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How to prove this ?

  1. Jul 27, 2006 #1
    How to prove that [tex]2^n > n^2[/tex] when n>4 ?
  2. jcsd
  3. Jul 27, 2006 #2


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    You could use the fact that both functions are continuous and never cross after that point. If you are only interested in integer n then you could use induction.
  4. Jul 27, 2006 #3
    *Waves hand*
    An exponential function grows faster than a polynomial.
  5. Jul 27, 2006 #4


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    That is not sufficient--it does not rule out the possibility of some point before infinity but after 4 where the inequality does not hold.

    For integer n, induction is the way to go. Otherwise, you can show directly that the second derivative of 2^n is greater than that of n^2 for n > 4, and the first derivative of 2^n is greater than that of n^2 for n = 4. From this it is possible to infer the conclusion by integration.
  6. Jul 28, 2006 #5
    If I don't want to use induction but want to prove it mathematically ,how to do it ?
  7. Jul 28, 2006 #6


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    What do you mean by mathematically? What's wrong with using induction, by the way? As a matter of fact, it's completely valid!
  8. Jul 28, 2006 #7


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    Induction would be simpler for this case if you want to prove it for integers. The other way I mentioned is using the function [tex]f(x) = 2^x - x^2[/tex]. You should be able to show that [tex]f'(4) > 0[/tex] and also that [tex]f''(x) > 0[/tex] for any x larger than 4. And you know that f(4) = 0. Then use
    [tex]\int_a^t g'(x) dx + g(a) = g(t)[/tex]
    to show that [tex]f'(x) > 0[/tex] for x > 4 and then that [tex]f(x) > 0[/tex] for x > 4.
    Last edited: Jul 28, 2006
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