Question

Fuliptic integrals The period of an undamped pendulum is given by $$ T=4 \sqrt{\frac{\ell}{g}} \int_{0}^{\pi / 2} \frac{d \theta}{\sqrt{1-k^{2} \sin ^{2} \theta}}=4 \sqrt{\frac{\ell}{g}} F(k) $$ where \(\ell\) is the length of the pendulum, \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, \(k=\sin \frac{\theta_{0}}{2},\) and \(\theta_{0}\) is the initial angular displacement of the pendulum (in radians). The integral in this formula \(F(k)\) is called an elliptic integral, and it cannot be evaluated analytically. Approximate \(F(0.1)\) by expanding the integrand in a Taylor (binomial) series and integrating term by term.

Short answer

Answer: The approximate value of the elliptic integral F(0.1) using a Taylor series expansion is F(0.1) ≈ (π/2) - (π/40) + (9π/320).

Step-by-step solution

Write the integrand as a binomial series

To expand the denominator, we will write it in terms of a binomial series. Observe that the denominator has the form \((1-u)^{-\frac{1}{2}}\) where \(u=k^{2} \sin ^{2} \theta\). We will now expand \((1-u)^{-\frac{1}{2}}\) using a binomial series: $$ (1-u)^{-\frac{1}{2}} = \sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} (-u)^{n} $$ Substitute \(u=k^{2}\sin^2\theta\) in the above expression, we have: $$ \frac{1}{\sqrt{1-k^{2} \sin ^{2} \theta}} = \sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} (-k^{2} \sin ^{2} \theta)^{n} $$

Integrate term by term

Now, we will integrate term by term over the interval \([0,\frac{\pi}{2}]\) to approximate \(F(k)\): $$ F(k) = \int_{0}^{\pi / 2} \sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} (-k^{2} \sin ^{2} \theta)^{n} d\theta = \sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} (-k^{2})^{n} \int_{0}^{\pi / 2} (\sin ^{2} \theta)^{n} d\theta $$

Approximate F(0.1)

To approximate \(F(0.1)\), we will calculate the first few terms of the sum and the integral. We can stop at any term after which the value of the series is converging. Let's consider the first 3 terms for approximation: $$ F(0.1) \approx \sum_{n=0}^{2} \binom{-\frac{1}{2}}{n} (-0.1^{2})^{n} \int_{0}^{\pi / 2} (\sin ^{2} \theta)^{n} d\theta $$ Now calculate the binomial coefficients and integrals: $$ \binom{-\frac{1}{2}}{0}=1, \binom{-\frac{1}{2}}{1}=-\frac{1}{2}, \binom{-\frac{1}{2}}{2}=\frac{3}{8} $$ $$ \int_{0}^{\pi / 2} (\sin ^{2} \theta)^{0} d\theta=\int_{0}^{\pi / 2} d\theta=\frac{\pi}{2} $$ $$ \int_{0}^{\pi / 2} (\sin ^{2} \theta)^{1} d\theta=\int_{0}^{\pi / 2} \sin^2\theta d\theta=\frac{\pi}{4} $$ $$ \int_{0}^{\pi / 2} (\sin ^{2} \theta)^{2} d\theta=\int_{0}^{\pi / 2} \sin^4\theta d\theta=\frac{3\pi}{8} $$ Substitute these values in the approximate expression of \(F(0.1)\): $$ F(0.1) \approx 1\cdot\frac{\pi}{2} - \frac{1}{2}\cdot0.1^2\cdot\frac{\pi}{4} + \frac{3}{8}\cdot0.1^4\cdot\frac{3\pi}{8} $$ $$ F(0.1) \approx \frac{\pi}{2} - \frac{\pi}{40} + \frac{9\pi}{320} $$ Therefore, the approximate value of the elliptic integral \(F(0.1)\) is given by: $$ F(0.1) \approx \frac{\pi}{2} -\frac{\pi}{40} + \frac{9\pi}{320} $$

Key concepts

Pendulum Period Formula

The period of a simple pendulum is a classical problem in physics, and it's given by the formula:
\[ T = 4\sqrt{\frac{l}{g}} \int_{0}^{\pi / 2} \frac{d \theta}{\sqrt{1 - k^{2} \sin^{2} \theta}} \]where \(l\) represents the length of the pendulum, \(g\) is the acceleration due to gravity, usually approximated as \(9.8 \text{m/s}^2\), and \(k = \sin{\frac{\theta_0}{2}}\), with \(\theta_0\) being the initial angular displacement. The integral part of this equation, denoted as \(F(k)\), is an elliptic integral, which arises in various scientific and engineering contexts. Since exact analytical solutions for elliptic integrals are not available, they are often evaluated through approximation methods. The significance of the pendulum period formula lies in its depiction of the relationship between the physical parameters of the pendulum and its oscillatory motion, which is foundational in mechanics and timekeeping devices.

Binomial Series Expansion

A binomial series expansion allows mathematicians and scientists to express functions that contain a square root or other exponents in terms of an infinite series of terms. The general form of a binomial series for \(1 - u)^\text{n}\) is given by:
\[ (1-u)^n = \sum_{n=0}^{\infty} \binom{n}{k} (-u)^k \]The expansion is particularly useful when \(n\) is not a positive integer, which makes the function more complicated to deal with. In the case of the pendulum period problem, the binomial series is used to expand the integrand \(\sqrt{1 - k^2 \sin^2 \theta}\) to approximate the value of the elliptic integral. This series is powerful because it allows term-by-term integration, transforming an otherwise intractable integral into a sum of simpler integrals that can be calculated to approximate the function's value.

Term-by-Term Integration

Term-by-term integration is a technique used to integrate series of functions. If a function can be represented as a power series, it can often be integrated by integrating each term of the series individually. This method is valid under certain conditions, such as the series being uniformly convergent on the interval of integration.
In the context of the pendulum period, after expanding the integrand using a binomial series, the term-by-term integration is applied:
\[ F(k) = \sum_{n=0}^{\infty} \binom{-\frac{1}{2}}{n} (-k^{2})^{n} \int_{0}^{\pi / 2}(\sin^{2} \theta)^{n} d\theta \]Each resulting integral involves powers of \(\sin^{2} \theta\) and is easier to evaluate. This approach is immensely helpful for approximating values of complex integrals such as elliptic integrals where direct methods fail. By only taking into account a finite number of terms from the series, an approximate value for the integral can be obtained, facilitating practical calculations in physics and engineering.