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Use the remainder term to find a bound on the error in the following approximations on the given interval. Error bounds are not unique. $$\sqrt{1+x}=1+\frac{x}{2} \text {on } [-0.1,0.1]$$

Short Answer

Expert verified
Answer: The maximum error is \(\frac{3}{1600}\).

Step by step solution

01

Identify the function and its Taylor series

We're given a function \(\sqrt{1+x}\) and its Taylor series approximation as \(1+\frac{x}{2}\) on the interval \([-0.1, 0.1]\). This approximation corresponds to the first two terms of the Taylor series of the function.
02

Find the next term and the remainder term of the Taylor series

Using the binomial theorem, the square root can be expanded as: $$\sqrt{1+x} = \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{2k}}{2k \choose k} x^k$$ The approximation given includes terms up to the first power of x (i.e., the \(k = 0\) and \(k = 1\) terms). The next term (i.e., the \(k = 2\) term) is $$R_2(x) = \frac{(-1)^2}{2^{4}}{4 \choose 2} x^2 = \frac{3}{16}x^2$$ The error in the approximation is therefore bounded by the magnitude of this term: $$|E(x)| = |f(x) - (1 + \frac{x}{2})| \leq |R_2(x)| = \frac{3}{16}|x^2|$$
03

Find the maximum error on the given interval

To find the maximum error on the interval \([-0.1, 0.1]\), we need to maximize the error term \((\frac{3}{16}|x^2|)\) on that interval. Since the term is always nonnegative, the maximum occurs at the boundaries. In this case, both boundaries produce the same maximum error: $$\max_{[-0.1,0.1]} |E(x)| = \frac{3}{16}|-0.1|^2 = \frac{3}{16} \times 0.01 = \frac{3}{1600}$$
04

State the error bound

Based on our analysis, we can conclude that the error in approximating \(\sqrt{1+x}\) with \(1+\frac{x}{2}\) on the interval \([-0.1, 0.1]\) is bounded by \(\frac{3}{1600}\). In other words, $$|f(x) - (1 + \frac{x}{2})| \leq \frac{3}{1600} \text{ for } x \in [-0.1, 0.1]$$

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Key Concepts

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

Taylor Series Approximation
A Taylor series approximation is a way to represent a function as the sum of an infinite series of terms. These terms are derived from the function's derivatives at a single point. This approach is powerful because it allows us to estimate complex functions using simpler polynomial expressions.
In the context of the given exercise, we're working with the function \( \sqrt{1+x} \). Its Taylor series approximation on the interval \([-0.1, 0.1]\) is expressed as \( 1 + \frac{x}{2} \). This approximation includes just the first two terms of the series, which gives a simplified yet effective representation of the function near the point of expansion (usually \(x=0\)).
Key benefits of the Taylor series include the ability to:
  • Estimate functions for small values of \(x\) accurately
  • Simplify complex functions into polynomials
  • Provide insight into the behavior of functions near specific points
For many problems, considering only the first few terms of a Taylor series approximation is enough to achieve a satisfactory accuracy.
Remainder Term
The remainder term in a Taylor series is crucial in determining how accurate an approximation is. In essence, it represents the difference between the true function value and the value obtained from the polynomial approximation.
The exercise required finding the remainder term or error bound for the approximation \( \sqrt{1+x} = 1 + \frac{x}{2} \). This involves computing the next term in the series that was not included in the approximation, which is known as the remainder. For the given problem, this next term is \( R_2(x) = \frac{3}{16}x^2 \).
The error bound is then derived from the maximum possible value of the remainder term within the given interval \([-0.1, 0.1]\). Here, the maximum error is calculated as:
  • \( |E(x)| \leq \frac{3}{16}|x^2| \)
  • The maximum occurs at the boundaries: \( x = -0.1 \) or \( x = 0.1 \)
  • This results in a maximum error of \( \frac{3}{1600} \)
Understanding the remainder term is essential for assessing the reliability of Taylor series approximations in practical applications.
Binomial Theorem
The Binomial Theorem is a fundamental mathematical principle used to expand expressions that are raised to a power. It provides a formula for expanding expressions of the form \((a + b)^n\).
For functions like \( \sqrt{1+x} \), which can be thought of as \((1 + x)^{1/2}\), the binomial theorem helps us derive series expansions. More concretely, it allows us to express the square root of \(1 + x\) as a series:
  • \[ \sqrt{1+x} = \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{2k}} {2k \choose k} x^k \]
  • This series gives us a systematic way of obtaining both the coefficients of terms and their sequence in a polynomial representation
In the exercise, the binomial theorem assists us in finding the terms of the series beyond what is used in the initial approximation \((1 + \frac{x}{2})\). This understanding allows us to compute the remainder term necessary for a precise error bound assessment.
By leveraging the binomial theorem, one can explore a variety of situations where series expansions offer practical approximations in fields as diverse as engineering, physics, and computer science.

