Mixing
Continuous paths joining positive x-axis to negative x-axis, through upper half plane
$begingroup$
I need a way to parametrise all continuous paths from the positive to the negative x-axis, which go through the upper half plane (in $2$ dimensions).
I do not care about the speed of the parametrization, just as long as I can describe the set of points the curve passes through.
I have two parametrisations in mind and 2 questions about them.
$1.$ ($f(t)cos(t),g(t)sin(t))$, where $f,g$ are continuous with $|f|,|g|>0$ and $tin [0,pi]$.
$2.$ $r(t)e^{it}$, where $r$ is continuous with $|r|>0$ and $tin [0,pi]$.
First question are these parametrizations equivalent? The second one seems easier to work with and is a special case of the first one where $f=g$.
Is there a known way of doing this, based on some known result?
parametrization
asked Jan 25 at 13:37
pureundergradpureundergrad
547211
$endgroup$
add a comment |
$begingroup$
I need a way to parametrise all continuous paths from the positive to the negative x-axis, which go through the upper half plane (in $2$ dimensions).
I do not care about the speed of the parametrization, just as long as I can describe the set of points the curve passes through.
I have two parametrisations in mind and 2 questions about them.
$1.$ ($f(t)cos(t),g(t)sin(t))$, where $f,g$ are continuous with $|f|,|g|>0$ and $tin [0,pi]$.
$2.$ $r(t)e^{it}$, where $r$ is continuous with $|r|>0$ and $tin [0,pi]$.
First question are these parametrizations equivalent? The second one seems easier to work with and is a special case of the first one where $f=g$.
Is there a known way of doing this, based on some known result?
parametrization
asked Jan 25 at 13:37
pureundergradpureundergrad
547211
$endgroup$
$begingroup$
Am I correct in believing that you only care about paths that take on each $x$-value at most once? Otherwise, both these approaches miss out on almost all paths. Think, for example, of a random walk in the open upper half-plane, but starting at, say $(1,0)$ and ending at $(-1, 0)$. It's likely to cross the $y$-axis a great many times, but in both your first and second parameterizations, there's only one point with $t = pi/2$.
$endgroup$
– John Hughes
Jan 25 at 13:56
$begingroup$
Good point, I would need to think about that, but for now lets rule that possibility out
$endgroup$
– pureundergrad
Jan 25 at 14:02
$begingroup$
I was mistaken about the first parameterization --- it DOES allow some kinds of backtracking, although not across the $y$-axis. More details in my answer below.
$endgroup$
– John Hughes
Jan 25 at 14:11
add a comment |
$begingroup$
I need a way to parametrise all continuous paths from the positive to the negative x-axis, which go through the upper half plane (in $2$ dimensions).
I do not care about the speed of the parametrization, just as long as I can describe the set of points the curve passes through.
I have two parametrisations in mind and 2 questions about them.
$1.$ ($f(t)cos(t),g(t)sin(t))$, where $f,g$ are continuous with $|f|,|g|>0$ and $tin [0,pi]$.
$2.$ $r(t)e^{it}$, where $r$ is continuous with $|r|>0$ and $tin [0,pi]$.
First question are these parametrizations equivalent? The second one seems easier to work with and is a special case of the first one where $f=g$.
Is there a known way of doing this, based on some known result?
parametrization
asked Jan 25 at 13:37
pureundergradpureundergrad
547211
$endgroup$
I need a way to parametrise all continuous paths from the positive to the negative x-axis, which go through the upper half plane (in $2$ dimensions).
I do not care about the speed of the parametrization, just as long as I can describe the set of points the curve passes through.
I have two parametrisations in mind and 2 questions about them.
$1.$ ($f(t)cos(t),g(t)sin(t))$, where $f,g$ are continuous with $|f|,|g|>0$ and $tin [0,pi]$.
$2.$ $r(t)e^{it}$, where $r$ is continuous with $|r|>0$ and $tin [0,pi]$.
First question are these parametrizations equivalent? The second one seems easier to work with and is a special case of the first one where $f=g$.
