To Reconnect Natural Logarithm and Unit Circle

The tangent to the logarithm curve $y = \tfrac{\ln x}{b}$ at the point $x=\alpha$ has equation $\begin{aligned} y - \tfrac{\ln \alpha}{b} & = \; \tfrac{1}{b\alpha}(x-\alpha) \\ b\alpha y - x & = \; \alpha(\ln\alpha - 1) \end{aligned}$ and this will be tangent to the unit circle around the origin provided that the distance from it to the origin is $1$ , so that $\frac{|\alpha(\ln \alpha - 1)|}{\sqrt{1 + b^2\alpha^2}} \; = \; 1$ which implies that $\alpha^2(\ln \alpha - 1)^2 \; = \; 1 + b^2\alpha^2$ Regarding $b$ as a function of $\alpha$ , we see that $b'(\alpha) > 0$ for $\alpha > 3$ , with $b(3) = -0.101$ . Moreover $b(\alpha) \to \infty$ as $\alpha \to \infty$ . Thus we can certainly find a unique value $\alpha(b) > 3$ for any $b\ge 1$ . Since $(\ln\alpha(b) - 1)^2 \; = \; b^2 + \alpha(b)^{-2}$ we deduce that $b^2 \; < \; (\ln\alpha(b) - 1)^2 \; < \; b^2 + 1 \; < \; (b+1)^2 \hspace{2cm} b \in \mathbb{N}$ and hence $b < \ln\alpha(b) - 1 < b+1$ for all positive integers $b$ , so that $\lfloor \ln\alpha(b)\rfloor - 1 = b$ for all positive integers $b$ . Thus $\left\lfloor 315\sum_{b=1}^\infty \frac{1}{(\lfloor\ln\alpha(b)\rfloor-1)^2}\right\rfloor \; = \; \left\lfloor 315 \sum_{b=1}^\infty b^{-2}\right\rfloor \; = \; \left\lfloor 315 \times \tfrac16\pi^2\right\rfloor \; = \; \boxed{518}$