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How one can write the delta Dirac function using Fourier series expansion such that summation is performed over odd indices only?
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The delta Dirac function can be presented in the form of Fourier series expansion as
$$
delta(x)=frac12pi
+frac1pi sum_nge 1 cos (nx) , .
$$
The proof is straightforward and can be found e.g. in this Mathworld Wolfram article.
In a mathematical physics problem, I would like to make use of the Fourier series technique in order to solve a system of differential equations involving a delta Dirac function, such that the solution has the form
$$
f(x) = sum_nge 1 f_n cos left( (2n-1)x right) , .
$$
That is, only the odd indices have to be need to be included in the Fourier series representation of the delta Dirac function.
What I tries is:
$$
delta(x) = frac2pi sum_nge 1 cos left( (2n-1)x right) , ,
$$
which seems to lead to correct final results.
I am wondering whether this can be proved in a rigorous way and/or whether this is true in the first place. Any help or comment would be very helpful.
Mickhausen
real-analysis sequences-and-series trigonometry fourier-analysis fourier-series
asked Jul 18 at 10:52
Mickhausen St Ilgen Sandhausen
92419
add a comment |Â
up vote
1
down vote
favorite
The delta Dirac function can be presented in the form of Fourier series expansion as
$$
delta(x)=frac12pi
+frac1pi sum_nge 1 cos (nx) , .
$$
The proof is straightforward and can be found e.g. in this Mathworld Wolfram article.
In a mathematical physics problem, I would like to make use of the Fourier series technique in order to solve a system of differential equations involving a delta Dirac function, such that the solution has the form
$$
f(x) = sum_nge 1 f_n cos left( (2n-1)x right) , .
$$
That is, only the odd indices have to be need to be included in the Fourier series representation of the delta Dirac function.
What I tries is:
$$
delta(x) = frac2pi sum_nge 1 cos left( (2n-1)x right) , ,
$$
which seems to lead to correct final results.
I am wondering whether this can be proved in a rigorous way and/or whether this is true in the first place. Any help or comment would be very helpful.
Mickhausen
real-analysis sequences-and-series trigonometry fourier-analysis fourier-series
asked Jul 18 at 10:52
Mickhausen St Ilgen Sandhausen
92419
1
How rigorous do you mean? Are you willing to delve into the theory of distributions?
– Jason Born
Jul 18 at 12:06
@JasonBorn A basic demonstration using the standard analytical tools would be highly appreciated. The expression seems to work out nicely but i unfortunately do not know why..
– Mickhausen St Ilgen Sandhausen
Jul 18 at 12:14
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
The delta Dirac function can be presented in the form of Fourier series expansion as
$$
delta(x)=frac12pi
+frac1pi sum_nge 1 cos (nx) , .
$$
The proof is straightforward and can be found e.g. in this Mathworld Wolfram article.
In a mathematical physics problem, I would like to make use of the Fourier series technique in order to solve a system of differential equations involving a delta Dirac function, such that the solution has the form
$$
f(x) = sum_nge 1 f_n cos left( (2n-1)x right) , .
$$
That is, only the odd indices have to be need to be included in the Fourier series representation of the delta Dirac function.
What I tries is:
$$
delta(x) = frac2pi sum_nge 1 cos left( (2n-1)x right) , ,
$$
which seems to lead to correct final results.
I am wondering whether this can be proved in a rigorous way and/or whether this is true in the first place. Any help or comment would be very helpful.
Mickhausen
real-analysis sequences-and-series trigonometry fourier-analysis fourier-series
asked Jul 18 at 10:52
Mickhausen St Ilgen Sandhausen
92419
The delta Dirac function can be presented in the form of Fourier series expansion as
$$
delta(x)=frac12pi
+frac1pi sum_nge 1 cos (nx) , .
$$
The proof is straightforward and can be found e.g. in this Mathworld Wolfram article.
In a mathematical physics problem, I would like to make use of the Fourier series technique in order to solve a system of differential equations involving a delta Dirac function, such that the solution has the form
$$
f(x) = sum_nge 1 f_n cos left( (2n-1)x right) , .
$$
That is, only the odd indices have to be need to be included in the Fourier series representation of the delta Dirac function.
What I tries is:
$$
delta(x) = frac2pi sum_nge 1 cos left( (2n-1)x right) , ,
$$
which seems to lead to correct final results.
I am wondering whether this can be proved in a rigorous way and/or whether this is true in the first place. Any help or comment would be very helpful.
Mickhausen
real-analysis sequences-and-series trigonometry fourier-analysis fourier-series
asked Jul 18 at 10:52
Mickhausen St Ilgen Sandhausen
92419
edited Jul 18 at 11:57
asked Jul 18 at 10:52
Mickhausen St Ilgen Sandhausen
92419
asked Jul 18 at 10:52
Mickhausen St Ilgen Sandhausen
92419
asked Jul 18 at 10:52
Mickhausen St Ilgen Sandhausen
92419
92419
1
How rigorous do you mean? Are you willing to delve into the theory of distributions?