Is there a known way of doing this, based on some known result?
parametrization
parametrization
asked Jan 25 at 13:37
pureundergradpureundergrad
547211
asked Jan 25 at 13:37
pureundergradpureundergrad
547211
asked Jan 25 at 13:37
pureundergradpureundergrad
547211
asked Jan 25 at 13:37
pureundergradpureundergrad
547211
asked Jan 25 at 13:37
pureundergradpureundergrad
547211
547211
$begingroup$
Am I correct in believing that you only care about paths that take on each $x$-value at most once? Otherwise, both these approaches miss out on almost all paths. Think, for example, of a random walk in the open upper half-plane, but starting at, say $(1,0)$ and ending at $(-1, 0)$. It's likely to cross the $y$-axis a great many times, but in both your first and second parameterizations, there's only one point with $t = pi/2$.
$endgroup$
– John Hughes
Jan 25 at 13:56
$begingroup$
Good point, I would need to think about that, but for now lets rule that possibility out
$endgroup$
– pureundergrad
Jan 25 at 14:02
$begingroup$
I was mistaken about the first parameterization --- it DOES allow some kinds of backtracking, although not across the $y$-axis. More details in my answer below.
$endgroup$
– John Hughes
Jan 25 at 14:11
add a comment |
$begingroup$
Am I correct in believing that you only care about paths that take on each $x$-value at most once? Otherwise, both these approaches miss out on almost all paths. Think, for example, of a random walk in the open upper half-plane, but starting at, say $(1,0)$ and ending at $(-1, 0)$. It's likely to cross the $y$-axis a great many times, but in both your first and second parameterizations, there's only one point with $t = pi/2$.
$endgroup$
– John Hughes
Jan 25 at 13:56
$begingroup$
Good point, I would need to think about that, but for now lets rule that possibility out
$endgroup$
– pureundergrad
Jan 25 at 14:02
$begingroup$
I was mistaken about the first parameterization --- it DOES allow some kinds of backtracking, although not across the $y$-axis. More details in my answer below.
$endgroup$
– John Hughes
Jan 25 at 14:11
$begingroup$
Am I correct in believing that you only care about paths that take on each $x$-value at most once? Otherwise, both these approaches miss out on almost all paths. Think, for example, of a random walk in the open upper half-plane, but starting at, say $(1,0)$ and ending at $(-1, 0)$. It's likely to cross the $y$-axis a great many times, but in both your first and second parameterizations, there's only one point with $t = pi/2$.
$endgroup$
– John Hughes
Jan 25 at 13:56
$begingroup$
Am I correct in believing that you only care about paths that take on each $x$-value at most once? Otherwise, both these approaches miss out on almost all paths. Think, for example, of a random walk in the open upper half-plane, but starting at, say $(1,0)$ and ending at $(-1, 0)$. It's likely to cross the $y$-axis a great many times, but in both your first and second parameterizations, there's only one point with $t = pi/2$.
$endgroup$
– John Hughes
Jan 25 at 13:56
$begingroup$
Good point, I would need to think about that, but for now lets rule that possibility out
$endgroup$
– pureundergrad
Jan 25 at 14:02
$begingroup$
Good point, I would need to think about that, but for now lets rule that possibility out
$endgroup$
– pureundergrad
Jan 25 at 14:02
$begingroup$
I was mistaken about the first parameterization --- it DOES allow some kinds of backtracking, although not across the $y$-axis. More details in my answer below.
$endgroup$
– John Hughes
Jan 25 at 14:11
$begingroup$
I was mistaken about the first parameterization --- it DOES allow some kinds of backtracking, although not across the $y$-axis. More details in my answer below.
$endgroup$
– John Hughes
Jan 25 at 14:11
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The parameterizations are not equivalent. Consider
$$
g(t) = 1
$$
and
$$
f(t) = 1 + sin(100 t)
$$
For $t$ near $0$, this looks like a rapid left-right zigzag rising from the point $(1, 0)$. Because of this, the ray from the origin in the direction $(1,1)$ (shown in magenta in the image below) intersects it multiple times
.
