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Magazine Chapter 8: Problem 12
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 \(R_{n}(x) \rightarrow 0.1]\) $$f(x)=x-x^{3}, \quad a=-2$$
Short Answer
Expert verified
The Taylor series is \( 6 - 11(x+2) + 6(x+2)^2 - (x+2)^3 \).
Step by step solution
01
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 \( x = -2 \).
02
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 \( 0 \).
03
Evaluate derivatives at the center
Evaluate these derivatives at \( x = -2 \):- \( f(-2) = -2 - (-2)^3 = 6 \).- \( f'(-2) = 1 - 3(-2)^2 = -11 \).- \( f''(-2) = -6(-2) = 12 \).- \( f'''(-2) = -6 \).
04
Write the Taylor series formula
The Taylor series expansion of a function \( f(x) \) about \( x = a \) is given by the formula:\[T(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots\]
05
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:\[T(x) = 6 - 11(x+2) + 6(x+2)^2 - (x+2)^3\]
06
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.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.
\[T(x) = 6 - 11(x+2) + 6(x+2)^2 - (x+2)^3\] This polynomial functions as a close approximation to \(x - x^3\) around \(x = -2\), making complex calculations simpler and efficient.
- 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.
\[T(x) = 6 - 11(x+2) + 6(x+2)^2 - (x+2)^3\] 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.
