Next, evaluate the derivatives at the center x = f(x) = cos(x) = f(n/4) = 2 √2 f'(x)=-sin(x) = f'(n/4) = f(x) = cos(x) = f(n/4) = 22 22 √2 f(x)=sin(x) => ""(n/4) = 22 (4)(x) = cos(x) -> (4)(π/4) √2 2 Π 1 √2 1 1 √2 Step 3 Substitute the derivative values in the Taylor series centered at x= cos(x)= XT 2 πT and simplify. (Hint: Note that (-1)(n+1)/2 = 1, −1, −1, 1, 1, -1, -1, 1 ...) 4 7) - √2 (1x-(1/4)]²) + √2 (Ex - (n/4)]³) 3! [x-(n/4)]4 = -ΣΟ 7=0 Submit Skip (you cannot come back) Submit Answer DETAILS LARCALC12 9.10.009.
Next, evaluate the derivatives at the center x = f(x) = cos(x) = f(n/4) = 2 √2 f'(x)=-sin(x) = f'(n/4) = f(x) = cos(x) = f(n/4) = 22 22 √2 f(x)=sin(x) => ""(n/4) = 22 (4)(x) = cos(x) -> (4)(π/4) √2 2 Π 1 √2 1 1 √2 Step 3 Substitute the derivative values in the Taylor series centered at x= cos(x)= XT 2 πT and simplify. (Hint: Note that (-1)(n+1)/2 = 1, −1, −1, 1, 1, -1, -1, 1 ...) 4 7) - √2 (1x-(1/4)]²) + √2 (Ex - (n/4)]³) 3! [x-(n/4)]4 = -ΣΟ 7=0 Submit Skip (you cannot come back) Submit Answer DETAILS LARCALC12 9.10.009.
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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Question
![Next, evaluate the derivatives at the center x =
f(x) = cos(x)
= f(n/4)
=
2
√2
f'(x)=-sin(x) = f'(n/4)
=
f(x) = cos(x) = f(n/4) =
22 22
√2
f(x)=sin(x) => ""(n/4) =
22
(4)(x)
= cos(x) ->
(4)(π/4)
√2
2
Π
1
√2
1
1
√2
Step 3
Substitute the derivative values in the Taylor series centered at x=
cos(x)=
XT
2
πT and simplify. (Hint: Note that (-1)(n+1)/2 = 1, −1, −1, 1, 1, -1, -1, 1 ...)
4
7) - √2 (1x-(1/4)]²) + √2 (Ex - (n/4)]³)
3!
[x-(n/4)]4
=
-ΣΟ
7=0
Submit
Skip (you cannot come back)
Submit Answer
DETAILS
LARCALC12 9.10.009.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F87ea163f-98f9-4b5a-bc49-88ec1dbff3d5%2Fd17eb696-27c9-4701-be34-07cf19adcb02%2Fldvzaqx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Next, evaluate the derivatives at the center x =
f(x) = cos(x)
= f(n/4)
=
2
√2
f'(x)=-sin(x) = f'(n/4)
=
f(x) = cos(x) = f(n/4) =
22 22
√2
f(x)=sin(x) => ""(n/4) =
22
(4)(x)
= cos(x) ->
(4)(π/4)
√2
2
Π
1
√2
1
1
√2
Step 3
Substitute the derivative values in the Taylor series centered at x=
cos(x)=
XT
2
πT and simplify. (Hint: Note that (-1)(n+1)/2 = 1, −1, −1, 1, 1, -1, -1, 1 ...)
4
7) - √2 (1x-(1/4)]²) + √2 (Ex - (n/4)]³)
3!
[x-(n/4)]4
=
-ΣΟ
7=0
Submit
Skip (you cannot come back)
Submit Answer
DETAILS
LARCALC12 9.10.009.
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