Showing that complex exponentials of the Fourier Series are an orthonormal basis
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I am revisiting the Fourier transform and I found great lecture notes by Professor Osgood from Standford (pdf ~30MB).
On page 30 and 31 he show that the complex exponentials form an orthonormal basis. I understand the result, but not his calculation. He shows that the inner product of two different exponentials $(e_n (t) = e^{2pi int}, e_m(t)= e^{2pi imt})$ with $m neq n$ is $0$ (He uses round parenthesis to denote the inner product). So, he does the calculation:
$$ (e_n, e_m) = int_0^1 e^{2pi int} overline{e^{2pi imt}}dt = dots = frac{1}{2pi i(n -m)} (e^{2pi i(n - m)} - e^0) = frac{1}{2pi i(n -m)} (1 - 1) = 0$$
So why is $e^{2pi i(n - m)} = 1$? And why do I have to look at the case $ m = n$ separately and cannot also just plug it into the last step? (I am aware that I would not get a sensible result then, but it still seems strange). Does this have anything to do because I am using Lebesgue integration and complex numbers? I suppose I have to review some math basics ...
fourier-analysis fourier-series orthonormal orthogonality
asked Jun 8 '14 at 19:05
Lucas
143115
add a comment |
up vote
7
down vote
favorite
I am revisiting the Fourier transform and I found great lecture notes by Professor Osgood from Standford (pdf ~30MB).
On page 30 and 31 he show that the complex exponentials form an orthonormal basis. I understand the result, but not his calculation. He shows that the inner product of two different exponentials $(e_n (t) = e^{2pi int}, e_m(t)= e^{2pi imt})$ with $m neq n$ is $0$ (He uses round parenthesis to denote the inner product). So, he does the calculation:
$$ (e_n, e_m) = int_0^1 e^{2pi int} overline{e^{2pi imt}}dt = dots = frac{1}{2pi i(n -m)} (e^{2pi i(n - m)} - e^0) = frac{1}{2pi i(n -m)} (1 - 1) = 0$$
So why is $e^{2pi i(n - m)} = 1$? And why do I have to look at the case $ m = n$ separately and cannot also just plug it into the last step? (I am aware that I would not get a sensible result then, but it still seems strange). Does this have anything to do because I am using Lebesgue integration and complex numbers? I suppose I have to review some math basics ...
fourier-analysis fourier-series orthonormal orthogonality
asked Jun 8 '14 at 19:05
Lucas
143115
What is the anti-derivative of $e^{2pi i (n-m)t}$? Does the formula hold for $n = m$?
– Christopher A. Wong
Jun 8 '14 at 19:16
$e^{itheta} = cos theta + isin theta$
– Paul Hurst
Jun 8 '14 at 19:16
Ok, so if understand you correctly it's much easier. They become zero for all exponents because of the 2 pi and the integral for m=n is not defined because otherwise the denominator isn't defined anymore?
– Lucas
Jun 8 '14 at 19:50
The Integral is defined. But the antiderivative of $e^{2pi i (n-m) x} = 1$ is the identity, so that the definite (normalized) integral is $1$ instead of $0$.
– PhoemueX
Jun 8 '14 at 20:27
You seems to have some trouble with the exponential function. For a crash course on it I recommend reading the introduction to Rudin's Real and complex analysis. This Q&A, more advanced, might also be interesting.
– Giuseppe Negro
Jun 8 '14 at 20:38
add a comment |
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I am revisiting the Fourier transform and I found great lecture notes by Professor Osgood from Standford (pdf ~30MB).
On page 30 and 31 he show that the complex exponentials form an orthonormal basis. I understand the result, but not his calculation. He shows that the inner product of two different exponentials $(e_n (t) = e^{2pi int}, e_m(t)= e^{2pi imt})$ with $m neq n$ is $0$ (He uses round parenthesis to denote the inner product). So, he does the calculation:
$$ (e_n, e_m) = int_0^1 e^{2pi int} overline{e^{2pi imt}}dt = dots = frac{1}{2pi i(n -m)} (e^{2pi i(n - m)} - e^0) = frac{1}{2pi i(n -m)} (1 - 1) = 0$$
So why is $e^{2pi i(n - m)} = 1$? And why do I have to look at the case $ m = n$ separately and cannot also just plug it into the last step? (I am aware that I would not get a sensible result then, but it still seems strange). Does this have anything to do because I am using Lebesgue integration and complex numbers? I suppose I have to review some math basics ...
fourier-analysis fourier-series orthonormal orthogonality
asked Jun 8 '14 at 19:05
Lucas
143115
I am revisiting the Fourier transform and I found great lecture notes by Professor Osgood from Standford (pdf ~30MB).
