Find the T, polynomial of a Taylor series polynomial for f(x) = cos(x)centered at a =Recall: the definition of a Taylor series of the function f centered at a is" (a),S (x) = ÷(x-a)" .п!On the same set of axes, sketch a graph of:1. The Taylor polynomial2. The function f(x) = cos(x) and373. Label the pointcos4

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Find the T, polynomial of a Taylor series polynomial for f(x) = cos(x) centered at a == Recall: the definition of a Taylor series of the function f centered at a is S(x) = * (@)(x-a)" . '(a), n! On the same set of axes, sketch a graph of: 1. The Taylor polynomial 2. The function f(x) = cos(x) and

Find the T, polynomial of a Taylor series polynomial for f(x) = cos(x) centered at a == Recall: the definition of a Taylor series of the function f centered at a is S(x) = * (@)(x-a)" . '(a), n! On the same set of axes, sketch a graph of: 1. The Taylor polynomial 2. The function f(x) = cos(x) and

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter10: Sequences, Series, And Probability

Section10.1: Infinite Sequences And Summation Notation

Problem 72E

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Question

Find the T, polynomial of a Taylor series polynomial for f(x) = cos(x)
centered at a =
Recall: the definition of a Taylor series of the function f centered at a is
" (a),
S (x) = ÷
(x-a)" .
п!
On the same set of axes, sketch a graph of:
1. The Taylor polynomial
2. The function f(x) = cos(x) and
37
3. Label the point
cos
4

Expert Solution

Step 1

Given :

The function $f (x) = \cos x$ at $x = a$ centered at $a = \frac{3 π}{4}$ .

To find: The $T_{3}$ polynomial of a Taylor series for $f (x) = \cos x$ at $a = \frac{3 π}{4}$ .

Solution :

The Taylor series for $f (x)$ centered at $x = a$ is given as,

$f (x) = \sum_{n = 0}^{\infty} \frac{f^{n} (a) {(x - a)}^{n}}{n!}$

Now, Find $T_{3}$ polynomial for $f (x)$ ,

Step 2

$\begin{array}{rcl} T_{3} & = & \sum_{n = 0}^{3} \frac{f^{n} (a) {(x - a)}^{n}}{n!} \\ T_{3} & = & f (a) + f^{1} (a) (x - a) + \frac{f^{2} (a) {(x - a)}^{2}}{2!} + \frac{f^{3} (a) {(x - a)}^{3}}{3!} \dots (1) \end{array}$

Since, $f (x) = \cos x$ ,

$f^{1} (x) = - \sin x$
$f^{2} (x) = - \cos x$ and
$f^{3} (x) = \sin x$

Also, the point $a$ is $\frac{3 π}{4}$ ,

Therefore, $f (\frac{3 π}{4}) = \cos \frac{3 π}{4} = - \frac{1}{\sqrt{2}}$

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