Show ${int_{0}^{1}|e^{-2picdot int} f(t)| dt} = {int_{0}^{1}|f(t)| dt}$
$begingroup$
The expressions below are from page 36 of Brad Osgood's "The Fourier Transform and its Applications". I don't understand how to establish the equality between the second and third expressions below (the text doesn't expand on this).
For the inequality between the first and second expressions, I can tell that it is using the idea that the absolute value of an integral is less than or equal to the integral of the absolute value of the same integrand. I also recognize the first expression as the absolute value of the n-th Fourier coefficient. The context of the previous pages was on the Cauchy-Schwartz inequality and the Triangle Inequality; however, it doesn't look like either of these inequalities are being used to establish this equality. Any hints on how to get at this?
fourier-analysis fourier-series
edited Nov 26 '18 at 2:37
Tianlalu
3,08621038
asked Nov 26 '18 at 1:59
hgilhgil
226
$endgroup$
add a comment |
$begingroup$
The expressions below are from page 36 of Brad Osgood's "The Fourier Transform and its Applications". I don't understand how to establish the equality between the second and third expressions below (the text doesn't expand on this).
For the inequality between the first and second expressions, I can tell that it is using the idea that the absolute value of an integral is less than or equal to the integral of the absolute value of the same integrand. I also recognize the first expression as the absolute value of the n-th Fourier coefficient. The context of the previous pages was on the Cauchy-Schwartz inequality and the Triangle Inequality; however, it doesn't look like either of these inequalities are being used to establish this equality. Any hints on how to get at this?
fourier-analysis fourier-series
edited Nov 26 '18 at 2:37
Tianlalu
3,08621038
asked Nov 26 '18 at 1:59
hgilhgil
226
$endgroup$
add a comment |
$begingroup$
The expressions below are from page 36 of Brad Osgood's "The Fourier Transform and its Applications". I don't understand how to establish the equality between the second and third expressions below (the text doesn't expand on this).
For the inequality between the first and second expressions, I can tell that it is using the idea that the absolute value of an integral is less than or equal to the integral of the absolute value of the same integrand. I also recognize the first expression as the absolute value of the n-th Fourier coefficient. The context of the previous pages was on the Cauchy-Schwartz inequality and the Triangle Inequality; however, it doesn't look like either of these inequalities are being used to establish this equality. Any hints on how to get at this?
fourier-analysis fourier-series
edited Nov 26 '18 at 2:37
Tianlalu
3,08621038
asked Nov 26 '18 at 1:59
hgilhgil
226
$endgroup$
The expressions below are from page 36 of Brad Osgood's "The Fourier Transform and its Applications". I don't understand how to establish the equality between the second and third expressions below (the text doesn't expand on this).
For the inequality between the first and second expressions, I can tell that it is using the idea that the absolute value of an integral is less than or equal to the integral of the absolute value of the same integrand. I also recognize the first expression as the absolute value of the n-th Fourier coefficient. The context of the previous pages was on the Cauchy-Schwartz inequality and the Triangle Inequality; however, it doesn't look like either of these inequalities are being used to establish this equality. Any hints on how to get at this?
fourier-analysis fourier-series
fourier-analysis fourier-series
edited Nov 26 '18 at 2:37
Tianlalu
3,08621038
asked Nov 26 '18 at 1:59
hgilhgil
226
edited Nov 26 '18 at 2:37
Tianlalu
3,08621038
asked Nov 26 '18 at 1:59
hgilhgil
226
edited Nov 26 '18 at 2:37
Tianlalu
3,08621038
edited Nov 26 '18 at 2:37
Tianlalu
3,08621038
edited Nov 26 '18 at 2:37
Tianlalu
3,08621038
3,08621038
asked Nov 26 '18 at 1:59
hgilhgil
226
asked Nov 26 '18 at 1:59
hgilhgil
226
asked Nov 26 '18 at 1:59
hgilhgil
226
226
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that
$$|e^{it}|=1 hspace{4mm}forall tin mathbb{R}.$$
It is because
$$|e^{it}|=|cos(t)+isin(t)|=sqrt{cos^2(t)+sin^2(t)}=1.$$
Thus,
$$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$
answered Nov 26 '18 at 2:11
LeBLeB
1,080317
$endgroup$
$begingroup$
Perfect! Thanks!
