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I think I barely understand the usefulness of asymptotic notations. And I came across small o and I want to prove that statement is true. But I have no idea how to start.

Those are 2 extra problems with the same idea : Prove that 1000.n^2+ 1000.n ≠ o(n^2) And Prove that 5.n + 10 ≠ o(n)

Note: someone marked this question as repeated by D.W., I looked D.W. all his questions and I didn't find any similarity.

John Sall
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    The first step is: write the definition of $f(n) = o(g(n))$. The second step is: specialize to the case $f(n) = g(n) = n^2$. The third step is: prove the statement in the second step. – Yuval Filmus Jul 06 '18 at 13:54
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    I don't think you're asking the right question, here. $n^2=o(n^2)$ would hold only if the value of a certain limit was zero, but that limit obviously has value one. So your problem must be with the definition of little-$o$, or with limits, or something else. But without knowing what that problem is, we can't really help you. – David Richerby Jul 06 '18 at 15:42
  • @DavidRicherby , this question was given to me by the lab instructor. How is it wrong? – John Sall Jul 07 '18 at 10:53
  • @YuvalFilmus Are you saying if I proved that f(n) = g(n) then I have proved that f(n) = o(g(n))? – John Sall Jul 07 '18 at 10:57
  • @tmc While what you wrote is true, this is very far from what I'm saying. I suggest you use the definition of little o. – Yuval Filmus Jul 07 '18 at 10:58
  • @YuvalFilmus this is the definition of little o : for any positive constant c>0, there exists a constant n_0 > 0 such that 0 <= f (n) < c.g(n) for all n >= n_0. and I cannot make head or tail of it – John Sall Jul 07 '18 at 11:00
  • @tmc If $f(n)=g(n)$ then you can take any $c<1$ to show that the definition cannot hold. – Yuval Filmus Jul 07 '18 at 11:08
  • @YuvalFilmus This is little o, not little omega. in little o : f(n) < c.g(n) which end up with c > 1. not c < 1. – John Sall Jul 07 '18 at 11:10
  • @tmc Your definition is wrong. Instead of "for any positive constant $c > 0$", it should be "for all positive constants $c > 0$". (Actually it isn't the definition but rather your interpretation of it which is wrong.) – Yuval Filmus Jul 07 '18 at 11:15
  • @YuvalFilmus I took this from Intro To Algorithms by Thomas H. Corman, page 50. Oh god, I'm so confused. – John Sall Jul 07 '18 at 11:20
  • @tmc As I wrote, the definition is fine, but your interpretation of the word "any" is wrong. It refers to universal quantification. – Yuval Filmus Jul 07 '18 at 11:21
  • @YuvalFilmus oh, when I said "f(n) < c.g(n) which end up with c >1" I didn't mean the definition, that's actually the proof. so : n^2 < c.n^2 and by factoring we left with 1 < c – John Sall Jul 07 '18 at 11:25
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    @tmc If you can't make head or tail of the definition of little-o then you need to go back to an introductory maths course where they explain limits. Maybe your problem comes from the fact this question is ridiculously simple. Try proving that sqrt (n) = o (n) using the definition. For example, if c = 0.1 which is the n_0 such that sqrt (n) < 0.1 n for all n > n_0? – gnasher729 Jul 07 '18 at 11:26
  • @tmc We are trying to show that the definition doesn't hold, that is, to prove its negation. So we get to choose $c$. – Yuval Filmus Jul 07 '18 at 11:27
  • @gnasher729 Yes, I know about limits, can you tell me how are limits useful here? I don't see any relation. Also, I just proved the equation you mentioned sqrt(n) = o(n), because o(n) has a higher order than sqrt(n). but the problem arise when they have the same order of growth, I don't understand how to prove it when they have the same order. – John Sall Jul 07 '18 at 11:29
  • @tmc This is not a reply to a comment of mine. – Yuval Filmus Jul 07 '18 at 11:30
  • @YuvalFilmus how about if I say because c can be any value, and I assigned c = 1. then c > 1 won't hold? is that a proof? – John Sall Jul 07 '18 at 11:32
  • @tmc Yes, this is the idea. I suggest reading an "intro to proofs" textbook for more information on what constitutes a mathematical proof. – Yuval Filmus Jul 07 '18 at 11:34
  • @YuvalFilmus Okay, thank you, I just solved the examples that I have using this method. However, I'm only following the definition, but what I'm seeking is understanding the big picture and I didn't get there yet. – John Sall Jul 07 '18 at 11:50
  • @sosscs I'm not saying that the question is wrong. I'm saying that the answer follows quickly from the defintion of little-$o$. – David Richerby Jul 07 '18 at 17:57
  • @sosscs "Because n has a higher order than sqrt(n)" isn't a proof. Write down the definition. Which n_0 do you have to choose if c = 0.1? Which n_0 if c = 0.01? It's some simple concrete maths. Which n_0 do you have to choose if I give you an arbitrary c > 0, depending on c? Write the proof down. Then try the same for n = o(n) and see why the proof fails for any c < 1. – gnasher729 Jul 07 '18 at 18:07
  • Connection to limits: How do you prove that the limit of 1/n for n->inf is zero? – gnasher729 Jul 07 '18 at 18:08
  • "someone marked this question as repeated by D.W., I looked D.W. all his questions and I didn't find any similarity." You've misunderstood. It was D.W. who marked the question as a repeat. It's a repeat of the question linked in the yellow box above your question. – David Richerby Jul 08 '18 at 08:06

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