How to Prove That $O(n) Cdot O(n^2) = O(n^3)$?

First, there seems to be some confusion about what you are asked to prove: what does $$tag1 O(n) cdot O(n^2) = O(n^3)$$ mean, exactly?The interpretation I am familiar with, in the words of Cormen et al., is the following:(Cormen et al., "Introduction to Algorithms", 3rd Ed., section 3.1, page 50.)Note how this makes the $;=;$ operator asymmetrical. Note also that this makes $;f(n) = O(g(n));$ mean $;f in O(g(n));$.$%

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ewcommandwhenLeftarrow %$This interpretation says that $Ref1$ is equivalent to: $$ beginalign& langle forall f,g :: f in O(n) ;land; g in O(n^2) ;then; &phantomlangle forall f,g ::; langle exists h :: h in O(n^3) ;land; langle forall n :: f(n) cdot g(n) = h(n)



angle endalign $$ or simplified: for all $;f,g;$, $$ tag1a f(n) = O(n) ;land; g(n) = O(n^2) ;then; f(n) cdot g(n) = O(n^3) $$Now, how would we prove $Ref1a$? You have the correct idea essentially, and let's try to write this down as cleanly as possible, using the definition that you seem to be using, which is $$ tag2 f(n) = O(g(n)) ;equiv; langle exists c : c > 0 : langle forall_textlarge enough n :: f(n) le c cdot g(n)


angle $$It seems simplest to start with $;f(n) cdot g(n);$, and rewrite that using the assumptions in the left hand side of $Ref1a$:$$calc f(n) cdot g(n) ople_forall_textlarge enough nhintsusing the LHS of $Ref1a$ and definition $Ref2$ twice,hintchoose $;c_1,c_2 > 0;$ c_1 cdot n cdot c_2 cdot n^2 op=hintarithmetic (c_1 cdot c_2) cdot n^3 op=hintusing definition $Ref2$ with $;c := c_1 cdot c_2;$ O(n^3) endcalc$$By transitivity of the three steps, this proves $Ref1a$.A side note: The above proof of course does not just work for $;n;$ and $;n^2;$, but for any functions $;F(n),G(n);$: in the same way we can prove the more general $$tag3 O(F(n)) cdot O(G(n)) = O(F(n) cdot G(n))$$Finally, what about proving the other direction? I would write this as $$tag4 O(n^3) = O(n) cdot O(n^2)$$ and Cormen et al. says this means we need to prove beginalign tag4a &langle forall f :: f in O(n^3) ;then; &phantomlangle forall f :: ; langle exists g,h :: g in O(n) land h in O(n^2) land &phantomlangle forall f :: langle exists g,h :: ; langle forall n :: f(n) = g(n) cdot h(n)



angle endalignSo given an $;f in O(n^3);$, there seem to be two simple choices, one of which is $;g(n) = f(n) / n^2;$, and $;h(n) = n^2;$. All that is then left to prove is $;f(n)/n^2 = O(n);$ and $;n^2 = O(n^2);$, which are easy exercises from definition $Ref2$.(Note that I am assuming here that $;n;$ ranges over positive integers, to avoid technical manipulations to avoid division by zero.)And again, of course this proof generalizes immediately to a proof of $$O(F(n) cdot G(n)) = O(F(n)) cdot O(G(n))$$The only qualm I have with this proof is that it is not really symmetrical in $;g;$ and $;h;$: we could have equally well have chosen $;g(n) = n;$ and $;h(n) = f(n) / n;$. I tried to find a more symmetrical choice for $;g,h;$, but do not see it yet.$% endgroup %$

1. Is the chemical form of rust Fe2O3.xH2O?

Yes, chemical form of rust is Fe2O3.XH2O, where the number of water molecules is variable.Is the chemical form of rust Fe2O3.XH2O?

2. ( o _ 0 ) What is Shaquille O'Neal smoking these days?

I've seen him practice BJJ and he LOVES MMA but I think he is kinda nuts if he does this. I would hate to see him raped but I do like seeing these athletes thinking they can fight PROFESSIONAL MMA FIGHTERS and then die trying

3. I feel.. *big* and uncomfortable at school? Same weight as Debby Ryan o.O?

First of all I like how you are admitting a problem you have. Anyway, I like to take fresh walks every now and then. Eating a snack such as a piece of fruit helps too. Just watch what you eat, and keep it steady. The more patience, the better. Wish you very good luck :) Joel

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