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Chapter 9: Problem 6

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) \(\cos 2 x\) (b) \(x^{2} e^{x}\) (c) \(x e^{-x}\) (d) \(\sin \left(x^{2}\right)\)

Short Answer

Expert verified
(a) \(1 - 2x^2 + \frac{2}{3}x^4 - \frac{1}{45}x^6\); (b) \(x^2 + x^3 + \frac{x^4}{2} + \frac{x^5}{6}\); (c) \(x - x^2 + \frac{x^3}{2} - \frac{x^4}{6}\); (d) \(x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}\); all with radius \(R = \infty\).

Step by step solution

01

Recall Maclaurin Series for Known Functions

The Maclaurin series for known functions are useful for substitution. For cosines and exponentials, we recall:- \[ \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots \]- \[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]- \[ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots \]
02

Step 2a: Substitute in Known Series for \(\cos 2x\)

Substitute \(2x\) for \(x\) in the known series for \(\cos(x)\):\[\cos(2x) = 1 - \frac{(2x)^2}{2!} + \frac{(2x)^4}{4!} - \frac{(2x)^6}{6!} + \ldots\]Simplifying the terms:- First term: \(1\)- Second term: \(-\frac{4x^2}{2} = -2x^2\)- Third term: \(\frac{16x^4}{24} = \frac{2}{3}x^4\)- Fourth term: \(-\frac{64x^6}{720} = -\frac{1}{45}x^6\)
03

Step 3a: Write the First Four Terms for \(\cos 2x\)

The first four nonzero terms of the Maclaurin series for \(\cos 2x\) are:\[\cos(2x) \approx 1 - 2x^2 + \frac{2}{3}x^4 - \frac{1}{45}x^6\]
04

Step 4a: Determine the Radius of Convergence of \(\cos 2x\)

The radius of convergence for the powers of cosine in terms of \(x\), \(R = \infty\), since cosine has a series expansion for all real numbers.
05

Step 2b: Substitute in Known Series for \(x^2 e^x\)

Start with the Maclaurin series for \(e^x\) and multiply each term by \(x^2\):\[x^2 e^x = x^2 \left( 1 + x + \frac{x^2}{2} + \frac{x^3}{6} + \ldots \right)\]Distributing \(x^2\):- First term: \(x^2\)- Second term: \(x^3\)- Third term: \(\frac{x^4}{2}\)- Fourth term: \(\frac{x^5}{6}\)
06

Step 3b: Write the First Four Terms for \(x^2 e^x\)

The first four nonzero terms are:\[x^2 e^x \approx x^2 + x^3 + \frac{x^4}{2} + \frac{x^5}{6}\]
07

Step 4b: Determine the Radius of Convergence of \(x^2 e^x\)

The radius of convergence for \(x^2 e^x\) is the same as for \(e^x\), which is \(R = \infty\).
08

Step 2c: Substitute in Known Series for \(x e^{-x}\)

Start with the Maclaurin series for \(e^{-x}\), negating \(x\):\[ e^{-x} = 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \ldots \]Multiply by \(x\):\[x e^{-x} = x \left( 1 - x + \frac{x^2}{2} - \frac{x^3}{6} + \ldots \right)\]Distributing \(x\):- First term: \(x\)- Second term: \(-x^2\)- Third term: \(\frac{x^3}{2}\)- Fourth term: \(-\frac{x^4}{6}\)
09

Step 3c: Write the First Four Terms for \(x e^{-x}\)

The first four nonzero terms are:\[x e^{-x} \approx x - x^2 + \frac{x^3}{2} - \frac{x^4}{6}\]
10

Step 4c: Determine the Radius of Convergence of \(x e^{-x}\)

The radius of convergence for \(x e^{-x}\) is the same as for \(e^x\), which is \(R = \infty\).
11

Step 2d: Substitute in Known Series for \(\sin x^2\)

Substitute \(x^2\) for \(x\) in the known series for \(\sin(x)\):\[\sin(x^2) = x^2 - \frac{(x^2)^3}{3!} + \frac{(x^2)^5}{5!} - \ldots \]Simplifying:- First term: \(x^2\)- Second term: \(-\frac{x^6}{6}\)- Third term: \(\frac{x^{10}}{120}\)- Fourth term: Further terms are not needed for the first four nonzero terms.
12

Step 3d: Write the First Four Terms for \(\sin x^2\)

The first four nonzero terms of \(\sin(x^2)\) are:\[\sin(x^2) \approx x^2 - \frac{x^6}{6} + \frac{x^{10}}{120}\]
13

Step 4d: Determine the Radius of Convergence of \(\sin x^2\)

The radius of convergence for \(\sin(x^2)\) is the same as for \(\sin(x)\), which is \(R = \infty\).

