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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.
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