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I came across a problem that I tried to formalize as follows:

Let say i have two functions $x(t)$ and $y(t)$ such that for $t \rightarrow t_0$ $$ \left\{ \begin{array} \;y(t) \rightarrow -\infty \\ x(t) \rightarrow -\infty \\ y(t) = o(x(t)) \end{array} \right. $$

Shouldn't be in such case $ 2^{y(t)} = o(2^{x(t)})$ ? If yes I struggle to prove it is true... otherwise what can we say instead? What about if we don't assume $y(t),x(t) \rightarrow - \infty$?

PS. Just to be clear.. by $y(t) = o(x(t))$ I mean that $\frac{y(t)}{x(t)} = 0$, for $t \rightarrow t_0$.

user8469759
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1 Answers1

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No, $ 2^{y(t)} = O(2^{x(t)})$ is false.

Consider the example

$$ \left\{ \begin{array} y(t) = -\frac1{|t-t_0|} \\ x(t) = -\frac2{|t-t_0|} \\ y(t) = \frac12 x(t) = O(x(t)) \end{array} \right. $$ But if we let $u = \frac1{|t-t_0|}$, then $2^{-u}$ is not $O(2^{-2u})$, although it is $O(\sqrt{2^{-2u}})$: as $u\to\infty$, the ratio of $2^{-u}$ to $2^{-2u}$ exceeds any constant $C$.

Mark Fischler
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  • I'm sorry but why in your example $y = O(x)$? should the ratio $y / x$ be 0 for $t \rightarrow t_0$? – user8469759 Mar 30 '16 at 16:14
  • CHeck your definition of the "big-O" notation -- it means that there is a constant $C$ and some distance $\epsilon>0$ such that $|t-t_0|<\epsilon \implies y(t) < Cx(t)$. And that is obviously true; for example, take $C=3$. – Mark Fischler Mar 31 '16 at 22:49
  • What I meant was $\frac{y(t)}{x(t)} \rightarrow 0 ;; t \rightarrow t_0$ – user8469759 Apr 04 '16 at 14:11