asymptotics

Feb 2016
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If $$ \lim\limits_{n \to \infty} \dfrac{f(n)}{g(n)} = \dfrac{a}{b}$$ with $a \ne b$, $b \ne 0$ then $$b f(n) \sim a g(n) $$ ?....
 
Dec 2015
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If you set \(\displaystyle a/b\) inside the limit then it is done .
 
Feb 2016
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no but then why define $\sim$ the way it is
 

v8archie

Math Team
Dec 2013
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If $$ \lim\limits_{n \to \infty} \dfrac{f(n)}{g(n)} = \dfrac{a}{b}$$ with $a \ne b$, $b \ne 0$ then $$b f(n) \sim a g(n) $$ ?....
$a=0$, $f(x)=x$, $g(x)=x^2$
 
Feb 2016
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No I get it's a stupid question but I meant $a \ne 0$ also. Basically I have a function with known limit which is not unity and I want to infer how it grows without having to juggle terms in the limit
 
Feb 2016
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The point is that the limit either a) converges to a constant greater than 0, diverges to $\pm \infty$ or converges to 0. If a) then $f \sim g$, b) $f$ grows faster, c) $g$ grows faster.
 

skeeter

Math Team
Jul 2011
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If $$ \lim\limits_{n \to \infty} \dfrac{f(n)}{g(n)} = \dfrac{a}{b}$$ with $a \ne b$, $b \ne 0$ then $$b f(n) \sim a g(n) $$ ?....
$\displaystyle \dfrac{b}{a} \lim_{n \to \infty} \dfrac{f(n)}{g(n)}= \dfrac{b}{a} \cdot \dfrac{a}{b}$

$\displaystyle \lim_{n \to \infty} \dfrac{b \cdot f(n)}{a \cdot g(n)}= 1$
 
Feb 2016
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$\displaystyle \dfrac{b}{a} \lim_{n \to \infty} \dfrac{f(n)}{g(n)}= \dfrac{b}{a} \cdot \dfrac{a}{b}$

$\displaystyle \lim_{n \to \infty} \dfrac{b \cdot f(n)}{a \cdot g(n)}= 1$
Yes.. just in the asymptotic sense, the constants $a,b$ don't factor into the expression $f \sim g$ because they're negligible in the limit.
 
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