Problem 13 - Calculus: Early Transcendentals

Question

(a) Approximate \( f \) by a Taylor polynomial with degree \( n \) at the number \( a. \) (b) Use Taylor's Inequality to estimate the accuracy of the approximation \( f(x) \approx T_n(x) \) when \( x \) lies in the given interval. (c) Check you result in part (b) by graphing \( \mid R_n(x) \mid . \) \( f (x) = 1/x, \) \( a = 1, \) \( n = 2, \) \( 0.7 \le x \le 1.3 \)

Step-by-Step Solution

Answer

The Taylor polynomial is \( T_2(x) = 1 - (x-1) + (x-1)^2 \) and the error \( |R_2(x)| \) is estimated as \( \leq 20.874|x-1|^3 \).

Step 1: Find the Taylor polynomial

The Taylor polynomial of degree \( n \) for a function \( f(x) \) about \( a \) is given by: \[ T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x-a)^n \] For \( f(x) = \frac{1}{x} \), calculate the derivatives at \( a = 1 \):- \( f(x) = x^{-1} \Rightarrow f'(x) = -x^{-2} \Rightarrow f'(1) = -1 \)- \( f''(x) = 2x^{-3} \Rightarrow f''(1) = 2 \)The degree 2 Taylor polynomial is:\[ T_2(x) = f(1) + f'(1)(x-1) + \frac{f''(1)}{2}(x-1)^2 = 1 - (x-1) + \frac{2}{2}(x-1)^2 = 1 - (x-1) + (x-1)^2 \] Thus, \[ T_2(x) = 1 - (x-1) + (x-1)^2 \]

Step 2: Use Taylor's Inequality to estimate accuracy

Taylor's Inequality states:\[ |R_n(x)| \leq \frac{M}{(n+1)!}|x-a|^{n+1} \] where \( M \) is an upper bound on the absolute value of the \( (n+1)\)-th derivative on the interval. - The third derivative for \( f(x) = x^{-1} \) is \( f'''(x) = -6x^{-4} \). - On the interval \( 0.7 \le x \le 1.3 \), the maximum value of \( |f'''(x)| \) is \( |-6 \times (0.7)^{-4}| \). Calculating gives \( M = 125.24 \).Using this in Taylor's Inequality:\[ |R_2(x)| \leq \frac{125.24}{3!}|x-1|^3 = \frac{125.24}{6}|x-1|^3 \] So, \[ |R_2(x)| \leq 20.874|x-1|^3 \]

Step 3: Visualize error using Remainder Graph

To check the accuracy estimation graphically, compute and plot the remainder |Rn(x)|:- Calculate \( R_2(x) = f(x) - T_2(x) \).- Use a graphing tool to plot \( |R_2(x)| \) for \( 0.7 \le x \le 1.3 \).- The graph should show values of \( |R_2(x)| \) as small since the Taylor polynomial approximates the function well within the interval.

Key Concepts

Derivative

To create a Taylor polynomial, it's essential to understand the derivatives of a function. Derivatives help us understand how a function changes. For a function like \( f(x) = \frac{1}{x} \), derivatives reveal the rate at which \( f(x) \) changes at any point. To form a Taylor polynomial, we need these rates of change at a specific point \( a \), which in this case is 1.

• The first derivative, \( f'(x) = -x^{-2} \), tells us the opposite of how steep the slope is at \( x \).
• The second derivative, \( f''(x) = 2x^{-3} \), indicates how the slope itself changes.

Plugging \( x = 1 \) into \( f'(x) \) and \( f''(x) \), we find that \( f'(1) = -1 \) and \( f''(1) = 2 \). These calculations are crucial to building the Taylor polynomial, showing behavior near \( x = 1 \). At the heart of the Taylor polynomial is this derivative calculation because it forms the backbone of the approximation near the point \( a \). By understanding derivatives, you can grasp how closely the Taylor polynomial mimics the original function.

Taylor's Inequality

Taylor's Inequality is a powerful tool for estimating how well a Taylor polynomial approximates a function. It helps determine the range of the remainder, \( R_n(x) \), which represents the error in approximation. This inequality provides an upper bound to quantify the difference between the function and its polynomial approximation. The inequality formula is:\[|R_n(x)| \leq \frac{M}{(n+1)!}|x-a|^{n+1}\]where \( M \) is the maximum possible value of the absolute value of the \((n+1)\)-th derivative on the interval considered. Finding \( M \) involves evaluating the (n+1)-th derivative within a specific range, ensuring you capture the worst-case scenario of how much the function deviates from the approximation.In our case, for the function \( f(x) = \frac{1}{x} \), the third derivative \( f'''(x) = -6x^{-4} \) was used to find \( M \), resulting in the estimation:\[|R_2(x)| \leq 20.874|x-1|^3\]This formula is pivotal in checking how small\( |R_2(x)|\) can get in the interval, indicating the quality of the approximation.

Approximation Error

The approximation error is the difference between the actual function and its approximation using a Taylor polynomial. When we approximate a function like \( f(x) = \frac{1}{x} \) using a Taylor polynomial, there's always some level of error, especially as you move further from the point \( a \). This error is represented by \( R_n(x) \).Understanding this error is crucial for determining the accuracy of the approximation. The Taylor polynomial gives us a simplified version of a function that is easier to work with. However, by knowing the error, you can judge how good the approximation really is.To see this error visually, you can plot \( |R_n(x)| \) over the desired interval, such as from 0.7 to 1.3 in this exercise. This plot shows how well the polynomial aligns with the actual function over the interval. Observing it graphically helps solidify the concept that, although the Taylor polynomial is a good estimate, it can't exactly match the real function over the entire range. Instead, it provides a very close approximation, particularly near the point \( a \). Understanding and estimating the approximation error lets you gauge how reliable your polynomial approximation is in solving a problem.