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Hi I need help figuring these out:

True or False: $\log_2 n$ is $O(\log_3 n)$

I used the definition of Big O in Dasgupta's book: ${f(n)}\over g{(n)}$ $\leq c$

So I used the base transformation rule to get: ${(\log_2 3)(\log_3 n)}\over \log_3 n$ $\leq c$

Which just gives me $\log_2 3 \leq c$

So I got false as the answer, what I'm not sure about is if I did this correctly and if it makes sense in terms of Big-O notation, because I still have trouble with Big O.

For another question: $2^{(\log_2 n)} = O(2^{(\log_3 n)})$

I simplfied this to $2^{(\log_2 n)}\over 2^{(\log_3 n)}$ to $n\over n 2^{\log_3 2}$ to $1 \over 2^{\log_3 2)}$ but I'm not sure that I did that correctly when I simplified it down.

And for this question

True or False: $(\log_2 n)^2$ is $O(\log_3 n)^2$

I'm not even sure how I would begin this problem, any hints?

One last quick question about sum of series for the series $S(n) = 1^c + 2^c + 3^c + \cdots + n^c$

I get $\sum_{i=1}^n i^c$ Does that equal the equation $n^{c+1} -1 \over n-1$?

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Forget about $n=1$, the (not so good) definition has problems there. One log is a constant multiple of another. So the big O relationship trivially holds. – André Nicolas Oct 3 '12 at 4:06

$$ \log_2 n = \frac{\log_3 n}{\log_3 2} = c\log_3 n\text{ where } c = \frac{1}{\log_3 2}=\log_2 3. $$ If $\log_2 n = c\log_3 n$ then $\log_2 n\le c\log_3 n$. If $\log_2 n\le c\log_3 n$ then $\log_2 n = O(\log_3 n)$.

(A crucial point here is that the quantity called "$c$" does not depend on $n$. "Constant" always means "not depending on something". What "something" is depends on the context.)

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I'm not sure how showing c = $1\over log_3 2$ $= log_2 3$ fits into showing that it is $O(log_3 n)$. Is it just showing that in this context that is what c is? And if that is the value of c the inequalities follow and that it is $O(log_3 n)$? Or well actually they would be equal in this situation correct? not less than, and is that why the answer is true? – Kevin Oct 3 '12 at 4:28
The constant is either $1/\log_3 2$ or $\log_2 3$; it's the same number either way. – Michael Hardy Oct 3 '12 at 4:47

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