Problem 39 - Calculus Early Transcendentals

Question

Residuals and errors Approximate the root of \(f(x)=x^{10}\) at \(x=0\) using Newton's method with an initial approximation of \(x_{0}=0.5 .\) Make a table showing the first 10 approximations, the error in these approximations (which is \(\left|x_{n}-0\right|=\left|x_{n}\right|\) ), and the residual of these approximations (which is \(f\left(x_{n}\right)\) ). Comment on the relative size of the errors and the residuals and give an explanation.

Step-by-Step Solution

Answer

The relationship between the errors and residuals in this case is that as the errors decrease, the residuals decrease at a faster rate. This is because Newton's method decreases the errors in an almost linear manner, whereas the residuals decrease in a polynomial manner (\(x_n^{10}\)). The rapidly decreasing residuals indicate that Newton's method converges quickly for this function.

Step 1: Newton's method formula

To approximate the root using Newton's method, we use the formula: \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\) Let's first find the derivative of \(f(x) = x^{10}\).

Step 2: Calculate the derivative of the function

The derivative of \(f(x) = x^{10}\) is: \(f'(x) = 10x^9\)

Step 3: Generate the first 10 approximations using Newton's method

Now, we can apply Newton's method iteratively with the initial approximation \(x_0 = 0.5\). The formula becomes: \(x_{n+1} = x_n - \frac{x_n^{10}}{10x_n^9} = x_n - \frac{1}{10}x_n = \frac{9}{10}x_n\) We can use this formula to find the first 10 approximations: 1. \(x_1 = \frac{9}{10}(0.5) = 0.45\) 2. \(x_2 = \frac{9}{10}(0.45) = 0.405\) 3. \(x_3 = \frac{9}{10}(0.405) = 0.3645\) 4. \(x_4 = \frac{9}{10}(0.3645) = 0.32805\) 5. \(x_5 = \frac{9}{10}(0.32805) = 0.295245\) 6. \(x_6 = \frac{9}{10}(0.295245) = 0.2657205\) 7. \(x_7 = \frac{9}{10}(0.2657205) = 0.23914845\) 8. \(x_8 = \frac{9}{10}(0.23914845) = 0.215233605\) 9. \(x_9 = \frac{9}{10}(0.215233605) = 0.1937102445\) 10. \(x_{10} = \frac{9}{10}(0.1937102445) = 0.17433922005\)

Step 4: Calculate the errors and residuals

For each approximation \(x_n\), the error is given by \(|x_n - 0| = |x_n|\) and the residual is given by \(f(x_n) = x_n^{10}\). | Approximation | Error | Residual | |---------------|--------------|------------------------------| | \(x_1 = 0.45\) | \(0.45\) | \(3.405\times10^{-11}\) | | \(x_2 = 0.405\) | \(0.405\) | \(9.648\times10^{-13}\) | | \(x_3 = 0.3645\)| \(0.3645\) | \(2.733\times10^{-14}\) | | \(x_4 = 0.32805\)| \(0.32805\) | \(7.747\times10^{-16}\) | | \(x_5 = 0.295245\)| \(0.295245\) | \(2.190\times10^{-17}\) | | \(x_6 = 0.2657205\)|\(0.2657205\)| \(6.198\times10^{-19}\) | | \(x_7 = 0.23914845\)|\(0.23914845\)|\(7.562\times10^{-21}\) | | \(x_8 = 0.215233605\)|\(0.215233605\)|\(2.148\times10^{-22}\) | | \(x_9 = 0.1937102445\)|\(0.1937102445\)|\(6.084\times10^{-24}\) | | \(x_{10} = 0.17433922005\)|\(0.17433922005\)|\(1.725\times10^{-25}\)|

Step 5: Analyze the results

We can see that as we iterate through the approximations, both the errors and residuals decrease in value. As the errors decrease, the residuals decrease at a faster rate. This is because Newton's method decreases the errors in an almost linear manner, whereas the residuals decrease in a polynomial manner (\(x_n^{10}\)). The rapidly decreasing residuals indicate that Newton's method converges quickly for this function.

Key Concepts

Root Approximation

Newton's Method is a powerful iterative approach for approximating the roots of a function. Starting with an initial guess, the method uses the derivative of the function to converge rapidly to the actual root. In this particular exercise, we're trying to approximate the root of the function \(f(x) = x^{10}\) at \(x = 0\). Given the initial guess of \(x_0 = 0.5\), Newton's formula, \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\), simplifies in this context to \(x_{n+1} = \frac{9}{10}x_n\) after computing the derivative, \(f'(x) = 10x^9\). This simplification allows for straightforward calculations and demonstrates how each iteration "pushes" the approximation closer to the true root.

Over ten iterations, the approximations progressively move towards zero: \(x_1 = 0.45\), \(x_2 = 0.405\), and so on, until \(x_{10} = 0.17433922005\). Even though we're initially quite far from the root, each calculation gets us significantly closer, effectively approximating the root with remarkable precision. This process exemplifies how Newton's Method accelerates towards the root with each step.

Error Analysis

Error analysis in Newton's Method gives us insight into how close our approximations are to the actual root. For this exercise, the error at each step is calculated as the absolute value of the current approximation. So, the error for each iteration is simply \(|x_n|\). This is because our target root is at zero, making the calculations straightforward.

When examining the errors from \(x_1\) through \(x_{10}\), we see a consistent decrease: starting from \(0.45\) and ending at \(0.17433922005\). This decrease highlights the method’s efficiency in narrowing down on the root. The reduction in error at each step reinforces that although the initial approximation is far from ideal, the quick narrowing of error margin suggests how iterative methods like Newton's are advantageous for refining guesses towards precision.

As students, understanding how errors shrink can guide expectations on how quickly a method converges, and also highlights the importance of selecting good initial guesses to ensure quicker convergence.

Residual Analysis

Residual analysis focuses on the difference between what the function outputs compared to the expected value, which is zero for this root-finding scenario. Examined as \(f(x_n) = x_n^{10}\), residuals further reveal the precision of approximations.

The rapid decrease in residual values, from \(3.405 \times 10^{-11}\) at \(x_1\) to a breathtaking \(1.725 \times 10^{-25}\) at \(x_{10}\), indicates that as the method moves closer to the root, the function value itself becomes negligible. This subtler form of precision is critical in numerical methods, where getting close means not only hitting the right number, but producing function values that get closer to zero.

The impressive rate at which the residuals decrease compared to the linear error response means the function is improving quadratically, a property of Newton's Method when the guess is sufficiently close to the root. For students, recognizing the difference between error and residual decay patterns is key in understanding why certain methods yield faster results and why residuals are vital for assessing the success of root approximations.