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Chapter 9: Problem 62

a. Use any analytical method to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. You do not need to use the definition of the Taylor series coefficients. b. Determine the radius of convergence of the series. $$f(x)=\cos 2 x+2 \sin x$$

Short Answer

Expert verified
To find the first four nonzero terms of the Taylor series for the function \(f(x)=\cos 2x + 2\sin x\), we first expanded the Taylor series for both \(\cos 2x\) and \(2\sin x\). Then, we added their respective series to create the Taylor series for the given function. After obtaining the first four nonzero terms, we found that \(f(x) \approx 1 - 2x^2 + 2x^3 + \frac{1}{3}x^4\). The radius of convergence for this Taylor series is infinite (鈭�) since it is the sum of the convergent series for the standard sin(x) and cos(x).

Step by step solution

01

Find Taylor series for \(\cos 2x\) and \(2\sin x\)

First, we need to find the Taylor series expansions for \(\cos 2x\) and \(2\sin x\). We can use the well-known expansions for sin(x) and cos(x): $$\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}x^{2n+1}$$ $$\cos x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}x^{2n}$$ For \(\cos 2x\), replace \(x\) with \(2x\) in the cos(x) expansion: $$\cos 2x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}(2x)^{2n}$$ For \(2\sin x\), multiply the sin(x) expansion by 2: $$2\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}(2x^{2n+1})$$
02

Add the series

As the given function is the sum of \(\cos 2x\) and \(2\sin x\), we can add the respective Taylor series: $$f(x) = \cos 2x + 2\sin x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}(2x)^{2n} + \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}(2x^{2n+1})$$
03

Find the first four nonzero terms

Now, let's find the first four nonzero terms of the Taylor series for f(x): When n=0: $$\frac{(-1)^0}{(2\cdot0)!}(2x)^{2\cdot0} = 1$$ $$\frac{(-1)^0}{(2\cdot0+1)!}(2x^{2\cdot0+1}) = 0$$ When n=1: $$\frac{(-1)^1}{(2\cdot1)!}(2x)^{2\cdot1} = -2x^2$$ $$\frac{(-1)^1}{(2\cdot1+1)!}(2x^{2\cdot1+1}) = 2x^3$$ When n=2: $$\frac{(-1)^2}{(2\cdot2)!}(2x)^{2\cdot2} = \frac{2^3}{4!}x^4 = \frac{1}{3}x^4$$ $$\frac{(-1)^2}{(2\cdot2+1)!}(2x^{2\cdot2+1}) = 0$$ When n=3: $$\frac{(-1)^3}{(2\cdot3)!}(2x)^{2\cdot3} = \frac{-2^5}{6!}x^6 = -\frac{2}{45}x^6$$ $$\frac{(-1)^3}{(2\cdot3+1)!}(2x^{2\cdot3+1}) = \frac{-2^7}{7!}x^7 = -\frac{2}{315}x^7$$ First four nonzero terms are obtained: $$f(x) \approx 1 - 2x^2 + 2x^3 + \frac{1}{3}x^4$$
04

Determine the radius of convergence

The radius of convergence for the standard sin(x) and cos(x) Taylor series is infinity. Therefore, since f(x) is the sum of these well-behaved series, the radius of convergence will be the smallest of these two which is 鈭�: $$\text{Radius of Convergence} = \infty$$

