Is there a difference between $g(x,y)$ and $g(x;y)$?
Is there a reason that the second notation uses a semicolon?
Here's the definition:
we say $g(x;y)$ is a Green's function
$$g(x;y) = left{
begin{array}{lr}
sin(kx)sin(k(y-1)/ksin(k) & : x lt y\
sin(ky)sin(k(x-1)/ksin(k) & : y lt x\ end{array}
right.$$
notation
add a comment |
Is there a reason that the second notation uses a semicolon?
Here's the definition:
we say $g(x;y)$ is a Green's function
$$g(x;y) = left{
begin{array}{lr}
sin(kx)sin(k(y-1)/ksin(k) & : x lt y\
sin(ky)sin(k(x-1)/ksin(k) & : y lt x\ end{array}
right.$$
notation
6
There is no difference, but the notation $g(x;y)$ suggests that we are going to think of $y$ as a parameter. For a fixed value of $y$, we will be interested in the function $x mapsto g(x; y)$. It would have been ok to use the notation $g(x,y)$ instead, and some authors do this when discussing Green's functions.
– littleO
Jan 4 at 14:03
1
@littleO it seems like you could make that comment verbatim into an answer
– Mark S.
Jan 4 at 14:15
add a comment |
Is there a reason that the second notation uses a semicolon?
Here's the definition:
we say $g(x;y)$ is a Green's function
$$g(x;y) = left{
begin{array}{lr}
sin(kx)sin(k(y-1)/ksin(k) & : x lt y\
sin(ky)sin(k(x-1)/ksin(k) & : y lt x\ end{array}
right.$$
notation
Is there a reason that the second notation uses a semicolon?
Here's the definition:
we say $g(x;y)$ is a Green's function
$$g(x;y) = left{
begin{array}{lr}
sin(kx)sin(k(y-1)/ksin(k) & : x lt y\
sin(ky)sin(k(x-1)/ksin(k) & : y lt x\ end{array}
right.$$
notation
notation
edited Jan 4 at 14:48
Le Anh Dung
1,0531521
1,0531521
asked Jan 4 at 13:59
user29418user29418
444513
444513
6
There is no difference, but the notation $g(x;y)$ suggests that we are going to think of $y$ as a parameter. For a fixed value of $y$, we will be interested in the function $x mapsto g(x; y)$. It would have been ok to use the notation $g(x,y)$ instead, and some authors do this when discussing Green's functions.
– littleO
Jan 4 at 14:03
1
@littleO it seems like you could make that comment verbatim into an answer
– Mark S.
Jan 4 at 14:15
add a comment |
6
There is no difference, but the notation $g(x;y)$ suggests that we are going to think of $y$ as a parameter. For a fixed value of $y$, we will be interested in the function $x mapsto g(x; y)$. It would have been ok to use the notation $g(x,y)$ instead, and some authors do this when discussing Green's functions.
– littleO
Jan 4 at 14:03
1
@littleO it seems like you could make that comment verbatim into an answer
– Mark S.
Jan 4 at 14:15
6
6
There is no difference, but the notation $g(x;y)$ suggests that we are going to think of $y$ as a parameter. For a fixed value of $y$, we will be interested in the function $x mapsto g(x; y)$. It would have been ok to use the notation $g(x,y)$ instead, and some authors do this when discussing Green's functions.
– littleO
Jan 4 at 14:03
There is no difference, but the notation $g(x;y)$ suggests that we are going to think of $y$ as a parameter. For a fixed value of $y$, we will be interested in the function $x mapsto g(x; y)$. It would have been ok to use the notation $g(x,y)$ instead, and some authors do this when discussing Green's functions.
– littleO
Jan 4 at 14:03
1
1
@littleO it seems like you could make that comment verbatim into an answer
– Mark S.
Jan 4 at 14:15
@littleO it seems like you could make that comment verbatim into an answer
– Mark S.
Jan 4 at 14:15
add a comment |
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I'll post my comment here so that the question receives an answer:
There is no difference, but the notation $g(x;y)$ suggests that we are going to think of $y$ as a parameter. For a fixed value of $y$, we will be interested in the function $x mapsto g(x;y)$. It would have been ok to use the notation $g(x,y)$ instead, and some authors do this when discussing Green's functions.
add a comment |
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I'll post my comment here so that the question receives an answer:
There is no difference, but the notation $g(x;y)$ suggests that we are going to think of $y$ as a parameter. For a fixed value of $y$, we will be interested in the function $x mapsto g(x;y)$. It would have been ok to use the notation $g(x,y)$ instead, and some authors do this when discussing Green's functions.
add a comment |
I'll post my comment here so that the question receives an answer:
There is no difference, but the notation $g(x;y)$ suggests that we are going to think of $y$ as a parameter. For a fixed value of $y$, we will be interested in the function $x mapsto g(x;y)$. It would have been ok to use the notation $g(x,y)$ instead, and some authors do this when discussing Green's functions.
add a comment |
I'll post my comment here so that the question receives an answer:
There is no difference, but the notation $g(x;y)$ suggests that we are going to think of $y$ as a parameter. For a fixed value of $y$, we will be interested in the function $x mapsto g(x;y)$. It would have been ok to use the notation $g(x,y)$ instead, and some authors do this when discussing Green's functions.
I'll post my comment here so that the question receives an answer:
There is no difference, but the notation $g(x;y)$ suggests that we are going to think of $y$ as a parameter. For a fixed value of $y$, we will be interested in the function $x mapsto g(x;y)$. It would have been ok to use the notation $g(x,y)$ instead, and some authors do this when discussing Green's functions.
answered Jan 4 at 14:23
littleOlittleO
29.3k644109
29.3k644109
add a comment |
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There is no difference, but the notation $g(x;y)$ suggests that we are going to think of $y$ as a parameter. For a fixed value of $y$, we will be interested in the function $x mapsto g(x; y)$. It would have been ok to use the notation $g(x,y)$ instead, and some authors do this when discussing Green's functions.
– littleO
Jan 4 at 14:03
1
@littleO it seems like you could make that comment verbatim into an answer
– Mark S.
Jan 4 at 14:15