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Most popular questions from this chapter

Find power series representations centered at 0 for the following functions using known power series. $$f(x)=\frac{3}{3+x}$$

Use the methods of this section to find the power series centered at 0 for the following functions. Give the interval of convergence for the resulting series. $$f(x)=e^{2 x}$$

Bessel functions Bessel functions arise in the study of wave propagation in circular geometries (for example, waves on a circular drum head). They are conveniently defined as power series. One of an infinite family of Bessel functions is $$ J_{0}(x)=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{2^{2 k}(k !)^{2}} x^{2 k} $$ a. Write out the first four terms of \(J_{0}\) b. Find the radius and interval of convergence of the power series for \(J_{0}\) c. Differentiate \(J_{0}\) twice and show (by keeping terms through \(x^{6}\) ) that \(J_{0}\) satisfies the equation \(x^{2} y^{\prime \prime}(x)+x y^{\prime}(x)+x^{2} y(x)=0\)

Symmetry a. Use infinite series to show that \(\cos x\) is an even function. That is, show \(\cos (-x)=\cos x\) b. Use infinite series to show that \(\sin x\) is an odd function. That is, show \(\sin (-x)=-\sin x\)

Proof of Taylor's Theorem There are several proofs of Taylor's Theorem, which lead to various forms of the remainder. The following proof is instructive because it leads to two different forms of the remainder and it relies on the Fundamental Theorem of Calculus, integration by parts, and the Mean Value Theorem for Integrals. Assume \(f\) has at least \(n+1\) continuous derivatives on an interval containing \(a\). a. Show that the Fundamental Theorem of Calculus can be written in the form $$f(x)=f(a)+\int_{a}^{x} f^{\prime}(t) d t$$ b. Use integration by parts \(\left(u=f^{\prime}(t), d v=d t\right)\) to show that \right. $$f(x)=f(a)+(x-a) f^{\prime}(a)+\int_{a}^{x}(x-t) f^{\prime \prime}(t) d t$$ c. Show that \(n\) integrations by parts give $$\begin{aligned} f(x)=& f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots \\ &+\frac{f^{(n)}(a)}{n !}(x-a)^{n}+\underbrace{\int_{a}^{x} \frac{f^{(n+1)}(t)}{n !}(x-t)^{n} d t}_{R_{n}(x)} \end{aligned}$$ d. Challenge: The result in part (c) has the form \(f(x)=p_{n}(x)+R_{n}(x),\) where \(p_{n}\) is the \(m\) th-order Taylor polynomial and \(R_{n}\) is a new form of the remainder, known as the integral form of the remainder. Use the Mean Value Theorem for Integrals (Section 5.4 ) to show that \(R_{n}\) can be expressed in the form $$R_{n}(x)=\frac{f^{(n+1)}(c)}{(n+1) !}(x-a)^{n+1}$$ where \(c\) is between \(a\) and \(x\)

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