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I need to prove $\sin(1/n)<1/n$ for all $n \in \Bbb N$ using mathematical induction.

Dont know how to start. Please help!

Stahl
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mopdf
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    Starting is the easy part! Base case: $n=1$ $$\sin \left(1\over 1\right)<1 , \color{green}\checkmark$$ – Git Gud Apr 21 '13 at 20:33
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    Why did you take it so literally? – mopdf Apr 21 '13 at 20:44
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    Because he is a mathematician. – Fixed Point Apr 21 '13 at 21:02
  • What makes you think induction is a good approach to this problem? – Potato Apr 21 '13 at 21:03
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    @mopdf Boys will be boys. – Git Gud Apr 21 '13 at 21:05
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    (You can easily show $\sin(x)<x$ for all positive $x$ using the Taylor series for $\sin$.) – Potato Apr 21 '13 at 21:07
  • @GitGud +1 for the comic and +1 to potato...exactly what I was going to say but you beat me to it ;-) – Fixed Point Apr 21 '13 at 21:08
  • @mopdf Are you allowed to use both induction and calculus? – Git Gud Apr 21 '13 at 21:10
  • Or you could show that $(x-\sin(x))' = 1-\cos(x) \geq 0$ and $>0$ in some neighborhood of $0$... It is not induction of course. – Myself Apr 21 '13 at 21:24
  • I'd like to see how to use both induction and calculus on this problem. If appropiate, a hint would be fine. Thank you all. – mopdf Apr 21 '13 at 21:25
  • @mopdf Well, just use Potato's comment above. You don't need induction. But since you're required to use it, use it! I mean, do the base case, then assume the induction hypothesis and prove what you want. You won't need to use the induction hypothesis, though. – Git Gud Apr 21 '13 at 21:30
  • Ok, I'll do it that way. Thanks Git Gud. – mopdf Apr 21 '13 at 21:41
  • @FixedPoint It may not have been clear, but that comic is part of a book called Logicomix: An Epic Search for Truth. – Git Gud Apr 21 '13 at 21:42
  • @Potato "(You can easily show sin(x)<x for all positive x using the Taylor series for sin.)"... If $x\leqslant\sqrt{20}$, I see how to do it. But for higher values of $x$... – Did May 18 '13 at 20:14
  • @Did Well, considering $|\sin(x)|\le 1$, that doesn't seem to present much of a problem. – Potato May 18 '13 at 20:27
  • @Did (I think I'm missing some extremely dry humor here. If so, my apologies.) – Potato May 18 '13 at 20:28
  • @Potato No humor here, neither dry nor wet, just wondering what you meant by "using the Taylor series". Can you explain the argument showing that sin x < x using the Taylor series? The argument which uses |sin x| < 1 to get rid of higher values of x is different. – Did May 19 '13 at 06:41
  • @Did I'm sorry if I was unclear. Small values of $x$ can be eliminated using the Taylor series as in Dominic's answer below. The bound for larger values follows from $\sin$ being bounded in magnitude by $1$. – Potato May 19 '13 at 07:25
  • @Potato OK. Then two different arguments it is (as hypothesized in my first comment), the one in your comment for small values and another one for large values. Makes sense now. – Did May 19 '13 at 10:42

1 Answers1

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Because of \[ \sin(x)=x-\frac{x^3}{3!}+ \frac{x^5}{5!}-\frac{x^7}{7!} +\dots \] As $x^n$ is monotone decreasing on $[0,1]$ in respect to $n$. Hence \[ \sin(x)\leq x \] for $x$ in $[0,1]$. As $\sin(x)\leq 1$ the inequality is even true for all $x\in [0,\infty)$.

We can see that when $x-\sin(x)> 0$ holds for $x\in (0,1]$ this implies that $\sin\big(\frac{1}{n}\big) < \frac{1}{n}$ holds for all $n \in \mathbb{N}$. At first we note that for $x=0$ $\sin(x)-x=0$. The derivative of \[ x - \sin(x)\] is $1-\cos(x)$. With the fundamental theoroem of calculus we know that \[ x-\sin(x) = (1-\cos(\xi)) \cdot x \] where $\xi \in (0,x)$. As $\cos (x) \leq 1$ and $x>0$ we have that $x-\sin(x)$ is positive and hence our inequality holds.

Dominic Michaelis
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