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I have an inequality consisted only of seven parameters as follows:

$(1 + g)^{-(m+w)}\big[\big\{a+(1-a)(1 + g)^m\big\}(1 + p q)Q - (1 + g)^w \big\{(1 + g)^m (1 + q)-(1-p)q\big\} Q\big] < 0$

Parameters, $g$, $w$, and $m$, are subject to the following restrictions:

$0<g$, $0<w$, $0<m$, $0<Q$

That is, these four parameters are all positive.

Now, we can manually verify that this inequality holds under the above parameter restrictions. And I would like to confirm this with Mathematica.

My codes for the assumptions:

$Assumptions = 0 < g && 0 < w && 0 < m && 0 < Q

And as for checking if the inequality holds, I used Refine and If as follows:

Refine[If[(1 + g)^(-m - w) ((a - (-1 + a) (1 + g)^m) (1 + p q) Q - (1 + g)^w ((-1 + p) q + (1 + g)^m (1 + q)) Q)<0, Print[True], Print[False]]]

Since the condition is true, the result I expect is True. But what I get is a simple repetition of the If bracket:

If[(1 + g)^(-m - w) ((a - (-1 + a) (1 + g)^m) (1 + p q) Q - (1 + g)^w ((-1 + p) q + (1 + g)^m (1 + q)) Q) < 0, Print[Yes], Print[No]]

When I use a little simpler, but similar, type of inequality, it gives me either True or False depending on the specific form of inequality I use. But it seems Mathematica is not able to figure out whether the above inequality, which is too complex (maybe?), is true or false.

Or am I missing something? Can anyone help? Thank you so much.

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  • $\begingroup$ I couldn't get it to Refine either, but I expect that part of the answer will be that you need to also tell Mma that pk, q, and Q are Real (and not Complex), since comparison with zero isn't well-defined for complex numbers. $\endgroup$ – evanb Sep 13 '14 at 0:26
  • $\begingroup$ PS. You don't need the If and Print statements. If Refine works it will return True! $\endgroup$ – evanb Sep 13 '14 at 0:27
  • $\begingroup$ I don't think this inequality is true. Try, for example, the values {g -> 0.001, m -> 0.001, w -> 0.001, Q -> -10^7, a -> 1, p -> 1, q -> 1}. Then the LHS gives 39.98, which is more than 0. So I think you have some unstated assumptions. $\endgroup$ – evanb Sep 13 '14 at 0:39
  • $\begingroup$ @evanb Oh yes, you are right. One missing assumption is $Q>0$. I will add this right away! $\endgroup$ – jim Sep 13 '14 at 1:03
  • $\begingroup$ Sorry, that's still not enough. Try {g -> 0.01, m -> 0.01, w -> 0.01, Q -> 1, p -> -10^8, a -> 0.01, q -> 0.01} $\endgroup$ – evanb Sep 13 '14 at 1:20
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I would recommend using Reduce instead of Refine. Also, you might as well cancel the factors (1+g)^(-m-w) and Q which are positive. Then, since g > 0 and m and w are independent positive real numbers, we may as well replace

{(1 + g)^m -> u, (1 + g)^w -> v}

to simplify its human readability. The conditions translate to u > 1 and v > 1. The inequality becomes

expr = ((a - (-1 + a) (1 + g)^m) (1 + p q) - (1 + g)^w ((-1 + p) q + (1 + g)^m (1 + q)))  /.
  {(1 + g)^m -> u, (1 + g)^w -> v}
(* (1 + p q) (a - (-1 + a) u) - ((-1 + p) q + (1 + q) u) v *)

Then we can use Reduce, with or without specifying the order of the variables.

