Q. 41 P4(x)In Exercises 41鈥�48 find t... [FREE SOLUTION]

Chapter 8: Q. 41 (page 680)

$P_{4} (x)$ In Exercises 41鈥�48 find the fourth Taylor polynomial for the specified function and the given value of $x_{0}$ .
$\cos x, \frac{蟺}{2}$

Short Answer

Expert verified

Ans: The fourth Taylor polynomial for the specified function is $= - (x - \frac{蟺}{2}) + \frac{1}{6} {(x - \frac{蟺}{2})}^{3}$ .

Step by step solution

Step 1. Given information:

$f (x) = \cos x$

Step 2. The fourth Taylor polynomial:

Since for any function $f$ with a derivative of order 4 at $x = \frac{蟺}{2}$ , the fourth Taylor polynomial for $x = \frac{蟺}{2}$ is given by

$P_{4} (x) = f (\frac{蟺}{2}) + f^{'} (\frac{蟺}{2}) (x - \frac{蟺}{2}) + \frac{f^{''} (\frac{蟺}{2})}{2!} {(x - \frac{蟺}{2})}^{2} + \frac{f^{''} (\frac{蟺}{2})}{3!} {(x - \frac{蟺}{2})}^{3} + \frac{f^{'''} (\frac{蟺}{2})}{4!} {(x - \frac{蟺}{2})}^{4}$

Therefore, first, find the value of the function along with $f^{'} (x), f^{''} (x), f^{'''} (x)$ and $f^{''''} (x)$ at $x = \frac{蟺}{2}$

Step 3. Finding the fourth Taylor polynomial through derivative of order 4:

$T h u s, t h e v a l u e o f t h e t h e f u n c t i o n s t x = \frac{蟺}{2} i s f (\frac{蟺}{2}) = \cos (\frac{蟺}{2}) = 0 T h e d e r i v a t i v e s o f t h e f u n c t i o n f (x) = \cos (x) a r e f' (x) = \frac{d}{d x} [\cos x] = - \sin x S o, a t x = \frac{蟺}{2} f' (\frac{蟺}{2}) = - \sin (\frac{蟺}{2}) = - 1 A l s o, f'' (x) = \frac{d}{d x} [- \sin x] = - \cos x S o, a t x = \frac{蟺}{2} f' (\frac{蟺}{2}) = - \cos (\frac{蟺}{2}) = 0 A g a i n, f'''' (x) = \frac{d}{d x} [\sin x] = \cos x S o, a t x = \frac{蟺}{2} f' (\frac{蟺}{2}) = \cos (\frac{蟺}{2}) = 0$

Step 4. Substituting the derivative of order 4 in the fourth Taylor polynomial :

Therefore, the fourth Taylor polynomial for the function $f (x) = \cos x$ is

P_{4} (x) = 0 + - 1 路 (x - \frac{蟺}{2}) + \frac{0}{2!} {(x - \frac{蟺}{2})}^{2} + \frac{1}{3!} {(x - \frac{蟺}{2})}^{3} + \frac{0}{4!} {(x - \frac{蟺}{2})}^{4} = - (x - \frac{蟺}{2}) + \frac{1}{3!} {(x - \frac{蟺}{2})}^{3} = - (x - \frac{蟺}{2}) + \frac{1}{6} {(x - \frac{蟺}{2})}^{3}

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$P_{4} (x)$ In Exercises 41鈥�48 find the fourth Taylor polynomial for the specified function and the given value of $x_{0}$ .
$\cos x, \frac{蟺}{2}$

Short Answer

Step by step solution

Step 1. Given information:

Step 2. The fourth Taylor polynomial:

Step 3. Finding the fourth Taylor polynomial through derivative of order 4:

Step 4. Substituting the derivative of order 4 in the fourth Taylor polynomial :

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91影视

P4(x)In Exercises 41鈥�48 find the fourth Taylor polynomial for the specified function and the given value of x0.cosx,蟺2

Short Answer

Step by step solution

Step 1. Given information:

Step 2. The fourth Taylor polynomial:

Step 3. Finding the fourth Taylor polynomial through derivative of order 4:

Step 4. Substituting the derivative of order 4 in the fourth Taylor polynomial :

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Decision Maths

Applied Mathematics

Calculus

Geometry

Logic and Functions

Theoretical and Mathematical Physics

Study anywhere. Anytime. Across all devices.

$P_{4} (x)$ In Exercises 41鈥�48 find the fourth Taylor polynomial for the specified function and the given value of $x_{0}$ .
$\cos x, \frac{蟺}{2}$