Asymptotic estimates for the Exponential - MathOverflow [closed] most recent 30 from http://mathoverflow.net 2013-06-19T13:13:15Z http://mathoverflow.net/feeds/question/104951 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/104951/asymptotic-estimates-for-the-exponential Asymptotic estimates for the Exponential Lasagna 2012-08-17T22:59:33Z 2012-08-18T05:17:27Z <p>Consider the function $z(n) = (1-f(n))^{g(n)}$. For $f(n) = \frac 1n, g(n) = n$ we have that $\lim z(n) = e^{-1}$; more generally, when $f(n) = \frac cn$ for any constant $c$, we have $\lim z(n) = e^{-c}$. In each of these cases we note that the limit is equal to $e^{-f(n)g(n)}$.</p> <p>Here's my question: Under what conditions (on $f,g$) can we claim that $(1-f(n))^{g(n)} \sim e^{-f(n)g(n)}$? Unless $f(n)g(n)$ is a constant, the RHS also depends on $n$, which is why I'm only asking about asymptotic equality. My original guess was that it is sufficient to have $f(n) \to 0$ and $g(n) \to \infty$ but I haven't been able to get anywhere with a proof.</p> <p>My context: I have the quantity $( 1 - \frac 1n)^{cn \log n}$ and I'd like desperately for this to be asymptotically equal to $e^{-c\log n} = n^{-c}$.</p> http://mathoverflow.net/questions/104951/asymptotic-estimates-for-the-exponential/104956#104956 Answer by Marc Chamberland for Asymptotic estimates for the Exponential Marc Chamberland 2012-08-17T23:42:33Z 2012-08-17T23:42:33Z <p>This is a standard calculus question. Let $z(n) = (1−f(n))^{g(n)}$. Then $\log z(n) = g(n) \log( 1 - f(n))$. Assuming $f(n)\rightarrow 0$, one has $\log z(n)/( f(n)g(n) ) \rightarrow -1$. For your particular problem, all works as hoped for and your limit is $n^{-c}$.</p> http://mathoverflow.net/questions/104951/asymptotic-estimates-for-the-exponential/104964#104964 Answer by Latro for Asymptotic estimates for the Exponential Latro 2012-08-18T02:32:30Z 2012-08-18T02:32:30Z <p>An easyish way to see that the first comment is valid:</p> <p>If f(n) tends to 0, then</p> <p>$\ln (1-f(n))$ tends to $-f(n)$ (take the first term of the Taylor series.)</p> <p>So</p> <p>$\ln (1-f(n))/f(n)$ tends to $-1$</p> <p>So</p> <p>$g(n) \ln (1-f(n))$ tends to $- g(n) f(n)$</p> <p>Then exponentiate and the result follows.</p> http://mathoverflow.net/questions/104951/asymptotic-estimates-for-the-exponential/104968#104968 Answer by Steeler for Asymptotic estimates for the Exponential Steeler 2012-08-18T05:17:27Z 2012-08-18T05:17:27Z <p>You need some sort of control over the $g(n)$:</p> <p>Consider $f(n)=1/n$</p> <p>$g(n)=n^2$</p> <p>Then $(1-f(n))^{g(n)} \over \exp(-f(n)g(n))$ does not go to 1 as $n \rightarrow \infty$ (goes to maybe $\exp(-1/2)$ instead)</p> <p>Ultimately the original functions are nice enough to work though.</p>