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Chapter 9: Problem 7

Find the first four nonzero terms of the Maclaurin series for the function by making an appropriate substitution in a known Maclaurin series and performing any algebraic operations that are required. State the radius of convergence of the series. (a) \(\frac{x^{2}}{1+3 x}\) (b) \(x \sinh 2 x\) (c) \(x\left(1-x^{2}\right)^{3 / 2}\)

Short Answer

Expert verified
(a) Terms: \(x^2 - 3x^3 + 9x^4 - 27x^5\), Radius: \(\frac{1}{3}\) (b) Terms: \(2x^2 + \frac{4x^4}{3}\), Radius: Infinite (c) Terms: \(x - \frac{3}{2}x^3\), Radius: 1

Step by step solution

01

Identify the Known Maclaurin Series

Let's start by identifying the Maclaurin series for functions that resemble parts of the given function.For part (a), we look at the geometric series \( \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots \) for \(|x| < 1\). This suggests substituting \( 3x \) for \( x \) to find the series for \( \frac{1}{1+3x}\).For part (b), the Maclaurin series for \( \sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots \). Substitute \( 2x \) for \( x \) to find \( \sinh(2x)\).For part (c), the function \( (1-x^2)^{3/2} \) resembles a binomial series expansion.
02

Substitute and Simplify for Part (a)

For the function \( f(x) = \frac{x^2}{1+3x} \):First, substitute \( 3x \) in the geometric series.\[ \frac{1}{1+3x} = 1 - 3x + (3x)^2 - (3x)^3 + \ldots = 1 - 3x + 9x^2 - 27x^3 + \ldots \]Now multiply each term by \( x^2 \):\[ x^2 - 3x^3 + 9x^4 - 27x^5 + \ldots \]The first four nonzero terms: \( x^2 - 3x^3 + 9x^4 - 27x^5 \).Radius of convergence: \( \frac{1}{3} \) from where the original series \( |3x| < 1 \).
03

Substitute and Simplify for Part (b)

For the function \( g(x) = x \sinh(2x) \):Start with:\[ \sinh(2x) = 2x + \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} + \ldots = 2x + \frac{8x^3}{6} + \frac{32x^5}{120} + \ldots = 2x + \frac{4x^3}{3} + \frac{4x^5}{15} + \ldots \]Now multiply each term by \( x \):\[ 2x^2 + \frac{4x^4}{3} + \frac{4x^6}{15} + \ldots \]The first four nonzero terms: \( 2x^2 + \frac{4x^4}{3} + \ldots \) (collecting only terms up to the fourth power).Radius of convergence: Infinite, because \( \sinh(x) \) converges for all \( x \).
04

Expand and Simplify for Part (c)

For the function \( h(x) = x(1-x^2)^{3/2} \), use the binomial series:The binomial series for \( (1-x)^{n} \) is:\[ (1-x)^{n} = 1 + nx + \frac{n(n-1)x^2}{2!} + \ldots \]Apply \( n = \frac{3}{2} \) and \( x \to x^2 \):\[ (1-x^2)^{3/2} = 1 - \frac{3}{2}x^2 + \frac{3 \cdot 1}{2 \cdot 2}x^4 + \ldots = 1 - \frac{3}{2}x^2 + \frac{3}{8}x^4 + \ldots \]Now multiply by \( x \):\[ x - \frac{3}{2}x^3 + \frac{3}{8}x^5 + \ldots \]The first four nonzero terms: \( x - \frac{3}{2}x^3 \).Radius of convergence: 1, as determined from the binomial pattern \(|x^2| < 1\).

