Problem 12 - Essential Calculus Early Transcendentals

Question

Find the Taylor series for \(f(x)\) centered at the given value of \(a\) . [Assume that \(f\) has a power series expansion. Do not show that \(R_{n}(x) \rightarrow 0.1]\) $$f(x)=x-x^{3}, \quad a=-2$$

Step-by-Step Solution

Answer

The Taylor series is \( 6 - 11(x+2) + 6(x+2)^2 - (x+2)^3 \).

Step 1: Identify the function and center

The given function is \( f(x) = x - x^3 \), and the Taylor series is to be centered at \( a = -2 \). We need to determine the Taylor series expansion of \( f(x) \) about \( x = -2 \).

Step 2: Find the derivatives

To construct the Taylor series, calculate the derivatives of \( f(x) \).- The first derivative is \( f'(x) = 1 - 3x^2 \).- The second derivative is \( f''(x) = -6x \).- The third derivative is \( f'''(x) = -6 \).- Higher derivatives: Since the third derivative and all subsequent ones are constant, \( f^{(n)}(x) \) for \( n \geq 3 \) is \( 0 \).

Step 3: Evaluate derivatives at the center

Evaluate these derivatives at \( x = -2 \):- \( f(-2) = -2 - (-2)^3 = 6 \).- \( f'(-2) = 1 - 3(-2)^2 = -11 \).- \( f''(-2) = -6(-2) = 12 \).- \( f'''(-2) = -6 \).

Step 4: Write the Taylor series formula

The Taylor series expansion of a function \( f(x) \) about \( x = a \) is given by the formula:\[T(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots\]

Step 5: Substitute into the Taylor series formula

By substituting our specific function \( f(x) \) and its derivatives into the Taylor series formula, we have:\[T(x) = 6 - 11(x+2) + \frac{12}{2}(x+2)^2 - \frac{6}{6}(x+2)^3\]Thus, simplifying results in:\[T(x) = 6 - 11(x+2) + 6(x+2)^2 - (x+2)^3\]

Step 6: Write out the Taylor series

The Taylor series expanded form:\[6 - 11(x+2) + 6(x+2)^2 - (x+2)^3 + \cdots\]This series represents \( f(x) = x - x^3 \) about \( x = -2 \). Since derivatives higher than the third derivative are zero, this simplification effectively captures the full behavior of \( f(x) \) in this case.

Key Concepts

Power Series Expansion

Power series expansion allows us to represent a function as an infinite sum of terms in the form of a power series. It breaks down complex functions into easier, polynomial-like expressions that are simple to work with. In the context of a Taylor series, the power series is centered around a specific point, which in this exercise, is given as \(a = -2\).

• A function \(f(x)\) is expressed as \(f(x) = c_0 + c_1(x-a) + c_2(x-a)^2 + \ldots\).
• Each term has a coefficient derived from the derivatives of the function at the center \(a\).
• Such a series can approximate complex functions to varying degrees of accuracy.

The idea is to approximate \(f(x)\) in such a way that even without computing the entire infinite series, we can still have a very accurate representation of the original function.

Derivative Evaluation

Derivative evaluation is a key step in forming a Taylor series. The derivatives of the function at the center \(a\) are used to determine the coefficients of the series.

• First, determine the base function \(f(x) = x - x^3\).
• Calculate consecutive derivatives: the first derivative \(f'(x) = 1 - 3x^2\), the second \(f''(x) = -6x\), and the third \(f'''(x) = -6\).
• Evaluate each derivative at the particular center point \(x = -2\) to gain \(f(a)\), \(f'(a)\), \(f''(a)\), etc.

In our example: - \(f(-2) = 6\), - \(f'(-2) = -11\), - \(f''(-2) = 12\), - and \(f'''(-2) = -6\). These values are crucial because they directly become the coefficients in the series expansion, contributing to its accuracy.

Polynomial Approximation

Polynomial approximation uses derivatives to approximate a function by a polynomial, specifically around the center point \(a\). The Taylor polynomial is a finite sum that can approximate the original function. The degree of the polynomial directly affects the approximation's accuracy.

• Begin by substituting the evaluated derivatives into the Taylor series formula.
• Assemble terms with increasing powers of \((x-a)\).
• The polynomial is truncated after a certain degree to simplify the representation.

In this exercise, the polynomial derived from the Taylor series is:
\[T(x) = 6 - 11(x+2) + 6(x+2)^2 - (x+2)^3\] This polynomial functions as a close approximation to \(x - x^3\) around \(x = -2\), making complex calculations simpler and efficient.

Series Centering

Series centering is about selecting a point around which the Taylor series is expanded. This center, referred to as \(a\), affects the convergence and accuracy of the series within a region.

• The choice of \(a\) should be relevant; commonly points where function behavior changes or needs closer examination are chosen.
• The entire expansion depends on accurately evaluated derivatives at this point \(a\).
• For our specific function, being centered at \(a = -2\) shapes the entire convergence of the series in that immediate area.

This centered approach ensures that the generated polynomial is a true reflection of the function in the vicinity of \(a\). With functions having symmetric behavior, centering at key points maximizes the simplicity and accuracy of the approximation.