Chapter 2: Problem 10
Konstruieren Sie mit dem NEWTON-Verfahren eine rekursive Folge zur näherungsweisen Berechnung von \(\sqrt{5}\).
Short Answer
Expert verified
Use recursive formula: \(x_{n+1} = \frac{x_n + \frac{5}{x_n}}{2}\) starting with \(x_0 = 2.5\).
Step by step solution
01
Define the Function
To use Newton's method to approximate \(\sqrt{5}\), first consider the function \(f(x) = x^2 - 5\). We seek a root of this function, which corresponds to \(\sqrt{5}\) because if \(x = \sqrt{5}\), then \(f(x) = (\sqrt{5})^2 - 5 = 0\).
02
Compute the Derivative
Calculate the derivative of the function. For \(f(x) = x^2 - 5\), the derivative is \(f'(x) = 2x\).
03
Set Up Newton's Iterative Formula
Newton's method uses the formula \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\). Substituting \(f(x)\) and its derivative, we get: \(x_{n+1} = x_n - \frac{x_n^2 - 5}{2x_n}\).
04
Simplify the Formula
Simplify the iterative formula: \(x_{n+1} = x_n - \frac{x_n^2 - 5}{2x_n} = x_n - \frac{x_n^2}{2x_n} + \frac{5}{2x_n} = \frac{2x_n^2 + 5}{2x_n}\).
05
Choose an Initial Guess
Select an initial guess \(x_0\). A reasonable starting point for \(\sqrt{5}\) could be 2.5, since \(2^2 = 4\) and \(3^2 = 9\), and \(\sqrt{5}\) is between these values.
06
Iterate the Formula
Starting from the initial guess, apply the iterative formula to calculate successive approximations: \(x_1 = \frac{2 \times 2.5 + 5}{2 \times 2.5} = 2.25\), then continue with further iterations until the result stabilizes.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Approximate Square Roots
Finding square roots can sometimes be tricky, especially without a calculator by your side. However, Newton's Method offers a useful way to approximate them using a systematic process. In this context, we focus on approximating \(\sqrt{5}\).
To start, we define a function related to the square root: \(f(x) = x^2 - 5\). The goal is to find \(x\) such that \(f(x) = 0\), which implies \(x = \sqrt{5}\). Since \(x^2 - 5\) describes a curve that intersects the x-axis at \(\sqrt{5}\), our job using this method is to zoom into that intersection point with ever-closer guesses.
The benefit of this approach is that it can be refined to be as precise as needed. Initially estimating that \(\sqrt{5}\) is somewhere between 2 and 3, Newton's Method can help pinpoint this number to as many decimal places as required by using iterative refinements.
To start, we define a function related to the square root: \(f(x) = x^2 - 5\). The goal is to find \(x\) such that \(f(x) = 0\), which implies \(x = \sqrt{5}\). Since \(x^2 - 5\) describes a curve that intersects the x-axis at \(\sqrt{5}\), our job using this method is to zoom into that intersection point with ever-closer guesses.
The benefit of this approach is that it can be refined to be as precise as needed. Initially estimating that \(\sqrt{5}\) is somewhere between 2 and 3, Newton's Method can help pinpoint this number to as many decimal places as required by using iterative refinements.
Numerical Methods
Numerical methods are tools used to solve mathematical problems that don't have easy symbolic answers.
Newton's Method, specifically, is a numerical method that provides a way to iteratively approximate the roots of a real-valued function, such as finding square roots.
Here’s how Newton’s Method shines:
By setting \(f(x) = x^2 - 5\) and calculating its derivative \(f'(x) = 2x\), we informed the iterative formula: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\), which acts as a stepping stone to a better approximation of \(x\).
Newton's Method, specifically, is a numerical method that provides a way to iteratively approximate the roots of a real-valued function, such as finding square roots.
Here’s how Newton’s Method shines:
- It transforms complex algebraic problems into a series of simpler arithmetic operations.
- The method leverages calculus, specifically derivatives, to improve an initial guess.
- It can solve problems with high precision, especially important where exact solutions are too cumbersome to acquire.
By setting \(f(x) = x^2 - 5\) and calculating its derivative \(f'(x) = 2x\), we informed the iterative formula: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\), which acts as a stepping stone to a better approximation of \(x\).
Iterative Process
The heart of Newton's Method lies in its iterative nature. It takes a guess and improves it repeatedly to zero in on the function's root.
This pattern follows a cycle, where each approximation builds upon the previous one. Starting with an initial guess, say \(x_0 = 2.5\), we apply the iterative formula:
\[ x_{n+1} = \frac{2x_n^2 + 5}{2x_n} \]
Subsequently, this formula is used repeatedly to produce a new guess, which is more accurate with each cycle until results stabilize.
This iteration is
This pattern follows a cycle, where each approximation builds upon the previous one. Starting with an initial guess, say \(x_0 = 2.5\), we apply the iterative formula:
\[ x_{n+1} = \frac{2x_n^2 + 5}{2x_n} \]
Subsequently, this formula is used repeatedly to produce a new guess, which is more accurate with each cycle until results stabilize.
This iteration is
- Efficient: Each step converges rapidly toward the actual root, often in just a few iterations.
- Adaptive: Modifying the initial guess directly impacts the speed and accuracy of convergence.
- Precise: The process can meet high precision by simply continuing the iterations until the desired accuracy is achieved.