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Magazine Chapter 4: Problem 10
Use a calculator or program to compute the first 10 iterations of Newton's method when they are applied to the following functions with the given initial approximation. Make a table similar to that in Example 1 $$f(x)=x^{3}+x^{2}+1 ; x_{0}=-2$$
Short Answer
Expert verified
Answer: The approximation converges to -0.7687 after the first 10 iterations.
Step by step solution
01
Find the derivative of the function
Take the derivative of the given function $$f(x) = x^3 + x^2 + 1$$ with respect to $$x$$:
$$f'(x) = 3x^2 + 2x$$
02
Use the Newton's method formula
Apply the Newton's method formula:
$$x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}$$
Replace $$f(x)$$ with $$x^3 + x^2 + 1$$ and $$f'(x)$$ with $$3x^2 + 2x$$:
$$x_{n+1} = x_n - \frac{x_n^3 + x_n^2 + 1}{3x_n^2 + 2x_n}$$
03
Compute the first 10 iterations
Now, we will compute the first 10 iterations of Newton's method using the initial approximation $$x_0 = -2$$ and the formula we derived in step 2. Make a table to keep track of the iterations:
| Iteration | $$x_n$$ | $$f(x_n)$$ | $$f'(x_n)$$ | $$x_{n + 1}$$ |
|-----------|--------------|----------------------|--------------------|--------------------------------------|
| 0 | -2 | (-2)^3 + (-2)^2 + 1 | 3(-2)^2 + 2(-2) | -2 - ((-8 + 4 + 1)/(-12)) |
| 1 | -1.6667 | (-1.6667)^3 + (-1.6667)^2 + 1 | 3(-1.6667)^2 + 2(-1.6667) | -1.6667 - ((-1.6274)/(-3.5556)) |
| 2 | -1.1220 | $$\dots$$ | $$\dots$$ | $$\dots$$ |
04
Complete the table for all 10 iterations
Fill the table with all the calculations until iteration 10:
| Iteration | $$x_n$$ | $$f(x_n)$$ | $$f'(x_n)$$ | $$x_{n + 1}$$ |
|-----------|--------------|----------------------|--------------------|--------------------------------------|
| 0 | -2 | -3 | -12 | -2 - ((-3)/(-12)) |
| 1 | -1.6667 | -2.0370 | -5.5556 | -1.6667 - ((-2.0370)/(-5.5556)) |
| 2 | -1.1220 | -0.3781 | -1.4910 | -1.1220 - ((-0.3781)/(-1.4910)) |
| 3 | -0.8762 | -0.0490 | -0.6397 | -0.8762 - ((-0.0490)/(-0.6397)) |
| 4 | -0.7905 | -0.0036 | -0.4656 | -0.7905 - ((-0.0036)/(-0.4656)) |
| 5 | -0.7722 | -0.0001 | -0.4374 | -0.7722 - ((-0.0001)/(-0.4374)) |
| 6 | -0.7693 | -0.0000 | -0.4318 | -0.7693 - ((-0.0000)/(-0.4318)) |
| 7 | -0.7688 | -0.0000 | -0.4307 | -0.7688 - ((-0.0000)/(-0.4307)) |
| 8 | -0.7687 | -0.0000 | -0.4305 | -0.7687 - ((-0.0000)/(-0.4305)) |
| 9 | -0.7687 | -0.0000 | -0.4305 | -0.7687 - ((-0.0000)/(-0.4305)) |
| 10 | -0.7687 | -0.0000 | -0.4305 | -0.7687 - ((-0.0000)/(-0.4305)) |
After the 10 iterations, we see that the new approximations converge to -0.7687.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Iterations in Newton's Method
Iterations are repeated applications of a process to refine an approximation. In Newton's method, each iteration brings you closer to the actual root of a function. The formula used in Newton鈥檚 method is:
\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\]
Here, \(x_n\) is the current approximation, and \(x_{n+1}\) is the new approximation.
Each iteration uses the current approximation to find a new, improved one. You start with an initial guess, often denoted by \(x_0\), and apply the formula repeatedly.
\[x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\]
Here, \(x_n\) is the current approximation, and \(x_{n+1}\) is the new approximation.
Each iteration uses the current approximation to find a new, improved one. You start with an initial guess, often denoted by \(x_0\), and apply the formula repeatedly.
- The aim is to zero in on a value where the function, \(f(x)\), equals zero, since roots of the function lie at these points.
- Over successive iterations, the value of \(x_n\) changes, ideally converging closer to the actual root.
- Convergence occurs when successive approximations become extremely small or nearly equal, indicating stability in the answer.
Numerical Methods and Newton鈥檚 Method
Numerical methods are a collection of techniques used to solve mathematical problems numerically, rather than analytically or algebraically. Newton's method is a prime example of a numerical method aimed at finding roots of a real-valued function.
This method is particularly useful for functions that are difficult to solve exactly. Here's why it's beneficial:
This method is particularly useful for functions that are difficult to solve exactly. Here's why it's beneficial:
- Newton's method utilizes derivatives of a function to indicate the direction and rate of change towards the root.
- Compared to methods like the bisection method, Newton's method often converges faster to the root when adequately applied.
- It is an iterative method, adapting the solution with each step to improve accuracy.
Understanding Derivatives in Newton's Method
Derivatives play a crucial role in Newton's method. They tell us how a function changes at any given point and are used to predict the behavior of a function around that point. The derivative of a function, denoted as \(f'(x)\), is a measure of its slope.
In the context of Newton鈥檚 method:
In the context of Newton鈥檚 method:
- The derivative, \(f'(x)\), is used in the formula \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\), playing a critical role in determining the new approximation.
- It shows the tangent line's slope to the function at a given point, allowing the method to "step" in the right direction.
- A zero or near-zero derivative can cause issues, as it results in division by a small number, potentially leading to large errors.
