1. ## Big-Oh Proofs

Last time I had some questions on Big-Oh proofs, this set of problems is similar, I was wondering if the different notation still means the same thing, except now f(n) is always above 0? So "there exists some c so that if n is above the threshold B, then no matter what n is cg(n) will always be equal or greater than f(n)"?

I'm not really familiar with the new notation.
Thanks.

2. Which exactly notation is new?

3. Oh the |-> (arrow) and the R>0.

4. Originally Posted by Selena
Oh the |-> (arrow) and the R>0.
the bar is read as "in which" and the arrow is "if" and the R>0 is real numbers that are greater than 0.

5. Ok, so it translates to "if f(n) results in a real number greater to or equal to 0, then ... ..."?

6. Ok, so it translates to "if f(n) results in a real number greater to or equal to 0, then ... ..."?
It's not clear what n you are referring to.

The notations here have nothing to do with big-Oh; they are standard mathematical notations.

$\{x:A\mid P(x)\}$ -- the set of elements $x$ of $A$ for which $P(x)$ holds. This is set-builder notation.

$f:\mathbb{N}\mapsto\mathbb{R}^{\ge0}$ -- $f$ is a function with domain $\mathbb{N}$ and codomain $\mathbb{R}^{\ge0}$ (the set of nonnegative real numbers). Actually, I think $\to$ should be used instead of $\mapsto$. One writes $x\mapsto y$ to mean a function maps $x$ (an element of the domain) into $y$.

$A\Rightarrow B$ -- if $A$, then $B$. See the list of logic symbols.

7. Hmm so it has a domain of natural numbers, and codomain of non-negative real numbers... isn't that just the same as saying it's in a set of non-negative real numbers? Don't they overlap?

8. isn't that just the same as saying it's in a set of non-negative real numbers?
What is in the set of non-negative real numbers?