$endgroup$
– hgil
Nov 26 '18 at 2:13
add a comment |
$begingroup$
For any two complex numbers $z_1$ and $z_2$ we have $$|z_1z_2|=|z_1||z_2|$$ so here
$$left|e^{-2pi i n t}f(t)right|=left|e^{-2pi i n t}right|left|f(t)right|=left|f(t)right|$$
if $ntin mathbb{R}$.
answered Nov 26 '18 at 2:13
M1183M1183
943
$endgroup$
add a comment |
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2 Answers
2
active
oldest
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2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that
$$|e^{it}|=1 hspace{4mm}forall tin mathbb{R}.$$
It is because
$$|e^{it}|=|cos(t)+isin(t)|=sqrt{cos^2(t)+sin^2(t)}=1.$$
Thus,
$$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$
answered Nov 26 '18 at 2:11
LeBLeB
1,080317
$endgroup$
$begingroup$
Perfect! Thanks!
$endgroup$
– hgil
Nov 26 '18 at 2:13
add a comment |
$begingroup$
Note that
$$|e^{it}|=1 hspace{4mm}forall tin mathbb{R}.$$
It is because
$$|e^{it}|=|cos(t)+isin(t)|=sqrt{cos^2(t)+sin^2(t)}=1.$$
Thus,
$$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$
answered Nov 26 '18 at 2:11
LeBLeB
1,080317
$endgroup$
$begingroup$
Perfect! Thanks!
$endgroup$
– hgil
Nov 26 '18 at 2:13
add a comment |
$begingroup$
Note that
$$|e^{it}|=1 hspace{4mm}forall tin mathbb{R}.$$
It is because
$$|e^{it}|=|cos(t)+isin(t)|=sqrt{cos^2(t)+sin^2(t)}=1.$$
Thus,
$$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$
answered Nov 26 '18 at 2:11
LeBLeB
1,080317
$endgroup$
Note that
$$|e^{it}|=1 hspace{4mm}forall tin mathbb{R}.$$
It is because
$$|e^{it}|=|cos(t)+isin(t)|=sqrt{cos^2(t)+sin^2(t)}=1.$$
Thus,
$$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$
answered Nov 26 '18 at 2:11
LeBLeB
1,080317
answered Nov 26 '18 at 2:11
LeBLeB
1,080317
answered Nov 26 '18 at 2:11
LeBLeB
1,080317
answered Nov 26 '18 at 2:11
LeBLeB
1,080317
1,080317
$begingroup$
Perfect! Thanks!
$endgroup$
– hgil
Nov 26 '18 at 2:13
add a comment |
$begingroup$
Perfect! Thanks!
$endgroup$
– hgil
Nov 26 '18 at 2:13
$begingroup$
Perfect! Thanks!
$endgroup$
– hgil
Nov 26 '18 at 2:13
$begingroup$
Perfect! Thanks!
$endgroup$
– hgil
Nov 26 '18 at 2:13
add a comment |
$begingroup$
For any two complex numbers $z_1$ and $z_2$ we have $$|z_1z_2|=|z_1||z_2|$$ so here
$$left|e^{-2pi i n t}f(t)right|=left|e^{-2pi i n t}right|left|f(t)right|=left|f(t)right|$$
if $ntin mathbb{R}$.
answered Nov 26 '18 at 2:13
M1183M1183
943
$endgroup$
add a comment |
$begingroup$
For any two complex numbers $z_1$ and $z_2$ we have $$|z_1z_2|=|z_1||z_2|$$ so here
$$left|e^{-2pi i n t}f(t)right|=left|e^{-2pi i n t}right|left|f(t)right|=left|f(t)right|$$
if $ntin mathbb{R}$.
answered Nov 26 '18 at 2:13
M1183M1183
943
$endgroup$
add a comment |
$begingroup$
For any two complex numbers $z_1$ and $z_2$ we have $$|z_1z_2|=|z_1||z_2|$$ so here
$$left|e^{-2pi i n t}f(t)right|=left|e^{-2pi i n t}right|left|f(t)right|=left|f(t)right|$$
if $ntin mathbb{R}$.
answered Nov 26 '18 at 2:13
M1183M1183
943
$endgroup$
For any two complex numbers $z_1$ and $z_2$ we have $$|z_1z_2|=|z_1||z_2|$$ so here
$$left|e^{-2pi i n t}f(t)right|=left|e^{-2pi i n t}right|left|f(t)right|=left|f(t)right|$$
if $ntin mathbb{R}$.
answered Nov 26 '18 at 2:13
M1183M1183
943
answered Nov 26 '18 at 2:13
M1183M1183
943
answered Nov 26 '18 at 2:13
M1183M1183
943
answered Nov 26 '18 at 2:13
M1183M1183
943
943
add a comment |
add a comment |
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