Answer
$T \approx 2.09894$ seconds.
Work Step by Step
Recall the elliptic integral $E\left( k \right)$ in Example 12:
$E\left( k \right) = \dfrac{\pi }{2} + \dfrac{\pi }{2}\mathop \sum \limits_{n = 1}^\infty {\left( {\dfrac{{1\cdot 3\cdot\cdot\cdot{{\left( {2n - 1} \right)}^2}}}{{2\cdot 4\cdot\cdot\cdot\left( {2n} \right)}}} \right)^2}{k^{2n}}$
1. Evaluate the first five terms of the Maclaurin series for the elliptic integral $E\left( k \right)$:
$E\left( k \right) \approx \dfrac{\pi }{2} + \dfrac{\pi }{2}\left[ {{{\left( {\dfrac{1}{2}} \right)}^2}{k^2} + \left( {\dfrac{{1\cdot 3}}{{2\cdot 4}}} \right){k^4} + \left( {\dfrac{{1\cdot 3\cdot 5}}{{2\cdot 4\cdot 6}}} \right){k^6} + \left( {\dfrac{{1\cdot 3\cdot 5\cdot 7}}{{2\cdot 4\cdot 6\cdot 8}}} \right){k^8}} \right]$
For $\theta = \dfrac{\pi }{4}$, we have $k = \sin \dfrac{1}{2}\theta = \sin \dfrac{\pi }{8} \approx 0.382683$. So,
$E\left( {0.382683} \right) \approx \dfrac{\pi }{2} + \dfrac{\pi }{2}\left[ {{{\left( {\dfrac{1}{2}} \right)}^2}{{\left( {0.382683} \right)}^2} + \left( {\dfrac{{1\cdot 3}}{{2\cdot 4}}} \right){{\left( {0.382683} \right)}^4} + \left( {\dfrac{{1\cdot 3\cdot 5}}{{2\cdot 4\cdot 6}}} \right){{\left( {0.382683} \right)}^6} + \left( {\dfrac{{1\cdot 3\cdot 5\cdot 7}}{{2\cdot 4\cdot 6\cdot 8}}} \right){{\left( {0.382683} \right)}^8}} \right]$
$E\left( {0.382683} \right) \approx 1.64268$
2. Estimate the period $T$ of a $1$-meter pendulum using the first five terms of the Maclaurin series
We have $T = 4\sqrt {\dfrac{L}{g}} E\left( k \right)$.
For $L=1$, $g=9.8$ and $E\left( {0.382683} \right) \approx 1.64268$, we obtain
$T = 4\sqrt {\dfrac{1}{{9.8}}} E\left( {0.382683} \right) \approx 2.09894$
So, $T \approx 2.09894$ seconds.
