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Re: FindROot and substitutions

  • To: mathgroup at smc.vnet.net
  • Subject: [mg89831] Re: FindROot and substitutions
  • From: Szabolcs Horvát <szhorvat at gmail.com>
  • Date: Sun, 22 Jun 2008 03:22:23 -0400 (EDT)
  • References: <g3ihp3$fd5$1@smc.vnet.net>

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;

----------^
You used a semicolon here, so I suppose that you do not understand the 
difference between SetDelayed and Set.  Please read about that:

tutorial/ImmediateAndDelayedDefinitions

Later also read these:
tutorial/TheStandardEvaluationProcedure
tutorial/Evaluation

> bad[z_] := FindRoot[ g == 5, {x, -10, 10 }];
> bad[5]

There is not "z" in "FindRoot[ g == 5, {x, -10, 10 }]" so bad[5] 
evaluates to FindRoot[ g == 5, {x, -10, 10 }] in the first step.  Use 
the function Trace to see this.

> good[k_] := FindRoot[ g == 5 /. z -> k, {x, -10, 10 }];
> good[6]

good[6] evaluates to FindRoot[ g == 5 /. z -> 6, {x, -10, 10 }] in the 
first step because it does contain k.


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