How to prove that [tex]2^n > n^2[/tex] when n>4 ?
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.
An exponential function grows faster than a polynomial.
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.
If I don't want to use induction but want to prove it mathematically ,how to do it ?
What do you mean by mathematically? What's wrong with using induction, by the way? As a matter of fact, it's completely valid!
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.
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