On page 30 and 31 he show that the complex exponentials form an orthonormal basis. I understand the result, but not his calculation. He shows that the inner product of two different exponentials $(e_n (t) = e^{2pi int}, e_m(t)= e^{2pi imt})$ with $m neq n$ is $0$ (He uses round parenthesis to denote the inner product). So, he does the calculation:
$$ (e_n, e_m) = int_0^1 e^{2pi int} overline{e^{2pi imt}}dt = dots = frac{1}{2pi i(n -m)} (e^{2pi i(n - m)} - e^0) = frac{1}{2pi i(n -m)} (1 - 1) = 0$$
So why is $e^{2pi i(n - m)} = 1$? And why do I have to look at the case $ m = n$ separately and cannot also just plug it into the last step? (I am aware that I would not get a sensible result then, but it still seems strange). Does this have anything to do because I am using Lebesgue integration and complex numbers? I suppose I have to review some math basics ...
fourier-analysis fourier-series orthonormal orthogonality
fourier-analysis fourier-series orthonormal orthogonality
asked Jun 8 '14 at 19:05
Lucas
143115
asked Jun 8 '14 at 19:05
Lucas
143115
asked Jun 8 '14 at 19:05
Lucas
143115
asked Jun 8 '14 at 19:05
Lucas
143115
asked Jun 8 '14 at 19:05
Lucas
143115
143115
What is the anti-derivative of $e^{2pi i (n-m)t}$? Does the formula hold for $n = m$?
– Christopher A. Wong
Jun 8 '14 at 19:16
$e^{itheta} = cos theta + isin theta$
– Paul Hurst
Jun 8 '14 at 19:16
Ok, so if understand you correctly it's much easier. They become zero for all exponents because of the 2 pi and the integral for m=n is not defined because otherwise the denominator isn't defined anymore?
– Lucas
Jun 8 '14 at 19:50
The Integral is defined. But the antiderivative of $e^{2pi i (n-m) x} = 1$ is the identity, so that the definite (normalized) integral is $1$ instead of $0$.
– PhoemueX
Jun 8 '14 at 20:27
You seems to have some trouble with the exponential function. For a crash course on it I recommend reading the introduction to Rudin's Real and complex analysis. This Q&A, more advanced, might also be interesting.
– Giuseppe Negro
Jun 8 '14 at 20:38
add a comment |
What is the anti-derivative of $e^{2pi i (n-m)t}$? Does the formula hold for $n = m$?
– Christopher A. Wong
Jun 8 '14 at 19:16
$e^{itheta} = cos theta + isin theta$
– Paul Hurst
Jun 8 '14 at 19:16
Ok, so if understand you correctly it's much easier. They become zero for all exponents because of the 2 pi and the integral for m=n is not defined because otherwise the denominator isn't defined anymore?
– Lucas
Jun 8 '14 at 19:50
The Integral is defined. But the antiderivative of $e^{2pi i (n-m) x} = 1$ is the identity, so that the definite (normalized) integral is $1$ instead of $0$.
– PhoemueX
Jun 8 '14 at 20:27
You seems to have some trouble with the exponential function. For a crash course on it I recommend reading the introduction to Rudin's Real and complex analysis. This Q&A, more advanced, might also be interesting.
– Giuseppe Negro
Jun 8 '14 at 20:38
What is the anti-derivative of $e^{2pi i (n-m)t}$? Does the formula hold for $n = m$?
– Christopher A. Wong
Jun 8 '14 at 19:16
What is the anti-derivative of $e^{2pi i (n-m)t}$? Does the formula hold for $n = m$?
– Christopher A. Wong
Jun 8 '14 at 19:16
$e^{itheta} = cos theta + isin theta$
– Paul Hurst
Jun 8 '14 at 19:16
$e^{itheta} = cos theta + isin theta$
– Paul Hurst
Jun 8 '14 at 19:16
Ok, so if understand you correctly it's much easier. They become zero for all exponents because of the 2 pi and the integral for m=n is not defined because otherwise the denominator isn't defined anymore?
