I was looking to see if curved asymptotes were possible and came across an answer that referred to an end behavior of a function as being $f(x)$ ~ $x^2$. I'm assuming this either means the end behavior of a function or a generalization of what a function does given a set of large x values, but I don't want to simply assume something I don't know. Is there a formal definition for what $f(x)$ ~ $g(x)$ is?
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2Yes. $$lim _{x\rightarrow \infty}\frac{f(x)}{x^2}=1$$ – Phicar Apr 15 '19 at 19:52
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2Sure is - $f(x) \sim g(x)$ as $x \to a$ if $\lim_{x \to a} f(x)/g(x) =1 $. More generally, check out https://en.wikipedia.org/wiki/Big_O_notation . $\sim$ is a refined version of $\Theta$. – stochasticboy321 Apr 15 '19 at 19:53
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@stochasticboy321 So then it's just different notation, like f'(x) and d/dx? – jstowell Apr 15 '19 at 19:55
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@Phicar could you explain that a little more? – jstowell Apr 15 '19 at 19:55
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4@jstowell with respect to $\Theta$? No, it's more than that. $f = \Theta(g)$ only demands that $f(x)/g(x) \to C$ for some constant $C$. $\sim$ demands that this constant is $1$. So for instance, $x/2 = \Theta(x)$ but $x/2 \not\sim x$. – stochasticboy321 Apr 15 '19 at 19:57
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1@stochasticboy321 that makes more sense, if you put it as an answer I will accept it – jstowell Apr 15 '19 at 19:58
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4@stochasticboy321 I think $f=\Theta(g)$ does not require the limit $\lim_{x \to a} f(x)/g(x)$ to exist; only that the ratio is bounded between two constants. – angryavian Apr 15 '19 at 20:01
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1I am surprised so many people upvoted the incorrect comment by @stochasticboy321. – Antonio Vargas Apr 15 '19 at 20:36
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@angryavian Right you are. Thanks. – stochasticboy321 Apr 16 '19 at 02:49
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$f(x) \sim g(x)$, I believe, is formally defined as $$\lim_{x \to \infty} \frac{f(x)}{g(x)} = 1$$ Take $f(x) = x^2$ and $g(x) = x^2 - x$ for example. $f(x) \sim g(x)$ because $$\lim_{x \to \infty} \frac{x^2}{x^2-x} = \lim_{x \to \infty} \frac{x}{x-1} = 1$$
Ryan Shesler
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