How to prove expontential = O (n)

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Discussion Overview

The discussion revolves around determining whether the statement \(2^n = O(n!)\) is true or false, with participants exploring how to prove this relationship. The focus is on mathematical reasoning and the application of big O notation.

Discussion Character

  • Mathematical reasoning
  • Homework-related
  • Exploratory

Main Points Raised

  • One participant suggests that \(2^n = O(n!)\) might be true based on their understanding of growing order functions but admits they cannot prove it.
  • Another participant proposes showing that \(n! > 2^n\) for all \(n\) above a certain value as a way to prove the statement.
  • A participant expresses difficulty in establishing the link between factorial and exponential functions, specifically questioning how to choose values for \(k\) and \(n_0\) in the big O definition.
  • Further hints are provided about examining specific values of \(n\) to identify a pattern, indicating that \(2^n\) is greater than \(n!\) for small \(n\) but less for larger \(n\).
  • Another participant emphasizes that it is sufficient to find a specific \(k\) (suggesting \(k=1\)) to prove the inequality holds for sufficiently large \(n\).

Areas of Agreement / Disagreement

Participants do not reach a consensus on the proof of the statement. There are multiple viewpoints on how to approach the proof, and uncertainty remains regarding the specific values of \(k\) and \(n_0\) needed.

Contextual Notes

Participants express limitations in their understanding of the relationship between factorial and exponential growth, and there are unresolved questions about the choice of constants in the big O notation.

Physicsnuubie
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Hi experts,

I came across a question to determine whether 2^n = O(n!) is true or false. If its true, i need to prove it.

I am aware of the list of growing order functions. So i suppose the statement is true.

However, I can't prove it.
anyone can help me with this? thank you very much!
 
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Can you show that n! > 2^n for all n above some specific value?
This is easy, once you found the starting point (just calculate the first few values and you will see it).
 
Hi!
Thanks for replying.

I couldn't establish the link between factorial and expontential.

2^n < kn! = kn(n-1)(n-2)...1 for n>n0

this would be the definition of big O. but how can i prove it? what values of k and n0 do i have to take?? thank you very much!
 
Physicsnuubie said:
I couldn't establish the link between factorial and expontential.

2^n < kn! = kn(n-1)(n-2)...1 for n>n0

this would be the definition of big O. but how can i prove it? what values of k and n0 do i have to take?? thank you very much!
It looks like you're on the right track.

I don't want to answer your question for you but I'll lay some pretty big hints since being stuck doesn't really help you...

k and n0 don't take on any specific values. You just have to show that for any k, there exists an n0 such that your inequality will be true. As to how to show this, just take a look at your inequality. Maybe examining some concrete values will help you see a pattern:

n = 1 => 2^n = 2 > 1 = n!
n = 2 => 2^n = 2 * 2 > 2 * 1 = n!
n = 3 => 2^n = 2 * 2 * 2 > 3 * 2 * 1 = n!
n = 4 => 2^n = 2 * 2 * 2 * 2 < 4 * 3 * 2 * 1 = n!

As you can see, 2^n > n! until n >= 4. Do you think there exists an n > 4 such that 2^n > n! again? Why or why not?

If you can answer this, all you have to do in incorporate the constant k in your argument and you will have proven your claim...

Good luck!
 
Jocko Homo said:
You just have to show that for any k, there exists an n0 such that your inequality will be true.
Even better: It is sufficient to find some specific k.
k=1 works here, and you can prove it. That was the hint in my post:

mfb said:
Can you show that n! > 2^n for all n above some specific value?
 

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