Re: Using FindRoot with free parameters

*To*: mathgroup at smc.vnet.net*Subject*: [mg132595] Re: Using FindRoot with free parameters*From*: Bill Rowe <readnews at sbcglobal.net>*Date*: Wed, 16 Apr 2014 03:39:46 -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

On 4/14/14 at 11:00 PM, ssplotkin at gmail.com (steviep) wrote: >Thanks, I had actually come up this solution in the meantime: >xf[alpha_, n_] = Function[{alpha, n}, x /. FindRoot[f[n, x] - alpha, >{x, 2}]][alpha,n] >which gives the same answer. It also gives a bunch of warnings, as >does Module when "=" is used instead of ":=". >Is one method preferred over another? Thanks in any event. -S Yes, you originally posted >basically using FindRoot but holding off on substituting in the >parameters until later. Is there a simple solution? Delaying evaluation is the precise reason for using := (SetDelayed) rather than = (Set). Set evaluates its arguments immediately.