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April 7th, 2019, 03:15 AM   #1
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Prove inequality

Prove inequality : $\displaystyle \underbrace{\sin \sin ... \sin }_{N-times} (N)\leq \sin \frac{1}{N}\; $ , $\displaystyle N\in \mathbb{N}$.

To write it better in short-terms : $\displaystyle \sin_{N} (N)\leq\sin \frac{1}{N}$.
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April 7th, 2019, 01:36 PM   #2
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if this is a product chain it should be written as

$\sin^N(N) \leq \sin\left(\dfrac 1 N \right),~N \in \mathbb{N}$

if it's a function composition chain I've seen it written as

$\underset{\text{N times}}{\underbrace{\sin\circ \sin \circ \dots \sin(N)} }=\sin^{\circ N}(N) \leq \sin\left(\dfrac 1 N \right)$

which is it?
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April 7th, 2019, 01:38 PM   #3
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The composition N-times of sin function.
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April 7th, 2019, 02:00 PM   #4
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Wolfram Alpha gives $\sin(\sin(2)) \approx 0.78907 \dots > \frac{1}{2}$. Of course, that's in radians. However, it's $< \frac{1}{2}$ for $2$ degrees. Just wanted to note that. OP must mean degrees.
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Last edited by skipjack; April 7th, 2019 at 03:21 PM.
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April 7th, 2019, 02:01 PM   #5
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For $x \gt 0$ $\sin(x)\lt x$. Therefore $\sin(\sin(1/N))\lt \sin(1/N)$..Repeat N times to get what you want.
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Last edited by skipjack; April 7th, 2019 at 03:14 PM.
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April 7th, 2019, 02:18 PM   #6
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Tried all methods posted above but got no result .
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April 8th, 2019, 01:18 PM   #7
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Quote:
Originally Posted by idontknow View Post
Tried all methods posted above but got no result .
Did you understand my reply?
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April 8th, 2019, 01:19 PM   #8
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Quote:
Originally Posted by mathman View Post
Did you understand my reply?
No.
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April 9th, 2019, 12:11 PM   #9
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Quote:
Originally Posted by idontknow View Post
No.
Sorry - I misread your question.
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April 9th, 2019, 06:06 PM   #10
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Math Focus: Dynamical systems, analytic function theory, numerics
Use induction and the fact that $\sin$ is entire and its Taylor expansion can be written as
\[ \sin(N) = \sin(N-1) + \frac{1}{2} \cos(N-1) - \frac{1}{6}\sin(N-1) + \dotsc \].
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