테일러 급수의 활용

Use the approximation $T\approx 2\pi \sqrt{\frac{L}{g}}\left(1+\frac{{k}^{2}}{4}\right)$ to approximate the period of a pendulum having length $10$ meters and maximum angle ${\theta }_{\max }=\frac{\pi }{6}$ where $k=\sin \left(\frac{{\theta }_{\max }}{2}\right).$ Compare this with the small angle estimate $T\approx 2\pi \sqrt{\frac{L}{g}}.$

Suppose that a pendulum is to have a period of $2$ seconds and a maximum angle of ${\theta }_{\max }=\frac{\pi }{6}.$ Use $T\approx 2\pi \sqrt{\frac{L}{g}}\left(1+\frac{{k}^{2}}{4}\right)$ to approximate the desired length of the pendulum. What length is predicted by the small angle estimate $T\approx 2\pi \sqrt{\frac{L}{g}}?$

[T] An equivalent formula for the period of a pendulum with amplitude ${\theta }_{\max }$ is $T({\theta }_{\max })=2\sqrt{2}\sqrt{\frac{L}{g}}{\int }_{0}^{{\theta }_{\max }}\frac{d\theta }{\sqrt{\cos \theta }-\cos ({\theta }_{\max })}$ where $L$ is the pendulum length and $g$ is the gravitational acceleration constant. When ${\theta }_{\max }=\frac{\pi }{3}$ we get $\frac{1}{\sqrt{\cos t-1/2}}\approx \sqrt{2}\left(1+\frac{{t}^{2}}{2}+\frac{{t}^{4}}{3}+\frac{181{t}^{6}}{720}\right).$ Integrate this approximation to estimate $T\left(\frac{\pi }{3}\right)$ in terms of $L$ and $g.$ Assuming $g=9.806$ meters per second squared, find an approximate length $L$ such that $T\left(\frac{\pi }{3}\right)=2$ seconds.