Re: rootsearch in a piecewise function

• To: mathgroup at smc.vnet.net
• Subject: [mg60257] Re: rootsearch in a piecewise function
• From: Peter Pein <petsie at dordos.net>
• Date: Thu, 8 Sep 2005 06:47:52 -0400 (EDT)
• References: <dfovrd\$feo\$1@smc.vnet.net>
• Sender: owner-wri-mathgroup at wolfram.com

```juejung schrieb:
> hi group,
>
> why does the root search in the following piecewise function not work,
> when the plot function before evaluates just fine.
> if i use the /.para already at the point where i define f2 and f3 then the
> root command works. however, the actual functions f2 and f3 are much
> longer and i would like to replace parametervalues only at the point where
> i do in the example below. it seems that findroot is evaluated before the
> replacement takes place??
>
> thanks
> juergen
>
> para = {a -> 3, b -> 4, c -> 20, d -> 1.5};
> f2[x_] := 10 + a* x^(1/2) - x - b;
> f3[x_] := -6 + c/x + d^2;
> f1[x_] := Which[0 < x < b, f2[x], b <= x < 10, f3[x],10 <= x, x^(1/2)]
> /.para;
> df1[x_] = D[f1[x], x];
>
> Plot[f1[x] /.para, {x, 0.1, 15}]
> Plot[df1[x] /.para, {x,0.1, 15}]
>
> FindRoot[{f1[x] == 0} /.para, {x, 2}]
> FindRoot[{df1[x] == 0} /.para, {x, 2}]
>
Hi Juergen,

you'll have to map Evaluate at f1[x], because Which has Attribute
HoldAll. This means, without Evaluate, f1[x] contains only calls to f2
and f3. And the expression f2[x] does not contain any of your parameters.

Compare:

f1[x]

Which[0 < x < 4, f2[x], 4 <= x < 10], f3[x], 10 <= x, Sqrt[x]]

with

Evaluate /@ f1[x]

Which[0 < x < 4, 10 - b + a*Sqrt[x] - x, 4 <= x < 10], -6 + d^2 + c/x,
10 <= x, Sqrt[x]]

FindRoot[Evaluate /@ f1[x]==0 /. para, {x, 5}] will find the root at
16/3. (The start value 2 leads the algorithm towards negative x-values,
where f is not defined).

and FindRoot[Evaluate /@ df1[x]==0 /. para, {x, 2}] gives you the
location of the maximum at 9/4.

--
Peter Pein, Berlin
GnuPG Key ID: 0xA34C5A82
http://people.freenet.de/Peter_Berlin/

```

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