Question

A well-known method for approximating \(\sqrt{c}\) for positive real numbers \(c\) consists of the following recurrence relation (based on Newton's method). Let \(a_{0}=c\) and $$a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{c}{a_{n}}\right), \quad \text { for } n=0,1,2,3, \dots$$ a. Use this recurrence relation to approximate \(\sqrt{10} .\) How many terms of the sequence are needed to approximate \(\sqrt{10}\) with an error less than \(0.01 ?\) How many terms of the sequence are needed to approximate \(\sqrt{10}\) with an error less than \(0.0001 ?\) (To compute the error, assume a calculator gives the exact value.) b. Use this recurrence relation to approximate \(\sqrt{c},\) for \(c=2\) \(3, \ldots, 10 .\) Make a table showing how many terms of the sequence are needed to approximate \(\sqrt{c}\) with an error less than \(0.01.\)

Short answer

Answer: 3 terms.

Step-by-step solution

Set up the initial values

As the problem states, for the given number \(c\), we start with \(a_{0}=c\). For example, for part (a), we will start with \(a_0=10\). ##Step 2: Apply the recurrence relation##

Apply the recurrence relation

To find the next terms in the sequence, we will use the mentioned recurrence relation: $$a_{n+1}=\frac{1}{2}\left(a_{n}+\frac{c}{a_{n}}\right)$$ Repeating this process, we will generate terms in the sequence which will approximate \(\sqrt{c}\). ##Step 3: Check the error##

Check the error

To check the error, we will use the following formula: $$\text{Error} = \left|a_n - \sqrt{c}\right|$$Keep calculating \(a_n\) until we reach the desired error as required by the question. The question demands us to find out for two different error levels, \(0.01\) and \(0.0001\). ##Step 4: Make a table for multiple values of c##

Make a table for multiple values of c

For \(c=2, 3, \ldots , 10\), we will use the same four steps as above and find the number of terms needed to approximate \(\sqrt{c}\) with an error less than \(0.01\). #Here is the implementation for the given exercise: a. To find how many terms of the sequence are needed to approximate \(\sqrt{10}\) with an error less than \(0.01\) and \(0.0001\), perform the following steps: Step 1: \(a_0 = 10\) Step 2: Apply the recurrence relation for next terms: \(a_1 = \frac{1}{2}(a_0 + \frac{10}{a_0}) = \frac{1}{2}(10 + \frac{10}{10}) = 5.5\) \(a_2 = \frac{1}{2}(a_1 + \frac{10}{a_1}) = \frac{1}{2}(5.5 + \frac{10}{5.5}) \approx 3.2782\) \(a_3 = \frac{1}{2}(a_2 + \frac{10}{a_2}) = \frac{1}{2}(3.2782 + \frac{10}{3.2782}) \approx 3.1623\) Step 3: The errors between true and approximation are: \(|a_1 - \sqrt{10}| = |5.5 - \sqrt{10}| \approx 1.3862 > 0.01\) \(|a_2 - \sqrt{10}| = |3.2782 - \sqrt{10}| \approx 0.1158 > 0.01\) \(|a_3 - \sqrt{10}| = |3.1623 - \sqrt{10}| \approx 0.0032 < 0.01\) Two terms are needed to achieve an error less than \(0.01\). To approximate \(\sqrt{10}\) with an error less than \(0.0001\), keep calculating: \(a_4 = \frac{1}{2}(a_3 + \frac{10}{a_3}) = \frac{1}{2}(3.1623 + \frac{10}{3.1623})\approx 3.1623 \) \(|a_4 - \sqrt{10}| = |3.1623 - \sqrt{10}| \approx 0.000015 < 0.0001\) Three terms are needed to achieve an error less than \(0.0001\). b. Calculate \(a_n\) for \(c=2, 3, 4, 5, 6, 7, 8, 9\) and find the number of terms needed to approximate \(\sqrt{c}\) with an error less than \(0.01\) by following the same procedure. We obtain the following table: | \(c\) | \(\sqrt{c}\) | Number of Terms Required for Error \(<0.01\) | |-----|------------|-------------------------------------------| | 2 | 1.4142 | 2 | | 3 | 1.7321 | 2 | | 4 | 2.0000 | 1 | | 5 | 2.2361 | 2 | | 6 | 2.4495 | 2 | | 7 | 2.6458 | 2 | | 8 | 2.8284 | 2 | | 9 | 3.0000 | 2 | For each value of c, the table indicates the number of terms needed to approximate \(\sqrt{c}\) with an error less than \(0.01\).

Key concepts

Newton's Method

Newton's Method is a powerful technique used to approximate solutions to equations, especially useful for finding square roots. It's based on the idea of linear approximation, providing an iterative formula to solve problems more precisely over time.
Newton's Method involves the following steps:

• Start with an initial guess, denoted as \(a_0\).
• Use the recurrence relation \(a_{n+1} = \frac{1}{2}(a_n + \frac{c}{a_n})\) to find subsequent terms closer to the actual root.

Each iteration brings us closer to the true value by refining our approximation. The beauty of this method lies in its simplicity and efficiency, especially as it converges quite rapidly for well-behaved functions.
It's important to choose a good initial guess, as it affects the speed of convergence. For square roots, starting with \(a_0 = c\) is common practice. With fewer iterations, Newton's Method ensures we zero in on the root with remarkable precision.

Approximating Square Roots

Approximating square roots is a common problem where Newton's Method comes highly beneficial. Square roots can be tricky to calculate directly without a calculator, hence the need for approximation methods.
By using the recurrence relation, \(a_{n+1} = \frac{1}{2}(a_n + \frac{c}{a_n})\), each term \(a_n\) becomes increasingly closer to \(\sqrt{c}\). This approach allows easy computation

• Begin by setting \(a_0 = c\), assuming you want to find \(\sqrt{c}\).
• Apply the recurrence relation iteratively.

For instance, in the case of calculating \(\sqrt{10}\), you start with \(a_0 = 10\) and apply the formula repeatedly to bind the approximation closely to the actual square root. This iterative process is fast enough for most practical purposes and provides satisfactory precision, depending on how tightly you need the approximation to be.

Error Analysis

Error Analysis in the context of approximating square roots is crucial to understand how close our approximations really are to the true values. It involves estimating the difference between the calculated approximation and the actual square root.
The error at each step can be calculated using:
\[ \text{Error} = |a_n - \sqrt{c}| \]

• Calculate \(a_n\) using the recurrence relation until your error is below the desired threshold.
• Two common thresholds are \(0.01\) and \(0.0001\).

These thresholds denote acceptable errors, wherein an approximation is considered good enough for practical use. Error analysis ensures that our approximations aren't just numerically close, but effectively useful for decision-making in practical scenarios.
By repeatedly calculating and checking the error, we can decide how many steps or iterations we require to meet our precision needs.