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This question already has an answer here:

can someone pls help

How do I prove

2⌊lg n⌋ = Θ(2⌈lg n⌉)

2 ⌈lg n⌉+⌊lg n⌋ = Θ(n2)

I'm not too good at maths. I know,

lim ( n -> infinty) = f(x)/g(x) if we get a real constant the statement is true.

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marked as duplicate by David Richerby, Raphael algorithms Jan 28 '18 at 20:56

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    $\begingroup$ You prove the first one by using the definition of $\Theta$; you don't prove the second one at all, because it isn't true. $\endgroup$ – David Richerby Jan 28 '18 at 17:17
  • $\begingroup$ @DavidRicherby i just wanted to know how does limit for floor and ceiling function work as by defn. if limit is (c,infinity) the it is big theta $\endgroup$ – Sandeep Ranjan Jan 28 '18 at 18:00
  • $\begingroup$ You might need to ask a more specific question, then. If it's only about the mathematics, you should ask it on Mathematics. $\endgroup$ – Raphael Jan 28 '18 at 20:58
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First, observe that $\lfloor \log N \rfloor < 2\lceil \log N \rceil$ holds for all $N \geq 2$.
Let $n_0 = 2$, $c_1=2$, $f(N)=2\lfloor \log N \rfloor$, $g(N) = 2\lceil \log N \rceil$.
So $$f(N) \leq c_1 \cdot g(N)$$ holds for all $N \geq n_0$.
Next, observe that $\lfloor \log N \rfloor > \frac{1}{2}\lceil \log N \rceil$ is true for all $N \geq 2$.
Let $c_2 = \frac{1}{2}$.
Thus, $$f(N) \geq c_2 \cdot g(N)$$ Therefore, $$f(N) \in \Theta(g(N))$$ But the second question, it is obvious that the statement is incorrect, so I'll not be proving here.

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