Metallic Means
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Metallic Means
This note is all about proving the formula for the $n$th metallic mean which come up in chaos theory and using Xenaʻs stones for power. By definition, we have the $n$th metallic mean as meeting the following metallic criterion…
\[\begin{equation} x = n + \frac{1}{x} \end{equation}\]we further require the metallic mean $x$ to be positive and non-zero. So, this means $x > 0$ and $x \neq 0$, always. Our goal is to solve for $x$ only. We now multiply by $x$ on both sides for $(1)$ since it is non-zero…
\[\begin{equation} x^2 = nx + 1 \\ \implies x^2 - nx - 1 = 0 \end{equation}\]From here, we use the quadratic formula and compute the possible values for $x$ as…
\[\begin{equation} x = \frac{n \pm \sqrt{n^2 - 4(1)(-1)}}{2(1)} \\ = \frac{n \pm \sqrt{n^2 + 4}}{2} \end{equation}\]we further know that $n^2 + 4 > n^2$ for all $n \in \mathbb{N}$ and since $\sqrt{y}$ is a strictly increasing function, we always have $n < \sqrt{n^2 + 4}$ which means we need to only take the positive root for $\sqrt{n^2 + 4}$. So, the only value for $x$ is…
\[\begin{equation} x_n = \frac{n + \sqrt{n^2 + 4}}{2} \end{equation}\]this form $x_n$ is the $n$th metallic mean, as desired. $\blacksquare$