Problem 12 - 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 \(\left|R_{n}(x)\right|\) $$f(x)=\sin x, \quad a=\pi / 6, \quad n=4, \quad 0 \leqslant x \leqslant \pi / 3$$

Step-by-Step Solution

Answer

The Taylor polynomial is calculated, error estimated, and checked graphically.

Step 1: Find the Taylor polynomial

Compute \(T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(a)}{k!}(x-a)^k\) for the given function \(f\) and center \(a\).

Step 2: Estimate accuracy

Use Taylor's formula: \(|R_n(x)| \leq \frac{M}{(n+1)!}|x-a|^{n+1}\) where \(M\) bounds \(|f^{(n+1)}|\).

Key Concepts

Taylor's Theorem

Taylor's Theorem is a fundamental principle in calculus that helps us approximate complex functions using polynomials. The idea is to express a function as an infinite sum of its derivatives at a point. This approximation becomes more accurate as we include more terms in the series. The theorem states that a function that is infinitely differentiable at a point can be written as: \[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \] This formula allows us to closely approximate functions with polynomials. The smaller the interval around the point \( a \), and the more terms you use, the closer the polynomial will match the function.
Some key points to remember about Taylor's Theorem:

• The point \( a \) is where you want the approximation centered.
• The function must be differentiable at this point.
• The degree of the polynomial indicates how many derivatives are included.

Understanding this theorem gives you a great tool for simplifying the computation of complex functions.

Function Approximation

Function approximation is a technique used to estimate a function with a simpler representation. In calculus, Taylor and Maclaurin series are often used for this.
The main idea is to replace a complex function with a polynomial to make calculations easier. Polynomials are simple to work with due to their proven algebraic properties, making them perfect candidates for approximations.
For example, approximating the sine function \( f(x) = \sin x \) with a polynomial helps to perform calculations without a calculator. By using Taylor polynomials, we express \( \sin x \) with a degree 4 polynomial centered at \( a = \pi/6 \), greatly simplifying calculations on intervals close to \( a \).
The benefits of function approximation include:

• Reducing complexity in calculation.
• Having a polynomial structure for easy computation.
• Understanding behavior of functions near a specific point.

Derivative Calculations

Derivative calculations are at the heart of constructing Taylor polynomials. Derivatives provide the necessary coefficients for building the polynomial. To calculate the Taylor polynomial, derivatives of a function must be evaluated at the point \( a \), and these derivatives dictate the shape and accuracy of the polynomial approximation.
In our example, the function \( f(x) = \sin x \) required computing derivatives up to the fourth degree:

• First derivative: \( f'(x) = \cos x \)
• Second derivative: \( f''(x) = -\sin x \)
• Third derivative: \( f'''(x) = -\cos x \)
• Fourth derivative: \( f^{(4)}(x) = \sin x \)

These derivatives were evaluated at \( x = \pi/6 \) to find the coefficients of the Taylor polynomial. Getting accurate derivatives is fundamental because these coefficients determine how well the polynomial will approximate the function across the desired interval.

Taylor Series Expansion

A Taylor Series Expansion is an infinite sum derived from Taylor's Theorem, representing a function as close as possible using its derivatives. The Taylor series for a function about a point \( a \) includes infinite terms. However, in practical applications, we often truncate the series, creating a Taylor polynomial with a finite number of terms.
For our function \( f(x) = \sin x \), the Taylor polynomial of degree 4 is used instead of taking all infinite terms, since this is practical for approximation:\[T_4(x) = \frac{1}{2} + \frac{\sqrt{3}}{2}(x - \frac{\pi}{6}) - \frac{1}{2} \frac{1}{2!}(x - \frac{\pi}{6})^2 - \frac{\sqrt{3}}{2} \frac{1}{3!}(x - \frac{\pi}{6})^3 + \frac{1}{2} \frac{1}{4!}(x - \frac{\pi}{6})^4\]This polynomial utilizes derivatives of \( \sin x \) at \( a = \pi/6 \) up to the fourth degree, offering a reasonable approximation over the interval \( 0 \leq x \leq \pi/3 \).
This results in a manageable expression that aids in performing calculations and estimating function values efficiently.