The question is to prove - $2^n>n^2$ (n is natural number >4) How do I prove it?
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2possible duplicate of Prove that $ n < 2^{n}$ for all natural numbers $n$. – Najib Idrissi Sep 12 '15 at 13:44
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1Why go to the trouble of posting a self-answered question without even checking if there's a duplicate first? – Najib Idrissi Sep 12 '15 at 13:45
3 Answers
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Prove it true for 5 and assume it true for some k.
So, $2^k>k^2$
Not prove it true for k+1.
We observe,$[2^{k+1}=2^k\cdot2]>[2\cdot k^2=k^2+k^2]$ (as assumed above)
Also, $2^{k+1}>k^2+3k$ (from the previous statement above as $k>4$ and so $k^2>3k$)
So, $2^{k+1}>k^2+2k+k$
So, $2^{k+1}>k^2+2k+1$ (as $k>1$)
So, $2^{k+1}>(k+1)^2$
Hence,proved
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3it would be good to mention this as the "proof by (mathematical) induction technique" – Nikos M. Sep 12 '15 at 12:39
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Alternatively, show by differentiation that the gradient of $f(x) = 2^x - x^2$ is always positive for $n \ge 4$. Hence conclude using the implication that $f(x)$ is strictly increasing (for $n \ge 4$), that
$$\begin{align}n > 4 &\implies f(n) > f(4) = 0\\&\implies 2^n - n^2 > 0\\&\implies 2^n > n^2\end{align}$$
Yiyuan Lee
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Of course induction is the best choice here, but if you want to use calculus, then you can discover a deeper result:
After squaring both sides, it's equivalent to proving $4^n>n^4$ for $n>4$ or equivalently $\sqrt[4]{4}>\sqrt[n]{n}$ for $n>4$. This is the famous result that the function $f(x)=x^{\frac{1}{x}}$ attains its minimum at $x=e$ and is strictly increasing for $x>e$. This is equivalent to $a^b>b^a$ for all $a,b$ (not necessarily integers!) with $a>b>e$
Tintarn
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