91Ó°ÊÓ

To prove the equivalence, let's first define the function \(f(x) = x^2 - a\). We want to find the value of \(x\) such that \(f(x) = 0\). If \(a > 0\), then the function \(f(x) = x^2 - a\) can be written as \(x^2 = a\). Taking the square root of both sides gives us \(x = \sqrt{a}\). Therefore, finding the positive root of the function \(f(x) = x^2 - a\) is equivalent to finding the square root \(\sqrt{a}\).

Now, we will show that using Newton's method on the function \(f(x) = x^2 - a\) results in the Babylonian method. Recall that Newton's method is defined as: $$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)},$$ where \(f'(x)\) is the derivative of \(f(x)\). In our case, \(f(x) = x^2 - a\) so the derivative \(f'(x) = 2x\). Now we can plug the function and its derivative into Newton's method: $$x_{n+1} = x_n - \frac{x_n^2 - a}{2x_n}$$ By simplification, we obtain the Babylonian method: $$x_{n+1} = \frac{1}{2}(x_n + \frac{a}{x_n}).$$

Using the Babylonian method, we can find approximations for \(\sqrt{13}\) and \(\sqrt{73}\). We need a good initial approximation \(x_0\) to start with. Usually, we can choose an integer close to the square root or simply guess a positive number. For \(\sqrt{13}\), we can choose \(x_0 = 3\), and for \(\sqrt{73}\), we can choose \(x_0 = 8\). These initial approximations will help us converge to the correct values more quickly.

Now, we can use the Babylonian method to iteratively find square roots with at least 10 significant digits. Given the initial approximations, we can continue to calculate with the following recursions: $$x_{n+1} = \frac{1}{2}(x_n + \frac{13}{x_n})$$ for \(\sqrt{13}\), and $$x_{n+1} = \frac{1}{2}(x_n + \frac{73}{x_n})$$ for \(\sqrt{73}\). To obtain at least 10 significant digits, we can keep track of the relative error between each iteration and stop when the relative error is less than \(10^{-10}\). For approximating \(\sqrt{13}\): 1. \(x_0 = 3\) 2. \(x_1 = \frac{1}{2}(3 + \frac{13}{3}) \approx 3.6666666667\) 3. \(x_2 = \frac{1}{2}(3.6666666667 + \frac{13}{3.6666666667}) \approx 3.6055513361\) 4. \(x_3 = \frac{1}{2}(3.6055513361 + \frac{13}{3.6055513361}) \approx 3.6055509726\) Thus, the approximation for \(\sqrt{13}\) is \(3.6055509726\). For approximating \(\sqrt{73}\): 1. \(x_0 = 8\) 2. \(x_1 = \frac{1}{2}(8 + \frac{73}{8}) \approx 6.8125\) 3. \(x_2 = \frac{1}{2}(6.8125 + \frac{73}{6.8125}) \approx 6.5445640134\) 4. \(x_3 = \frac{1}{2}(6.5445640134 + \frac{73}{6.5445640134}) \approx 6.5413819775\) Thus, the approximation for \(\sqrt{73}\) is approximately \(6.5413819775\).