Is there a formula or algorithm that can find the power ${^x}$ that returns $0$ for $n$ ${\mathbb N} $?
Asked
Active
Viewed 59 times
-1
-
2Do you have any example of a power of an integer that returns zero? – Hagen von Eitzen Aug 18 '16 at 10:10
-
Only by trial and error I can return an approximate such that $2$ to the power $-13.287$ is $0.00010049$. I know that $0$ to the power of any number is also $0$. I am looking for the approach to solve this question? – ctrl-alt-delete Aug 18 '16 at 10:13
-
$[n^x=0]\implies[n=0]\wedge[x\neq0]$. – barak manos Aug 18 '16 at 10:20
2 Answers
2
No such thing exists. Taking the example of $n = 2$, then $f(x) = 2^{-x}$ is a strictly decreasing function on $(0, \infty)$. You can see: $f(1) = \frac{1}{2}$, $f(2) = \frac{1}{4}$ etc etc, but only: $\lim\limits_{x \rightarrow \infty} f(x) = 0$.
Piotr Benedysiuk
- 1,622
2
No number, other than zero itself, has a solution to what you want. What I mean is that $n^x \neq 0$ unless $n = 0$. What you can do is just raising it to a very large negative power, e.g. with your example $2^{-13}\approx 0$, but $2^{-100}$ is even closer to zero, and $2^{-1000}$ is even closer etc. This is true for any $n > 0$.
Eff
- 12,989