Re: Rookie questions about solving for small numbers and others

*To*: mathgroup at smc.vnet.net*Subject*: [mg131039] Re: Rookie questions about solving for small numbers and others*From*: Peter Klamser <klamser at googlemail.com>*Date*: Wed, 5 Jun 2013 03:30:01 -0400 (EDT)*Delivered-to*: l-mathgroup@mail-archive0.wolfram.com*Delivered-to*: l-mathgroup@wolfram.com*Delivered-to*: mathgroup-outx@smc.vnet.net*Delivered-to*: mathgroup-newsendx@smc.vnet.net*References*: <20130604055823.C8D676A27@smc.vnet.net>

You have two problems: 1. Finding the root of a expression in an area with very small numbers. 2. In that area with very small numbers you have a very flat gradient. Because I can not read the expressions you posted with Mathematica 9 I can not help you further on. Peter 2013/6/4 Samuel Mark Young <sy81 at sussex.ac.uk>: > > Hello, > > I'm trying to solve a problem involving various integrals. Essentially, I= 'm perturbing a gaussian distribution, and then trying to find a value for = sigma (the standard deviation) for which there is a 10^-5 chance of being g= reater than 1 (i.e. What value of sigma gives a value of 10^-5 when the pdf= is integrated from 1 to infinity). The aim is to find how sigma changes wi= th the different perturbations. The code below is a shortened version of wh= at I'm currently using - you may not need to know all the relevant details. > > > Explanation: Here I'm adding a quadratic perturbation with coefficient f = to the Gaussian variable x to make a new variable zeta. The aim here is to = plot a graph (eventually) showing how sigma changes with f. Because I want = to use this code to work with higher order equations (for which there is no= analytic solution), I use a temporary value for sigma (=1B$B&R=1B(Btem= p - provided that the value of sigma found is close to =1B$B&R=1B(Btemp= , this works well) and solve numerically. Similarly, the easiest way to han= dle the in tegrations is just to find the relevant values of x for which ze= ta>1, and integrate the Gaussian distribution over those values. I then use= FindRoot to find a value of sigma which satisfies the equation. > > > f = 0.5; > > =1B$B&R=1B(Btemp = 0.2; The values here have been picked arbitraril= y > > > zeta = x + f (x^2 - =1B$B&R=1B(B^2); > xCritical = x /. NSolve[1 == zeta /. =1B$B&R=1B(B -> =1B$B&R= =1B(Btemp, x, Reals]; > yCritical = xCritical/=1B$B&R=1B(B; This calculates the critical va= lues to integrate between > > > =1B$B&R=1B(Btemp = =1B$B&R=1B(B /. > FindRoot[ > > Sum[(1/ > Sqrt[2*Pi]) (If[(Abs[D[zeta, x]] /. x -> xCritical[[n]]) > > 0, If[(D[zeta, x] /. x -> xCritical[[n]]) > 0, 1, -1], > 0]) (Integrate[Exp[-(y^2)/2], {y, yCritical[[n]], =1B$B!g=1B= (B}]), {n, > Length[xCritical]}] The Sum[] command generates a series of ERFC'= s (this code will throw up errors for certain values of f, but the full cod= e doesn't) > > > + If[(Simplify[D[zeta, x] /. x -> xCritical[[1]]]) < 0, 1, 0] = = 10^(-5), {=1B$B&R=1B(B, 0.00001, 2}] FindRoot searches for a value of s= igma between 0 and 2 (the solution should always lie in this range - though= putting in zero exactly results in divide by zero errors > > > There are currently 3 problems I'm having: > > 1) Underflow occurs in the computation a lot for certain values of f > > 2) I want to be able to solve for when the integrals equal 10^-20 (as opp= os= > e to 10^-5) - which is greater than machine precision. I've tried fiddlin= g = > with settings like AccuracyGoal, PrecisionGoal and WorkingPrecision but c= an= > 't find anything that makes it work reliably (instead of a smooth curve, = I = > end up with jagged spikes. Its entirely possible, even likely, that I'm m= is= > sing something obvious. > > 3) For negative values of f, there are no solutions to x + f (x^2 - =1= B$B&= > R=1B(Btemp^2)=1 if =1B$B&R=1B(Btemp is too small. The problem the= n, is th= > at, when =1B$B&R=1B(Btemp is increased, unless there is remarkable fi= ne tun= > ing, the final value of sigma found is not similar to =1B$B&R=1B(Btem= p. The= > only way I've found to handle this is to very slowly increment =1B$B&R= =1B(= > Btemp until there are real solutions, then use FindRoot to find a value f= or= > sigma, and compare sigma to =1B$B&R=1B(Btemp to see if they match (t= o 4 s.= > f. is fine). However, this takes a very long time when I want to do this = re= > peatedly. > > > > Many thanks in advance for taking the time to read this, and any help is = ve= > ry well appreciated. I think I have included all the information needed, = bu= > t please ask if you need more. Please feel free to contact me directly at= s= > y81 at sussex.ac.uk with any questions. > > > Regards, > > Sam Young > >

**References**:**Rookie questions about solving for small numbers and others***From:*Samuel Mark Young <sy81@sussex.ac.uk>

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