I am a little confused on how to prove/disprove Big O.

For the problem, $2^{n+3}= O(2^n)$, I did the following:

$$2^{n+3} \leq K \times 2^n$$

Set $K = 1$

$$2^{n+3} \leq 2^n$$

Test for large values of n (so I plugged in n = 100)

$2^{103} \leq 2^{100}$ --- which is false therefore Big O is disproven

Is this process correct?

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    $\begingroup$ No, it's not correct. $2^{n+3}$ is $O(2^n)$ if there exists a constant $c$ such that $2^{n+3}\le c2^n$ for all large enough $n$. All you have shown is that $c\neq 1$; you need $c=8$. $\endgroup$ – David Richerby Feb 3 '14 at 23:28
  • $\begingroup$ Thanks. That makes sense. Is there a method to prove them without picking values for 'c' and 'n' or is that the only way? $\endgroup$ – user13371 Feb 3 '14 at 23:33
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    $\begingroup$ "which is false therefore Big O is disproven" -- wrong. Wrong! Landau notation says something about behaviour in the limit; you can not disprove such a relation with a finite sample! $\endgroup$ – Raphael Feb 4 '14 at 9:52
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    $\begingroup$ @Raphael, the "it fails for this single case, therefore it is false" is sadly widespread. Here it even adds a parameter $K$, arbitrarily set to 1... $\endgroup$ – vonbrand Feb 4 '14 at 16:23
  • $\begingroup$ @vonbrand Sad, but probably won't change as long as we keep telling people to just plot it without warning. $\endgroup$ – Raphael Feb 4 '14 at 17:21

Just, replace the constant $K = 2^3$, and you have to use this constant for your proof.

$$ 2^{n+3} = K\times 2^n$$

So, $O(2^{n+3}) = O(2^n)$.

  • $\begingroup$ This says nothing that isn't already contained in my comment to the question. $\endgroup$ – David Richerby Feb 4 '14 at 8:00
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    $\begingroup$ @DavidRicherby So why did you not post an answer? $\endgroup$ – Raphael Feb 4 '14 at 9:50
  • $\begingroup$ Why we have to use constant K=8? I am not getting it $\endgroup$ – rohan-patel Feb 11 '14 at 0:04

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