6  
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}×11×227 Posts 
Incidentally, I have just heard from much more knowledgeable people, and I will simply quote:
Quote:


20170725, 16:41  #7 
"Serge"
Mar 2008
Phi(4,2^7658614+1)/2
2^{2}·11·227 Posts 
And for the Inverse of Smoothness Probability question, you want a reasonable numerical estimate the inverse of Dickman function.
So you want to take Newton method on the inverse of Dickman ρ  because if you know the derivative of Dickman ρ by definition, then you know the derivative of the inverse of Dickman ρ. 
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