Need help disproving this

  • Thread starter Melodia
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  • #1
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Homework Statement



Thanks for the help on the last problem. Here is the final problem set I'm stuck on:


SYPzV.jpg


Homework Equations





The Attempt at a Solution



To me it seems that there will always be a positive c so that cg(n) is greater or equal to f(n). No matter how large n is, since there's no limit to how large c can be (can even be a decimal), wouldn't that always be possible?

Thanks.
 

Answers and Replies

  • #2
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First off, g is not a polynomial, as it is not the sum of multiples of integral powers of n. It might be helpful for you to graph the two functions, because you would see that for n larger than about 20, the linear function dominates the other function. Although f(n) = 2n + 3 is a linear function and grows at only a constant rate, that rate is larger than that of the other function, for large enough n.
 
  • #3
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Oooh I see. Since c is fixed and chosen before the n, then there will always be an n that contradicts the statement right?

I'm guessing the same thing applies to this other problem right?

8cBSM.jpg
 
Last edited:
  • #4
vela
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Yes, essentially. For the first problem, there's no fixed value of c that works because you can always make n large enough so that the inequality doesn't hold. In other words, no such c exists.
 

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