1
$\begingroup$

I try to solve the following inequation.

Reduce[
  (k^-n (k (-k + k^n) y + (-1 + k) x (-1 + k + y)))/((-1 + k) (x + y*n)) < 1/n && 
   x > 0 && y > 0 && n ∈ Integers &&k ∈ Integers, 
  {x, y, k, n}]

Unfortunately,Reduce does not get a result. Any suggestion on how to solve it?

$\endgroup$
  • 1
    $\begingroup$ Although the documentation does not explicitly mention it, Reduce[] is very lame when dealing with exponentials. For example, it can't do Reduce[2 == 2^n + 2^m, {n, m}, Integers] which should be easy. Try FindInstance[] instead which is more capable. $\endgroup$ – Somos Mar 30 at 18:28
  • $\begingroup$ As I read this, you are trying to find a solution for 4 variables from one equation. Is this what you want? Do you really want an inequality for x? $\endgroup$ – mikado Mar 30 at 18:28
  • $\begingroup$ Something can be done by specification of $n$ and $k$, e.g. Reduce[(n*((k^-n (k (-k + k^n) y + (-1 + k) x (-1 + k + y)))/((-1 + k) (x + y*n))) < 1) /. {n -> 1, k -> 2} && x > 0 && y > 0, {x, y}, Reals] $\endgroup$ – user64494 Mar 30 at 18:31
  • 1
    $\begingroup$ If you are looking for an upper bound, please place that important informaion in the body of your question instead of a comment. $\endgroup$ – Somos Mar 30 at 22:27
  • 2
    $\begingroup$ By the way, the inequality is false in general. Try n=k=x=y=2. $\endgroup$ – Somos Mar 30 at 22:38
1
$\begingroup$

Although the answer provided by bill-s provides ten instances violating the original inequality,

FindInstance[(k^-n (k (-k + k^n) y + (-1 + k) x (-1 + k + y)))/((-1 + k) (x + y*n)) < 1/n 
    && x > 0 && y > 0 && n \[Element] Integers && k \[Element] Integers, {x, y, k, n}, 10]

gives ten instances satisfying the original inequality. An example is

(* {x -> 5683/5, y -> 2331/10, k -> 10084, n -> 1132} *)
((k^-n (k (-k + k^n) y + (-1 + k) x (-1 + k + y)))/((-1 + k) (x + y*n)) < 1/n) /. %
(* True *)

It is, therefore, natural to ask what portion of random sets of {x, y, k, n} satisfy the inequality. The expression,

Count[ParallelTable[{x = RandomReal[10^5], y = RandomReal[10^5], 
    k = RandomInteger[{-10^5, 10^5}], n = RandomInteger[{-10^5, 10^5}],
    tst = (k^-n (k (-k + k^n) y + (-1 + k) x (-1 + k + y)))/((-1 + k) (x + y*n)), 
    tst < 1/n}, 100000], False, Infinity]

suggests that about 70% do. Replacing the right side of the inequality to 2/n increases the percentage to about 82%, but further increasing the right side moderately yields no improvement. An example of a random set for which the inequality is far from satisfied is

(* {99896.2, 4488.5, -40330, -43962, 7.317356679375576*10^202473, False} *)

Such instances arise when k and n both are large negative integers, with k or n even, and also occasionally when k is positive and n is negative and even. Further, the left side of inequality can be arbitrarily large amount in such instances.

To assure that the inequality always is satisfied, require that both k > 1 and n > 0, and increase the right side of the inequality to 2/n. (A tighter bound, say 1.6/n, also works.)

$\endgroup$
3
$\begingroup$

Taking a hint from Somos, you can see that your inequality is false in general. Use FindInstance with the opposite sign on the inequality:

FindInstance[(k^-n (k (-k + k^n) y + (-1 + k) x (-1 + k + y)))/((-1 + k) (x + y*n)) > 1/n 
    && x > 0 && y > 0 && n \[Element] Integers && k \[Element] Integers, {x, y, k, n}, 10]

This returns 10 cases where the original inequality, with less then 1/n, is false.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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