Assuming[a > 0 && p > 0 && q > 0 && u > 1 && v > 1,
 Reduce[$Assumptions && expr < 0]
 ]
(*
  v > 1 && (
    (0 < a < 1 && (
       (1 < u <= (a - v)/(-1 + a) && p > 0 && q > 0) ||
       (u > (a - v)/(-1 + a) && (
          (0 < p <= (v - u v)/(-a - u + a u + v) && q > 0) ||
          (p > (v - u v)/(-a - u + a u + v) && 
             0 < q < (a + u - a u - u v)/(-a p - p u + a p u - v + p v + u v)))))
     ) ||
    (a >= 1 && u > 1 && p > 0 && q > 0))
*)

The order of the variables matters (because Reduce will return a cylindrical decomposition). You can play with it to see if one makes the problem clearer.

Assuming[a > 0 && p > 0 && q > 0 && u > 1 && v > 1,
 Reduce[$Assumptions && expr < 0, {p, q, a, u, v}]
 ]
(*
  (0 < p <= 1 && q > 0 && a > 0 && u > 1 && v > 1) ||
  (p > 1 && q > 0 && (
     (0 < a < (-q + p q)/(1 + p q) && u > 1 &&
          v > (a + a p q + u - a u + p q u - a p q u)/(-q + p q + u + q u)) ||
     (a >= (-q + p q)/(1 + p q) && u > 1 && v > 1)))
*)

One can also play with the alternative inequality to understand why it is not simply true or false.

Assuming[a > 0 && p > 0 && q > 0 && u > 1 && v > 1,
 red = Reduce[$Assumptions && expr > 0]
 ]
(*
  v > 1 && 0 < a < 1 && u > (a - v)/(-1 + a) && 
    p > (v - u v)/(-a - u + a u + v) && 
    q > (a + u - a u - u v)/(-a p - p u + a p u - v + p v + u v)
*)

Interpreting this, it says we can pick any v > 1 and 0 < a < 1. Let's do this and examine red:

red /. {v -> 2, a -> 1/2} // Simplify
(* u > 3 && 4 + p (-3 + u) > 4 u && 1 + q (4 + p (-3 + u) - 4 u) > 3 u *)

Now we can pick any u > 3:

red /. {v -> 2, a -> 1/2, u -> 4} // Simplify
(* p > 12 && q > 11/(-12 + p) *)

Next, pick p > 12:

red /. {v -> 2, a -> 1/2, u -> 4, p -> 13} // Simplify
(* q > 11 *)

And finally pick q > 12:

red /. {v -> 2, a -> 1/2, u -> 4, p -> 13, q -> 12} // Simplify
(* True *)

So the settings {v -> 2, a -> 1/2, u -> 4, p -> 13, q -> 12} make the expression positive:

expr /. {v -> 2, a -> 1/2, u -> 4, p -> 13, q -> 12}
(* 1/2 *)

Similarly one can find numbers that make the expression negative. @evanb has pointed some of this out in comments.

| improve this answer | |
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  • $\begingroup$ thank you so much for your help. I wasn't able to interpret the results of the third and fourth code. Do you know what these mean? It seems, with these you are saying that we cannot tell whether my inequality is true or not. Thanks a lot! $\endgroup$ – jim Sep 17 '14 at 14:16
  • $\begingroup$ @jim I think that very last command gives conditions for which the inequality is false. Just find values of the variables which make the reduced inequalities true. From the second last command, the condition (0 < p <= 1 && q > 0 && a > 0 && u > 1 && v > 1) shows at least some range for the variables over which the inequality is true. I'll add something to the last one to clarify. $\endgroup$ – Michael E2 Sep 17 '14 at 14:22
  • $\begingroup$ It really help! Thank you! Can I ask you one more question? What do we mean by 'order' here? $\endgroup$ – jim Sep 17 '14 at 14:48
  • $\begingroup$ @jim Whether you list the variables in the order {p, q, a, u, v} or {a, q, u, p, v} or any other order. The order determines how the inequality is broken down. With {p, q, a, u, v}, you first have a p inequality, && then a q, && then a, etc. Different cases are connected with Or (||). With {a, q, u, p, v}, it would start with an a inequality. $\endgroup$ – Michael E2 Sep 17 '14 at 15:05
  • $\begingroup$ Got it. Thanks a lot! I really appreciate! $\endgroup$ – jim Sep 17 '14 at 15:09

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