       Re: Why doesn't this work?

• To: mathgroup at smc.vnet.net
• Subject: [mg20362] Re: [mg20352] Why doesn't this work?
• From: BobHanlon at aol.com
• Date: Sun, 17 Oct 1999 02:45:37 -0400
• Sender: owner-wri-mathgroup at wolfram.com

```Joe,

In the first case, you need to use Evaluate since FindRoot has the attribute
HoldAll.

Attributes[FindRoot]

{HoldAll, Protected}

Dog[t_] := (t/tau1)^6 * Exp[-6 * (t/tau1 - 1)] - (t/tau2)^6 *
Exp[-6 * (t/tau2 - 1)]

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

{t -> 18.1648}

Bob Hanlon

In a message dated 10/16/1999 5:26:04 AM, joe at strout.net writes:

>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?
>

```

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