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In my class, we have defined that $$ f(x) \ll g(x) $$ on $A$ if there exist a strictly positive c such that $$ |f(x)| \le cg(x) $$ for every $x$ on $A$.

I'm a bit confused. Say that $ f(x) = x$ and $ g(x) = \frac x2 $ We could take $c = 3$, and that would mean that $x\ll\frac x2$? I'm confused, it doesn't make sense to me.

Hagen von Eitzen
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math7dg7
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  • Yes, $x\ll x/2$. – Gerry Myerson Apr 12 '16 at 09:31
  • But you could also say that x/2 << x, with that definition. I'm not sure I actually understand what it means. I thought it meant that the function grew slower asymptotically. – math7dg7 Apr 12 '16 at 09:33
  • According to this definition, $2x << x$. In particular, this symbol does not denote "is much bigger than". The definition is equivalent to $$\inf_{x \in A} \frac{g(x)}{|f(x)|} > 0$$ – Crostul Apr 12 '16 at 09:34
  • Not slower; slower than some constant multiple. $x$ grows more slowly than $(73)(x/2)$, and $x/2$ grows more slowly than $(1)x$. – Gerry Myerson Apr 12 '16 at 09:37
  • According to my notes, the definition I used is equivalent to Big-O notation. Is this correct? – math7dg7 Apr 12 '16 at 09:41
  • @math7dg7 No. Big-Oh differs. In particular, Big-Oh is always relative to some boundary point of $A$, i.e., sometimes understood to refer to $x\to\infty$, sometimes for $x\to 0$, sometimes others. As even these variations differ ($x+1\in O(x^2)$ as $x\to\infty$, but $x+1\notin O(x^2)$ as $x\to 0$), they cannot coincide with your notation that is unaware of the boundary point picked. – Hagen von Eitzen Apr 12 '16 at 09:49
  • Well, $f\ll g$ as $x\to\infty$ is the same as $f$ is $O(g)$ as $x\to\infty$. They are used interchangeably, when the "boundary point" is understood from context. – Gerry Myerson Apr 12 '16 at 13:28

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