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Question Number 216830 by MrGaster last updated on 22/Feb/25

$$\mathrm{Prove}:\mathrm{\forall }x\in \mathbb{R},\mid \cos x\mid +\mid \cos 2x\mid +\dots +\mid \cos nx\mid \ge \frac{n-1}{2}(n\in {\mathbb{Z}}_{>0})$$

Commented by MathematicalUser2357 last updated on 25/Feb/25

$$\mathrm{Or}\mathrm{you}\mathrm{could}\mathrm{do}\{n\mid n\in \mathbb{Z}\wedge n>0\}\left(\mathrm{or}\{n\mid n\in \mathbb{N}\}\right)$$

Answered by MrGaster last updated on 23/Feb/25

$$\mathrm{Prove}:\underset{k=1}{\sum ^n}\mid \cos \left(k\pi \right)\mid \ge \frac{n-1}{2}(n\in {\mathbb{Z}}_{>0})$$$$\mathrm{Let}{S}_n\left(x\right)=\underset{k=1}{\sum ^n}\mid \cos \left(kx\right)\mid$$$$\Rightarrow S_n\left(x\right)=\mid \cos \left(x\right)\mid +\mid \cos \left(2x\right)\mid +\dots +\mid \cos \left(nx\right)\mid$$$$\mathrm{Consider}{T}_n\left(x\right)=\underset{k=1}{\sum ^n}\cos \left(k\pi \right)$$$$=\frac{\sin \left(\frac{nx}{2}\right)\cos \left(\frac{(n+1)x}{2}\right)}{\sin \left(\frac{x}{2}\right)}$$$$\Rightarrow \mid T_n\left(x\right)\mid \le \frac{1}{\mid \sin \left(\frac{x}{2}\right)\mid }$$$$\mathrm{Let}{U}_n\left(x\right)=\underset{k=1}{\sum ^n}{\cos }^2\left(k\pi \right)$$$$=\frac{n}{2}+\frac{1}{2}\underset{k=1}{\sum ^n}\cos \left(2kx\right)$$$$=\frac{n}{2}+\frac{1}{2}\left(\frac{\sin \left(nx\right)\cos \left((n+1)x\right)}{\sin \left(x\right)}\right)$$$$\Rightarrow U_n\left(x\right)\le \frac{n}{2}+\frac{1}{2}$$$$\because \mid \cos \left(kx\right)\mid \ge {\cos }^2\left(kx\right)$$$$\Rightarrow S_n\left(x\right)\ge U_n\left(x\right)$$$$\Rightarrow S_n\left(x\right)\ge \frac{n}{2}-\frac{1}{2}$$$$\therefore S_n\left(x\right)\ge \frac{n-1}{2}$$$$[\mathrm{Q}.\mathrm{E}.\mathrm{D}]$$

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