– Jason Born
Jul 18 at 12:06
@JasonBorn A basic demonstration using the standard analytical tools would be highly appreciated. The expression seems to work out nicely but i unfortunately do not know why..
– Mickhausen St Ilgen Sandhausen
Jul 18 at 12:14
add a comment |Â
1
How rigorous do you mean? Are you willing to delve into the theory of distributions?
– Jason Born
Jul 18 at 12:06
@JasonBorn A basic demonstration using the standard analytical tools would be highly appreciated. The expression seems to work out nicely but i unfortunately do not know why..
– Mickhausen St Ilgen Sandhausen
Jul 18 at 12:14
1
1
How rigorous do you mean? Are you willing to delve into the theory of distributions?
– Jason Born
Jul 18 at 12:06
How rigorous do you mean? Are you willing to delve into the theory of distributions?
– Jason Born
Jul 18 at 12:06
@JasonBorn A basic demonstration using the standard analytical tools would be highly appreciated. The expression seems to work out nicely but i unfortunately do not know why..
– Mickhausen St Ilgen Sandhausen
Jul 18 at 12:14
@JasonBorn A basic demonstration using the standard analytical tools would be highly appreciated. The expression seems to work out nicely but i unfortunately do not know why..
– Mickhausen St Ilgen Sandhausen
Jul 18 at 12:14
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
The details below are a bit of a quick and dirty (more physicist approach) and that it ought to be demonstrated properly by distribution theory.
The expansion that you use is not a delta function but a combination thereof. I.e., it is not to difficult to check that:
$$
delta(x) - delta(x-pi) = frac2pi sum_n ~textodd cos( n x)
$$
Note that the series expansion implies that you consider functions with a $2pi$-periodicity, and that one also could have written it in a more symmetric form as
$$
delta(x) - frac12 left[ delta(x-pi) + delta(x+pi)right]
$$
that again should be repeated periodically.
Checking the validity of the expansion needs a a little extra attention because the second delta peak is located at the boundary of the integration interval for the coefficients. One simple approach to circumvent this issue is to consider the complex series:
$$
f(x) = sum_n A_n e^imath n x
$$
with
$$
A_n = frac12 pi int_-pi/2^3 pi/2 f(x) e^-imath n x textd x
$$
which would not affect the usual definition for an arbitrary function $f(x)$, but here ensures that both delta functions are within the integration interval. As an alternative approach one could count the delta functions only half if they are exactly on the boundaries.
Whether your approach is valid or not depends on the actual problem. It could be accidental of course, but I would not be surprised if the the underlying symmetry of the problem indeed allows for it.
answered Jul 18 at 15:15
Ronald Blaak
1,69438
add a comment |Â
up vote
1
down vote
No, that can't be right. The Fourier series of a distribution is unique.
answered Jul 18 at 15:30
David C. Ullrich
54.3k33583
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The details below are a bit of a quick and dirty (more physicist approach) and that it ought to be demonstrated properly by distribution theory.
The expansion that you use is not a delta function but a combination thereof. I.e., it is not to difficult to check that:
$$
delta(x) - delta(x-pi) = frac2pi sum_n ~textodd cos( n x)
$$
Note that the series expansion implies that you consider functions with a $2pi$-periodicity, and that one also could have written it in a more symmetric form as
$$
delta(x) - frac12 left[ delta(x-pi) + delta(x+pi)right]
$$
that again should be repeated periodically.
Checking the validity of the expansion needs a a little extra attention because the second delta peak is located at the boundary of the integration interval for the coefficients. One simple approach to circumvent this issue is to consider the complex series:
$$
f(x) = sum_n A_n e^imath n x
$$
with
$$
A_n = frac12 pi int_-pi/2^3 pi/2 f(x) e^-imath n x textd x
$$
which would not affect the usual definition for an arbitrary function $f(x)$, but here ensures that both delta functions are within the integration interval. As an alternative approach one could count the delta functions only half if they are exactly on the boundaries.
Whether your approach is valid or not depends on the actual problem. It could be accidental of course, but I would not be surprised if the the underlying symmetry of the problem indeed allows for it.
answered Jul 18 at 15:15
Ronald Blaak
1,69438
add a comment |Â
up vote
1
down vote
accepted
The details below are a bit of a quick and dirty (more physicist approach) and that it ought to be demonstrated properly by distribution theory.
The expansion that you use is not a delta function but a combination thereof. I.e., it is not to difficult to check that:
$$
delta(x) - delta(x-pi) = frac2pi sum_n ~textodd cos( n x)
$$
Note that the series expansion implies that you consider functions with a $2pi$-periodicity, and that one also could have written it in a more symmetric form as
$$
delta(x) - frac12 left[ delta(x-pi) + delta(x+pi)right]
$$
that again should be repeated periodically.
