 Nylah Hendrix

2022-07-09

If a function f(x) is constant complexity $f\left(x\right)=O\left(1\right)$, describe C and k.
If a function f(x) is constant complexity $f\left(x\right)=O\left(1\right)$, describe C and k.
This needs to be described in terms of the relation of C and k.
There exists constants C and k such that $C|f\left(x\right)|\le |1|$, for all $x>k$.
The above cannot be true because using the explanation from the book of discrete math:
f(x) is O(g(x)) if there are constants C and k such that $\mathrm{\forall }x>k,|f\left(x\right)|\le C|g\left(x\right)|.$.
So if $f\left(x\right)=O\left(1\right)$, then it would mean $O\left(1\right)\le C|g\left(x\right)|$, or $C|g\left(x\right)|\ge O\left(1\right)$.
Instead the below statement is true.
There exists constants C and k such that $C|f\left(x\right)|\ge |1|$ for all $x>k$. Leslie Rollins

Expert

Explanation:
I submitted my work and got it wrong, just wanted to share with others who are looking for answers to this quesion. It is:
There exists constants C and k such that $|f\left(x\right)|\le C$ for all $x>k$.

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