Order Comparison of Functions: f(n) < THETA(g(n)) for O(2^n) and O(3^n)

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Discussion Overview

The discussion revolves around the comparison of the growth rates of two exponential functions, f(n) = 2^n and g(n) = 3^n, specifically examining their classifications within Big O notation and the implications for their relationship in terms of Θ notation. Participants explore whether f(n) is less than Θ(g(n)) and the correctness of the notation used.

Discussion Character

  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant asserts that since O(2^n) < O(3^n), it follows that f(n) < Θ(g(n)).
  • Another participant challenges this by stating that O(2^n) is equivalent to O(3^n) due to the nature of exponential functions differing only by a multiplicative constant.
  • A further reply emphasizes that there is no constant C such that C · 2^x > 3^x for all x, questioning the initial claim.
  • One participant acknowledges a misunderstanding regarding the comparison of functions and clarifies their thought process.
  • Another participant explains that while O(2^n) is often treated as equal to O(3^n) in informal contexts, the set of functions classified as O(2^n) is a proper subset of those classified as O(3^n). They suggest that the correct notation might be that 2^n is o(3^n) based on the limit comparison.
  • A participant expresses gratitude for the discussion and reflects on their understanding of Θ notation in relation to upper and lower bounds.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the relationship between O(2^n) and O(3^n), with some arguing for equivalence and others for a distinction. The discussion remains unresolved regarding the correct interpretation of the notation and the implications for the functions' growth rates.

Contextual Notes

There are limitations in the discussion regarding the definitions and interpretations of Big O and Θ notation, as well as the conditions under which these classifications hold. The use of informal notation and potential misunderstandings about the relationships between functions contribute to the complexity of the discussion.

Ad2d
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Hello All, I need help making sure I am right about this. Say for example I have these two functions: f(n) = 2^n and g(n) = 3^n.

Now the order is O(2^n) and O(3^n) respectfuilly. Now I need make sure I am right about this: based on the order, O(2^n) < O(3^n) therefore f(n) < THETA(g(n)).

Is this statement correct here when I say f(n) < THETA(g(n)) and that O(2^n) < O(3^n)?
 
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Isn't O(2^n) = O(3^n)? Exponential functions differ by a multiplicative constant.
 
Tac-Tics said:
Isn't O(2^n) = O(3^n)? Exponential functions differ by a multiplicative constant.

No?
The constant C does not exist such that [tex]C \cdot 2^x > 3^x[/tex] for all x.
 
Ack. My intuition failed me. I was thinking of 2x^n versus 3x^n for a constant n. Sry.
 
O(2^n) is O(3^n) in the usual abuse of notation sense. But the set of functions that are O(2^n) is a proper subset of the set of functions that are O(3^n). For example, 2.0001^n is not O(2^n).

"f(n) < THETA(g(n))" isn't defined -- O, o, Theta, and the others use only the element operator (or, by the abuse mentioned above, an equal sign). Do you perhaps mean that 2^n is o(3^n)? That would be correct, because lim 2^n/3^n = 0.
 
Thank You all. I was able do what CRGreathouse with the limits which made more since to me. From what the professor was saying, I am thinking that Theta is a value that is I am between the upper and lower bound, and I need to show whether or not the functions where less than the upperbond (f(n) < theta (g(n)) ), greater than the lower bound (f(n) > theta (g(n)) ), or equal ((f(n) = theta (g(n)) )). I hope that make sense (I'm not a genius when it comes to wording in math). Just like to say thanks again for all your input. It helped.
 

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