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Chapter 8: Q. 37 (page 692)

In Exercises 41鈥�48 in Section 8.2, you were asked to find the fourth Taylor polynomial $P_{4} (x)$ for the specified function and the given value of x 0. Here give Lagrange鈥檚 form for the remainder $R_{4} (x)$ .
$\cos x, \frac{蟺}{2}$

Short Answer

Expert verified

Ans: $R_{4} (x) = 鈭� \frac{\sin 鈦� c}{120} {(x 鈭� \frac{蟺}{2})}^{5}$

Step by step solution

01

Step 1. Given information.

given,

$\cos x, \frac{蟺}{2}$

02

Step 2. Consider the given function,

The derivatives of the function $f (x) = \cos x$ are

$\begin{aligned} f^{鈥�} (x) = \frac{d}{d x} [\cos 鈦� x] \\ = 鈭� \sin 鈦� x \end{aligned}$

Also,

$\begin{aligned} f^{鈥测赌�} (x) = \frac{d}{d x} [鈭� \sin 鈦� x] \\ = 鈭� \frac{d}{d x} [\sin 鈦� x] \\ = 鈭� \cos 鈦� x \end{aligned}$

Again

$\begin{aligned} f^{鈥测赌�'} (x) = 鈭� \frac{d}{d x} [\cos] \\ = 鈭� (鈭� \sin 鈦� x) \\ = \sin x \end{aligned}$

Also,

$\begin{aligned} f^{鈥测赌� 鈥测赌�} (x) = \frac{d}{d x} [\sin 鈦� x] \\ = \cos 鈦� x \end{aligned}$

Implies that

$f^{(4)} (x) = \cos x$

Finally,

$\begin{aligned} f^{(5)} (x) = \frac{d}{d x} [\cos 鈦� x] \\ = 鈭� \sin 鈦� x \end{aligned}$

03

Step 3. Now,

by the Lagrange's form for the remainder, if f is a function that can be differentiated n+1 times in some open interval / containing the point $x_{0}$ and $R_{n} (x)$ be the nth remainder for $f$ at $x = x_{0}$ . Then there exists at least one c between $x_{0}$ and x such that

$R_{n} (x) = \frac{f^{(n + 1)} (c)}{(n + 1)!} {(x 鈭� x_{0})}^{n + 1}$

Since $f^{(5)} (x) = 鈭� \sin 鈦� x and x_{0} = \frac{蟺}{2} then$

$R_{4} (x) = \frac{f^{5} (c)}{5!} {(x 鈭� \frac{蟺}{2})}^{5}$

That is,

$R_{4} (x) = 鈭� \frac{\sin 鈦� c}{120} {(x 鈭� \frac{蟺}{2})}^{5}$

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Most popular questions from this chapter

What is meant by the interval of convergence for a power series in $x$ ? How is the interval of convergence determined? If a power series in $x$ has a nontrivial interval of convergence, what types of intervals are possible?

The second-order differential equation

$x^{2} y^{''} + x y^{'} + (x^{2} - p^{2}) = 0$

where p is a non-negative integer, arises in many applications in physics and engineering, including one model for the vibration of a beaten drum. The solution of the differential equation is called the Bessel function of order p, denoted by $J_{p} (x)$ . It may be shown that $J_{p} (x)$ is given by the following power series in x :

$J_{p} (x) = {鈭�}_{k = 0}^{鈭�} \frac{(- 1)^{k}}{k! (k + p)! 2^{2 k + p}} x^{2 k + p}$

What is the interval of convergence for $J_{1} (x)$ ?

What is Lagrange鈥檚 form for the remainder? Why is Lagrange鈥檚 form usually more useful for analyzing the remainder than the definition of the remainder or the integral provided by Taylor theorem?

Prove that if the power series ${鈭�}_{k = 0}^{鈭�} 鈥� a_{k} x^{k}$ and ${鈭�}_{k = 0}^{鈭�} 鈥� a_{k} x^{2 k}$ have the same radius of convergence $p$ , then $p$ is $0, 1,$ or infinite.

What is a power series in x?

See all solutions

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