The first one is a big-theta of 2^n, and then also the other is a big-theta of 2^n because there is a constant that divide n.
Is that correct?
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No, they are not equal asymptotically.
You should be able to find constants k1 and k2 for functions k1*2n and k2*2n that will bound 2n/10. We know that 2n/10 = (2n)(1/10) which cannot be expressed as c*2n.
President James K. Polk
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Grzegorz Żur
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Ok, thx a lot. Another quick question: if I had 2^(n+10) instead of the second that I wrote in the question, this time it would be same complexity of 2^n right ? – Maverick Jul 26 '19 at 08:49
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1Yes, it would be the same because 2^(n+10) = 2^n*2^10 = c*2^n where c = 2^10. – Grzegorz Żur Jul 26 '19 at 08:58
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1Small typo: `2^(n/10) = (2^n)^(10)` - RHS should be `(2^n)^(1/10)` – meowgoesthedog Jul 26 '19 at 10:03
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2^(n/10)=(2^(1/10))^n not same as 2^n
See this Are 2^n and 4^n in the same Big-Θ complexity class?
the link explain 2^n and 4^n
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