2 Use the identity sin x= 1- cos (2x) 2 to obtain the Maclaurin series for sin ´x. Then differentiate this series to obtain the Maclaurin series for 2 sin x cos x. Check that this is the series for sin (2x). What is the Maclaurin series for sin ²x? 00 Ο Α. Σ k=0 00 Ο Β. Σ k=0 00 ⭑C. Σ k=0 -- (-1)*22k+1 (2k+1)! (-1)*22k (2k)! x 2k (-1)*22k+1 00 (2k+2)! 1 * D. Σ 2 k=0 (-1)*22k (2k + 1)! 25 27 2k+1 3 5 7 9 =2x x + x 5! x + 22 2 =1- x + 4 x 2k+2 2 23 4 = 1 2k+1 2º x + 6 8 8 x 25 27 x + x 6 8! 8 -X +... ==-x+ -x 3! 3 - 6! 24 5 X + 2° 7! What is the Maclaurin series for 2 sin x cos x = sin (2x)? 00 O A. Σ k=0 00 'B. Σ k = 0 00 ○ C. Σ k = 0 00 OD. Σ k=0 2k (-1)+12 (2k)! (-12-1 (2k+1)! (-12-1 (2k)! 2k x =-1+ x x 2k+1 22 24 2 4 x + 6 X 2! =2x 23 3 7' 3' 3! + 25 X 5 - 7! 2k 21 (-1)+122k+2 (2k+1)! 2k+1 + 25 * 6 + 7 7 28 9 28 음... 8! 8 + ... -- =- 3 --- 5! + 7 7!
2 Use the identity sin x= 1- cos (2x) 2 to obtain the Maclaurin series for sin ´x. Then differentiate this series to obtain the Maclaurin series for 2 sin x cos x. Check that this is the series for sin (2x). What is the Maclaurin series for sin ²x? 00 Ο Α. Σ k=0 00 Ο Β. Σ k=0 00 ⭑C. Σ k=0 -- (-1)*22k+1 (2k+1)! (-1)*22k (2k)! x 2k (-1)*22k+1 00 (2k+2)! 1 * D. Σ 2 k=0 (-1)*22k (2k + 1)! 25 27 2k+1 3 5 7 9 =2x x + x 5! x + 22 2 =1- x + 4 x 2k+2 2 23 4 = 1 2k+1 2º x + 6 8 8 x 25 27 x + x 6 8! 8 -X +... ==-x+ -x 3! 3 - 6! 24 5 X + 2° 7! What is the Maclaurin series for 2 sin x cos x = sin (2x)? 00 O A. Σ k=0 00 'B. Σ k = 0 00 ○ C. Σ k = 0 00 OD. Σ k=0 2k (-1)+12 (2k)! (-12-1 (2k+1)! (-12-1 (2k)! 2k x =-1+ x x 2k+1 22 24 2 4 x + 6 X 2! =2x 23 3 7' 3' 3! + 25 X 5 - 7! 2k 21 (-1)+122k+2 (2k+1)! 2k+1 + 25 * 6 + 7 7 28 9 28 음... 8! 8 + ... -- =- 3 --- 5! + 7 7!
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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Transcribed Image Text:2
Use the identity sin x=
1- cos (2x)
2
to obtain the Maclaurin series for sin ´x. Then differentiate this series to obtain the Maclaurin series for 2 sin x cos x. Check that this is the series for sin (2x).
What is the Maclaurin series for sin ²x?
00
Ο Α. Σ
k=0
00
Ο Β. Σ
k=0
00
⭑C. Σ
k=0
--
(-1)*22k+1
(2k+1)!
(-1)*22k
(2k)!
x
2k
(-1)*22k+1
00
(2k+2)!
1
* D.
Σ
2
k=0
(-1)*22k
(2k + 1)!
25
27
2k+1
3
5
7
9
=2x
x +
x
5!
x +
22
2
=1- x +
4
x
2k+2
2
23
4
=
1
2k+1
2º
x +
6
8
8
x
25 27
x + x
6
8!
8
-X +...
==-x+ -x
3!
3
-
6!
24
5
X +
2°
7!
What is the Maclaurin series for 2 sin x cos x = sin (2x)?
00
O A. Σ
k=0
00
'B. Σ
k = 0
00
○ C. Σ
k = 0
00
OD. Σ
k=0
2k
(-1)+12
(2k)!
(-12-1
(2k+1)!
(-12-1
(2k)!
2k
x =-1+
x
x
2k+1
22
24
2
4
x +
6
X
2!
=2x
23
3
7'
3'
3!
+
25
X
5
-
7!
2k
21
(-1)+122k+2
(2k+1)!
2k+1
+
25
*
6
+
7
7
28
9
28
음...
8!
8
+ ...
--
=-
3
---
5!
+
7
7!
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