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On the supremum and infimum of a sine sequence

Let \{x_n\}_{n \in \mathbb{N}} be a sequence defined as

    \[x_n = \sin 1 + \sin 3 + \sin 5 + \cdots + \sin (2n -1)\]

Find the supremum as well as the infimum of the sequence x_n.

Solution

Background: This problem was on the shortlist of the 2014 Olimpiada Nationala de Matematica de Romania and was suggested by Leo Giugiuc.

We begin by the very well known manipulation.

    \begin{align*} \sum_{k=1}^n\sin (2k-1) &= {\rm Im}\left[\sum_{k=1}^ne^{i(2k-1)}\right]\\ &= {\rm Im}\left[\frac{e^{i}(1-e^{2ni})}{1-e^{2i}}\right]\\ &= {\rm Im}\left[\frac{e^{i}}{1-e^{2i}}(1-e^{2ni})\right]\\ &= {\rm Im}\left[\frac{e^{i}}{1-e^{2i}}(1-\cos 2n+i\sin 2n)\right]\\ &= {\rm Im}\left[\frac{i}{2\sin 1}(1-\cos 2n+i\sin 2n)\right]\\ &=\frac{1-\cos 2n}{2\sin 1} \end{align*}

Thus \displaystyle x_n= \frac{1-\cos 2n}{2\sin 1} and we have to find the supremum and infimum of \cos 2n. Since the values n \mod 2\pi are dense on the unit circle , the same shall hold for 2 n \mod 2\pi implying that \inf \cos 2n =-1 and \sup \cos 2n = 1. Thus,

\displaystyle \inf\{x_n\}=\frac{1}{2\sin 1}(1-1)=0 \quad , \quad \sup\{x_n\}=\frac{1}{2\sin 1}(1+1)=\frac{1}{\sin 1}

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