I'm trying to prove that $2^{n+1} < n^2 + 2$ for $n \ge 0$ by use of mathematical induction, but I get to the inductive step and get lost. I don't know how to link my assumption to the proof.
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3maybe the problem is that your inequality is wrong, the left hand side grows exponential the right hand side polynomial – Dominic Michaelis Oct 02 '15 at 14:25
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Well, that was a final exam question, so I don't know if it is right or wrong – Mahmoud Dawlatly Oct 02 '15 at 14:31
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1well I told you this is wrong, it is even false for 1, and indeed for every larger number – Dominic Michaelis Oct 02 '15 at 14:34
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Ok thanks anyway. Its my fault that I didn't check – Mahmoud Dawlatly Oct 02 '15 at 14:35
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stuff like that happens, but you should be positive about it, this really explains why you got stuck at the inductive step – Dominic Michaelis Oct 02 '15 at 14:37
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The question was solved in comments. I am posting a CW-answer so that it does not remain unanswered.
You can check that the inequality in the question if false even for very small numbers: $$ \begin{array}{|c|c|c|} \hline n & 2^{n+1} & n^2+2 \\\hline 1 & 4 & 3 \\\hline 2 & 8 & 6 \\\hline 3 &16 &11 \\\hline 4 &32 &18 \\\hline \end{array} $$
However, if you turn the inequality sign, you should be able to show that it is true by induction. Have a look at proofs of very similar inequality given in answers here: Proof that $n^2 < 2^n$
Martin Sleziak
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Try writing the inequality the other way: $2^{n+1}>n^2+2$. It is clear this is not true for $n=0$, but try to prove it is true for $n\geq 1$. For $n=1$, we have that $2^{n+1}=2^{2}=4>3=1^2+2=n^2+2$. Now show the inequality holds for when $n=2$: we have $2^{3}=8>6=2^2+2$, and this is true. Now try a proof by induction. I'll provide a brief sketch below.
Claim: For all $n\geq 1, 2^{n+1}>n^2+2$. Let this claim be denoted by $S(n)$.
Proof. We checked the base cases for $n=1,2$. Now, for some fixed $k\geq2$, assume that $$ S(k) : 2^{k+1}>k^2+2 $$ holds. We must now use this assumption (called the inductive hypothesis) to prove that $$ S(k+1) : 2^{k+2}>(k+1)^2+2 $$ follows. Starting with the left-hand side of $S(k+1)$, \begin{align} 2^{k+2} &= 2(2^{k+1})\tag{exponent law}\\[0.5em] &> 2(k^2+2)\tag{by $S(k)$, the ind. hyp.}\\[0.5em] &= 2k^2+4\tag{simplify}\\[0.5em] &> k^2+2k+3\tag{since $k\geq 2$}\\[0.5em] &= (k^2+2k+1)+2\tag{rearrange}\\[0.5em] &= (k+1)^2+2,\tag{factor} \end{align} we end up at the right-hand side of $S(k+1)$, completing the inductive step.
By mathematical induction, the statement $S(n)$ is true for all $n\geq 1$.
Did all of that make sense? If a step was unclear, feel free to leave a comment.
Daniel W. Farlow
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