Show ${int_{0}^{1}|e^{-2picdot int} f(t)| dt} = {int_{0}^{1}|f(t)| dt}$
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 at 2:37
Tianlalu
3,05521038
asked Nov 26 at 1:59
hgil
226
add a comment |
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 at 2:37
Tianlalu
3,05521038
asked Nov 26 at 1:59
hgil
226
add a comment |
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 at 2:37
Tianlalu
3,05521038
asked Nov 26 at 1:59
hgil
226
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
3,05521038
asked Nov 26 at 1:59
hgil
226
edited Nov 26 at 2:37
Tianlalu
3,05521038
asked Nov 26 at 1:59
hgil
226
edited Nov 26 at 2:37
Tianlalu
3,05521038
edited Nov 26 at 2:37
Tianlalu
3,05521038
edited Nov 26 at 2:37
Tianlalu
3,05521038
3,05521038
asked Nov 26 at 1:59
hgil
226
asked Nov 26 at 1:59
hgil
226
asked Nov 26 at 1:59
hgil
226
226
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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
989217
Perfect! Thanks!
– hgil
Nov 26 at 2:13
add a comment |
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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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
989217
Perfect! Thanks!
– hgil
Nov 26 at 2:13
add a comment |
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
989217
Perfect! Thanks!
– hgil
Nov 26 at 2:13
add a comment |
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
989217
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
989217
answered Nov 26 at 2:11
LeB
989217
answered Nov 26 at 2:11
LeB
989217
answered Nov 26 at 2:11
LeB
989217
989217
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 |
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
add a comment |
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
add a comment |
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
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
answered Nov 26 at 2:13
M1183
943
answered Nov 26 at 2:13
M1183
943
answered Nov 26 at 2:13
M1183
943
943
add a comment |
add a comment |
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