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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 determines the interval in which a series converges. To find the radius of convergence for a power series centered at zero, analyze where the series sums to a finite number.
  • For part (a) of our problem, the series derived from the geometric series \( \frac{1}{1-x} \) has a radius of convergence where \( |3x| < 1 \), simplifying to \( |x| < \frac{1}{3} \). This means the series converges when the absolute value of \( x \) is less than one-third.
  • For part (b), the series \( \sinh(2x) \) is calculated using the hyperbolic sine function, which converges for all real values of \( x \). Therefore, it has an infinite radius of convergence.
  • For part (c), related to the binomial series, the radius of convergence is 1, obtained from the inequality \( |x^2| < 1 \).
Understanding the radius is crucial when determining where you can confidently use your series approximation without running into errors.
Geometric Series
The geometric series is a fundamental tool in calculus. It forms the backbone for numerous applications and series expansions. A basic geometric series has the form:\[ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots \]The series converges when \( |x| < 1 \). In our problem:
  • For part (a), we substitute \( 3x \) for \( x \) to create the series for \( \frac{1}{1+3x} \). After the substitution, the series becomes \( 1 - 3x + 9x^2 - 27x^3 + \ldots \).
Geometric series make complex functions simpler by converting them into an infinite sum of more manageable terms. This is particularly useful when applying operations like multiplication or division in series form.
Binomial Series
The binomial series represents expressions of the form \( (1-x)^n \) when \( n \) is not necessarily an integer. This series is given by:\[ (1-x)^n = 1 + nx + \frac{n(n-1)x^2}{2!} + \frac{n(n-1)(n-2)x^3}{3!} + \ldots \]It's essential for expanding binomials that involve fractional powers. For part (c) of our problem:
  • We have \( (1-x^2)^{3/2} \), which uses the binomial series for this non-integer exponent. Substituting \( x^2 \) for \( x \) gives \( 1 - \frac{3}{2}x^2 + \frac{3}{8}x^4 + \ldots \).
Multiplying this result by \( x \), the expression becomes \( x - \frac{3}{2}x^3 + \frac{3}{8}x^5 + \ldots \), showing the first few terms needed for precise approximations. Binomial expansions serve as a pivotal component in constructing power series expansions.
Hyperbolic Functions
Hyperbolic functions are analogs of trigonometric functions and often denoted similarly. They appear frequently in calculus and are defined using exponential functions. Here are the basics:
  • The hyperbolic sine function is given by \( \sinh(x) = \frac{e^x - e^{-x}}{2} \). Its Maclaurin series expansion is: \( x + \frac{x^3}{3!} + \frac{x^5}{5!} + \ldots \).
  • For part (b), this function is used in \( x \sinh(2x) \), where substituting \( 2x \) yields \( 2x + \frac{4x^3}{3} + \ldots \).
Hyperbolic functions are crucial when solving problems involving hyperbolic angles or areas. They extend the ideas of trigonometric functions into the realm of imaginary numbers and are essential tools in analysis and series expansions.

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Most popular questions from this chapter

(a) Use the Maclaurin series for \(1 /(1-x)\) to find the Maclaurin series for $$f(x)=\frac{x}{1-x^{2}}$$ (b) Use the Maclaurin series obtained in part (a) to find \(f^{(5)}(0)\) and \(f^{(6)}(0)\) (c) What can you say about the value of \(f^{(n)}(0) ?\)

Exercise will show how a partial sum can be used to obtain upper and lower bounds on the sum of a series when the hypotheses of the integral test are satisfied. This result will be needed in Exercises. It was stated in Exercise 35 that $$\sum_{k=1}^{\infty} \frac{1}{k^{4}}=\frac{\pi^{4}}{90}$$ (a) Let \(s_{n}\) be the \(n\) th partial sum of the series above. Show that $$ s_{n}+\frac{1}{3(n+1)^{3}} < \frac{\pi^{4}}{90} < s_{n}+\frac{1}{3 n^{3}} $$ (b) We can use a partial sum of the series to approximate \(\pi^{4} / 90\) to three decimal-place accuracy by capturing the sum of the series in an interval of length 0.001 (or less). Find the smallest value of \(n\) such that the interval containing \(\pi^{4} / 90\) in part (a) has a length of 0.001 or less. (c) Approximate \(\pi^{4} / 90\) to three decimal places using the midpoint of an interval of width at most 0.001 that contains the sum of the series. Use a calculating utility to confirm that your answer is within 0.0005 of \(\pi^{4} / 90\).

Find examples to show that if the series \(\sum u_{k}\) and \(\sum v_{k}\) both diverge, then the series \(\sum\left(u_{k}+v_{k}\right)\) and \(\sum\left(u_{k}-v_{k}\right)\) may either converge or diverge.

Use the Remainder Estimation Theorem to find an interval containing \(x=0\) over which \(f(x)\) can be approximated by \(p(x)\) to three decimal-place accuracy throughout the interval. Check your answer by graphing \(|f(x)-p(x)|\) over the interval you obtained. $$f(x)=\sin x ; p(x)=x-\frac{x^{3}}{3 !}$$

Find the Maclaurin polynomials of orders \(n=0,1,2,3\) and \(4,\) and then find the \(n\) th Maclaurin polynomials for the function in sigma notation. $$x e^{x}$$
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