Mixing
Regarding Fourier transform
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Let $fin L^1(mathbbT)$. Where $mathbbT$ is the unite cirle in the complex plane. Define $$g(e^it)=f(e^2it).$$ Show that $hatg(n)= hatf(n/2)$ if 2 divides $n$ and $hatg(n)= 0$ if 2 does not divide $n$. I understand how the first part comes clearly, can anyone help me with the second part?
fourier-analysis fourier-transform
edited Jul 19 at 9:50
Bernard
110k635103
asked Jul 19 at 9:44
user534666
202
add a comment |Â
up vote
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Let $fin L^1(mathbbT)$. Where $mathbbT$ is the unite cirle in the complex plane. Define $$g(e^it)=f(e^2it).$$ Show that $hatg(n)= hatf(n/2)$ if 2 divides $n$ and $hatg(n)= 0$ if 2 does not divide $n$. I understand how the first part comes clearly, can anyone help me with the second part?
fourier-analysis fourier-transform
edited Jul 19 at 9:50
Bernard
110k635103
asked Jul 19 at 9:44
user534666
202
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $fin L^1(mathbbT)$. Where $mathbbT$ is the unite cirle in the complex plane. Define $$g(e^it)=f(e^2it).$$ Show that $hatg(n)= hatf(n/2)$ if 2 divides $n$ and $hatg(n)= 0$ if 2 does not divide $n$. I understand how the first part comes clearly, can anyone help me with the second part?
fourier-analysis fourier-transform
edited Jul 19 at 9:50
Bernard
110k635103
asked Jul 19 at 9:44
user534666
202
Let $fin L^1(mathbbT)$. Where $mathbbT$ is the unite cirle in the complex plane. Define $$g(e^it)=f(e^2it).$$ Show that $hatg(n)= hatf(n/2)$ if 2 divides $n$ and $hatg(n)= 0$ if 2 does not divide $n$. I understand how the first part comes clearly, can anyone help me with the second part?
fourier-analysis fourier-transform
edited Jul 19 at 9:50
Bernard
110k635103
asked Jul 19 at 9:44
user534666
202
edited Jul 19 at 9:50
Bernard
110k635103
edited Jul 19 at 9:50
Bernard
110k635103
edited Jul 19 at 9:50
Bernard
110k635103
110k635103
asked Jul 19 at 9:44
user534666
202
asked Jul 19 at 9:44
user534666
202
asked Jul 19 at 9:44
user534666
202
202
add a comment |Â
add a comment |Â
2 Answers
2
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oldest
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2
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You have to show that $int_-pi ^pi e^-int f(e^2it) , dt =0$ if $n$ is odd. Just make the substitution $s=t+pi $ ad see what you get.
answered Jul 19 at 10:21
Kavi Rama Murthy
20.9k2830
Can you explain how it is happening?
– user534666
Jul 21 at 10:25
$int e^-intf(e^2it), dt =int (-1)^ne^-insf(e^2is), ds=-int e^-intf(e^2it), dt$. The important point in this proof is the integral of a $2pi $ periodic function over intervals of length $2pi $ are all the same, so integrating from $-2pi $ to $0$ gives the same value as integrating from $-pi $ to $pi $.
– Kavi Rama Murthy
Jul 21 at 11:51
add a comment |Â
up vote
0
down vote
This should be intuitively clear: If $g$ is only made up of even frequencies, then it should be orthogonal to the odd frequencies.
Suggestion: What happens if $f$ is a polynomial? What do you know about the subset of all polynomials inside $L^1(mathbbT)$?
answered Jul 19 at 9:56
user357980
1,556213
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
You have to show that $int_-pi ^pi e^-int f(e^2it) , dt =0$ if $n$ is odd. Just make the substitution $s=t+pi $ ad see what you get.
answered Jul 19 at 10:21
Kavi Rama Murthy
20.9k2830
Can you explain how it is happening?
– user534666
Jul 21 at 10:25
$int e^-intf(e^2it), dt =int (-1)^ne^-insf(e^2is), ds=-int e^-intf(e^2it), dt$. The important point in this proof is the integral of a $2pi $ periodic function over intervals of length $2pi $ are all the same, so integrating from $-2pi $ to $0$ gives the same value as integrating from $-pi $ to $pi $.
– Kavi Rama Murthy
Jul 21 at 11:51
add a comment |Â
up vote
2
down vote
accepted
You have to show that $int_-pi ^pi e^-int f(e^2it) , dt =0$ if $n$ is odd. Just make the substitution $s=t+pi $ ad see what you get.
answered Jul 19 at 10:21
Kavi Rama Murthy
20.9k2830
Can you explain how it is happening?
