Trouble understanding little-o notation example

Question

I'm having trouble with this one problem

9n <= cn^3

basically I can get down to

9/c <= n^2

But how do I solve the rest?

@nabla, It's actually 9n <= o(n^3) – Frightlin Feb 07 '14 at 11:01 — Frightlin, Feb 07 '14 at 11:01
@Frghtlin Yes reading it again I understand. – Feb 07 '14 at 11:02 — , Feb 07 '14 at 11:02

Tanmoy Banerjee · Accepted Answer · 2014-02-08T08:25:04.843

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definition of little o is

enter image description here we say f(x)=o(g(x)).

let f(x)=9*x and g(x)=c*x^3 where c is a positive constant. when x tends to infinity, f(x)/g(x) tends to 0.so we can say f(x)=o(g(x)).

asyptotic notations are applicable for sufficiently large value of n.so for large value of n

9n << cn^3

for all c>0.

edited Feb 08 '14 at 08:25

answered Feb 08 '14 at 08:17

Tanmoy Banerjee

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score -1 · Answer 2 · edited May 23 '17 at 11:56

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Read this link to undersatnd big-O and little-O link

see in case of your equation, when n=3 it becomes 9*3=23=3^3 so for n<3 9n > n^3. so if you choose c as any number to make 9n<=n^3 for n<3 then it can be in O(n).

edited May 23 '17 at 11:56

Community

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answered Feb 07 '14 at 11:02

LearningC

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This is not what the OP tries to proof. – Feb 07 '14 at 11:03
do you mean that it can be in little o(n)? – Frightlin Feb 07 '14 at 11:04
@Frigthlin It is O(n), but not o(n). – Feb 07 '14 at 11:05
@nabla, ah, I'm looking for the little oh, not big oh :( – Frightlin Feb 07 '14 at 11:06

Trouble understanding little-o notation example

2 Answers2