Re: Re: parameter restrictions
- To: mathgroup at smc.vnet.net
- Subject: [mg32464] Re: [mg32442] Re: [mg32428] parameter restrictions
- From: BobHanlon at aol.com
- Date: Tue, 22 Jan 2002 03:19:34 -0500 (EST)
- Sender: owner-wri-mathgroup at wolfram.com
Since your original question did not include an example I did not understand
your intent. I suggest that you use (Full)Simplify with assumptions to
evaluate inequalities.
Clear[f,x,a,b];
f[x_,a_,b_]:=-x^(a+b);
Note that you do not want to restrict the parameters values since this would
preclude evaluation for non-numeric values of the parameters. That is why
you obtained the unevaluated f^(2,0,0)[x,a,b]
FullSimplify[D[f[x,a,b],{x,2}]>0,{x>0,a>0,b>0,a+b-1<0}]
0 > (a + b - 1)*(a + b)*
x^(a + b - 2)
Unfortunately, this did not fully evaluate. The problem is the power of x.
FullSimplify[x^y > 0, {x > 0, Element[{x,y}, Reals]}]
x^y > 0
However, the difficulty of the evaluation can be eliminated by assuming a
positive value for x, say 1.
Simplify[D[f[x,a,b],{x,2}]>0,{x==1,a>0,b>0,a+b-1<0}]
True
Bob Hanlon
In a message dated 1/21/02 4:12:20 AM, sosolala at hotmail.com writes:
>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
>
>but I receive
>
>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
>To: mathgroup at smc.vnet.net
>>Subject: [mg32464] [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
>