       Re: parameter restrictions

• To: mathgroup at smc.vnet.net
• Subject: [mg32442] Re: [mg32428] parameter restrictions
• From: "soso lala" <sosolala at hotmail.com>
• Date: Mon, 21 Jan 2002 02:54:53 -0500 (EST)
• Sender: owner-wri-mathgroup at wolfram.com

```Hi

Thanks for your answer. Unfortunately, it does not work like this. Consider
the following example:

Clear[f,x,a,b]
f[x_,a_,b_]:= -x^(a+b)

The second derivation with respect to x is:
D[f[x,a,b],{x,2}]
-(-1+a+b)(a+b)x^(-2+a+b)

Given that x>0,a>0,b>0 and (a+b)<1 this derivation is unambiguously
positive. I have tried to show this with Mathematica in the manner proposed:

Clear[f,x,a,b]
f[x_?Positive,a_?Positive,b_?Positive]:= -x^(a+b) /; (a+b)<1

Now, if I write

D[f[x,a,b],{x,2}]>0

I would expect the output

TRUE

f^(2,0,0)[x,a,b]>0

Can somebody tell me where the mistake is or how I must define parameter
restrictions?

Thanks for your efforts

Jack

>From: BobHanlon at aol.com
To: mathgroup at smc.vnet.net
>Subject: [mg32442] Re: [mg32428] parameter restrictions
>Date: Sat, 19 Jan 2002 20:47:10 EST
>
>
>In a message dated 1/19/02 8:05:36 PM, sosolala at hotmail.com writes:
>
> >I am using Mathematica 4.0 and have a question about parameter
>restrictions.
> >
> >How can I define the range of values of a parameter, e.g. that alpha must
> >be
> >positive or that (alpha + beta) must be less than unity?
> >
> >Although it may be a very simple question (and/or answer) I would be
> >delighted if I get an answer.
> >
>
>f[x_, a_?Positive] := a*x;
>
>f[x_, a_?Positive, b_?NonNegative] := a^x*b  /; (a+b) < 1;
>
>
>Bob Hanlon
>Chantilly, VA  USA

_________________________________________________________________
Downloaden Sie MSN Explorer kostenlos unter http://explorer.msn.de/intl.asp.

```

• Prev by Date: Re: Simplify with assumptions
• Next by Date: How to separate imaginary unit inside an arbitrary function?
• Previous by thread: Re: parameter restrictions
• Next by thread: Re: parameter restrictions