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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 is an important concept when dealing with series, especially in the context of Maclaurin series. It describes the set of values for which the series converges (or sums up to a finite value) around a particular point. For the types of functions common in this context, like cosine, sine, and exponentials, the Maclaurin series typically has an infinite radius of convergence. This means the series converges for all real numbers.

To understand radius of convergence from a practical perspective, think of it as the "zone of reliability" for the series. For instance:
  • Cosine and Sine Functions: The series for \(\cos(x)\) and \(\sin(x)\) converge for all real values of \(x\), giving them an infinite radius of convergence.
  • Exponential Functions: Similarly, \(e^x\) also converges for all real numbers, resulting in \(R = \infty\).
The unlimited radius of convergence for these functions makes the Maclaurin series incredibly useful for approximations in various mathematical applications.
Algebraic Operations
When working with Maclaurin series, algebraic operations are often necessary to adjust the series for specific functions. Understanding these operations is crucial for solving problems that require modifying known series. Essentially, this involves either substituting values or multiplying series terms.

For example:
  • Substitution: When given a series for a function like \(\cos(x)\), to find the series for \(\cos(2x)\), you replace \(x\) with \(2x\) throughout the series.
  • Multiplication: Consider a function such as \(x^2e^x\). To determine its series, you need to first expand \(e^x\) and then multiply each term of that series by \(x^2\).
These operations allow us to modify the Maclaurin series to fit a variety of function forms, making them versatile tools for problem-solving.
Series Expansion
Series expansion is the process of expressing a function as an infinite sum of terms. Each of these terms is calculated based on derivatives of the function evaluated at a single point, often zero for Maclaurin series. This infinite sum gives us a polynomial representation that can approximate the function.

The beauty of series expansion lies in:
  • Simplicity: It transforms complex functions into simple polynomials, making them easier to analyze and use in calculations.
  • Flexibility: By expanding functions, you can perform operations like differentiation or integration directly on the polynomial terms.
  • Approximation: Even a finite number of terms can provide a powerful approximation of a function for values close to the expansion point.
For instance, in solving the problem:- Expanding \(\cos(x)\) or \(e^x\) already gives series that can be adapted for more complex functions through substitution or multiplication.Understanding series expansions allows us to break down real-world problems into manageable pieces, especially when exact solutions are difficult to come by.

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Most popular questions from this chapter

Determine whether the series converges. $$\sum_{k=1}^{\infty} k^{2} \sin ^{2}\left(\frac{1}{k}\right)$$

The limit of an indeterminate form as \(x \rightarrow x_{0}\) can sometimes be found by expanding the functions involved in Taylor series about \(x=x_{0}\) and taking the limit of the series term by term. Use this method to find the limits in these exercises. (a) \(\lim _{x \rightarrow 0} \frac{\sin x}{x}\) (b) \(\lim _{x \rightarrow 0} \frac{\tan ^{-1} x-x}{x^{3}}\)

Determine whether the series converges. $$\sum_{k=1}^{\infty} \operatorname{sech}^{2} k$$

(a) We will see later that the polynomial \(1-x^{2} / 2\) is the "local quadratic" approximation for \(\cos x\) at \(x=0.\) Make a conjecture about the convergence of the series $$\sum_{k=1}^{\infty}\left[1-\cos \left(\frac{1}{k}\right)\right]$$ by considering this approximation. (b) Try to confirm your conjecture using the limit comparison test.

Suppose that the values of a function \(f\) and its first three derivatives at \(x=1\) are $$f(1)=2, \quad f^{\prime}(1)=-3, \quad f^{\prime \prime}(1)=0, \quad f^{\prime \prime \prime}(1)=6$$ Find as many Taylor polynomials for \(f\) as you can about \(x=1\)
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