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Key Concepts

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

Radius of Convergence
When studying the Taylor series, a crucial concept to understand is the radius of convergence. It tells you how far out, from the center point of the series, you can go while still having the series sum up to the actual function. Essentially, it provides a boundary鈥攖he "safe zone"鈥攆or using your series approximation.
For functions such as sine and cosine, represented by the series expansions, this becomes even more interesting. These trigonometric functions have an infinite radius of convergence, which means their series representation is valid over the entire set of real numbers.
The infinite radius of convergence indicates that no matter how large the value of x, the series will be a reliable estimate of the function itself. This is one of the marvelous aspects of sine and cosine functions鈥攖heir predictability and stability across the real line. Thus, in analyzing the Taylor series of functions like \(\cos 2x\) and \(2\sin x\), the radius remains infinite, making them particularly useful in numerous mathematical applications.
Series Expansion
Series expansion is a powerful method in calculus that allows us to express a complex function as a sum of simpler polynomial terms. It transforms difficult or abstract functions into easier-to-manage sums of powers of x.
For instance, consider the functions \(\cos 2x\) and \(2\sin x\). The series expansion of these functions lets us break them down into a sequence of terms involving powers of x. This is done using the known series for \(\sin x\) and \(\cos x\). By altering these series slightly鈥攕ubstituting '2x' for 'x' in their expansions鈥攚e gain the required series for \(\cos 2x\). Similarly, multiplying the entire sine series by 2 gives us the expansion for \(2\sin x\).
  • For \(\cos 2x\): The expansion is derived from the cosine series by replacing x with 2x, resulting in terms like 1, \-2x^2\, \-\frac{2}{45}x^6\, and more.
  • For \(2\sin x\): The series is a straightforward amplification of the sine series, leading to terms like 0, \+2x^3\, and \-\frac{2}{315}x^7\.
Understanding these expansions simplifies the addition of these series, yielding a comprehensive representation of functions like \(f(x) = \cos 2x + 2\sin x\), which become manageable through their first few non-zero terms in a polynomial-like form.
Cosine and Sine Functions
The cosine and sine functions are foundational elements of trigonometry, and their behavior is crucial in both theoretical and applied mathematics. As periodic functions, they repeat their values over intervals of \(2\pi\) and have a wide range of applications from waves to oscillations.
Exploring these functions through their Taylor series gives us insights beyond their trigonometric identities. The Taylor series allows us to approximate these functions infinitely close by summing a sequence of terms, each a multiple of a power of x. For example:
  • The cosine function, \(\cos x\), can be expanded into a series starting with terms such as 1, \-\frac{x^2}{2!}\, \frac{x^4}{4!}\, and so on.
  • The sine function, \(\sin x\), has an alternate pattern where it begins with terms like \frac{x}{1!}\, \-\frac{x^3}{3!}\, \frac{x^5}{5!}\, and so forth.
These power series offer approximations that persist across all values of x due to their infinite radius of convergence. Particularly, when analyzing combinations like \(\cos 2x + 2\sin x\), understanding these expansions is crucial for accurate approximations and practical applications in fields ranging from engineering to physics to computer graphics.

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Most popular questions from this chapter

Assume that \(f\) has at least two continuous derivatives on an interval containing \(a\) with \(f^{\prime}(a)=0 .\) Use Taylor's Theorem to prove the following version of the Second Derivative Test. a. If \(f^{\prime \prime}(x)>0\) on some interval containing \(a,\) then \(f\) has a local minimum at \(a\) b. If \(f^{\prime \prime}(x)<0\) on some interval containing \(a,\) then \(f\) has a local maximum at \(a\)

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 that \(f\) 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 $$f(x)=f(a)+(x-a) f^{\prime}(a)+\int_{a}^{x}(x-t) f^{\prime \prime}(t) d t$$ c. Show that \(n\) integrations by parts gives $$ \begin{aligned} f(x)=& f(a)+\frac{f^{\prime}(a)}{1 !}(x-a)+\frac{f^{\prime \prime}(a)}{2 !}(x-a)^{2}+\cdots \\ &+\frac{f^{(n)}(a)}{n !}(x-a)^{n}+\underbrace{\int_{a}^{x} \frac{f^{(n+1)}(t)}{n !}(x-t)^{n} d t}_{R_{n}(x)} \end{aligned} $$ d. Challenge: The result in part (c) looks like \(f(x)=p_{n}(x)+\) \(R_{n}(x),\) where \(p_{n}\) is the \(n\) 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 \(x\)

Exponential function In Section 9.3, we show that the power series for the exponential function centered at 0 is $$e^{x}=\sum_{k=0}^{\infty} \frac{x^{k}}{k !}, \quad \text { for }-\infty< x <\infty$$ Use the methods of this section to find the power series for the following functions. Give the interval of convergence for the resulting series. $$f(x)=e^{2 x}$$

Choose a Taylor series and center point to approximate the following quantities with an error of \(10^{-4}\) or less. $$\sin (0.98 \pi)$$

Elliptic integrals The period of a pendulum is given by $$ T=4 \sqrt{\frac{\ell}{g}} \int_{0}^{\pi / 2} \frac{d \theta}{\sqrt{1-k^{2} \sin ^{2} \theta}}=4 \sqrt{\frac{\ell}{g}} F(k) $$ where \(\ell\) is the length of the pendulum, \(g \approx 9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, \(k=\sin \left(\theta_{0} / 2\right),\) and \(\theta_{0}\) is the initial angular displacement of the pendulum (in radians). The integral in this formula \(F(k)\) is called an elliptic integral, and it cannot be evaluated analytically. a. Approximate \(F(0.1)\) by expanding the integrand in a Taylor (binomial) series and integrating term by term. b. How many terms of the Taylor series do you suggest using to obtain an approximation to \(F(0.1)\) with an error less than \(10^{-3} ?\) c. Would you expect to use fewer or more terms (than in part (b)) to approximate \(F(0.2)\) to the same accuracy? Explain.
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