Re: Interval, Range of a function

• To: mathgroup at smc.vnet.net
• Subject: [mg81651] Re: Interval, Range of a function
• From: "David W.Cantrell" <DWCantrell at sigmaxi.net>
• Date: Sun, 30 Sep 2007 04:06:10 -0400 (EDT)
• References: <fdkrt6\$7s7\$1@smc.vnet.net>

```janos <janostothmeister at gmail.com> wrote:
> If you ask Mathematica what is the range of Sin restricted to e.g. [0,
> Pi/4] you will get a correct answer.
> This is not always the case, however. Let
> f[x_]:=5Exp[-2x]-3Exp[-x]
> f[Interval[{-Infinity,+Infinity}]]
>
> Then the answer Interval[{-Infinity,+Infinity}] is different from what
> you see on the figure of the function.
>
> Is there a bug here?

As Andrzej has already pointed out, there is no bug.

It should also be mentioned that one can, easily, get the desired range in
this example using f[Interval[{-Infinity, Infinity}]]. To do so, we must
avoid the so-called "problem of dependence" by being careful about exactly
how f is expressed. (Just think of "completing the square" in order to
rewrite f.)

In[1]:= f[x_] := 5*(Exp[-x] - 3/10)^2 - 9/20

In[2]:= f[Interval[{-Infinity, Infinity}]]

Out[2]= Interval[{-9/20, Infinity}]

But be aware that it is all too easy to find functions which cannot be
expressed in a way that avoids the problem of dependence.

David W. Cantrell

```

• Prev by Date: Re: [Mathematica 6] Editing 3D graphics? (Editing tool for 2D only)
• Next by Date: Plot3D with NDSolve
• Previous by thread: Re: Interval, Range of a function
• Next by thread: Exporting Arrays