Problem 9 - Calculus: Early Transcendentals

Question

Find the \(5^{\text {th }}\) degree Taylor polynomial for \(f(x)=\sin x\) around \(a=0\). (a) Use this Taylor polynomial to approximate \(\sin (0.1)\). (b) Use a calculator to find \(\sin (0.1) .\) How does this compare to our approximation in part \((a) ?\)

Step-by-Step Solution

Answer

The Taylor polynomial approximation is accurate and closely matches the calculator value for \(\sin(0.1)\).

Step 1: Understanding the Taylor Series

The Taylor series for a function \( f(x) \) centered at \( a \) is given by: \[ T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k \]. We will use this to find the 5th degree Taylor polynomial for \( f(x) = \sin x \) around \( a = 0 \).

Step 2: Determine the Derivatives

We need the derivatives of \( \sin x \) evaluated at \( x = 0 \):- \( f(x) = \sin x, \quad f(0) = 0 \)- \( f'(x) = \cos x, \quad f'(0) = 1 \)- \( f''(x) = -\sin x, \quad f''(0) = 0 \)- \( f'''(x) = -\cos x, \quad f'''(0) = -1 \)- \( f^{(4)}(x) = \sin x, \quad f^{(4)}(0) = 0 \)- \( f^{(5)}(x) = \cos x, \quad f^{(5)}(0) = 1 \)

Step 3: Construct the 5th Degree Taylor Polynomial

Using the derivatives, the 5th degree Taylor polynomial is:\[ T_5(x) = 0 + \frac{1}{1!}x - \frac{1}{3!}x^3 + \frac{1}{5!}x^5 \]Thus, \[ T_5(x) = x - \frac{x^3}{6} + \frac{x^5}{120} \].

Step 4: Approximate \( \sin(0.1) \) Using the Polynomial

Substitute \( x = 0.1 \) into the polynomial:\[ T_5(0.1) = 0.1 - \frac{(0.1)^3}{6} + \frac{(0.1)^5}{120} \]Calculate each term:- \( 0.1 = 0.1 \)- \( \frac{(0.1)^3}{6} = \frac{0.001}{6} = 0.0001667 \)- \( \frac{(0.1)^5}{120} = \frac{0.00001}{120} = 0.0000000833 \)Approximation: \( T_5(0.1) \approx 0.1 - 0.0001667 + 0.0000000833 = 0.0998334 \)

Step 5: Exact Value Using Calculator

Using a calculator to find \( \sin(0.1) \), we get approximately \(0.0998334\).

Step 6: Compare the Approximation and Exact Value

The approximation \( T_5(0.1) \approx 0.0998334 \) using the Taylor polynomial is very close to the exact value \( \sin(0.1) \approx 0.0998334 \) found using a calculator, showing that the approximation is accurate.

Key Concepts

Taylor Series

The Taylor series is a powerful tool in mathematics, used to represent functions as an infinite sum of terms calculated from its derivatives at a single point. This concept is particularly useful in approximating functions that are difficult to compute exactly. For the function \( f(x) \), its Taylor series centered at \( a \) is expressed as:

• \( T_n(x) = \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k!}(x-a)^k \)

Here, \( n \) denotes the degree of the polynomial, \( f^{(k)}(a) \) is the \( k^{th} \) derivative of \( f \) evaluated at \( a \), and \( k! \) is the factorial of \( k \). This expansion allows complex functions to be expressed in terms of polynomials, making them easier to compute, analyze, and understand. By increasing the degree \( n \), the polynomial can provide a more accurate approximation to the actual function. In the context of trigonometric functions like \( \sin x \), Taylor series are very useful because these functions often repeat in cycles, making them suitable candidates for polynomial approximations.

Derivative

The derivative of a function provides critical information about the rate at which the function value changes as the input changes. Calculating the derivatives is an essential step when forming a Taylor series, as each term in the series depends on the value of these derivatives. For the function \( f(x) = \sin x \), derivatives are cyclic due to its periodic nature within the trigonometric family.

• The first derivative, \( f'(x) = \cos x \), represents the instantaneous rate of change of \( \sin x \).
• The second derivative, \( f''(x) = -\sin x \), shows further change and is negative, indicating a downward slope.
• This cycle continues with subsequent derivatives alternating between \( \sin x \) and \( \cos x \) with changes in sign, as seen with \( f'''(x) = -\cos x \) and \( f^{(4)}(x) = \sin x \).

This alternating pattern helps construct the Taylor polynomial, particularly for cyclic or periodic functions like trigonometric ones.

Trigonometric Functions

Trigonometric functions, such as sine and cosine, are fundamental in mathematics, with applications ranging from geometry to complex system modeling. The sine function \( \sin x \) is especially important and behaves predictably through its cyclical nature. The use of Taylor polynomials allows these functions to be approximated near a point, typically when exact computation is cumbersome. For \( \sin x \), its Taylor series expansion around \( x = 0 \) provides a tool for approximating values that can be tedious to calculate directly. In our example, the Taylor polynomial for \( \sin x \) includes only odd-powered terms due to the derivatives of \( \cos x \) and its relation with sine:

• \( T_5(x) = x - \frac{x^3}{6} + \frac{x^5}{120} \)

This approximation reveals how polynomials can mimic the complexities of the sine function through derivatives, offering insight into behavior patterns within trigonometric graphs.

Approximation

Approximating a function using a Taylor polynomial means representing the function with a finite number of terms from its Taylor series. This approximation is highly dependent on the degree of the polynomial chosen. In our example, we built a 5th degree Taylor polynomial for \( \sin x \) and used it to approximate \( \sin(0.1) \). The result was compared to the exact value obtained via a calculator.

• With \( x = 0.1 \), substituting into \( T_5(x) \) resulted in an approximate value of 0.0998334.
• The calculator value for \( \sin(0.1) \) was the same, demonstrating the polynomial's accuracy.

The proximity of these results illustrates the efficiency of Taylor polynomials in producing reliable approximations for small values of \( x \). Such approximations are vital in scientific and engineering applications, where exact values are often unnecessary or impractical to compute.