– Lucas
Jun 8 '14 at 19:50
Ok, so if understand you correctly it's much easier. They become zero for all exponents because of the 2 pi and the integral for m=n is not defined because otherwise the denominator isn't defined anymore?
– Lucas
Jun 8 '14 at 19:50
The Integral is defined. But the antiderivative of $e^{2pi i (n-m) x} = 1$ is the identity, so that the definite (normalized) integral is $1$ instead of $0$.
– PhoemueX
Jun 8 '14 at 20:27
The Integral is defined. But the antiderivative of $e^{2pi i (n-m) x} = 1$ is the identity, so that the definite (normalized) integral is $1$ instead of $0$.
– PhoemueX
Jun 8 '14 at 20:27
You seems to have some trouble with the exponential function. For a crash course on it I recommend reading the introduction to Rudin's Real and complex analysis. This Q&A, more advanced, might also be interesting.
– Giuseppe Negro
Jun 8 '14 at 20:38
You seems to have some trouble with the exponential function. For a crash course on it I recommend reading the introduction to Rudin's Real and complex analysis. This Q&A, more advanced, might also be interesting.
– Giuseppe Negro
Jun 8 '14 at 20:38
add a comment |
2 Answers
2
active
oldest
votes
up vote
9
down vote
If $l=n-mneq 0$ you have
$$e^{2pi i(n-m)}=e^{2pi i l}=cos(2pi l)+isin(2pi l)=1+icdot 0=1$$
The case $m=n$ has to be considered separately because in the solution you showed you divide by $m-n$, which you can't do if $m-n=0$. For $m=n$ you simply get
$$int_0^1e^{2pi i nt}e^{-2pi i nt}dt=int_0^1dt=1$$
answered Jun 8 '14 at 22:28
Matt L.
8,821821
I know that the integral must be over $[0,1)$ in order to work, but do you know why is that? Is it because the circles have radius of 1?
– Ooker
Nov 6 '17 at 20:13
add a comment |
up vote
-1
down vote
You can of course consider $m=n$ case separately, as Matt suggested.
Alternatively, you can use L'Hospitals rule, in the expression you found, and see that in the limit $m to n,$ the integral indeed converges to $1.$
edited Nov 14 at 22:45
amWhy
191k27223437
answered Nov 14 at 21:55
Anyon
11
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
9
down vote
If $l=n-mneq 0$ you have
$$e^{2pi i(n-m)}=e^{2pi i l}=cos(2pi l)+isin(2pi l)=1+icdot 0=1$$
The case $m=n$ has to be considered separately because in the solution you showed you divide by $m-n$, which you can't do if $m-n=0$. For $m=n$ you simply get
$$int_0^1e^{2pi i nt}e^{-2pi i nt}dt=int_0^1dt=1$$
answered Jun 8 '14 at 22:28
Matt L.
8,821821
I know that the integral must be over $[0,1)$ in order to work, but do you know why is that? Is it because the circles have radius of 1?
– Ooker
Nov 6 '17 at 20:13
add a comment |
up vote
9
down vote
If $l=n-mneq 0$ you have
$$e^{2pi i(n-m)}=e^{2pi i l}=cos(2pi l)+isin(2pi l)=1+icdot 0=1$$
The case $m=n$ has to be considered separately because in the solution you showed you divide by $m-n$, which you can't do if $m-n=0$. For $m=n$ you simply get
$$int_0^1e^{2pi i nt}e^{-2pi i nt}dt=int_0^1dt=1$$
answered Jun 8 '14 at 22:28
Matt L.
8,821821
I know that the integral must be over $[0,1)$ in order to work, but do you know why is that? Is it because the circles have radius of 1?
– Ooker
Nov 6 '17 at 20:13
add a comment |
up vote
9
down vote
up vote
9
down vote
If $l=n-mneq 0$ you have
$$e^{2pi i(n-m)}=e^{2pi i l}=cos(2pi l)+isin(2pi l)=1+icdot 0=1$$
The case $m=n$ has to be considered separately because in the solution you showed you divide by $m-n$, which you can't do if $m-n=0$. For $m=n$ you simply get
$$int_0^1e^{2pi i nt}e^{-2pi i nt}dt=int_0^1dt=1$$
answered Jun 8 '14 at 22:28
Matt L.
