Show that the function Y₂ (0, 6) = (5/16m)¹/2 (3 cos² 0-1) is normalized over the interval 0 ≤0 ≤ and 0 ≤ ≤ 2π. Match the items in the left column to the appropriate blanks in the sentences and the equations on the right. Make certain each sentence and equation is complete before submitting your answer. 2πT is -9/5 cos 0+2 cos³ 0 - cos 0 3 cos² - 3 cos 0+1 -9 cos¹0+ 6 cos² 0 - cos 0 is not 7T 1 9 cos4 - 6 cos² 0+1 dr = dod cos 0 = sin 0 do de 2T ㅠ || [(Y₂°(0,4))² dr = (15m) [ dø [ sin 0(3 cos² 0 − 1)² do 0 0 5 T ) Ï 0 16T 5 16п Therefore, the function sin 0( ) de )= normalized over the interval 0 ≤0 ≤ T and 0 ≤ ≤ 2TT. Reset Help
Show that the function Y₂ (0, 6) = (5/16m)¹/2 (3 cos² 0-1) is normalized over the interval 0 ≤0 ≤ and 0 ≤ ≤ 2π. Match the items in the left column to the appropriate blanks in the sentences and the equations on the right. Make certain each sentence and equation is complete before submitting your answer. 2πT is -9/5 cos 0+2 cos³ 0 - cos 0 3 cos² - 3 cos 0+1 -9 cos¹0+ 6 cos² 0 - cos 0 is not 7T 1 9 cos4 - 6 cos² 0+1 dr = dod cos 0 = sin 0 do de 2T ㅠ || [(Y₂°(0,4))² dr = (15m) [ dø [ sin 0(3 cos² 0 − 1)² do 0 0 5 T ) Ï 0 16T 5 16п Therefore, the function sin 0( ) de )= normalized over the interval 0 ≤0 ≤ T and 0 ≤ ≤ 2TT. Reset Help
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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![Show that the function Y₂º(0, ¢) = (5/16π)¹/² (3 cos² 0 — 1) is normalized over the interval 0 ≤ 0 ≤ π and 0 ≤ ø ≤ 2π.
Match the items in the left column to the appropriate blanks in the sentences and the equations on the right. Make certain each sentence and equation is complete before submitting your answer.
2πT
is
-9/5 cos5 0+2 cos³ 0 - cos
0
3 cos² - 3 cos 0 + 1
-9 cos¹ 0 + 6 cos² 0 - cos
is not
ㅠ
1
9 cos4 - 6 cos² 0 + 1
dr = do d cos
sin e do de
2πT
5
[ [ (xº (0, 0))² &T = (187) † do] s sin 0(3 cos² - 1)² de
16TT
0
0
) de
||
5
16п
5
16T
Therefore, the function
0
π
sin (
)=
normalized over the interval 0 ≤ 0 ≤ π and 0 ≤ ≤ 2TT.
Reset Help](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Faa93ff2f-dc04-4549-9d2f-0f80a4020e3f%2F3040dc7a-0bd2-4635-8239-ae7c99aad208%2Fgid9nl_processed.png&w=3840&q=75)
Transcribed Image Text:Show that the function Y₂º(0, ¢) = (5/16π)¹/² (3 cos² 0 — 1) is normalized over the interval 0 ≤ 0 ≤ π and 0 ≤ ø ≤ 2π.
Match the items in the left column to the appropriate blanks in the sentences and the equations on the right. Make certain each sentence and equation is complete before submitting your answer.
2πT
is
-9/5 cos5 0+2 cos³ 0 - cos
0
3 cos² - 3 cos 0 + 1
-9 cos¹ 0 + 6 cos² 0 - cos
is not
ㅠ
1
9 cos4 - 6 cos² 0 + 1
dr = do d cos
sin e do de
2πT
5
[ [ (xº (0, 0))² &T = (187) † do] s sin 0(3 cos² - 1)² de
16TT
0
0
) de
||
5
16п
5
16T
Therefore, the function
0
π
sin (
)=
normalized over the interval 0 ≤ 0 ≤ π and 0 ≤ ≤ 2TT.
Reset Help
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