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
- References:
- FindROot and substitutions
- From: Aaron Fude <aaronfude@gmail.com>
- FindROot and substitutions