Groups | Search | Server Info | Login | Register

Groups > comp.soft-sys.math.mathematica > #16814

Re: Using FindRoot with free parameters

Show all headers | View raw

On 4/14/14 at 11:00 PM, ssplotkin@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.

Back to comp.soft-sys.math.mathematica | Previous | Next | Find similar

Thread

Re: Using FindRoot with free parameters Bill Rowe <readnews@sbcglobal.net> - 2014-04-16 07:39 +0000

csiph-web