Checking the validity of the expansion needs a a little extra attention because the second delta peak is located at the boundary of the integration interval for the coefficients. One simple approach to circumvent this issue is to consider the complex series:
$$
f(x) = sum_n A_n e^imath n x
$$
with
$$
A_n = frac12 pi int_-pi/2^3 pi/2 f(x) e^-imath n x textd x
$$
which would not affect the usual definition for an arbitrary function $f(x)$, but here ensures that both delta functions are within the integration interval. As an alternative approach one could count the delta functions only half if they are exactly on the boundaries.
Whether your approach is valid or not depends on the actual problem. It could be accidental of course, but I would not be surprised if the the underlying symmetry of the problem indeed allows for it.
answered Jul 18 at 15:15
Ronald Blaak
1,69438
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The details below are a bit of a quick and dirty (more physicist approach) and that it ought to be demonstrated properly by distribution theory.
The expansion that you use is not a delta function but a combination thereof. I.e., it is not to difficult to check that:
$$
delta(x) - delta(x-pi) = frac2pi sum_n ~textodd cos( n x)
$$
Note that the series expansion implies that you consider functions with a $2pi$-periodicity, and that one also could have written it in a more symmetric form as
$$
delta(x) - frac12 left[ delta(x-pi) + delta(x+pi)right]
$$
that again should be repeated periodically.
Checking the validity of the expansion needs a a little extra attention because the second delta peak is located at the boundary of the integration interval for the coefficients. One simple approach to circumvent this issue is to consider the complex series:
$$
f(x) = sum_n A_n e^imath n x
$$
with
$$
A_n = frac12 pi int_-pi/2^3 pi/2 f(x) e^-imath n x textd x
$$
which would not affect the usual definition for an arbitrary function $f(x)$, but here ensures that both delta functions are within the integration interval. As an alternative approach one could count the delta functions only half if they are exactly on the boundaries.
Whether your approach is valid or not depends on the actual problem. It could be accidental of course, but I would not be surprised if the the underlying symmetry of the problem indeed allows for it.
answered Jul 18 at 15:15
Ronald Blaak
1,69438
The details below are a bit of a quick and dirty (more physicist approach) and that it ought to be demonstrated properly by distribution theory.
The expansion that you use is not a delta function but a combination thereof. I.e., it is not to difficult to check that:
$$
delta(x) - delta(x-pi) = frac2pi sum_n ~textodd cos( n x)
$$
Note that the series expansion implies that you consider functions with a $2pi$-periodicity, and that one also could have written it in a more symmetric form as
$$
delta(x) - frac12 left[ delta(x-pi) + delta(x+pi)right]
$$
that again should be repeated periodically.
Checking the validity of the expansion needs a a little extra attention because the second delta peak is located at the boundary of the integration interval for the coefficients. One simple approach to circumvent this issue is to consider the complex series:
$$
f(x) = sum_n A_n e^imath n x
$$
with
$$
A_n = frac12 pi int_-pi/2^3 pi/2 f(x) e^-imath n x textd x
$$
which would not affect the usual definition for an arbitrary function $f(x)$, but here ensures that both delta functions are within the integration interval. As an alternative approach one could count the delta functions only half if they are exactly on the boundaries.
Whether your approach is valid or not depends on the actual problem. It could be accidental of course, but I would not be surprised if the the underlying symmetry of the problem indeed allows for it.
answered Jul 18 at 15:15
Ronald Blaak
1,69438
answered Jul 18 at 15:15
Ronald Blaak
1,69438
answered Jul 18 at 15:15
Ronald Blaak
1,69438
answered Jul 18 at 15:15
Ronald Blaak
1,69438
1,69438
add a comment |Â
add a comment |Â
up vote
1
down vote
No, that can't be right. The Fourier series of a distribution is unique.
answered Jul 18 at 15:30
David C. Ullrich
54.3k33583
add a comment |Â
up vote
1
down vote
No, that can't be right. The Fourier series of a distribution is unique.
answered Jul 18 at 15:30
David C. Ullrich
54.3k33583
add a comment |Â
up vote
1
down vote
up vote
1
down vote
No, that can't be right. The Fourier series of a distribution is unique.
answered Jul 18 at 15:30
David C. Ullrich
54.3k33583
No, that can't be right. The Fourier series of a distribution is unique.
answered Jul 18 at 15:30
David C. Ullrich
54.3k33583
answered Jul 18 at 15:30
David C. Ullrich
54.3k33583
answered Jul 18 at 15:30
David C. Ullrich
54.3k33583
answered Jul 18 at 15:30
David C. Ullrich
54.3k33583
54.3k33583
add a comment |Â
add a comment |Â
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1
How rigorous do you mean? Are you willing to delve into the theory of distributions?
– Jason Born
Jul 18 at 12:06
@JasonBorn A basic demonstration using the standard analytical tools would be highly appreciated. The expression seems to work out nicely but i unfortunately do not know why..
– Mickhausen St Ilgen Sandhausen
Jul 18 at 12:14