Problem 98 - College Algebra with Modeling and Visualization

Question

The following recursively defined sequence can be used to compute \(\sqrt{k}\) for any positive number \(k .\) \(a_{1}=k ; a_{n}=\frac{1}{2}\left(a_{n-1}+\frac{k}{a_{n-1}}\right)\) This sequence was known to Sumerian mathematicians 4000 years ago, but it is still used today. Use this sequence to approximate the given square root by finding a \(6 .\) Compare your result with the actual value. $$\sqrt{11}$$

Step-by-Step Solution

Answer

Using the sequence, \( a_6 \approx 3.31662479 \), which closely matches \( \sqrt{11} \approx 3.31662479 \).

Step 1: Identify Initial Value

We start the sequence with the initial value given by the problem. Here, it is \( a_1 = k \), where \( k = 11 \). Therefore, \(a_1 = 11\).

Step 2: Apply Recurrence Relation for a_2

The recurrence relation provided is \( a_n = \frac{1}{2} \left( a_{n-1} + \frac{k}{a_{n-1}} \right) \). Substitute for \( n = 2 \) to get:\[ a_2 = \frac{1}{2} \left( a_1 + \frac{11}{a_1} \right) = \frac{1}{2} \left( 11 + \frac{11}{11} \right) = \frac{1}{2} (11 + 1) = 6 \]

Step 3: Compute a_3 Using Recurrence Relation

Substitute \( a_2 \) into the recurrence formula to calculate \( a_3 \):\[ a_3 = \frac{1}{2} \left( a_2 + \frac{11}{a_2} \right) = \frac{1}{2} \left( 6 + \frac{11}{6} \right) = \frac{1}{2} \left( 6 + 1.8333 \right) \approx 3.9167 \]

Step 4: Compute a_4 Using Recurrence Relation

Substitute \( a_3 \) into the recurrence formula to calculate \( a_4 \):\[ a_4 = \frac{1}{2} \left( a_3 + \frac{11}{a_3} \right) \approx \frac{1}{2} \left( 3.9167 + \frac{11}{3.9167} \right) = \frac{1}{2} (3.9167 + 2.808) \approx 3.36235 \]

Step 5: Compute a_5 Using Recurrence Relation

Substitute \( a_4 \) into the recurrence formula to calculate \( a_5 \):\[ a_5 = \frac{1}{2} \left( a_4 + \frac{11}{a_4} \right) \approx \frac{1}{2} \left( 3.36235 + \frac{11}{3.36235} \right) = \frac{1}{2} (3.36235 + 3.271) \approx 3.316675 \]

Step 6: Compute a_6 Using Recurrence Relation

Substitute \( a_5 \) into the recurrence formula to calculate \( a_6 \):\[ a_6 = \frac{1}{2} \left( a_5 + \frac{11}{a_5} \right) \approx \frac{1}{2} \left( 3.316675 + \frac{11}{3.316675} \right) = \frac{1}{2} (3.316675 + 3.317) \approx 3.31662479 \]

Step 7: Compare Result with Actual Square Root

The actual square root, \( \sqrt{11} \), is approximately \( 3.31662479 \). The value of \( a_6 = 3.31662479 \) aligns closely with this value, confirming the result.

Key Concepts

Square Root Approximation

Square root approximation is a method to estimate the value of a square root without a calculator. One common way is by using recursive sequences. Such sequences help find increasingly accurate values by providing a formula to follow repeatedly.
In our case, we're using the sequence starting with an initial guess, say the number itself, and refining that guess gradually. By applying the given formula repeatedly, the sequence rapidly zeroes in on the actual square root.
The neat thing about this method is that even very early approximations can be surprisingly close. This accuracy improves with each step, allowing calculations to reach a high degree of precision. Importantly, this technique doesn't require complex calculations or tools, making it both powerful and accessible for educational purposes.

Recurrence Relation

A recurrence relation is a mathematical way of defining a sequence using preceding terms. By this, a complex pattern can be broken into smaller, repeatable steps.
In the context of square root estimation, the recurrence relation used is:\[ a_n = \frac{1}{2} \left( a_{n-1} + \frac{k}{a_{n-1}} \right) \]
- The first term, \( a_1 \), starts as the number whose square root we seek.- Each subsequent term refines the approximation using the result from the previous step.
This method efficiently narrows down toward the precise value by leveraging both the initial guess and continuous feedback from each calculation step. Recursive approaches, like this, simplify tackling otherwise challenging problems by breaking them into digestible parts.

Sumerian Mathematics

Sumerian mathematics refers to the mathematical methods practiced by the Sumerian civilization over 4000 years ago. Despite their ancient origins, these methods are surprisingly advanced and have laid the groundwork for modern mathematics.
The Sumerians, known for their innovations, were pioneers in developing methods for solving complex mathematical problems, including the approximation of square roots. Their understanding of recursive sequences, as seen in this very problem, showcases their capability in dealing with mathematical complexity.
They not only developed these sequences but also might have been among the first to use recurrence relations. Today, their legacy is appreciated as we continue to use modified versions of their techniques in various computational and educational contexts.

Mathematical Sequences

Mathematical sequences are ordered lists of numbers following a particular rule. They are everywhere in mathematics, from the simplest arithmetic series to complex Fibonacci sequences.
In a mathematical sequence, each term is generated based on a rule or a formula, which could be a simple addition or a more intricate recursion. Sequences form the basis for many areas of mathematics, including calculus, number theory, and computer science.
The sequence used for approximating the square root of a number is an example of a recursive sequence, where each term refines the understanding of the target number. Understanding sequences helps us appreciate the beauty of patterns in numbers and how they can solve real-world problems.