8,821821
If $l=n-mneq 0$ you have
$$e^{2pi i(n-m)}=e^{2pi i l}=cos(2pi l)+isin(2pi l)=1+icdot 0=1$$
The case $m=n$ has to be considered separately because in the solution you showed you divide by $m-n$, which you can't do if $m-n=0$. For $m=n$ you simply get
$$int_0^1e^{2pi i nt}e^{-2pi i nt}dt=int_0^1dt=1$$
answered Jun 8 '14 at 22:28
Matt L.
8,821821
answered Jun 8 '14 at 22:28
Matt L.
8,821821
answered Jun 8 '14 at 22:28
Matt L.
8,821821
answered Jun 8 '14 at 22:28
Matt L.
8,821821
8,821821
I know that the integral must be over $[0,1)$ in order to work, but do you know why is that? Is it because the circles have radius of 1?
– Ooker
Nov 6 '17 at 20:13
add a comment |
I know that the integral must be over $[0,1)$ in order to work, but do you know why is that? Is it because the circles have radius of 1?
– Ooker
Nov 6 '17 at 20:13
I know that the integral must be over $[0,1)$ in order to work, but do you know why is that? Is it because the circles have radius of 1?
– Ooker
Nov 6 '17 at 20:13
I know that the integral must be over $[0,1)$ in order to work, but do you know why is that? Is it because the circles have radius of 1?
– Ooker
Nov 6 '17 at 20:13
add a comment |
up vote
-1
down vote
You can of course consider $m=n$ case separately, as Matt suggested.
Alternatively, you can use L'Hospitals rule, in the expression you found, and see that in the limit $m to n,$ the integral indeed converges to $1.$
edited Nov 14 at 22:45
amWhy
191k27223437
answered Nov 14 at 21:55
Anyon
11
add a comment |
up vote
-1
down vote
You can of course consider $m=n$ case separately, as Matt suggested.
Alternatively, you can use L'Hospitals rule, in the expression you found, and see that in the limit $m to n,$ the integral indeed converges to $1.$
edited Nov 14 at 22:45
amWhy
191k27223437
answered Nov 14 at 21:55
Anyon
11
add a comment |
up vote
-1
down vote
up vote
-1
down vote
You can of course consider $m=n$ case separately, as Matt suggested.
Alternatively, you can use L'Hospitals rule, in the expression you found, and see that in the limit $m to n,$ the integral indeed converges to $1.$
edited Nov 14 at 22:45
amWhy
191k27223437
answered Nov 14 at 21:55
Anyon
11
You can of course consider $m=n$ case separately, as Matt suggested.
Alternatively, you can use L'Hospitals rule, in the expression you found, and see that in the limit $m to n,$ the integral indeed converges to $1.$
edited Nov 14 at 22:45
amWhy
191k27223437
answered Nov 14 at 21:55
Anyon
11
edited Nov 14 at 22:45
amWhy
191k27223437
edited Nov 14 at 22:45
amWhy
191k27223437
edited Nov 14 at 22:45
amWhy
191k27223437
191k27223437
answered Nov 14 at 21:55
Anyon
11
answered Nov 14 at 21:55
Anyon
11
answered Nov 14 at 21:55
Anyon
11
11
add a comment |
add a comment |
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What is the anti-derivative of $e^{2pi i (n-m)t}$? Does the formula hold for $n = m$?
– Christopher A. Wong
Jun 8 '14 at 19:16
$e^{itheta} = cos theta + isin theta$
– Paul Hurst
Jun 8 '14 at 19:16
Ok, so if understand you correctly it's much easier. They become zero for all exponents because of the 2 pi and the integral for m=n is not defined because otherwise the denominator isn't defined anymore?
– Lucas
Jun 8 '14 at 19:50
The Integral is defined. But the antiderivative of $e^{2pi i (n-m) x} = 1$ is the identity, so that the definite (normalized) integral is $1$ instead of $0$.
– PhoemueX
Jun 8 '14 at 20:27
You seems to have some trouble with the exponential function. For a crash course on it I recommend reading the introduction to Rudin's Real and complex analysis. This Q&A, more advanced, might also be interesting.
– Giuseppe Negro
Jun 8 '14 at 20:38