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Chapter 9: Problem 44

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. $$ f(x)=\cos x ; p(x)=1-\frac{x^{2}}{2 !}+\frac{x^{4}}{4 !} $$

Short Answer

Expert verified
The interval is \( -0.405 \leq x \leq 0.405 \).

Step by step solution

01

Identify the Degree of the Approximating Polynomial

The polynomial given is \( p(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} \). This polynomial is the Taylor polynomial of degree 4 for \( \cos(x) \), as it includes terms up to \( x^4 \).
02

Recall the Remainder Estimation Theorem

The Remainder Estimation Theorem states that the remainder \( R_n(x) \) of the Taylor series, when approximating a function \( f(x) \) by a Taylor polynomial of degree \( n \), is: \[ |R_n(x)| \leq \frac{M|x-a|^{n+1}}{(n+1)!} \] where \( a \) is the center of the Taylor series, \( n \) is the degree of the polynomial, and \( M \) is an upper bound on the \( (n+1)^{th} \) derivative of the function on the interval \[ |x-a| \leq d \].
03

Calculate the Derivative and Determine M

The \(5^{th}\) derivative of \( \cos(x) \) is either \( \sin(x) \) or \( -\sin(x) \), both of which are bounded by 1 on the whole real line because the largest value \( \sin(x) \) can achieve is 1 and the smallest is -1. Therefore, \( M = 1 \).
04

Set Up the Remainder Inequality for Three Decimal Places

We need \( |R_4(x)| \leq 0.0005 \) for three decimal place accuracy. Substituting the known values, we get: \[ \frac{|x|^5}{5!} \leq 0.0005 \] Which simplifies to: \[ |x|^5 \leq 0.06 \]
05

Solve for the Interval of x

Solve \( |x|^5 \leq 0.06 \) to find the interval:\[ |x| \leq (0.06)^{1/5} \] Calculate \( (0.06)^{1/5} \approx 0.405 \). Therefore, the interval is \( -0.405 \leq x \leq 0.405 \).
06

Check by Graphing the Difference

Graph \( |f(x) - p(x)| \) over the interval \( -0.405 \leq x \leq 0.405 \). Ensure that \( |f(x) - p(x)| \leq 0.0005 \) throughout this interval. This confirms that the approximation is within the desired accuracy.

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Key Concepts

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

Taylor Polynomial
A Taylor Polynomial is an approximation of a function using a finite sum of terms derived from the function's derivatives at a certain point. It's a way to approximate more complex functions with a polynomial which is often easier to handle.
In our example, the Taylor polynomial we discussed is of degree 4, approximating the function \( f(x) = \cos(x) \) at \( x = 0 \). This polynomial is:
  • \( p(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} \)
The terms in the polynomial stem from the Taylor Series expansion for cosine. Because it is a polynomial of degree 4, it accurately represents the function up to the \( x^4 \) term, ignoring higher degree terms.
Using this Taylor Polynomial, we can approximate \( \cos(x) \) within a small interval around \( x = 0 \) efficiently.
Taylor Series
A Taylor Series is a way to express a function as an infinite sum of terms calculated from its derivatives at a specific point. The series offers a precise approximation of the function within a certain range.
For any function \( f(x) \), the Taylor Series about \( x = a \) is:
  • \( f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \)
In our specific case, the function is \( \cos(x) \) and the series is centered at \( x = 0 \). The estimation uses the beginning of this infinite series, up to the \( x^4 \) term, making it a Taylor Polynomial. This truncation simplifies calculations and gives us a manageable approximation that is useful for values around the center, ensuring precision within a given range, as determined by the Remainder Estimation Theorem.
Derivative
Derivatives are fundamental in creating Taylor Polynomials and Taylor Series. They measure how a function changes as its input changes. Each derivative provides essential data on the function's slope at a specific point.
The nth derivative is crucial in determining the terms of the Taylor Series. For example, when approximating \( \cos(x) \) with a 4th-degree Taylor Polynomial, the derivatives at \( x = 0 \) helped form each term in \( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} \).
  • The 1st derivative of \( \cos(x) \) is \(-\sin(x)\)
  • The 2nd derivative is \(-\cos(x)\)
  • The 3rd derivative is \(\sin(x)\)
  • The 4th derivative returns to \(\cos(x)\), and this cycle continues.
These cycles showcase how patterns in derivatives inform the makeup of Taylor Polynomials. Using derivatives helps us approximate and understand complex continuous functions accurately.

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Most popular questions from this chapter

Let \(\sum u_{k}\) be a series and define series \(\sum p_{k}\) and \(\sum q_{k}\) so that $$p_{k}=\left\\{\begin{array}{ll}{u_{k},} & {u_{k}>0} \\ {0,} & {u_{k} \leq 0}\end{array} \quad \text { and } \quad q_{k}=\left\\{\begin{aligned} 0, & u_{k} \geq 0 \\\\-u_{k}, & u_{k}<0 \end{aligned}\right.\right.$$ (a) Show that \(\sum u_{k}\) converges absolutely if and only if \(\sum p_{k}\) and \(\sum q_{k}\) both converge. (b) Show that if one of \(\sum p_{k}\) or \(\sum q_{k}\) converges and theother diverges, then \(\sum u_{k}\) diverges. (c) Show that if \(\sum u_{k}\) converges conditionally, then both $$\sum p_{k} \text { and } \sum q_{k} \text { diverge. }$$

Apply the divergence test and state what it tells you about the series. $$ \begin{array}{ll}{\text { (a) } \sum_{k=1}^{\infty} \frac{k}{e^{k}}} & {\text { (b) } \sum_{k=1}^{\infty} \ln k} \\ {\text { (c) } \sum_{k=1}^{\infty} \frac{1}{\sqrt{k}}} & {\text { (d) } \sum_{k=1}^{\infty} \frac{\sqrt{k}}{\sqrt{k}+3}}\end{array} $$

Confirm the derivative formula by differentiating the appropriate Maclaurin series term by term. $$ \text { (a) } \frac{d}{d x}[\cos x]=-\sin x \quad \text { (b) } \frac{d}{d x}[\ln (1+x)]=\frac{1}{1+x} $$

(a) Find the Maclaurin series for \(e^{x^{4}} .\) What is the radius of convergence? (b) Explain two different ways to use the Maclaurin series for \(e^{x^{4}}\) to find a series for \(x^{3} e^{x^{4}} .\) Confirm that both methods produce the same series.

Use the root test to find the interval of convergence of $$ \sum_{k=2}^{\infty} \frac{x^{k}}{(\ln k)^{k}} $$
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