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So I'm studying for a midterm and my professor put out a sample exam with the answers, and I'm stuck on one of the questions.
The answer is Big-O(n^2 log n)
Could someone show me the steps necessary to go from regular time complexity to asymptotic time complexity?
Informally:
There are two applicable rules here, wrt. a polynomial:
So due to rule 1, $20n^2 + 2n^2\log n - 2n + 1$ becomes $n^2+n^2\log n - n + 1$.
Now, only the largest-growing term here will be significant. The other terms will be insignificant when $n \rightarrow \infty$. The two first terms are obviously much faster growing, since they have an exponent of $2$. The second term is faster than the first because it has a non-constant multiplier that is more than $1$ as $n$ approaches infinity; it will thus make the term grow faster. Thus the second term becomes the only one remaining.
Now for the math:
Because, let $f(x)$ be that term, then $af(x) \in \mathcal{O}\left(f(x)\right)$, where $a$ is a constant, not dependent on $x$. What this means, is that $f(x)$ will grow at the same rate as $af(x)$, (as $x \rightarrow \infty$), even if it will always be larger by a multiplicative constant. How can we prove this?
The definition of $f(x) \in g(x)$:
$$|f(x)| \le \; M |g(x)|\text{ for all }x>x_0.$$
Where $M$ is some positive constant, and there exists some $M$ and $x_0$ that makes this true.
So let's set $g(x)=f(x)$:
So the question becomes: Does there exist an $M > 0$ and $x_0$ such that this holds true?
$$|af(x)| \le \; M |f(x)|\text{ for all }x>x_0.$$
The answer is simple: set $M=|a|$, then we get $|af(x)| \le \; |af(x)|\text{ for all }x>x_0$, which is of course true, since they are actually equal. And since this is true, we get $af(x) \in \mathcal{O}\left(f(x)\right)$ as per the formal definition of big O notation.
See wikipedia section on properties of big O notation:
If a function f(n) can be written as a finite sum of other functions, then the fastest growing one determines the order of f(n). For example
$f(n) = 9 \log n + 5 (\log n)^3 + 3n^2 + 2n^3 = O(n^3) \,, \qquad\text{as } n\to\infty \,\!.$
In particular, if a function may be bounded by a polynomial in n, then as n tends to infinity, one may disregard lower-order terms of the polynomial.
To prove this, we need to prove that given two terms in a polynomial, if one grows as fast or faster than another, then the entire polynomial is as fast as the faster growing term. Formally:
$\left.f(x) + g(x) \in \mathcal{O}\left(f(x)\right) \vphantom{\Bigg|} \right| _{g(x)\in \mathcal{O}\left(f(x)\right)}$
Translation: term $f(x)$ plus term $g(x)$ is as fast as $f(x)$ if $g(x)$ is bounded by $f(x)$ (bounded means: grows at most as fast as).
Proof of this statement:
$$ \begin{eqnarray} &&&\hspace{2pt}&\hphantom{(1)}&\hspace{1pt}& \\ \exists_{x_0,M>0} ~ \left|f(x) + g(x)\right| &\stackrel{?}{\le}& M\left|f(x)\right| ~~~x>x_0 \text{ as } x\rightarrow \infty &&(1)&&\text{Reformulate question} \\ g(x) &\in& \mathcal{O}\left(f(x)\right) &&(2)&&\text{Given} \\ \exists_{x_{g0},M_g>0}~\left|g(x)\right|&\le&M_g\left|f(x)\right| &&(3)&&\text{Restatement of (2)} \\ &&~~~x>x_{g0} \text{ as } x\rightarrow \infty \\ f(x) &\in& \mathcal{O}\left(f(x)\right) &&(4)&&\text{Reflexive} \\ \exists_{x_{f0}=0,M_f=1}~\left|f(x)\right|&\le&M_f\left|f(x)\right| &&(5)&&\genfrac {}{}{0}{}{\text{Restatement of (4) via}}{\text{def., choosing values}} \\ &&~~~x>x_{f0} \text{ as } x\rightarrow \infty \\ \exists_{ \genfrac{}{}{0}{} {x_{f0},x_{g0}} {M_f=1,M_g>0} } ~|f(x)|+|g(x)|&\le& M_f\left|f(x)\right|+M_g\left|f(x)\right| &&(6)&&\text{Adding (3), (5)} \\ && ~~~x>\max\left(x_{f0},x_{g0}\right) \text{ as } x\rightarrow \infty \\ &\le& \left(M_f+M_g\right)\left|f(x)\right| &&(7)&&\text{Simplying (6)} \\ &&~~~x>\max\left(x_{f0},x_{g0}\right) \text{ as } x\rightarrow \infty \\ |f(x)+g(x)| &\le&|f(x)| + |g(x)| &&(7.1)&&\text{Math} \\ \exists_{ \genfrac{}{}{0}{} {x_{0}=\max\left(x_{g0},x_{f0}\right)} {M=M_f+M_g} }~\left|f(x)+g(x)\right|&\le& M\left|f(x)\right| ~~~x>x_0\text{ as } x\rightarrow \infty &&(8)&&\text{Simplifying (7,7.1)} \\ M_f &=& 1 &&(9)&&\text{Valid/chosen in (5)} \\ x_{f0} &=& 0 &&(10)&&\text{Valid/chosen in (5)} \\ \exists_{ \genfrac{}{}{0}{} {x_{0}=\max\left(x_{g0},0\right)} {M=1+M_g} }~\left|f(x)+g(x)\right|&\le& M\left|f(x)\right| ~~~x>x_0\text{ as } x\rightarrow \infty &&(11)&&\text{Simplifying (8,9,10)} \end{eqnarray} $$
Thus we can construct a valid $M$ and $x_0$ given a valid $M_g$, and $x_{g0}$, which we know exist, because $g(x)\in \mathcal{O}\left(f(x)\right)$ is given.
