Prove by induction that $$3^m > (m+1) \cdot \sin m,\quad\forall m \geq 0$$
I need to get to $> (m+2) \sin (m+1)$.
The sine $m$ and $m+1$ oscillate differently, I can't make it smaller.
Asked
Active
Viewed 113 times
1
-
2The thought process here should be something like: this inequality is massively, massively true. If I need to make it easier, I can afford to make some very crude approximations. What approximations can I make on $\sin$ and on $m+1$ to make them into more exponential- and induction-friendly expressions? – Patrick Stevens Sep 13 '21 at 21:39
-
@PatrickStevens In proofs by induction do you always need to use the induction hypothesis? Because that way it looks like you don't need it. – Lambert macuse Sep 14 '21 at 00:19
-
Yes, a proof by induction is characterised by using an induction hypothesis. However, you don't have to use the first induction hypothesis that springs to mind! You can prove something else by induction, and then show that what you proved implies what you were meant to prove. – Patrick Stevens Sep 14 '21 at 17:17
1 Answers
7
Since $-1\le \sin m\le 1$, let instead prove the stronger
$$3^m > m+1 \ge (m+1) \cdot \sin m$$
user
- 154,566
-
$3^k (k+1) \cdot \sin k$, $k>0$ I didn't figure out how to use sine to get to $3^{k+1} > (k+2) \sin (k+1)$ – Lambert macuse Sep 14 '21 at 00:04
-
@Behemooth We can prove by induction that $3^m>m+1$ since $m+1\ge (m+1)\sin m$, in this way we can ignore $\sin m$. – user Sep 14 '21 at 05:07
-
If you desire keep $\sin m$, in the induction hypothesis we have that $3^k>k+1\ge (k+1)\sin k$ and then $3^{k+1}=3\cdot 3^k>3(k+1)>(k+2)>(k+2)\sin(k+1)$ – user Sep 14 '21 at 05:17