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What is the minimum order of the Taylor polynomial required to approximate the following quantities with an absolute error no greater than \(10^{-3} ?\) (The answer depends on your choice of a center.) $$\cos (-0.25)$$

Short Answer

Expert verified
Answer: The minimum order of the Taylor polynomial required is 3.

Step by step solution

01

Identify the Taylor expansion for cosine function around the center (0)

The Taylor expansion for the cosine function \(\cos(x)\) around the center 0 (Maclaurin series) is given by: $$\cos(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots$$
02

Calculate the error bound using Taylor's theorem

According to Taylor's theorem, the error bound, \(R_n(x)\), for a Taylor polynomial is given by: $$|R_n(x)| = \frac{M}{(n+1)!}\, |x|^{n+1}$$ Where \(M\) is the maximum value of the absolute value of the \((n+1)\)-st derivative of the function on the interval containing the center (0 in our case) and the expansion point \(x=-0.25\). Since we have the cosine function, all of its derivatives are also bounded, with a maximum absolute value of 1. Now, we need to find the minimum order \(n\) for which the error bound is less than \(10^{-3}\). That is, we need to find the smallest integer \(n\) such that: $$\frac{1}{(n+1)!}\, |-0.25|^{n+1} < 10^{-3}$$
03

Find the minimum order

Now, we can start checking for different values of \(n\) in order to find the smallest integer that satisfies the error bound: \begin{align*} n=0: \quad &\frac{1}{(0+1)!}\, |-0.25|^{0+1} &= 0.25\\ n=1: \quad &\frac{1}{(1+1)!}\, |-0.25|^{1+1} &= 0.015625\\ n=2: \quad &\frac{1}{(2+1)!}\, |-0.25|^{2+1} &= 0.00260417\\ n=3: \quad &\frac{1}{(3+1)!}\, |-0.25|^{3+1} &= 0.00026042 \end{align*} It is evident that the error bound is less than \(10^{-3}\) when \(n=3\). Thus, the minimum order of the Taylor polynomial required to approximate \(\cos(-0.25)\) with an absolute error no greater than \(10^{-3}\) is 3.

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Key Concepts

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

Maclaurin series
The Maclaurin series is a special type of Taylor series that expands a function into an infinite sum of terms calculated from the derivatives of the function at a single point, specifically at zero. This series can approximate functions to a high degree of accuracy within the radius of convergence. For many functions, including the cosine function which we see in our exercise, the Maclaurin series offers a straightforward and familiar form.

For instance, the Maclaurin series for \(\cos(x)\) is expressed as:\[\cos(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n}}{(2n)!} = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \cdots\]Each term in the series involves higher even-powered derivatives of the cosine function evaluated at zero. For our exercise, students need to understand that by using enough terms from this series, one can approximate \(\cos(x)\) for small values of \(x\) with a high degree of accuracy, which is key to solving the problem.
Cosine function
The cosine function is an even, periodic function that is fundamental in trigonometry and appears in many different areas of mathematics and physics. Its wave-like properties and symmetry make it important in modeling oscillations and waves. The graph of \(\cos(x)\) shows a peak at \(x=0\) and troughs at \(x=\pm\pi\), repeating every \(2\pi\) radians.

Using the Maclaurin series, the cosine function can be approximated by polynomials, and the more terms are included, the closer the polynomial will resemble the actual cosine curve near \(x=0\). However, as \(x\) gets farther from zero, more terms are generally needed to maintain an accurate approximation. The approximation of \(\cos(x)\) with a Taylor polynomial is especially useful for small \(x\), such as the \(\cos(-0.25)\) value from our exercise.
Error bound
The error bound in the context of Taylor and Maclaurin series is the maximum distance, or 'error', that the polynomial approximation is from the actual function value. It offers students a quantitative measure of how accurate their approximation is. According to Taylor's theorem, the remainder or error term for a Taylor polynomial is denoted by \(R_n(x)\).

The error bound for the \(n\)-th order Taylor polynomial is given by:\[|R_n(x)| = \frac{M}{(n+1)!} |x|^{n+1}\]where \(M\) is the maximum value of the absolute value of the \(n+1\)-st derivative of the function over the interval of interest. For the cosine function, all its derivatives are bounded by 1. To keep the approximation error within \(10^{-3}\), one needs to find \(n\) such that the above expression is less than \(10^{-3}\). By testing successive values of \(n\), we determined from our exercise that a third-order polynomial is needed to achieve the desired precision for \(\cos(-0.25)\).
Taylor's theorem
Taylor's theorem is a cornerstone of calculus that provides the foundation for polynomial approximation of functions. The theorem tells us that a smooth function can be represented by a Taylor polynomial plus a remainder term that becomes smaller as the polynomial's degree – or the number of terms – increases.

The formula involves derivatives up to a certain order and powers of \(x\) minus the center of the expansion. In our exercise where we're approximating \(\cos(-0.25)\), using Taylor's theorem helps us to understand how many terms we should keep to achieve a given level of accuracy. The theorem also gives the error bound formula used to ensure that the Taylor polynomial of a certain order will approximate the function within the specified error margin. For teaching, emphasize how this powerful theorem bridges the gap between theoretical functions and practical polynomial approximations, which are much easier to use and compute.

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Most popular questions from this chapter

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 \(x\)

Use composition of series to find the first three terms of the Maclaurin series for the following functions. a. \(e^{\sin x}\) b. \(e^{\tan x} \quad\) c. \(\sqrt{1+\sin ^{2} x}\)

Approximations with Taylor polynomials a. Approximate the given quantities using Taylor polynomials with \(n=3\) b. Compute the absolute error in the approximation, assuming the exact value is given by a calculator. $$e^{0.12}$$

Use the power series representation $$f(x)=\ln (1-x)=-\sum_{k=1}^{\infty} \frac{x^{k}}{k}, \quad \text { for }-1 \leq x<1$$ to find the power series for the following functions (centered at 0 ). Give the interval of comvergence of the new series. $$p(x)=2 x^{6} \ln (1-x)$$

What is the minimum order of the Taylor polynomial required to approximate the following quantities with an absolute error no greater than \(10^{-3} ?\) (The answer depends on your choice of a center.) $$e^{-0.5}$$

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