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Re: FindROot and substitutions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg89846] Re: [mg89811] FindROot and substitutions
  • From: Andrzej Kozlowski <akoz at mimuw.edu.pl>
  • Date: Sun, 22 Jun 2008 03:25:15 -0400 (EDT)
  • References: <200806210930.FAA15731@smc.vnet.net>

On 21 Jun 2008, at 18:30, Aaron Fude wrote:

> Hi,
>
> I have found a workaround for this problem, but I would like to
> understand how Mathematica wants you to think so I'm asking the
> question anyway.
>
> Why does "bad" not work, while "good" works?
>
> g := x z;
> bad[z_] := FindRoot[ g == 5, {x, -10, 10 }];
> bad[5]
> good[k_] := FindRoot[ g == 5 /. z -> k, {x, -10, 10 }];
> good[6]
>
> Thanks!
>
> Aaron
>

This has little to do with FindRoot but a lot to do with how :=  
(SetDelayed) works. When you evaluate bad[5], Mathematica tries to  
substitute 5 for z in the RHS without evaluating it, but as there is  
no z there, nothing is substituted. You then end up with FindRoot[x z  
==5, {x,-10,10}], which clearly is not going to work. In your second  
exmple, the right hand side of the definition of good involves k, so 6  
is substituted for k and you end up with
FindRoot[x z == 5 /. z -> 6, {x, -10, 10}], which is fine.

Andrzej Kozlowski





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