I'm new to calculus and i found a function expressed as :

$f(x)= x + O(x)$

And after studying some material i understood that $O(x)$ simply represent an upper bound, telling us that this $O(x)$ represent an unspecified function that from a certain $x_{0}$ grows at max as $cx$ where c > 0. Is it correct ?

In the definition it's said that c must be > 0, so i deduce that if i want to express a growth of $-x$ i simply write $-O(x)$. Correct ?

Thanks for your help :)

  • $\begingroup$ $O(x)$ involves also negative values, so $-O(x)$ is not different from $O(x)$ $\endgroup$ – Peter Jul 15 '18 at 14:14
  • $\begingroup$ If i want to express that as $x\rightarrow\infty$ it goes to $-\infty$ as $-x$ ? $\endgroup$ – themagiciant95 Jul 15 '18 at 14:23
  • $\begingroup$ Proper use of "Big-$O$" and "Little-$o$" notation, similarly to the "$\lim$" notation. should state where the variable is tending. ...BTW $f(x)=x+O(x)$ as $x\to \infty$ is equivalent to $ f(x)-x=O(x)$ as $x \to \infty,$ which means means that there exists $K>0$ and $r\in \Bbb R$ such that $ \forall x>r\; (|f(x)-x|<K|x)$|... Which is equivalent to $f(x)=O(x)$ as $x\to \infty.$ $\endgroup$ – DanielWainfleet Jul 16 '18 at 1:27

Usually $O(x)$ denotes any function whose absolute value grows at most as $cx$ for some $c>0$.

So each of the functions $x^2+x$ and $x^2-x$ is both $x^2+O(x)$ and $x^2-O(x)$.


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