How to Prove 2^n > n^2 for n>4 without using induction?

[ngkamsengpeter]

How to prove that $$2^n > n^2$$ when n>4 ?

 

[0rthodontist]

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.

 

[Corneo]

*Waves hand*
An exponential function grows faster than a polynomial.

 

[0rthodontist]

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.

 

[ngkamsengpeter]

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.

If I don't want to use induction but want to prove it mathematically ,how to do it ?

 

[VietDao29]

If I don't want to use induction but want to prove it mathematically ,how to do it ?

?
Huh?
What do you mean by mathematically? What's wrong with using induction, by the way? As a matter of fact, it's completely valid!

 

[0rthodontist]

If I don't want to use induction but want to prove it mathematically ,how to do it ?

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