1
$\begingroup$

Say I want to define a function fwith input x but also involving parameter p not yet specified:

f[x_] := p^x

Now I want to verify some inequalities with additional assumptions. A (hopefully) minimal example, showing my problem:

If, for some for input y in the function and for some paremeter value q>0, it holds that f[y] >= 2 q, then I would expect f[y] >= q to yield True.

However,

FullSimplify[f[y] >= q, {f[y] >= 2 q, q > 0}]

yields p^y >= q instead. Even if I add q \[Element] Reals as an assumption.

Do I overlook an obvious mathematical (non-Mathematica) issue, or how I can get Mathematica to "fully" simplify the expression? I have also looked at this issue with FullSimplify, but the solution provided did not help.

Improve this question
$\endgroup$
3
  • $\begingroup$ Have you tried the functions ForAll and Resolve? $\endgroup$
    – Mr Puh
    May 13, 2020 at 10:56
  • $\begingroup$ I rather seek to verify whether for some specific (not all/arbitrary) values of the parameters certain statements hold true. So as in the example, I have a general function f, "local" assumption f[y]>=2q with q>0 and the statement I want to verify, f[y] >= q, which I expected to turn into True. ForAll and Resolve don't seem to fare any bettr. I hoped my example was minimalistic enough, I can add more if required. $\endgroup$
    – Bernd
    May 13, 2020 at 11:21
  • $\begingroup$ I am not sure how capable Mathematica is in solving these kind of analytical questions. It might also be due to the complexity of your function f. Try using f[x_] := Hold[E^(Log[p] x)] as your equation, this seems to give the desired result. $\endgroup$
    – Mr Puh
    May 13, 2020 at 12:07

1 Answer 1

Reset to default
1
$\begingroup$

I am not certain that this is what you are looking for?

f[x_] := p^x
Reduce[f[x] > q && q > 0, p, Reals]

Visual:

With[
     {f = p^x},
     Manipulate[
      Plot[{f, q}, {x, 0, 2}, Filling -> {1 -> {{2}, {Orange, Blue}}}],
      {{p, 1/2}, -1, 1},
      {{q, 1/2}, -4, 4}
      ]
     ]
Improve this answer
$\endgroup$
1
  • $\begingroup$ I am not so much interested in the requirements/ranges for the parameters for certain (in)equalities to be true (which, I presume, Reduce would give me to some extend), but rather in having local restrictions/assumptions on certain values, then trying to check how these simplify my computations. I.e. "if some statement x holds, can I then infer y"? So I would want to "reduce" w.r.t. all parameters, i.e. some sort of "simplifying". $\endgroup$
    – Bernd
    May 13, 2020 at 10:28

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.