Show that the given set of functions is orthogonal on the indicated interval. Find the norm of each function in the set. {sin x, sin 3x, sin 5x, ...}; [0, x/2] For m + n /2 x/2 2/ (cos(2(n – m)x) – cos (2(n + m + 1)x) sin((2n + 1)x) sin((2m + 1)x) dx = dx 1 1 5(sin(2(n– m)x)) – sin((2(n+m+ 1)x)) 4(п - т) 4(n + m + 1) For m = n sin (2(2n + 1)x) 4(2n + 1) /2 R/2 | sin?((2n + 1)x) dx = dx sin (2(2n + 1)x) 4(2n +1) /2 To 4
Show that the given set of functions is orthogonal on the indicated interval. Find the norm of each function in the set. {sin x, sin 3x, sin 5x, ...}; [0, x/2] For m + n /2 x/2 2/ (cos(2(n – m)x) – cos (2(n + m + 1)x) sin((2n + 1)x) sin((2m + 1)x) dx = dx 1 1 5(sin(2(n– m)x)) – sin((2(n+m+ 1)x)) 4(п - т) 4(n + m + 1) For m = n sin (2(2n + 1)x) 4(2n + 1) /2 R/2 | sin?((2n + 1)x) dx = dx sin (2(2n + 1)x) 4(2n +1) /2 To 4
Linear Algebra: A Modern Introduction
4th Edition
ISBN:9781285463247
Author:David Poole
Publisher:David Poole
Chapter6: Vector Spaces
Section6.5: The Kernel And Range Of A Linear Transformation
Problem 30EQ
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This is practice and doesn't break honor code. I just need the one that is wrong.
![Show that the given set of functions is orthogonal on the indicated interval. Find the norm of each function in the set.
{sin x, sin 3x, sin 5x, ...}; [0, a/2]
For m # n
T/2
T/2
cos(2(n – m)x) - (
cos (2(n + m + 1)x)
sin((2n + 1)x) sin((2m + 1)x) dx =
dx
1
1
| T/2
sin(2(n –
m)x)) –
sin((2(n+ m+ 1)x))
4(п — т.
4(n + m + 1)
For m = n
sin (2(2n +1)x)
4(2n + 1)
T/2
T/2
| sin?((2n + 1)x) dx
%3D
dx
2
sin (2(2n + 1)x)
4(2n + 1)
| T/2
2
4
so that
||sin((2n + 1)x)|| =
2
II](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6f51a47d-2c3a-4935-8f76-d9a574028b9a%2F54cae1f5-7c32-4d60-bc0b-c7f6536e2e28%2Fryk07j1_processed.png&w=3840&q=75)
Transcribed Image Text:Show that the given set of functions is orthogonal on the indicated interval. Find the norm of each function in the set.
{sin x, sin 3x, sin 5x, ...}; [0, a/2]
For m # n
T/2
T/2
cos(2(n – m)x) - (
cos (2(n + m + 1)x)
sin((2n + 1)x) sin((2m + 1)x) dx =
dx
1
1
| T/2
sin(2(n –
m)x)) –
sin((2(n+ m+ 1)x))
4(п — т.
4(n + m + 1)
For m = n
sin (2(2n +1)x)
4(2n + 1)
T/2
T/2
| sin?((2n + 1)x) dx
%3D
dx
2
sin (2(2n + 1)x)
4(2n + 1)
| T/2
2
4
so that
||sin((2n + 1)x)|| =
2
II
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