May 27th, 2008, 05:46 AM  #1 
Newbie Joined: May 2008 Posts: 4 Thanks: 0  Is this in anyway solvable?
Hi, For my Bsc Thesis, i'm trying to create random variables for a custom pdf (probability denisty function). The final step involves solving the cdf function: y = 1(exp(a*x)*(1+a*x) So what i'm trying to create is a function x = f(y) from y = f(x) However is this possible? I have been trying to crack it all day, maybe I'm just not seeing it. Thanks a million! 
May 27th, 2008, 06:07 AM  #2 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
That looks like it could probably be solved with Lambert's W function. Would it help you to have a solution in terms of a special function? I don't suspect there's a closedform solution in +, , *, /, ^, log, and exp. 
May 27th, 2008, 10:11 AM  #3  
Newbie Joined: May 2008 Posts: 4 Thanks: 0  Quote:
I was frustrated and therefore used mathematica to solve it and indeed fould a solution that involved the productlog. However once I tested it the mathematica generated function did not return correct values! It would help me to have a solution, so if you are able to solve I would greatly appreciate this! would it be possible to do in steps so I can follow and learn? Ps. Should I have know how to do this? I have only had introductionary calculas. Right now I'm going to read up on the Lambert's W/ Product log. Thanks a million..  
May 27th, 2008, 11:03 AM  #4 
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms 
I'll look into some material on the function myself to see if I can help. I've used it to solve problems, but I'll admit its use does not come fully naturally to me yet. As an aside: How can it be that you're writing your thesis, but have only taken introductory calculus? My college required introductory/advanced/multivariable calculus and real analysis (total 15 semester hours) for any math major, and even that seems flimsy to me since it avoids complex analysis.* * As I was not required, I never did take complex analysis. I still feel the lack. 
May 27th, 2008, 11:20 AM  #5  
Newbie Joined: May 2008 Posts: 4 Thanks: 0  Quote:
I really wish I had the opportunity... cause I would of jumped on it!  
May 27th, 2008, 12:32 PM  #6  
Member Joined: Aug 2007 Posts: 93 Thanks: 0  Re: Is this in anyway solvable? Quote:
i.e. let a= discount rate and x = time then forgetting about the 1 term it is (1+ax) / exp(ax) so if a = 5% and t=2 it is 1.1 / 1.105  
May 27th, 2008, 12:50 PM  #7  
Global Moderator Joined: Nov 2006 From: UTC 5 Posts: 16,046 Thanks: 938 Math Focus: Number theory, computational mathematics, combinatorics, FOM, symbolic logic, TCS, algorithms  Quote:
Of course, because you're in biology not math. Sorry. I'll test the function a but when I get home.  
May 28th, 2008, 02:48 AM  #8  
Newbie Joined: May 2008 Posts: 4 Thanks: 0  Re: Is this in anyway solvable? Quote:
I am modeling (pollutant) dispersal from multiple sources but to start only from a single point source. I will use different probability density functions to model this. Though I thought I might start with the exponential: http://mathworld.wolfram.com/Exponentia ... ution.html Cause it looked simple enough. I suppose that this demonstrates dispersal in one single direction; however I am interested in dispersal along all directions (for now Iâ€™m assuming isotropy). So I arcwise integrated it and renormalized it (= ensuring that the area under the curve is 1). That resulted in this curve: (a/(2*pi))*2*pi*x*a*exp(a*x) A second step would be to calculate its cumulative density function (cdf), I got this one: y = 1(exp(a*x)*(1+a*x) The last and final step would be to solve the x=cdf(y) function so I can use a random number generator to start stochastic â€œsimulationâ€? of pollutant dispersal. And that is where I got stuck! So thatâ€™s the history and todayâ€™s a new day.. so Iâ€™ll work on it again. Thanks for the clues so far!  

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