Question

Approximate the function value with the indicated Taylor polynomial and give approximate bounds on the error. Approximate \(\sqrt{10}\) with the Taylor polynomial of degree 2 centered at \(x=9\).

Short answer

The approximation of \( \sqrt{10} \) is 3.1620, with an error bound of ±0.00017.

Step-by-step solution

Understand the Problem

We need to approximate \( \sqrt{10} \) using a Taylor polynomial approximation of degree 2 centered at \( x = 9 \). The Taylor polynomial of a function \( f \) about the point \( a \) is \( T_n(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 \).

Differentiate the Function

The function we are considering is \( f(x) = \sqrt{x} = x^{1/2} \). The first derivative is \( f'(x) = \frac{1}{2}x^{-1/2} \), and the second derivative is \( f''(x) = -\frac{1}{4}x^{-3/2} \).

Evaluate the Function and Its Derivatives

Evaluate \( f(9) \), \( f'(9) \), and \( f''(9) \). We find that \( f(9) = 3 \), \( f'(9) = \frac{1}{6} \), and \( f''(9) = -\frac{1}{108} \).

Construct the Taylor Polynomial

Using the derivatives, construct the Taylor polynomial: \[ T_2(x) = f(9) + f'(9)(x-9) + \frac{f''(9)}{2}(x-9)^2 \] \[ T_2(x) = 3 + \frac{1}{6}(x-9) - \frac{1}{216}(x-9)^2 \]

Approximate \( \sqrt{10} \)

Substitute \( x = 10 \) into the Taylor polynomial: \[ T_2(10) = 3 + \frac{1}{6}(10-9) - \frac{1}{216}(10-9)^2 \] \[ T_2(10) = 3 + \frac{1}{6}(1) - \frac{1}{216}(1) \] \[ T_2(10) = 3 + \frac{1}{6} - \frac{1}{216} = 3.1620 \]

Estimate the Error Bound

The error in a Taylor polynomial is bounded by the remainder term: \[ R_3(x) = \frac{f'''(c)}{3!}(x-9)^3 \] Determine \( f'''(x) = \frac{3}{8}x^{-5/2} \). Evaluate this at some \( c \) between 9 and 10. At most, \( f'''(c) \approx \frac{3}{8}\cdot \frac{1}{243}, \) providing an approximate maximum error of \( R_3(10) \approx \frac{1}{5832} \approx 0.00017 \). This suggests an accurate approximation of 3.1620 ± 0.00017.

Key concepts

Polynomial Approximation

Polynomial approximation is a powerful mathematical tool that helps us estimate the values of functions in the vicinity of a point. A Taylor polynomial is a specific type of polynomial approximation. It represents a function as a sum of derivatives at a particular point, scaled by powers of the distance from this point. This method is especially useful for approximating complex functions with simpler polynomials.
The exercise demonstrates this by approximating the function \( \sqrt{10} \) using a Taylor polynomial centered at \( x = 9 \). By constructing a Taylor polynomial of degree 2 using derivatives, we can approximate the square root function near 9. This approach simplifies the calculation, making it more accessible and manageable, especially when an exact computation is difficult or impossible.

Derivative Calculations

Derivative calculations are crucial when constructing a Taylor polynomial. Derivatives provide information about the function's rate of change and curvature, which are used to build the polynomial approximation. Think of them as the building blocks of the polynomial.
In our problem, the function is \( f(x) = \sqrt{x} \), which means we need its derivatives evaluated at \( x = 9 \). The first derivative, \( f'(x) = \frac{1}{2}x^{-1/2} \), gives the slope, while the second derivative, \( f''(x) = -\frac{1}{4}x^{-3/2} \), provides the concavity.

• First, we find \( f(9) = 3 \), representing the function value at \( x = 9 \).
• Next, \( f'(9) = \frac{1}{6} \) describes how quickly the function is changing at this point.
• Finally, \( f''(9) = -\frac{1}{108} \) offers insight into the bending of the function.

These derivatives help create the Taylor polynomial, directly influencing the approximation's accuracy.

Error Estimation

Error estimation assesses how accurate our polynomial approximation is. When using a Taylor polynomial, the error is related to the next derivative beyond those used in the polynomial, which is part of what's called the remainder term.
The remainder term provides a bound on the error for approximating a function using a Taylor polynomial. In this exercise, we estimate the error for the approximation of \( \sqrt{10} \) with a second-degree polynomial.
The error term is given by:\[ R_3(x) = \frac{f'''(c)}{3!}(x-9)^3 \]Evaluating the third derivative \( f'''(x) = \frac{3}{8}x^{-5/2} \) at a point \( c \) between 9 and 10, we ensure that the error is as small as possible. In this case, the error is approximately 0.00017, indicating a high level of precision for our approximation.

Function Evaluation

Function evaluation is the final check to ensure that our Taylor polynomial gives a reasonable approximation of our target value. After deriving and constructing the polynomial, we substitute the desired point into the polynomial to get an approximate value.
For \( \sqrt{10} \), we substitute \( x = 10 \) into the Taylor polynomial derived in previous steps:

• The polynomial is: \( T_2(x) = 3 + \frac{1}{6}(x-9) - \frac{1}{216}(x-9)^2 \).
• Substituting \( x = 10 \), we perform the calculations to find \( T_2(10) = 3.1620 \).

This step confirms that \( \sqrt{10} \approx 3.1620 \), demonstrating how the Taylor polynomial provides an accessible method to approximate more complicated function values close to a chosen point. The process efficiently bridges tough calculations with simpler arithmetic, making it easier to achieve a satisfactory result without extensive resources.