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Help with comparsion

  1. Oct 24, 2008 #1
    1. The problem statement, all variables and given/known data

    k is a constant where k1 < k2

    Is the relation between the two <. > or = to:

    A = nk1 + k2n

    B = nk2 + k1n

    3. The attempt at a solution

    I did this problem said that A > B, and I got it wrong. I am having a hard time telling if nki or kin is greater.
     
  2. jcsd
  3. Oct 24, 2008 #2
    Do you mean *n* is a constant? I don't see a k. Also, you have given us insufficient information to solve the problem because depending on k1 and k2 relative to n the answer will change. For example, n=10, k1=1, k2=2 and n=2, k1=10, k2=11.
     
    Last edited: Oct 24, 2008
  4. Oct 24, 2008 #3

    Dick

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    If n=1 then A>B. If n=2, k1=1 and k2=2 then A>B. If n=2, k1=3 and k2=4, then A<B. Do you know something about n you aren't telling us? What makes you think there is a definite relation between the two?
     
  5. Oct 26, 2008 #4
    Sorry about that, n can be any number, does not have to be fixed like the k's. So, for example nk1 < nk2.
     
  6. Oct 26, 2008 #5

    Dick

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    That doesn't help. You've already been given examples where A<B and B>A.
     
  7. Oct 26, 2008 #6
    I think that k's are fixed value for a give problem, and that as n-> infinite that one of those will be >, <, or =. I think that is what gear2d is asking.
     
  8. Oct 26, 2008 #7

    Dick

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    You might be right. If that's the real question, I wish Gear2d would clarify.
     
  9. Oct 27, 2008 #8
    Thanks Ad2d you said what I wanted to say. So how would one approach this problem?
     
  10. Oct 27, 2008 #9

    Dick

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    Figure out which term is dominant. Take n^k1 and k2^n. The logs are k1*log(n) and n*log(k2). Which is larger as n gets large? I.e. what is lim n->infinity (k1*log(n))/(n*log(k2))? Is it zero or infinity? Think l'Hopital.
     
  11. Oct 27, 2008 #10
    Thank you, that makes sense. Don't know why I did not see that. Thanks again.
     
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