1

There is a function defined as follows:

formula

$x, y, w(u), w'(u)$ are all positive real values. $x > y \geq 2$.

Does this mean that $\Phi(x, y) > \frac{x}{\log{y}}\left(w(u)+\left(\frac{y}{2x^2}-w'(u)\right)\frac{1}{\log{y}}\right)$ ?

In other words, is the big O strictly greater than 0?

The reason I ask is that it doesn't seem to be true in the application I am using it for and I want to check if I am making a mistake.

Simd
  • 417
  • 2
    There are popular textbooks (especially in computer science) that define the big-O notation as always positive, but in analysis it is always allowed to be negative. – Brendan McKay Dec 26 '22 at 12:44
  • I agree with Ethan. In the definition shown here, the $O$ need not be positive. – GEdgar Dec 26 '22 at 12:58
  • @BrendanMckay Thank you. This is from number theory. So what is the full definition of big O where it is allowed to be negative? – Simd Dec 26 '22 at 13:05
  • 1
    "$f(x)=O(g(x))$ as $x\to x_0$" means that there are constants $\varepsilon,C$ such that $|f(x)|\leq C g(x)$ whenever $|x-x_0|\leq\varepsilon$. In the case of $x_0=\infty$, the condition $|x-x_0|\leq\varepsilon$ is replaced by $x\geq \varepsilon$. Some people write $|g(x)|$ in place of $g(x)$ but it doesn't matter as we can always choose $g(x)$ to be positive. – Brendan McKay Dec 27 '22 at 00:19

3 Answers3

4

No. For $|x| < 1$

$$ \frac{1}{1-x} = 1 +x + x^2 + O(x^3) $$ where the tail of the truncated series is negative when $x$ is negative.

Ethan Bolker
  • 95,224
  • 7
  • 108
  • 199
  • Sorry I should have put lower bounds on x and y. – Simd Dec 26 '22 at 12:25
  • @Simd I doubt that matters in the general case. In your particular problem it may be true but you will have to find a particular argument. Perhaps ask about what you are actually trying to show rather than this general assertion. – Ethan Bolker Dec 26 '22 at 12:27
1

If you “solve” the equation for the big-Oh term, $O(1/(\log y)^2)=something$, then all that is claimed is that there exists $C$ such that $|something|<\frac C{(\log y)^2}$ for $y\gg0$.

Hagen von Eitzen
  • 374,180
1

Here's wikipedia's definition:

One writes $f(x) = O(g(x))$ as $x\rightarrow\infty$ if the absolute value of $f(x)$ is at most a positive constant multiple of $g(x)$ for all sufficiently large values of $x$. That is, $f(x) = O(g(x))$ if there exists a positive real number $M$ and a real number $x_0$ such that $|f(x)| \le Mg(x)$ for all $x\ge x_0$

So $g(x)$ should be a positive function, but $f(x)$ is free to be negative. Big-O is a statement about its absolute value. Furthermore, big-O notation is about asymptotic behavior. $1+x$ is in $O(x)$ even though at $x=0$, $1+x$ is infinitely larger than $x$; what matters is what happens "infinitely out", not what happens for any finite value.

Acccumulation
  • 12,210