Home » Blog » Find the angle between the vectors (2,4,6, …, 2n) and (1,3,5,…,2n-1) as n goes to infinity

Find the angle between the vectors (2,4,6, …, 2n) and (1,3,5,…,2n-1) as n goes to infinity

Consider the vectors A = (2,4,6, \ldots, 2n) and B = (1,3,5, \ldots, 2n-1) in \mathbb{R}^n. Let \theta denote the angle between A and B. Find the value of \theta as n \to \infty.


If \theta is the angle between A and B then we have

    \[ \cos \theta &= \frac{A \cdot B}{\lVert A \rVert \lVert B \rVert}. \]

Computing each of these pieces we have

    \begin{align*}  A \cdot B &= \sum_{k=1}^n (2k)(2k-1) \\[9pt]  &= \sum_{k=1}^n (4k^2 - 2k) \\[9pt]  &= 4 \left( \frac{2n^3 + 3n^2 + n}{6} \right) - 2 \left( \frac{n^2+n}{2} \right)  \\[9pt]  &= \frac{4n^3 + 3n^2 - n}{3}.\\[9pt]  \lVert A \rVert &= \left( \sum_{k=1}^n (2k)^2 \right)^{\frac{1}{2}} \\[9pt]  &= \left( 4 \left( \frac{2n^3+3n^2 + n}{6} \right) \right)^{\frac{1}{2}} \\[9pt]  &= \left( \frac{4n^3 + 6n^2 + 2n}{3} \right)^{\frac{1}{2}}.\\[9pt]  \lVert B \rVert &= \left(\sum_{k=1}^n (2k-1)^2 \right)^{\frac{1}{2}} \\[9pt]  &= \left( \sum_{k=1}^n 4k^2 - \sum_{k=1}^n 4k + \sum_{k=1}^n 1 \right)^{\frac{1}{2}} \\[9pt]  &= \left( \frac{4n^3+3n^2+2n}{3} \right)^{\frac{1}{2}}. \end{align*}

Therefore,

    \[ \cos \theta = \frac{A \cdot B}{\lVert A \rVert \lVert B \rVert} = \left( \frac{4n^3+3n^2-n}{3} \right) \left( \frac{9}{16n^3 + o(n^2)} \right)^{\frac{1}{2}}. \]

Hence, \cos \theta \to 1 as n \to \infty which implies \theta \to 0 as n \to \infty.

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