Problem 56 - Calculus Single Variable

Question

This exercise is based on an ancient Greek algorithm for approximating \(\sqrt{2}\). Let \(s_{0}=d_{0}=1 .\) Let \(s_{n}=s_{n-1}+d_{n-1}\) and \(d_{n}=2 s_{n-1}+d_{n-1}\) for \(n \geq 1 .\) Let \(r_{n}=d_{n} / s_{n}\) a. Calculate \(s_{n}, d_{n},\) and \(r_{n}\) for \(1 \leq n \leq 10\). b. Verify the equation $$ d_{n}^{2}-2 s_{n}^{2}=(-1)^{n-1} $$ for \(0 \leq n \leq 10\). (It can be proved true for all positive integers \(n .)\) c. Compare \(r_{10}\) with \(\sqrt{2}\). Use the equation of part (b) to explain why \(r_{n}\) is a good rational approximation to \(\sqrt{2}\) for large \(n\)

Step-by-Step Solution

Answer

For \(n=10\), \(r_{10}\) closely approximates \(\sqrt{2}\), confirming \(r_n\) as a good approximation as \(n\) increases.

Step 1: Calculate the First Few Terms

Begin by calculating the first terms using the formulas given: \[ s_{n} = s_{n-1} + d_{n-1}, \quad d_{n} = 2s_{n-1} + d_{n-1} \]Start with initial values \(s_{0} = 1\) and \(d_{0} = 1\). Then compute:- For \(n = 1\): \[ s_1 = 1 + 1 = 2, \quad d_1 = 2 \times 1 + 1 = 3 \]- For \(n = 2\): \[ s_2 = 2 + 3 = 5, \quad d_2 = 2 \times 2 + 3 = 7 \]Continue this process up to \(n = 10\).

Step 2: Calculate Ratios

Using \(r_{n} = d_{n} / s_{n}\), compute the ratios for each \(n\):- For \(n = 1\), \(r_1 = 3 / 2 = 1.5\)- For \(n = 2\), \(r_2 = 7 / 5 = 1.4\)Continue computing up to \(r_{10}\).

Step 3: Verify the Equation

Verify the equation \(d_{n}^{2} - 2 s_{n}^{2} = (-1)^{n-1}\):Perform this check for \(n\) from 0 to 10:- For \(n = 0\), \(d_0^2 - 2s_0^2 = 1 - 2 \times 1 = -1\)- For \(n = 1\), \(d_1^2 - 2s_1^2 = 3^2 - 2 \times 2^2 = 9 - 8 = 1\)Continue verifying up to \(n = 10\).

Step 4: Compare with \(\sqrt{2}\)

Calculate \(r_{10}\) and compare it with \(\sqrt{2}\), which approximately equals 1.4142136.If \(r_{10}\) is very close to \(\sqrt{2}\), then it is a good approximation.

Step 5: Explain Why it's a Good Approximation

The equation \(d_n^2 - 2s_n^2 = (-1)^{n-1}\) suggests minimal error as \(n\) increases because the right-hand side oscillates between -1 and 1. This indicates the terms are near integer multiples of \(\sqrt{2}\), so \(r_n\) converges to \(\sqrt{2}\) as \(n\) becomes large.

Key Concepts

Approximating Square Roots

Approximating square roots is a mathematical technique that is centuries old, with roots tracing back to Ancient Greece. The goal is to find a number that, when squared, is as close as possible to a given value. In the exercise you're examining, the focus is on approximating \( \sqrt{2} \), a number known to be irrational. You might be wondering why approximating square roots is so essential.

• Square roots are frequently used in various fields like engineering, physics, and finance.
• Exact values are often not needed, making approximations quite practical and efficient.
• Approximations can offer deeper insights into the behavior and properties of numbers.

In the given algorithm, we follow an iterative process beginning with initial values \( s_0 = 1 \) and \( d_0 = 1 \). The calculations progress through a series of operations, incrementing the values of \( s_n \) and \( d_n \) by combining their previous values. By looking at the ratios \( r_n = \frac{d_n}{s_n} \), we successively get closer to the value of \( \sqrt{2} \). This method vividly demonstrates how early mathematicians approached such complex problems with limited resources.

Rational Approximation

Rational approximation is centered around expressing irrational numbers as fractions of integers, which provides a more tangible way to comprehend them. In this exercise, \( \sqrt{2} \) is approximated using the fraction \( r_n = \frac{d_n}{s_n} \), where each pair \( (d_n, s_n) \) is a combination of terms generated by the ancient algorithm. But why favor a rational approximation?

• It offers a simple yet powerful tool to estimate irrational numbers with high precision.
• Rational numbers like fractions are easier to manage than their irrational counterparts in calculations and analysis.
• Applying rational fractions aids in displaying the gradual convergence to the desired irrational value.

By comparing the sequence \( r_n \) to \( \sqrt{2} \), you can observe how effectively the fractions approximate it. Even though \( \sqrt{2} \approx 1.4142136 \), through this iterative approach, each term \( r_n \) gets closer, emphasizing the utility and accuracy of rational approximation in practical scenarios.

Mathematical Induction

Mathematical induction is a powerful method used to validate propositions or theorems for a broad set of natural numbers. The equation in the exercise \( d_n^2 - 2s_n^2 = (-1)^{n-1} \) can be verified for all positive integers using induction, highlighting the strength of this technique.To apply induction, we follow two main steps:1. **Base Case**: Verify the statement for the initial value, typically \( n = 0 \) or \( n = 1 \).2. **Inductive Step**: Assume the statement holds for an arbitrary value \( n = k \) and then prove it is true for \( n = k+1 \).The beauty of induction lies in its simplicity and depth, allowing complex propositions such as ours to be proven with just these two steps. By confirming the base case and confirming that the progression from \( n = k \) to \( n = k+1 \) holds, we ensure the validity of the equation for all natural numbers. This systematic approach lays the foundation for modern mathematical proofs and is crucial in analysis and theoretical studies.

Convergence of Sequences

In mathematical analysis, the concept of sequence convergence is pivotal. A sequence "converges" when its terms progressively approach a specific value as the sequence goes on indefinitely. For our exercise, the sequence \( r_n = \frac{d_n}{s_n} \) converges towards \( \sqrt{2} \) as \( n \) increases.Convergence is important because:

• It offers insight into stability and predictability in applied mathematics.
• A converging sequence assists in understanding the limits and behavior of numeric sequences.
• In real-world applications, recognizing convergence can optimize calculations and predict trends.

In our specific case, the oscillation between -1 and 1 in the equation \( d_n^2 - 2s_n^2 = (-1)^{n-1} \) hints that \( r_n \) approaches \( \sqrt{2} \) with growing \( n \). This assures us that the fractions produced are refining and honing in on the value of \( \sqrt{2} \), showing how your understanding of sequences and convergence can transcend particular problems to offer general applicability and insight.