What is the Big Theta of $(\log n)^2-9\log n+7$?

Question

How can I find the Big Theta of $(\log n)^2-9\log n+7$?

I thought of $(\log n)^2-9\log(n)+7 < c_1(\log n)^2 +7$ or something like this and can't find the right way.

Your thought is in the right direction. You can eliminate the "+7" on the right and pick $c_1=1$. Then for all $n$ for which $-9\log(n)+7\le 0$ you'll have what you need to show that your function is $O(\log^2(n))$. Finding a big-Omega bound is only slightly more difficult (you can use, for example, $c=1/2$). — Rick Decker, Feb 13 '16 at 18:58
"The Big Theta" of $(\log n)^2-9\log n+7$ if $\Theta((\log n)^2-9\log n+7)$. You'll have to phrase your question more precisely. — Raphael, Sep 20 '18 at 21:29

score 5 · Answer 1 · answered Feb 14 '16 at 09:46

5

There is no such thing as "the" big $\Theta$ of a function. For a given function $f(n)$, many functions $g(n)$ satisfy $f(n) = \Theta(g(n))$. For example, every function satisfied $f(n) = \Theta(f(n))$.

However, usually we are interested in a "succinct", "canonical" or "simple" expression $g(n)$, for example of the form $c^n n^a (\log n)^b$, for some $c \geq 1$. In your case $g(n) = (\log n)^2$: your function is $\Theta((\log n)^2)$, which is not too hard to show.

Setting the proof apart, how do I know that your function is $\Theta((\log n)^2)$? Since $(\log n)^2$ is the dominant term: all other terms ($\log n$ and $1$) grow slower than $(\log n)^2$. The constants in front of them don't matter asymptotically.

answered Feb 14 '16 at 09:46

Yuval Filmus

276,994
27
311
503

Why not $O(logn)^2$? – vivek Mar 07 '16 at 08:07
The OP asked for big Theta, not big O. – Yuval Filmus Mar 07 '16 at 08:12
True, but what if I say that its $O(logn)^2$? is it a valid statement? – vivek Mar 07 '16 at 08:15
It's much better to write it as $O(\log^2 n)$, with the square inside the big O. – Yuval Filmus Mar 07 '16 at 08:19
I am a beginner and finding much confusion, the problem is that I can't find a good reference. There seems a lot of confusions too in lot of text books. I recently saw this answer, can you please point me to some good reference? – vivek Mar 07 '16 at 08:23
I believe it is covered in CLRS. – Yuval Filmus Mar 07 '16 at 08:25
@vivek Ask yourself what $O(f)^2$ is supposed to mean. Unfold the definition of $O$. Compare with $O(f) \cdot O(f)$. See the problem? – Raphael Sep 20 '18 at 21:31

score 2 · Answer 2 · answered Feb 13 '16 at 14:30

2

Hint: what is the answer for $n^2-9n+7$ ?

answered Feb 13 '16 at 14:30

mrk

3,688
22
35

1

Maybe The answer is Θ(n^2)??... so what's your point.... I don't want the answer like Θ(( log(n)^2) ) I need to know the way to find it... like big-O big-Ω and then theta. I can't comment(need 50 rep), that's why I made an answer.... – Feb 13 '16 at 15:56

What is the Big Theta of $(\log n)^2-9\log n+7$?

2 Answers2

Linked