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Show that $2^n > n^2$ through induction and so far I got to the $k+1$ step, but I am stuck.

I have $2^{k+1} = 2 +2^k$, but I don`t know how the book turned it into $k^2 +k^2$. The book then follows to turn it into $k^2+4n$ and then $k^2+2k+1$. Was the $K$th step used because I cant understand how it was used in case.

Caleb Stanford
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Belphegor
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2 Answers2

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Inductive hypothesis: For $n = k$, $$\color{blue}{2^k \geq k^2},\quad k \geq 4.$$

$$2^{k+1} = 2\cdot \color{blue}{2^k} \geq 2(\color{blue}{k^2}) \geq k^2 + k^2 \geq k^2 + 2k + 1 \overset{\large k>2} = (k+1)^2$$

amWhy
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  • is 2^k supposed to equal n^2 in this case allowing us to.replace – Belphegor Jul 23 '14 at 18:01
  • Since we have assumed $2^k \geq k^2$ (inductive hypotheis), it follows that $2\cdot 2^k \geq 2\cdot k^2$. We simply have the inductive hypothesis (inequality), multiplied by $2$ on each side of the inequality. – amWhy Jul 23 '14 at 18:08
  • how.did k^2 turn into 2k +1 – Belphegor Jul 23 '14 at 18:09
  • $k^2 \gt 2k + 1$ for all $k \in \mathbb N$, $k\gt 2$. Since we are proving what holds for $k \geq 4$, it follows that we're safe in asserting that $k^2>2k + 1$. – amWhy Jul 23 '14 at 18:25
  • what i mean is, how did you figure out it was 2k +1. How did.you get it. – Belphegor Jul 23 '14 at 18:32
  • Remember our objective is to show $2^{k+1}\geq (n+1)^2$. And $(n+1)^2 = n^2 + 2n + 1$. So by showing that $k^2 + k^2 \geq k^2 + 2k + 1$, we proven our objective: $2^{k+1} \geq 2k^2 \geq (n+1)^2$. – amWhy Jul 23 '14 at 18:39
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Induction problems like this can be done mechanically by telescopy. Rewriting $\,f(n) = 2^n-n^2\,$ as a telescopic sum of its differences makes its positivity obvious, because each summand is $\color{#c00}{\ge 0}.$

$$\displaystyle\begin{eqnarray}n\ge 5\ \ \Rightarrow\ \ \ f(n) &=&\! f(5)\,+ \sum_{\large k\,=\,5}^{\large n-1}\ (f(k\!+\!1)-f(k))\\ \displaystyle\Rightarrow\ \ \ 2^n-n^2 &=& \ 7\ +\ \sum_{\large k\,=\,5}^{\large n-1}\, (\overbrace{2^{k}- (2k\!+\!1)}^{\large \color{#c00}{\ge 0}})\, >\, 0 \end{eqnarray}\qquad$$

That the overbraced term is $\color{#c00}{\ge 0}\,$ is obvious by induction (telescopic or not). Note that it fails for $\,n < 5,\,$ e.g. $\,f(4) = 0,\ f(3) = -1.\ $

For further discussion see my many posts on telescopic induction.

Community
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Bill Dubuque
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  • It'd be more complete if the overbraced term is shown to be greater or equal to zero by telescoping induction. This will highlight the fact that this method is powerful because it reduces a harder induction problem to smaller, simpler ones. – Maxis Jaisi Jul 27 '16 at 05:59