– user534666
Jul 21 at 10:25
$int e^-intf(e^2it), dt =int (-1)^ne^-insf(e^2is), ds=-int e^-intf(e^2it), dt$. The important point in this proof is the integral of a $2pi $ periodic function over intervals of length $2pi $ are all the same, so integrating from $-2pi $ to $0$ gives the same value as integrating from $-pi $ to $pi $.
– Kavi Rama Murthy
Jul 21 at 11:51
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
You have to show that $int_-pi ^pi e^-int f(e^2it) , dt =0$ if $n$ is odd. Just make the substitution $s=t+pi $ ad see what you get.
answered Jul 19 at 10:21
Kavi Rama Murthy
20.9k2830
You have to show that $int_-pi ^pi e^-int f(e^2it) , dt =0$ if $n$ is odd. Just make the substitution $s=t+pi $ ad see what you get.
answered Jul 19 at 10:21
Kavi Rama Murthy
20.9k2830
answered Jul 19 at 10:21
Kavi Rama Murthy
20.9k2830
answered Jul 19 at 10:21
Kavi Rama Murthy
20.9k2830
answered Jul 19 at 10:21
Kavi Rama Murthy
20.9k2830
20.9k2830
Can you explain how it is happening?
– user534666
Jul 21 at 10:25
$int e^-intf(e^2it), dt =int (-1)^ne^-insf(e^2is), ds=-int e^-intf(e^2it), dt$. The important point in this proof is the integral of a $2pi $ periodic function over intervals of length $2pi $ are all the same, so integrating from $-2pi $ to $0$ gives the same value as integrating from $-pi $ to $pi $.
– Kavi Rama Murthy
Jul 21 at 11:51
add a comment |Â
Can you explain how it is happening?
– user534666
Jul 21 at 10:25
$int e^-intf(e^2it), dt =int (-1)^ne^-insf(e^2is), ds=-int e^-intf(e^2it), dt$. The important point in this proof is the integral of a $2pi $ periodic function over intervals of length $2pi $ are all the same, so integrating from $-2pi $ to $0$ gives the same value as integrating from $-pi $ to $pi $.
– Kavi Rama Murthy
Jul 21 at 11:51
Can you explain how it is happening?
– user534666
Jul 21 at 10:25
Can you explain how it is happening?
– user534666
Jul 21 at 10:25
$int e^-intf(e^2it), dt =int (-1)^ne^-insf(e^2is), ds=-int e^-intf(e^2it), dt$. The important point in this proof is the integral of a $2pi $ periodic function over intervals of length $2pi $ are all the same, so integrating from $-2pi $ to $0$ gives the same value as integrating from $-pi $ to $pi $.
– Kavi Rama Murthy
Jul 21 at 11:51
$int e^-intf(e^2it), dt =int (-1)^ne^-insf(e^2is), ds=-int e^-intf(e^2it), dt$. The important point in this proof is the integral of a $2pi $ periodic function over intervals of length $2pi $ are all the same, so integrating from $-2pi $ to $0$ gives the same value as integrating from $-pi $ to $pi $.
– Kavi Rama Murthy
Jul 21 at 11:51
add a comment |Â
up vote
0
down vote
This should be intuitively clear: If $g$ is only made up of even frequencies, then it should be orthogonal to the odd frequencies.
Suggestion: What happens if $f$ is a polynomial? What do you know about the subset of all polynomials inside $L^1(mathbbT)$?
answered Jul 19 at 9:56
user357980
1,556213
add a comment |Â
up vote
0
down vote
This should be intuitively clear: If $g$ is only made up of even frequencies, then it should be orthogonal to the odd frequencies.
Suggestion: What happens if $f$ is a polynomial? What do you know about the subset of all polynomials inside $L^1(mathbbT)$?
answered Jul 19 at 9:56
user357980
1,556213
add a comment |Â
up vote
0
down vote
up vote
0
down vote
This should be intuitively clear: If $g$ is only made up of even frequencies, then it should be orthogonal to the odd frequencies.
Suggestion: What happens if $f$ is a polynomial? What do you know about the subset of all polynomials inside $L^1(mathbbT)$?
answered Jul 19 at 9:56
user357980
1,556213
This should be intuitively clear: If $g$ is only made up of even frequencies, then it should be orthogonal to the odd frequencies.
Suggestion: What happens if $f$ is a polynomial? What do you know about the subset of all polynomials inside $L^1(mathbbT)$?
answered Jul 19 at 9:56
user357980
1,556213
answered Jul 19 at 9:56
user357980
1,556213
answered Jul 19 at 9:56
user357980
1,556213
answered Jul 19 at 9:56
user357980
1,556213
1,556213
add a comment |Â
add a comment |Â
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