asymptotics and calculating it

Question

I have a task to show by definition that

$\ln n^{2} + 1 = \Theta (\ln n)$

by definition , we know that we have to find c1 and c2 such as

$c_{1} * \ln n<\ln n^{2} + 1 < c_{2} * (\ln n)$

we can find find c2 easily

$\ln n^{2} + 1 < c_{2} * (\ln n)$

$2*\ln n + 1 < c_{2} * (\ln n)$

so we know that c2 hast to be c2 >= 3.

c1 is the same story

$c_{1} * \ln n< 2*\ln n + 1 $

we see that c1 <= 2

My question is from students point of view. Is it legit to assume that c2 >= 3 and 0 < c1 <= 2 because we "see it" or do we have to prove it by some further calculation?

@AbdallahHammam I do not see anything wrong , what did u meant by it? I may be explaining it poorly , but i cannot think of another way of explaining it. — trolkura, Oct 29 '16 at 14:30
@AbdallahHammam now i see what you meant by it , it should be ln n , im going to edit question. — trolkura, Oct 29 '16 at 14:49

hamam_Abdallah · Answer 1 · 2016-10-29T15:39:38.363

1

As

$$\lim_{n\to +\infty}\frac{ln(n^2)+1}{ln(n)}=$$

$$\lim_{n\to+\infty}(2+\frac{1}{ln(n)})=\color{green}{2}$$

with$\;\;\epsilon=1$ for example,

$$\exists N>0: \forall n\geq N$$

$$1<\frac{ln(n^2)+1}{ln(n)}<3$$

So, you can take $c_1=1\;$ and$\;c_2=3$ .

edited Oct 29 '16 at 15:39

answered Oct 29 '16 at 14:56

hamam_Abdallah

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asymptotics and calculating it

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