Show ${int_{0}^{1}|e^{-2picdot int} f(t)| dt} = {int_{0}^{1}|f(t)| dt}$
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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
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up vote
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favorite
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
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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
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 at 2:37
Tianlalu
2,9801936
2,9801936
asked Nov 26 at 1:59
hgil
226
226
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2 Answers
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1
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accepted
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,
$$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$
Perfect! Thanks!
– hgil
Nov 26 at 2:13
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1
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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 $ntin mathbb{R}$.
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
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,
$$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$
Perfect! Thanks!
– hgil
Nov 26 at 2:13
add a comment |
up vote
1
down vote
accepted
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,
$$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$
Perfect! Thanks!
– hgil
Nov 26 at 2:13
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
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,
$$|e^{-2pi i n t}f(t)|=|e^{-2pi i n t}||f(t)|=|e^{(-2pi n t)i}||f(t)|=1|f(t)|=|f(t)|.$$
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 at 2:11
LeB
938217
938217
Perfect! Thanks!
– hgil
Nov 26 at 2:13
add a comment |
Perfect! Thanks!
– hgil
Nov 26 at 2:13
Perfect! Thanks!
– hgil
Nov 26 at 2:13
Perfect! Thanks!
– hgil
Nov 26 at 2:13
add a comment |
up vote
1
down vote
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 $ntin mathbb{R}$.
add a comment |
up vote
1
down vote
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 $ntin mathbb{R}$.
add a comment |
up vote
1
down vote
up vote
1
down vote
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 $ntin mathbb{R}$.
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 at 2:13
M1183
943
943
add a comment |
add a comment |
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