Problem 9 - Essential Calculus Early Transcendentals

Question

(a) Approximate \(f\) by a Taylor polynomial with degree \(n\) at the number \(a\) . (b) Use Taylor's Formula to estimate the accuracy of the approximation \(f(x) \approx T_{n}(x)\) when \(x\) lies in the given interval. (c) Check your result in part (b) by graphing \(\left|R_{n}(x)\right|\) $$f(x)=\sqrt{x}, \quad a=4, \quad n=2, \quad 4 \leqslant x \leqslant 4.2$$

Step-by-Step Solution

Answer

The approximation \\( f(x) \approx T_2(x) \\) is very accurate, with a maximum possible error of about 0.000053 for \\( 4 \leq x \leq 4.2 \\). Graphing confirms this accuracy.

Step 1: Find the Taylor Polynomial

To approximate the function using a Taylor polynomial, we need to compute the polynomial up to degree 2 centered at \( a = 4 \). The Taylor series is given by: \[ T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 \]First compute the derivatives of \( f(x) = \sqrt{x} \):- \( f'(x) = \frac{1}{2\sqrt{x}} \)- \( f''(x) = -\frac{1}{4x^{3/2}} \)Evaluate at \( a = 4 \):- \( f(4) = \sqrt{4} = 2 \)- \( f'(4) = \frac{1}{2\sqrt{4}} = \frac{1}{4} \)- \( f''(4) = -\frac{1}{4(4)^{3/2}} = -\frac{1}{32} \)The Taylor polynomial is:\[ T_2(x) = 2 + \frac{1}{4}(x-4) - \frac{1}{64}(x-4)^2 \]

Step 2: Estimate Remainder Using Taylor's Formula

Taylor's formula remainder term for degree 2 is: \[ R_n(x) = \frac{f'''(c)}{3!}(x-a)^3 \]We need \( f'''(x) \), which is \[ f'''(x) = \frac{3}{8x^{5/2}} \]For \( 4 \leq x \leq 4.2 \), choose \( c \) to be close to 4, as this maximizes \( |f'''(x)| \) in the interval:\[ |f'''(4.2)| = \frac{3}{8(4.2)^{5/2}} \approx 0.008 \]Estimate the remainder:\[ |R_2(x)| \leq \frac{0.008}{6}(0.2)^3 \approx 0.000053 \]

Step 3: Verify Using Graph

To verify the approximation's accuracy, graph \( |R_2(x)| \) over the interval \( [4, 4.2] \). The graph should show that the remainder is small, supporting our earlier estimate that \( |R_2(x)| \leq 0.000053 \). This confirms that the approximation \( f(x) \approx T_2(x) \) is very accurate in the specified interval.

Key Concepts

Taylor polynomial

The Taylor polynomial is a powerful mathematical tool used to approximate functions. It helps to represent a function as a polynomial where the core idea is to match as many derivatives of the function as possible at a particular point. In our exercise, we aim to approximate the function \( f(x) = \sqrt{x} \) centered at \( a = 4 \) using a Taylor polynomial of degree 2.

To construct the Taylor polynomial, we compute the necessary derivatives of the function and evaluate them at \( a = 4 \). The polynomial expansion we get is:

• \( T_2(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 \)
• Carrying out the calculations, we achieve: \( T_2(x) = 2 + \frac{1}{4}(x-4) - \frac{1}{64}(x-4)^2 \)

The result is a quadratic polynomial that closely approximates the function \( f(x) = \sqrt{x} \) near \( x = 4 \). It rests on making incremental changes to \( x \) into simple terms that we can handle easily in calculations.

remainder estimation

Remainder estimation tells us how accurate our Taylor polynomial is in approximating the original function over a specific interval. When using a Taylor polynomial to approximate a function, the remainder represents the error or the difference between the actual function and the polynomial approximation.

The remainder \( R_n(x) \) for a Taylor polynomial of degree \( n = 2 \) involves the third derivative of the function, as shown in the expression:

• \( R_n(x) = \frac{f'''(c)}{3!}(x-a)^3 \)

For \( 4 \leq x \leq 4.2 \), we choose \( c \) such that \( |f'''(c)| \) is maximized. In our exercise, this gives us:

• \( |f'''(4.2)| \approx 0.008 \)
• Thus, \( |R_2(x)| \leq \frac{0.008}{6}(0.2)^3 \approx 0.000053 \)

This small value hints nicely at how the quadratic Taylor polynomial is extraordinarily close to the actual function within the specified range.

accuracy of approximation

The accuracy of approximation tells us how closely a Taylor polynomial matches the function it represents in a specific interval. In this exercise, we check the accuracy of our approximation by considering the size of the remainder \( |R_2(x)| \).

Since our estimated maximum error is very small, \( |R_2(x)| \leq 0.000053 \), the Taylor polynomial is very accurate for values of \( x \) between 4 and 4.2.

• This means that any result calculated using the polynomial is expected to be exceedingly close to the true value of \( f(x) = \sqrt{x} \).
• The importance of understanding this estimation lies in knowing the limits of the polynomial's effectiveness and reliability in practice.

The fantastically minute remainder illustrates that the extension to real-world applications, such as predicting behavior of functions over small intervals, is genuinely potent with Taylor polynomials.

derivatives of functions

Understanding derivatives of functions is crucial in forming a Taylor polynomial. Derivatives represent the rate at which a function changes at any point and form the backbone of the Taylor series.

For our function \( f(x) = \sqrt{x} \), we had to calculate the first three derivatives for our approximation:

• First derivative: \( f'(x) = \frac{1}{2\sqrt{x}} \)
• Second derivative: \( f''(x) = -\frac{1}{4x^{3/2}} \)
• Third derivative: \( f'''(x) = \frac{3}{8x^{5/2}} \)

Evaluating these at \( a = 4 \) provided us with necessary values to construct the Taylor polynomial and estimate the remainder.

• At \( x = 4 \), these derivatives capture how the function behaves near this point.
• Through the use of derivatives, we ascertain how quickly and in what manner the function diverges from the polynomial approximation.

These computations begin with understanding derivatives and then simplifying them for easy use in constructing polynomials, ensuring that all approximations are as accurate as possible.