The statement in the question is obviously true. But when I tried
FullSimplify[(1 - x)^y <= 1, 1 > x > 0 && y >= 1]
I got
(*(1 - x)^y <= 1*)
Is there any way to get this done through Mathematica?
$\begingroup$
$\endgroup$
1
2 Answers
$\begingroup$
$\endgroup$
5
Try this:
Reduce[(1 - x)^y > 1 && 1 > x > 0 && y >= 1]
(* False *)
Have fun!
answered Nov 15, 2018 at 15:29
-
$\begingroup$ That is rather weird. How come
Reduce[(1 - x)^y <= 1 && 1 > x > 0 && y >= 1]does not work? $\endgroup$ Nov 15, 2018 at 15:32 -
2$\begingroup$ Come on, if one statement is false, the opposite one is true. What is your aim: to solve a problem, or to do it in one special way? $\endgroup$ Nov 15, 2018 at 16:18
-
$\begingroup$ @ablmf In what sense did your variant fail to work? $\endgroup$ Nov 15, 2018 at 16:29
-
$\begingroup$ @DanielLichtblau It does not return True. $\endgroup$ Nov 15, 2018 at 16:59
-
3$\begingroup$ @ablmf:
Reducedoesn't return a True or False value unless the listed statement is automatically true or automatically False. In your version, the first clause(1 - x)^y <= 1does follow automatically from the other two, which means if you runReduce[(1 - x)^y <= 1 && 1 > x > 0 && y >= 1, Reals], you get back1 > x > 0 && y >= 1. But there are values of $x$ and $y$ for which this latter statement isn't true, so it can't be reduced further. $\endgroup$ Nov 15, 2018 at 18:10
$\begingroup$
$\endgroup$
What about the following?
Reduce[(1 - x)^y <= 1 && 1 > x > 0 && y >= 1, {x, y}, Reals]
(* 0 < x < 1 && y >= 1 *)
In[126]:= FullSimplify[y*Log[(1 - x)] <= 0, 1 > x > 0 && y >= 1] Out[126]= True$\endgroup$