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\(\forall n \in \mathbf{N}^+, \forall x\in (0,\pi) \left(\displaystyle{\sum_{k=1}^n \frac {1}{k} \sin kx > 0} \right)\)
03-04-2010, 04:41 PM
Post: #1
\(\forall n \in \mathbf{N}^+, \forall x\in (0,\pi) \left(\displaystyle{\sum_{k=1}^n \frac {1}{k} \sin kx > 0} \right)\)
Show that [试证] \(\forall n \in \mathbf{N}^+, \forall x\in (0,\pi) \left(\displaystyle{\sum_{k=1}^n \frac {1}{k} \sin kx > 0} \right)\)
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03-05-2010, 12:25 PM
Post: #2
The proof of the inequality
We prove it by contradiction:
Let \(n \) be the smallest positive integer such that for some minimal point \(x_0 \in (0,\pi)\), \(\displaystyle{\sum_{k=1}^{n} \frac{1}{k}\sin kx_0 \leq 0}\)
Then clearly \(n > 1\) and \(\sin n x_0 < 0\)
But \(\displaystyle{\frac{d}{dx} \sum_{k=1}^{n}\frac{1}{k}\sin kx = \sum_{k=1}^{n} \cos kx = \frac{\cos \frac{(n+1)x}{2} \sin \frac{nx}{2}}{\sin \frac{x}{2}}}\) and \(\displaystyle{\left(\sin \frac{nx}{2} = 0 \right)\Rightarrow (\sin nx = 0)}\),
we see that \(\cos \frac{(n+1)x_0}{2} = 0\) and so for some integer \(m\),
\((n+1)x_0/2 = (m+1/2)\pi \Rightarrow x_0 \in \left\{\displaystyle{\frac{(2m+1)\pi}{n+1}: m = 0,\cdots,\left[\frac{n-1}{2} \right]}\right \}\) since \(x_0 \in (0,\pi)\)
This means that for some \(m\) we have \(x_0 = (2m+1)\pi /(n+1) \) hence \(\sin nx_0 = \sin \displaystyle{\left(2m\pi+\frac{(n-2m)\pi}{n+1} \right) = \sin \frac {(n-2m)\pi}{n+1}} > 0\)
and this contradicts the necessary condition for disproving the inequality.
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