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Chapter 10: Problem 23

Find the Taylor series generated by \(f\) at \(x=a.\) \(f(x)=x^{3}-2 x+4, \quad a=2\)

Short Answer

Expert verified
The Taylor series is \( f(x) = 8 + 10(x-2) + 6(x-2)^2 + (x-2)^3 \).

Step by step solution

01

Introduction to Taylor Series

A Taylor series generated by a function \( f \) at a point \( a \) is an infinite sum that represents \( f(x) \) as \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \ldots \). The goal is to calculate each derivative at \( a \) and evaluate the series.
02

Find \( f(a) \)

The first term in the Taylor series is the value of the function at \( a \). Here, \( f(x) = x^3 - 2x + 4 \), so \( f(2) = 2^3 - 2(2) + 4 = 8 - 4 + 4 = 8 \).
03

First Derivative \( f'(x) \)

Calculate the first derivative of \( f(x) = x^3 - 2x + 4 \). The derivative is \( f'(x) = 3x^2 - 2 \). Evaluate this at \( x = 2 \), so \( f'(2) = 3(2)^2 - 2 = 12 - 2 = 10 \). Thus, the term is \( 10(x-2) \).
04

Second Derivative \( f''(x) \)

Calculate the second derivative \( f''(x) = 6x \). Evaluate this at \( x = 2 \), so \( f''(2) = 6(2) = 12 \). Thus, the term is \( \frac{12}{2!}(x-2)^2 = 6(x-2)^2 \).
05

Third Derivative \( f'''(x) \)

Calculate the third derivative \( f'''(x) = 6 \), which is constant. Evaluate at \( x = 2 \), so \( f'''(2) = 6 \). Thus, the term is \( \frac{6}{3!}(x-2)^3 = 1(x-2)^3 \).
06

Construct The Taylor Series

Combine these results to construct the Taylor series: \[ f(x) = 8 + 10(x-2) + 6(x-2)^2 + (x-2)^3 \]. This represents the Taylor series generated by \( f \) at \( x=2 \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Derivative
Derivatives help in understanding how a function changes. They are fundamental for constructing Taylor series. In a Taylor series, we use derivatives to approximate a function as a polynomial.
  • First Derivative: Tells the slope or rate of change of the function. For the function \(f(x) = x^3 - 2x + 4\), the first derivative is \(f'(x) = 3x^2 - 2\), indicating how steep the function is at any point \(x\).
  • Second Derivative: Reflects how the slope itself is changing. The second derivative \(f''(x) = 6x\) for our example shows the curvature of the function. A positive second derivative means the function is curving upwards.
  • Higher Derivatives: Can be generalized as the slope of the slope, and so on. The third derivative \(f'''(x) = 6\) here is constant, meaning changes in the slope are consistent.
Derivatives give us the coefficients for the Taylor series expansion, allowing precise approximation of the function near a point \(a\). When calculated at \(x = a\), they form key parts of the Taylor polynomial.
Polynomial
Polynomials are expressions made up of variables and coefficients, combined using addition, subtraction, multiplication, and non-negative integer exponents.In the Taylor series context, a function can often be represented as a polynomial around a certain point \(a\). For the function \(f(x) = x^3 - 2x + 4\), it is initially a polynomial, which makes it directly suitable for series expansion.
  • Coefficients: These are determined by the derivatives of the function at \(x = a\). For example, the constant term in the Taylor polynomial comes from \(f(a)\), the first derivative gives the linear term coefficient, and so on.
  • Degree: The degree of the polynomial reflects the highest power of \(x\). In Taylor series, the degree is expanded period-wise with higher derivatives contributing further terms.
  • Flexibility: Polynomials have a straightforward structure, making them easy to evaluate and differentiate multiple times.
Using Taylor polynomials, complex functions can be approximated by simpler polynomials making calculations easier and intuitive especially near the base point \(a\).
Series Expansion
Series expansion is a method of expressing a function as an infinite sum of terms. The Taylor series is one of the primary ways to achieve this.
  • Infinite Sum: Although it is termed as 'infinite', often we only use a finite number of terms for practical calculations, especially if those terms efficiently estimate the function around a point.
  • Approximation: Taylor series allows us to approximate functions with polynomial expressions by utilizing their derivatives at a specific point \(a\). This is helpful for calculative convenience, especially near \(a\).
  • Convergence: For a Taylor series to effectively represent a function, it must converge, meaning the sum of terms should approach the function value as more terms are included.
In this exercise, the Taylor series expansion of \(f(x) = x^3 - 2x + 4\) at \(x=2\) results in a polynomial sum: \[ f(x) = 8 + 10(x-2) + 6(x-2)^2 + (x-2)^3 \]. This series, with its finite terms, provides a flexible approximation of the function within a range around \(x = 2\).

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Most popular questions from this chapter

In Exercises \(25-28\) , find a polynomial that will approximate \(F(x)\) throughout the given interval with an error of magnitude less than \(10^{-3} .\) \begin{equation} F(x)=\int_{0}^{x} \sin t^{2} d t, \quad[0,1] \end{equation}

Use series to approximate the values of the integrals in Exercises \(19-22\) with an error of magnitude less than \(10^{-8}\) . \begin{equation} \int_{0}^{0.1} \frac{\sin x}{x} d x \end{equation}

Show that if \(\sum_{n=1}^{\infty} a_{n}\) and \(\sum_{n=1}^{\infty} b_{n}\) both converge absolutely, then so do the following. $$\begin{array}{ll}{\text { a. }} & {\sum_{n=1}^{\infty}\left(a_{n}+b_{n}\right)} & {\text { b. } \sum_{n=1}^{\infty}\left(a_{n}-b_{n}\right)} \\ {\text { c. }} & {\sum_{n=1}^{\infty} k a_{n}(k \text { any number })}\end{array}$$

Show that \(\sum_{n=1}^{\infty} 2^{\left(n^{2}\right)} / n !\) diverges. Recall from the Laws of Exponents that \(2^{\left(n^{2}\right)}=\left(2^{n}\right)^{n}\)

Use series to evaluate the limits. \begin{equation} \lim _{t \rightarrow 0} \frac{1-\cos t-\left(t^{2} / 2\right)}{t^{4}} \end{equation}
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