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MathGroup Archive 1999

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Re: Why doesn't this work?

  • To: mathgroup at smc.vnet.net
  • Subject: [mg20376] Re: Why doesn't this work?
  • From: "Allan Hayes" <hay at haystack.demon.co.uk>
  • Date: Sun, 17 Oct 1999 02:45:47 -0400
  • References: <7u9098$qg5@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Joe,

FindRoot has the attribute HoldAll, so, usually, the evaluation of
FindRoot[ body, {t,n}] begins by matching  the unevaluated body to the first
possible one of  the forms {lhs==rhs},{expr},lhs == rhs, expr.
Whichever  one matches it will then proceed on the assumption that  lhs-rhs,
respectively expr, will give a numerical value when t is assigned a
numerical value.

In your case the match of
(Dog[t]==0)/.{tau1->13.7,tau2->24.8}
is with  {expr} (it is not an equation), so  FindRoot assumes that, for
example, Dog[t]==0)/.{tau1->13.7,tau2->24 will give a numerical value when t
is given the value 20 - and it does not -  it gives False.

We can get round this by, in effect, forcing FindRoot to evaluate body
before matching.

FindRoot[Evaluate[(Dog[t] == 0) /. {tau1 -> 13.7, tau2 -> 24.8}], {t, 20}]

{0.11 Second, {t -> 18.1648}}

The correct assumption is also made with the following  (note the
parentheses).

FindRoot[(Dog[t] /. {tau1 -> 13.7, tau2 -> 24.8}) == 0, {t, 20}]

{0.11 Second, {t -> 18.1648}}

Your method

Dog2[t_] = Dog[t] /. {tau1 -> 13.7, tau2 -> 24.8};
FindRoot[Dog2[t] == 0, {t, 20}]

{t -> 18.1648}

also works because the matching leads to the correct assumption is made:
that Dog2[t] will be numeric for numeric t.

Allan
---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565

Joe Strout <joe at strout.net> wrote in message news:7u9098$qg5 at smc.vnet.net...
> I'm trying to understand why a temporary is needed in the following
> case.  I've got a function of one parameter and two constants.  I
> define the constants on the fly with "/.".  I want to find a root of
> this equation.  When I use "/." in FindRoot, it doesn't work, but if I
> create another function that has these as true constants (rather than
> symbols), FindRoot works.  I don't understand why these two cases are
> different.
>
> Here's my function (a difference of gaussians):
>
> Dog[t_] := (t/tau1)^6 * Exp[-6 * (t/tau1 - 1)] - (t/tau2)^6 * Exp[-6 *
> (t/tau2 - 1)]
>
> Here's my first attempt to find a root, and the result:
>
> FindRoot[(Dog[t]==0) /. { tau1->13.7, tau2->24.8}, {t, 20}]
> FindRoot::frnum: Function {False} is not a length 1 list of numbers at
> {t} = {20.}.
>
> Here, I define a temporary function, and a FindRoot which should come
> out to the exact same thing as above -- but this one works:
>
> Dog2[t_] = Dog[t] /. { tau1->13.7, tau2->24.8}
> FindRoot[Dog2[t]==0, {t,20}]
> {t->18.1648}
>
> Why is this?  I've looked at it every way I can think of, thrown in
> extra parens for grouping, checked that Dog2[t]==0 is exactly the same
> as (Dog[t]==0) /. { tau1->13.7, tau2->24.8}, etc., but I can't figure
> it out.  Any clues?
>
> Thanks,
> -- Joe
>
> P.S. How do I copy a notebook in plain text format?  I tried "Copy As >
> Plain Text" (and many others in that submenu) but it always gives me
> text befuddled by lots of markup tags.  I had to paste the above line
> by line, and fix the "->" and "==" symbols which came through as
> foreign characters.
>
> --
> ,------------------------------------------------------------------.
> |    Joseph J. Strout           Biocomputing -- The Salk Institute |
> |    joe at strout.net             http://www.strout.net              |
> `------------------------------------------------------------------'
> Check out the Mac Web Directory!    http://www.strout.net/macweb.cgi
>



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