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Hello!:) How can I solve this inequality?

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Hello everybodyemoticon

Mathematica is really great! I am new here and I kindly wanted to ask you a question about solving an inequality with parameters.

So, I firstly solved the following expression (with all the parameters >0):
FullSimplify [ Reduce [ (Gu  A  h -  A  Gr - P  h +  r) / (2 b N + N T) > (Gu  A  h  -  A  Gr  -  P h) / (2 b N) 
&& h > 0 && b > 0 &&T > 0 && r > 0 && N > 0 && Gu > 0 && A > 0 && Gr > 0 ]]

which gave me as a result:
T > 0 && r > 0 && h > 0 && Gu > 0 && Gr > 0 && b > 0 && A > 0 && N > 0 && A Gu h T < 2 b r + A Gr T + h P T      
(and this was ok)

The problem is that now, I wanted to introduce an additional parameter (D<1)

Rewriting the expression:
FullSimplify [ Reduce [ (Gu  A  h^D -  A  Gr - P  h +  r) / (2 b N + N T) > (Gu  A  h^D  -  A  Gr  -  P h) / (2 b N)
&& h > 0 && b > 0 &&T > 0 && r > 0 && N > 0 && Gu > 0 && A > 0 && Gr > 0 && D<1 ]]

I get:

" Reduce::nsmet: This system cannot be solved with the methods available to Reduce. "

and I also get as Out:
Reduce[b N (2 b + T) (2 b r + (A (Gr - Gu h^D) + h P) T) > 0 &&h > 0 && b > 0 && T > 0 && r > 0 && N > 0 && Gu > 0 && A > 0 && Gr > 0 && D < 1]

So my question is:  if I cannot solve the system with this new parameter D<1 (but why????) what is this result?

I apologize if this question looks stupid but it is giving me a lot of problems and I don't know how to solve it!

Thank you very much, in the case someone would like to help me!emoticon

Kodi
POSTED BY: Kodi Hannon
Answer
2 months ago
In some situations, Reduce works better if it is used step-wise.  Since your first Reduce calculation works, I would try Reduce on the result of the first calculation and the added constraint.
POSTED BY: Frank Kampas
Answer
2 months ago
Hi Frank!

Thank you very much for your kind replyemoticon

Sorry..I am little bit confused..you're saying to use Reduce on the first correct result (belonging to the first correct inequality), adding the condition D<1? But how can I do that? I mean, D is not contained in the first inequality..:/

Sorry if I understood wrong, I'm a little bit confused..emoticon
POSTED BY: Kodi Hannon
Answer
2 months ago
Which version of Mathematica are you using?  If I run
Reduce[(Gu A h^D - A Gr - P h + r)/(2 b N + N T) > (Gu A h^D - A Gr -
      P h)/(2 b N) && h > 0 && b > 0 && T > 0 && r > 0 && N > 0 &&
  Gu > 0 && A > 0 && Gr > 0 && D < 1]
in Version 9, i get
h^D \[Element]
  Reals && ((h^(-1 + D) \[Element] Reals && A > 0 && b > 0 && D < 1 &&
      Gr > 0 && Gu > 0 && h > 0 && N > 0 && P < (-A Gr + A Gu h^D)/h &&
      r > 0 && 0 < T < -((2 b r)/(A Gr - A Gu h^D + h P))) || (A > 0 &&
      b > 0 && D < 1 && Gr > 0 && Gu > 0 && h > 0 && N > 0 &&
     P >= (-A Gr + A Gu h^D)/h && r > 0 && T > 0))
POSTED BY: Frank Kampas
Answer
2 months ago
Dear Frank,

thank you very much!emoticon

I am also using the version 9.

I apologize, I made a mistake, I forgot the condition P>0; if I clarify that condition, I can't get a result! emoticon

Moreover, in trying the calculations without P>0, the parameter I had introduced was originally not labled "D" but "alpha" (I didn't think that there could be a difference!:/  I just used D in this forum as I couldn't write alpha). I noticed that in doing the same procedure, with "D" I get your same result, while with "alpha" I cannot. How is this possible?..I saw that alpha appears labeled in blue, while D in black..is there a meaning??

Still, if I use "E" I obtain "False" ???

Why does the result change in changing the letter ???

Anyways the problem now is also due to the fact that if I insert P>0 (which I had forgotten) I can't get a result anymore, neither with D!

emoticon

emoticon
POSTED BY: Kodi Hannon
Answer
2 months ago
D (ref) and E (ref) and N (ref) are probably not the best choices for variable names.
POSTED BY: Ilian Gachevski
Answer
2 months ago
Hello Ilianemoticon

Yes, but, so there is, actually, a difference in using "D" or "alpha"...how can I know when I'm right and when I'm wrong?

:/

The problem is that I still cannot solve this inequality:

FullSimplify [ Reduce [ (Gu  A  h^(alpha) -  A  Gr - P  h +  r) / (2 b N + N T) > (Gu  A  h^(alpha)  -  A  Gr  -  P h) / (2 b N)
&& h > 0 && b > 0 &&T > 0 && r > 0 && N > 0 && Gu > 0 && A > 0 && Gr > 0 && P>0 && (alpha) < 1 ]]

emoticon
POSTED BY: Kodi Hannon
Answer
2 months ago
If you look up N or D or E in the help system you find those names have already been defined to mean something other than an unknown user variable.

As a general rule, if you want to use names that start with a capital letter then first look that name up in the help system and make sure that Mathematica hasn't already given that some meaning.

If you don't use capital N and don't constrain P then this appears to work.
In[1]:= FullSimplify[
Reduce[(Gu A h^alpha - A Gr - P h + r)/(2 b n + n T) > (Gu A h^alpha - A Gr - P h)/(2 b n) &&
   h > 0 && b > 0 && T > 0 && r > 0 && n > 0 && Gu > 0 && A > 0 && Gr > 0 && alpha < 1]]

Out[3]= A > 0 && b > 0 && Gr > 0 && Gu > 0 && h > 0 && n > 0 && r > 0 && alpha < 1 && T > 0 &&
  ((2 b r)/(A Gr - A Gu h^alpha + h P) + T < 0 ||  A Gu h^(-1 + alpha) <= (A Gr)/h + P)
But that doesn't really say anything more than slightly rearranging your input.

If you look in the help system for Reduce you will see it is Reduce[ expression, vars ] where you are asking it to reduce an expression for a particular variable or variables. Perhaps if you were looking for a particular variable then it might be able to be more helpful.

If you include the condition P > 0 but do not include any variable to reduce to then it complains it doesn't have a method which can do that.

Perhaps if you try to fit your problem into the form described for Reduce you might have better luck.
POSTED BY: Bill Simpson
Answer
2 months ago
Hi Bill!emoticon

thank you very much, this was really helpfulemoticon

Yeah, I just have to try different combination and see what happens..

Thank you!

emoticon
POSTED BY: Kodi Hannon
Answer
2 months ago