Problem 18 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the Maclaurin polynomial of order \(n(n\) odd \()\) for \(\sin x .\) Then use it with \(n=5\) to approximate each of the fol. lowing. (This example should convince you that the Maclaurin approximation can be exceedingly poor if \(x\) is far from zero. Compare your answers with those given by your calculator. What conclusion do you draw? (a) \(\sin (0.1)\) (b) \(\sin (0.5)\) (c) \(\sin (1)\) (d) \(\sin (10)\)

Step-by-Step Solution

Answer

The Maclaurin approximation is accurate for small \( x \); accuracy decreases with larger \( x \).

Step 1: Understanding the Maclaurin Series

The Maclaurin series expansion for a function is a type of Taylor series where the series is centered at 0. For a function \( f(x) \), the series is given by:\[f(x) = f(0) + f'(0)x + \frac{f''(0)x^2}{2!} + \frac{f'''(0)x^3}{3!} + \ldots\]For the specific case of \( \sin x \), the derivative pattern produces only non-zero terms for odd derivatives.

Step 2: Derivatives of \\( \\sin x \\)

Calculate the derivatives of \( \sin x \) and evaluate them at \( x = 0 \):- \( \sin x \) has derivatives \( \sin x, \cos x, -\sin x, -\cos x, \sin x, \ldots \)- Evaluating these at 0: \( \sin(0) = 0 \), \( \cos(0) = 1 \), so higher derivatives follow this pattern: 0, 1, 0, -1, 0, 1, 0, -1, \ldots.

Step 3: Constructing the Maclaurin Series for \\( \\sin x \\)

From Step 2, recognizing non-zero terms and their alternating sign, the series for \( \sin x \) becomes:\[\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots\]For \( n = 5 \), consider terms up to \( x^5 \).

Step 4: Approximating \\( \\sin(x) \\) \\text{ using } \\ n=5

Using the Maclaurin series for \( n = 5 \), approximate \( \sin x \) with:\[\sin x \approx x - \frac{x^3}{6} + \frac{x^5}{120}\]Use this expression to calculate for (a), (b), (c), and (d).

Step 5: Calculation for \\( \\sin(0.1) \\)

Plug \( x = 0.1 \) into the series:\[\sin(0.1) \approx 0.1 - \frac{0.1^3}{6} + \frac{0.1^5}{120} \approx 0.0998333\]Compare with calculator value: \( \sin(0.1) \approx 0.0998334 \).

Step 6: Calculation for \\( \\sin(0.5) \\)

Plug \( x = 0.5 \) into the series:\[\sin(0.5) \approx 0.5 - \frac{0.5^3}{6} + \frac{0.5^5}{120} \approx 0.4794255\]Compare with calculator value: \( \sin(0.5) \approx 0.4794255 \).

Step 7: Calculation for \\( \\sin(1) \\)

Plug \( x = 1 \) into the series:\[\sin(1) \approx 1 - \frac{1^3}{6} + \frac{1^5}{120} \approx 0.8416667\]Compare with calculator value: \( \sin(1) \approx 0.8414709 \).

Step 8: Calculation for \\( \\sin(10) \\)

Plug \( x = 10 \) into the series:\[\sin(10) \approx 10 - \frac{10^3}{6} + \frac{10^5}{120} = -160\]Compare with calculator value: \( \sin(10) \approx -0.5440211 \).

Step 9: Conclusion from the Approximation Results

The Maclaurin series is a good approximation for small \( x \) values close to zero, such as \( x = 0.1 \) and \( x = 0.5 \). It becomes less accurate as \( x \) increases. For \( x = 10 \), the approximation is very poor, which shows the limitation of the Maclaurin series for large \( x \).

Key Concepts

Taylor Series

A Taylor series is a powerful mathematical tool used to approximate complex functions with polynomials. This technique involves expanding a given function into an infinite sum of terms calculated from the values of its derivatives at a single point. The general formula for the Taylor series of a function \( f(x) \) centered at \( a \) is:\[f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)(x-a)^2}{2!} + \frac{f'''(a)(x-a)^3}{3!} + \ldots\]The idea is to represent a function that may not even be a polynomial as an infinite series where the polynomial terms increase in degree. The Taylor series offers a way to understand, analyze, and even compute functions that are complex or undefined in parts. When the center of the series, \( a \), is chosen as 0, the series is known as the Maclaurin series, which is particularly useful for approximating functions near zero. The coefficients of each term in the series are derived from the derivatives of the function at \( a \). By using more terms in the series, one can increase the accuracy of the approximation.

sin x approximation

Approximating the function \( \sin x \) using a Maclaurin or Taylor series expands our ability to calculate its value without a calculator. Sinusoidal functions are naturally periodic and can be tricky to approximate with standard polynomials. The Maclaurin series provides a practical method for creating a polynomial that can approximate \( \sin x \) quite accurately for small values of \( x \). The series for \( \sin x \) is:\[\sin x \approx x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots\]Each term in the series is determined by successive derivatives of \( \sin x \) at 0 and divided by the factorial of the term's degree, ensuring convergence for small \( x \). The pattern involves alternating signs, reflecting the nature of the sine wave's oscillation. The approximation tends to be very precise when \( x \) is close to zero, but it becomes less reliable as \( x \) grows larger. For example, for \( \sin(0.1) \) or \( \sin(0.5) \), the polynomial provides a close approximation, yet for \( \sin(10) \), the result significantly diverges from the true value. Approximations using the series highlight how polynomial terms contribute to capturing the waveform nature of the sine function.

Polynomial Approximation

Polynomial approximation involves using polynomials to estimate or model a more complicated function over certain intervals. In the context of Maclaurin and Taylor series, these approximations leverage derivatives to capture the behavior of the original function. By constructing a polynomial through an infinite series, one can estimate the value of functions such as \( \sin x \) with tailored precision.

• The degree of the polynomial reflects the number of terms used from the series.
• More terms generally lead to greater accuracy, especially near the center of the series.
• A Taylor polynomial of degree \( n \) includes terms up to \( (x-a)^n \), and the more terms you add, the more the polynomial behaves like the actual function.

These approximations are constrained by the range of \( x \) for which they remain accurate. The primary advantage of polynomial approximations is their simplicity and ability to estimate values without direct computation. For functions like \( \sin x \), employing a polynomial offers a quick and manageable way to approximate values, albeit with the caution that this approximation falters as \( x \) deviates substantially from the central point.

Derivatives of Functions

Derivatives are essential in calculus and are the stepping stones for comprehension of Taylor and Maclaurin series. They represent the rate of change of a function, providing insight into the function’s behavior at any given point. Each term in a Taylor series stems from the derivatives of the function, evaluated at the center of the series.When dealing with a function like \( \sin x \), the derivatives have a repeating pattern because the function is periodic:

• First derivative, \( \cos x \)
• Second derivative, \(-\sin x \)
• Third derivative, \(-\cos x \)
• Fourth derivative, \( \sin x \)

This cyclical nature aids in constructing the Maclaurin series. By determining these derivatives at \( x = 0 \), and noticing which derivatives vanish at this point (for \( \sin x \), every other derivative turns out zero), a simplified pattern emerges to form the polynomial approximation. Understanding derivatives and their evaluations is crucial because they dictate both the shape and behavior of the polynomial approximation, providing the subtle accuracy a calculation requires.