Re: The equivalent of FindRoot for an interpolating function
- To: mathgroup at smc.vnet.net
- Subject: [mg37559] Re: The equivalent of FindRoot for an interpolating function
- From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
- Date: Tue, 5 Nov 2002 04:59:33 -0500 (EST)
- Organization: Universitaet Leipzig
- References: <apve0r$s00$1@smc.vnet.net>
- Reply-to: kuska at informatik.uni-leipzig.de
- Sender: owner-wri-mathgroup at wolfram.com
Hi,
sol = NDSolve[{y'[t] == 1 - y[t], y[0] == 0}, y[t], {t, 0, 20}]
fy = y[t] /. sol[[1]];
FindRoot[fy == 0.99, {t, 1}]
work fine. And you should add an argument to a function
if you ask for an argument.
Regards
Jens
"Philip M. Howe" wrote:
>
> Hi Folks,
>
> I wish to find the value of the independent variable in an
> interpolating function that makes the dependent variable assume some
> value of interest. For example,
>
> sol = NDSolve[{y'[t] == 1-y[t], y[0]==0}, y, {t, 0, 20}]
> fy = y/.sol[[1]]
>
> produces an interpolating function. I would like to extract the
> value of t that yields a value of 0.99 or 0.9999 (say) for y. Is
> there a straightforward way of doing this?
>
> Thanks in advance for the help.
>
> Regards,
>
> Phil
> --
> Philip M. Howe
> Program Manager, Stockpile Surety
> Los Alamos National Laboratory
>
> (505) 665-5332
> (505) 667-9498
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