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how would I prove

$ 8^n = Θ(4^n)$ is either true or false.

I so far have attempted to prove big O but cant find the value of C1

Pythonlover
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  • This seems like a homework question. What values of $c_1$ have you tried? Remember it can also be less than 1 (but has to be greater than 0). – ryan Mar 24 '19 at 23:58
  • I did it using this method link and came to $3^n<=C1$ – Pythonlover Mar 25 '19 at 00:01
  • Oh wait, I misread. A good approach to this (when they are both exponential) is trying to convert them to a common base. For instance we can convert them both to base 2, meaning $4^n = 2^{2n}$ and $8^n = 2^{3n}$. Can you now prove if $2^{3n} = \Theta(2^{2n})$? – ryan Mar 25 '19 at 00:14
  • would I do this the same way ? and sorry I meant I has 2^n <= C1 – Pythonlover Mar 25 '19 at 00:16
  • You could draw a similar conclusion. If you know $c_1 \geq 2^n$ this would imply that $c_1$ must be greater than a non-constant value (because $2^n$ is a function of $n$). Therefore we have a contradiction that $c_1$ is not a constant, and thus no constant $c_1$ would exist for this inequality. Does this logic make sense? – ryan Mar 25 '19 at 00:17
  • that's what I was thinking so have I then proved the statement is false because in order for it to be true $ f(n)<=C1$ – Pythonlover Mar 25 '19 at 00:19
  • but if $c1>=1/2^n$ c1 is a half ? – Pythonlover Mar 25 '19 at 00:24
  • Correct, you can do this with a simple argument by contradiction. Assume a value $c_1$ exists that satisfies $c_1 8^n \leq 4^n$. Next show that this value $c_1$ must also satisfy $c_1 \geq 2^n$. This, then violates our assumption that $c_1$ is a constant, therefore we have a contradiction. – ryan Mar 25 '19 at 00:25
  • Yes, you do have to be careful because just saying $c_1 \geq f(n)$ does not give a contradiction necessarily. Consider your example or any monotone decreasing functions (I think). So you do have to be a little careful with your argument, but for your case it should be fine. Note, you can also prove it with limits, like is shown below. – ryan Mar 25 '19 at 00:26
  • why is this a contradiction is it because there is no maximum value for n? it just keeps going up to infinity? – Pythonlover Mar 25 '19 at 00:26
  • It's a contradiction because $2^n$ is monotone increasing. – ryan Mar 25 '19 at 00:27
  • right ok so then to prove big omega I have $c2<=3^n$ is this also a contradiction or? – Pythonlover Mar 25 '19 at 00:35
  • That is actually fine, because we just need that $4^n \leq c_2 8^n$. So the restriction $c_2 \leq 3^n$ is OK. For instance if we choose $c_2 = 1$, then this is fine because $c_2 = 1 \leq 3^n$ for all $n \geq 0$. – ryan Mar 25 '19 at 01:21
  • I would recommend reviewing the definitions of big-$O$, big-$\Theta$ and big-$\Omega$. The inequalities only need to hold for all $n \geq n_0$ for some constant $n_0$. You basically get a set of constraints with these definitions. So if you satisfy all constraints, you then satisfy the definition! – ryan Mar 25 '19 at 01:24

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Just compute $\lim_{n\to\infty} \frac{8^n}{4^n} = \infty$. Hence, $8^n = \Omega(4^n)$ and also $8^n = \omega(4^n)$.

OmG
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