Re: Rookie questions about solving for small numbers

*To*: mathgroup at smc.vnet.net*Subject*: [mg131092] Re: Rookie questions about solving for small numbers*From*: Samuel Mark Young <sy81 at sussex.ac.uk>*Date*: Tue, 11 Jun 2013 02:29:31 -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*: <66577FF8848F674883C6A3CCB603563DB7FE70@EX-SHA-MBX2.ad.susx.ac.uk>,<90764878.135010054.1370676546100.JavaMail.root@jaguar8.sfu.ca>

Thank you for the advice. What I need is something that will do that automatically (without, for example, plotting a log-log graph to find approximate solutions) - so that I can quickly run this for a large range of values off. I also need it to work for higher order equations (i.e. past 4th order, the equation cannot be solved symbolically). Essentially, I'm solving for a taylor series type expansion, where a variable (call it w) is expanded in terms of a gaussian variable x. i.e. w=x+f*(x^2-s^2)+g*x^3+h*(x^4 - 3*s^4)+... (Equation 1). And then solving Pr(w>1)=e (Equation 2) for s. The range of values of x which give w>1 depends on the coefficients f, g, h, etc, as well as s (the standard deviation of the Gaussian variable x). As seen for the quadratic case, x + f*(x^2 - s^2), if f is negative we have: Pr(w>1)=Pr(x>x1)-Pr(x>x2) I've written the RHS that way for 2 reasons: it was easier to write the code that way, and this gives a sum of 2 ERFCs to solve for. The code I included in the original email works for the majority of cases - but the problems I run into are the same as those which occur for just the quadratic case: 1) If the highest order term is even, then typically for negative values of the coefficients there are no solutions to Equation 1 unless s becomes large. For example, there may be no solutions unless s>0.7, so to solve I use a temporary value for s, say stemp=0.8, and I use NSolve to find the solutions for x. I then use find root to find a value for s by solving Equation 2 - which can come out as, for example, 0.1. However, s=0.1 would have meant there were no solutions for w=1. At the moment, my best method is to use a trial and error approach until stemp matches s to within 3sf (an approximation to 3sf is fine). This doesn't take too long to do once, around 10s or so on my laptop, but I want the code to run for multiple values of f, g, h etc, so even running for only 10 values of each coefficient quickly means I have to run this thousands of times. Ideally, I'd like Mathematica to have a streamlined way of doing the trial and error, or a way for FindRoot to solve the system of equations simultaneously (so that I don't need to use a temporary value for s to solve equation 1). 2) Underflow occurs a lot - you seem to have got around that using WorkingPrecision? Where do I need to place that in the code? Many thanks, Sam ________________________________________ From: Ray Koopman [koopman at sfu.ca] Sent: 08 June 2013 08:29 To: Samuel Young Cc: mathgroup at smc.vnet.net Subject: [mg131092] Re: Rookie questions about solving for small numbers Writing s instead of sigma, and letting e denote the probability that is variously specified as 10^-5 or 10^-20, we have Pr[x + f*(x^2 - s^2) > 1] = e, where x is Normal with mean 0 and standard deviation s. Rewrite that as Pr[z + f*s*(z^2 - 1) > 1/s] = e, where z is standard normal. If f = 0 then we have Pr[z > 1/s] = e. Plot log e as a function of log s to identify approximate solutions, then use FindRoot. f = 0, e = 10^-5 gives s -> .234473 f = 0, e = 10^-20 gives s -> .107964 If f > 0 then z + f*s*(z^2 - 1) = 1/s has two real roots, say z1 < z2. The coefficient of z is > 0, so we want Pr[z < z1] + Pr[z > z2]. Again do a log-log plot, then use FindRoot. f = .5, e = 10^-5 gives s -> .173683 f = .5, e = 10^-20 gives s -> .0792307 If -1/4 < f < 0 then the roots are real for all s > 0. The coefficient of z^2 is < 0, so we want Pr[z1 < z < z2]. f = -.125, e = 10^-5 gives s -> .271650 f = -.125, e = 10^-20 gives s -> .126184 If f < -1/4 then the roots are real only for s > sqrt[-1-4f]/(-2f). f = -1/2 requires s > 1. f = -1/2, e = 10^-5, WorkingPrecision->30 gives s -> 1.000000000213493355543893763569933088415336334927`30. f = -1/2, e = 10^-20, WorkingPrecision->60 gives s -> 1.0000000000000000000000000000000000000002134933555514410720359161572336783883`60. In general, s -> smin as e -> 0 with f < -1/4. How much precision do you need? Would an approximation suffice? ----- Samuel Mark Young <sy81 at sussex.ac.uk> wrote: > Yes that's correct. The code I've included does this fine for positive values of f, but can't handle negative values very well. I also run into problems when I want to solve for Pr[x + f*(x^2 - sigma^2) > 1] = 10^-20 instead. > > Sam > ________________________________________ > From: Ray Koopman [koopman at sfu.ca] > Sent: 07 June 2013 04:19 > To: mathgroup at smc.vnet.net > Subject: Re: Rookie questions about solving for small numbers > > It's still not clear exactly what you want. Is this it? > > X is a random variable whose distribution is normal with mean 0 > and standard deviation sigma. Given a constant f, solve for sigma > such that Pr[x + f*(x^2 - sigma^2) > 1] = 10^-5.

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**Re: Rookie questions about solving for small numbers**

**Re: Rookie questions about solving for small numbers**