Perl: the Markov chain saw  
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Re: [OT] Normalizing the return result of an exponential formulaby SimonClinch (Deacon) 
on Apr 18, 2011 at 14:43 UTC ( #899953=note: print w/replies, xml )  Need Help?? 
What I am missing:  how the yvalue should tend to a limit as the distance approaches the radius (quadratically? hyperbolically? exponentially? linearly?.  Should the function be continuous or discontinuous at the exact radius point.  What should be the slope of the curve at that point? (or the slope as it tends to it on either side if the function should be discontinuous)  How should the yvalue decay to 0 as distance tends to infinity (same options as my first question. Update: I'll rephrase this all in plainer English: If you were to draw a 2D graph of this with x is distance on the horizontal axis and y is boost on the vertical axis, and that graph represents a function f, so that y=f(x), there will be a starting value for boost where distance is 0 let's say Z = f(0). From your description, this should decay firstly to f(R), so f(R) < Z. After that it drops more rapidly towards zero. So to describe an example formula for f(x), a mathematician needs to know the shape of the curve before and after the point x=R and whether the point at x=R is smooth or a corner or whether it jumps to a different value at that point (otherwise known as a point of discontinuity). Given this information it then possible to translate the function into Perl. One world, one people
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