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Assume $ a $ is real, and $ a^n > -1 $ for all positive integers $n$. It follows that $ a > -1 $. How can I do this simplification? More generally, I would like to find restrictions on $a$ implied by $ f(a, n)> -1 $ for all positive integer $ n $.

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  • $\begingroup$ How about Reduce[{a^n > -1, n > 0}, {a, n}, Reals]? $\endgroup$ – Αλέξανδρος Ζεγγ Dec 15 '18 at 6:58
  • $\begingroup$ Hi - Thanks, that works by assuming that $a$ and $n$ are real, which "coincidentally" solves the specific problem. I think there will be difficulties with the general problem in which $f(a,n)$ may be real and less than -1 for some value of positive $n$ which is not a positive integer. In that case, only the assumption that $n$ is a positive integer will give the correct results. $\endgroup$ – Paul R. Dec 15 '18 at 13:47
  • $\begingroup$ Reduce[ForAll[n, n \[Element] Integers && n > 0, a^n > -1], {a, n}, Reals] --> "Reduce::nsmet: This system cannot be solved with the methods available to Reduce." :/ $\endgroup$ – Michael E2 Dec 15 '18 at 14:05
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I'm not sure your first premise is strictly true.

a^n > -1;

Log[%[[1]]] > Log[%[[2]]] // PowerExpand
(*n Log[a] > I π*)

%[[1]]/n > %[[2]]/n;

Exp[%[[1]]] > Exp[%[[2]]]
(*a > E^((I π)/n)*)

a > (E^(I π))^(1/n)
(*a > (-1)^(1/n)*)

Table[{n, %}, {n, 1, 5}]
(*Greater::nord: Invalid comparison with I attempted.*)
(*{{1, a > -1}, {2, a > I}, {3, a > (-1)^(1/3)}, {4, a >(-1)^(1/4)}, {5, a > (-1)^(1/5)}}*)

Now for specific n we can use reduce

eq = a^n > -1

Table[Reduce[eq /. n -> b, a], {b, 1, 5}]
(*{a>-1,True,a>-1,True,a>-1}*)

It looks like MMa is saying a > -1 for odd n and for even n the eq is true for any a, but a warning in the first part indicates that inequalities may not be valid comparing complex numbers.

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  • $\begingroup$ Thanks for your input - To restate the problem more clearly: Assume $a$ is real, and that $a_n>−1$ for all positive integers $n$. It follows that $a>-1$. $\endgroup$ – Paul R. Dec 15 '18 at 12:46
  • $\begingroup$ (-2)^2 > -1 but -2 is not > -1 $\endgroup$ – Bill Watts Dec 15 '18 at 20:41
  • $\begingroup$ Hi Bill - There was some confusion in my original post, I was using $a_n>-1$ as shorthand for $a_n>-1$ for all $n$, where $n$ is a positive integer. (Ok for my notes, but not good for communication) So yes, (-2)^2>-1, but (-2)^3 is not. I have since edited the OP to clarify this. $\endgroup$ – Paul R. Dec 16 '18 at 0:12

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