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Chapter 11: Problem 13

Find the Maclaurin series for \(f(x)\) using the definition of a Maclaurin series. [Assume that \(f\) has a power series expansion. Do not show that \(\left.R_{n}(x) \rightarrow 0 .\right]\) Also find the associated radius of convergence. $$ f(x)=\cos x $$

Short Answer

Expert verified
The Maclaurin series for \( \cos x \) is \( \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \) with radius of convergence \( \infty \).

Step by step solution

01

Understand the Maclaurin Series

The Maclaurin series for a function \( f(x) \) is a power series expansion of \( f(x) \) around \( x = 0 \). It is given by the formula: \[ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n \]Here, \( f^{(n)}(0) \) represents the \( n \)-th derivative of the function evaluated at \( x = 0 \).
02

Calculate Derivatives of \( f(x) = \cos x \)

Calculate the first few derivatives of \( f(x) = \cos x \) and evaluate them at \( x = 0 \):- \( f(x) = \cos x \), \( f(0) = 1 \)- \( f'(x) = -\sin x \), \( f'(0) = 0 \)- \( f''(x) = -\cos x \), \( f''(0) = -1 \)- \( f'''(x) = \sin x \), \( f'''(0) = 0 \)- \( f^{(4)}(x) = \cos x \), \( f^{(4)}(0) = 1 \)Observe the pattern: derivatives at \( x = 0 \) repeat every four terms.
03

Write the Maclaurin Series

Substitute the evaluated derivatives into the Maclaurin series formula:\[ f(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \]This series represents \( \cos x \) around \( x = 0 \).
04

Determine the Radius of Convergence

The Maclaurin series for \( \cos x \) is similar to the exponential series based on even powers. The general form of the term is given by \( \frac{(-1)^n x^{2n}}{(2n)!} \). The series converges for all \( x \), meaning the radius of convergence is unlimited. Therefore, the radius of convergence is \( \infty \).

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Key Concepts

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

Radius of Convergence
The radius of convergence tells us within which interval a power series like the Maclaurin series effectively represents a function. For a general power series \( \sum_{n=0}^{\infty} a_n x^n \), this radius can be determined using the ratio test. However, in certain cases like the Maclaurin series of \( \cos x \) and other entire functions, the series converges for all real numbers. Here's why:
  • In the case of \( \cos x \), each term in the series diminishes rapidly since both the factorial in the denominator and the alternating sign lead the terms to have smaller magnitudes.
  • Thus, the radius of convergence \( R \) is infinite, meaning the series converges for every real number \( x \).
Understanding the radius of convergence helps not only in determining where a series represents a function but also in understanding where it converges and diverges.
Power Series Expansion
A power series expansion is a representation of a function as an infinite sum of terms based on powers of its variable, typically around a central point \( a \). The Maclaurin series is a specific case of a power series expansion where the expansion is centered at \( x = 0 \). The general formula is:\[f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n\]Here's why this is useful:
  • It allows complex functions to be expressed as sums of polynomial terms, which can be easier to work with.
  • The power series can provide approximate solutions by truncating the series to only a few terms, which is especially useful in numerical analysis.
For \( \cos x \), we used this tool to express it as \( \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \), showing the pattern of even-powered terms.
Derivatives of Trigonometric Functions
Finding derivatives is crucial when constructing a Maclaurin series because derivatives appear directly in the formula. For trigonometric functions like \( \cos x \), their derivatives symbolize a repetitive cycle:
  • The first derivative of \( \cos x \) is \( -\sin x \).
  • Continuing, the second derivative, \( -\cos x \), and the third, \( \sin x \), reveal a pattern.
  • Every four derivatives, the cycle repeats, returning back to \( \cos x \).
This cyclical pattern simplifies calculating higher-order derivatives, allowing us to see the full layout of the series and to recognize term repetition every four derivatives, which provides an added layer of understanding when evaluating such trigonometric functions.
Convergence of Series
The convergence of a series like the Maclaurin series determines the values of \( x \) where the series correctly represents the given function. It's important to establish whether a series converges, meaning its terms approach a fixed value as more terms are added.
  • The convergence behavior depends on the specific function and its expansion.
  • For \( \cos x \), known to have an infinite radius of convergence, the series converges over all real numbers.
  • The alternating signs and factorial growth ensure that each subsequent term becomes significantly smaller, tightening the series around the true function value.
This convergence property underlies much of calculus, allowing powerful approximations and understanding of functions through their series expansions.

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Most popular questions from this chapter

A uniformly charged disk has radius \(R\) and surface charge density \(\sigma\) as in the figure. The electric potential \(V\) at a point \(P\) at a distance \(d\) along the perpendicular central axis of the disk is \(V=2 \pi k_{*} \sigma(\sqrt{d^{2}+R^{2}}-d)\) where \(k_{x}\) is a constant (called Coulomb's constant). Show that $$ V=\frac{\pi k_{r} R^{2} \sigma}{d} \quad \text { for large } d $$

Use multiplication or division of power series to ind the first three nonzero terms in the Maclaurin series for each function. $$ y=(\arctan x)^{2} $$

Evaluate the indefinite integral as an infinite series. $$ \int \frac{\cos x-1}{x} d x $$

The resistivity \(\rho\) of a conducting wire is the reciprocal of the conductivity and is measured in units of ohm-meters \((\Omega-\mathrm{m}) .\) The resistivity of a given metal depends on the temperature according to the equation \(\rho(t)=\rho_{20} e^{\alpha(t-2 \omega)}\) where \(t\) is the temperature in \(^{\circ} \mathrm{C}\). There are tables that list the values of \(\alpha\) (called the temperature coefficient) and \(\rho_{20}\) (the resistivity at \(20^{\circ} \mathrm{C}\) ) for various metals. Except at very low temperatures, the resistivity varies almost linearly with temperature and so it is common to approximate the expression for \(\rho(t)\) by its first-or second-degree Taylor polynomial at \(t=20 .\) $$ \begin{array}{l}{\text { (a) Find expressions for these linear and quadratic }} \\ {\text { approximations. }} \\ {\text { (b) For copper, the tables give } \alpha=0.0039 /^{\circ} \mathrm{C} \text { and }} \\ {\rho_{20}=1.7 \times 10^{-8} \Omega \text { -m. Graph the resistivity of copper }} \\\ {\text { and the linear and quadratic approximations for }} \\ {-250^{\circ} \mathrm{C} \leqslant t \leqslant 1000^{\circ} \mathrm{C} \text { . }} \\\ {\text { (c) For what values of } t \text { does the linear approximation agree }} \\ {\text { with the exponential expression to within one percent? }}\end{array} $$

(a) By completing the square, show that $$ \int_{0}^{1 / 2} \frac{d x}{x^{2}-x+1}=\frac{\pi}{3 \sqrt{3}} $$ (b) By factoring \(x^{3}+1\) as a sum of cubes, rewrite the integral in part (a). Then express \(1 /\left(x^{3}+1\right)\) as the sum of a power series and use it to prove the following formula for \(\pi:\) $$ \pi=\frac{3 \sqrt{3}}{4} \sum_{n=0}^{\infty} \frac{(-1)^{n}}{8^{n}}\left(\frac{2}{3 n+1}+\frac{1}{3 n+2}\right) $$
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