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ngkamsengpeter
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How to prove that [tex]2^n > n^2[/tex] when n>4 ?
If I don't want to use induction but want to prove it mathematically ,how to do it ?0rthodontist said: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.
?ngkamsengpeter said: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 [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 usengkamsengpeter said:If I don't want to use induction but want to prove it mathematically ,how to do it ?
To prove this inequality, we can use mathematical induction. First, we will show that the statement is true for n=5. Then, we will assume it is true for some arbitrary value k and prove that it is also true for k+1. This will show that the statement is true for all values of n greater than 4.
This inequality is important because it helps us understand the growth rate of exponential and polynomial functions. It also has many applications in fields such as computer science, economics, and physics.
Sure, for n=6, 2^6 = 64 and 6^2 = 36. Since 64 is greater than 36, the inequality holds true for n=6.
Yes, this inequality can be represented geometrically by plotting the graphs of 2^n and n^2. The graph of 2^n will always be above the graph of n^2 for n>4, showing that 2^n grows faster than n^2.
No, this inequality holds true for all values of n greater than 4. This can be shown through the mathematical induction proof, which covers all possible values of n.