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June 4th, 2019, 06:29 AM   #1
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asymptotics

If $$ \lim\limits_{n \to \infty} \dfrac{f(n)}{g(n)} = \dfrac{a}{b}$$ with $a \ne b$, $b \ne 0$ then $$b f(n) \sim a g(n) $$ ?....
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June 4th, 2019, 07:16 AM   #2
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If you set $\displaystyle a/b$ inside the limit then it is done .
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June 4th, 2019, 07:38 AM   #3
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no but then why define $\sim$ the way it is
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June 4th, 2019, 08:09 AM   #4
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Quote:
Originally Posted by Joppy View Post
If $$ \lim\limits_{n \to \infty} \dfrac{f(n)}{g(n)} = \dfrac{a}{b}$$ with $a \ne b$, $b \ne 0$ then $$b f(n) \sim a g(n) $$ ?....
$a=0$, $f(x)=x$, $g(x)=x^2$
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June 6th, 2019, 11:02 PM   #5
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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
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June 12th, 2019, 12:47 AM   #6
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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.
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June 12th, 2019, 05:21 AM   #7
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Quote:
Originally Posted by Joppy View Post
If $$ \lim\limits_{n \to \infty} \dfrac{f(n)}{g(n)} = \dfrac{a}{b}$$ with $a \ne b$, $b \ne 0$ then $$b f(n) \sim a g(n) $$ ?....
$\displaystyle \dfrac{b}{a} \lim_{n \to \infty} \dfrac{f(n)}{g(n)}= \dfrac{b}{a} \cdot \dfrac{a}{b}$

$\displaystyle \lim_{n \to \infty} \dfrac{b \cdot f(n)}{a \cdot g(n)}= 1$
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June 12th, 2019, 05:40 AM   #8
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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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