Hence in the second parameterization, you'd need multiple values for $r(pi/4)$, which is impossible if $r$ is a function.
On the other hand, if you're willing to work with "at most one $y$-value for each $x$-value," then you can parameterize all such paths with the following:
begin{align}
x(t) = (1-t)a + t b & text{where $a > 0, b < 0$}\
y(t) = h(t) * sin(pi t)
end{align}
where $h: [0, 1] to Bbb R^{+}$ is continuous.
This parameterization has the charm that for any continuous curve from the point $a$ (on the positive $x$-axis) to $b$ (on the negative $x$-axis) that meets each $x$-coordinate at most once, there is a unique function $h$, and vice versa.
[I'm assuming that by "upper half plane" you mean the open upper half plane, i.e., points with $y > 0$. If you want to allow the closed upper half plane ($y ge 0$), then the target for $h$ should be the nonnegative reals rather than the positive reals. ]
answered Jan 25 at 14:25
John HughesJohn Hughes
64.6k24191
$endgroup$
$begingroup$
Excellent! Is there anywhere I can read more about that last parametrization? Does it have a name?
$endgroup$
– pureundergrad
Jan 25 at 14:28
$begingroup$
Well...the $x$-part is just "linear interpolation from $a$ to $b$." The $y$-part was inspired by your sine-and-cosine parameterization. So...to be honest, it was just invented. But roughly speaking, it's almost exactly the graph of $h$, except for the sine factor that makes it touch the $x$-axis at the ends. There's nothing really clever or special about it, and it surely doesn't have a name.
$endgroup$
– John Hughes
Jan 25 at 14:39
add a comment |
Your Answer
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1 Answer
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1 Answer
1
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oldest
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active
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oldest
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$begingroup$
The parameterizations are not equivalent. Consider
$$
g(t) = 1
$$
and
$$
f(t) = 1 + sin(100 t)
$$
For $t$ near $0$, this looks like a rapid left-right zigzag rising from the point $(1, 0)$. Because of this, the ray from the origin in the direction $(1,1)$ (shown in magenta in the image below) intersects it multiple times.
Hence in the second parameterization, you'd need multiple values for $r(pi/4)$, which is impossible if $r$ is a function.
On the other hand, if you're willing to work with "at most one $y$-value for each $x$-value," then you can parameterize all such paths with the following:
begin{align}
x(t) = (1-t)a + t b & text{where $a > 0, b < 0$}\
y(t) = h(t) * sin(pi t)
end{align}
where $h: [0, 1] to Bbb R^{+}$ is continuous.
This parameterization has the charm that for any continuous curve from the point $a$ (on the positive $x$-axis) to $b$ (on the negative $x$-axis) that meets each $x$-coordinate at most once, there is a unique function $h$, and vice versa.
[I'm assuming that by "upper half plane" you mean the open upper half plane, i.e., points with $y > 0$. If you want to allow the closed upper half plane ($y ge 0$), then the target for $h$ should be the nonnegative reals rather than the positive reals. ]
answered Jan 25 at 14:25
John HughesJohn Hughes
64.6k24191
$endgroup$
$begingroup$
Excellent! Is there anywhere I can read more about that last parametrization? Does it have a name?
$endgroup$
– pureundergrad
Jan 25 at 14:28
$begingroup$
Well...the $x$-part is just "linear interpolation from $a$ to $b$." The $y$-part was inspired by your sine-and-cosine parameterization. So...to be honest, it was just invented. But roughly speaking, it's almost exactly the graph of $h$, except for the sine factor that makes it touch the $x$-axis at the ends. There's nothing really clever or special about it, and it surely doesn't have a name.
$endgroup$
– John Hughes
Jan 25 at 14:39
add a comment |
$begingroup$
The parameterizations are not equivalent. Consider
$$
g(t) = 1
$$
and
$$
f(t) = 1 + sin(100 t)
$$
For $t$ near $0$, this looks like a rapid left-right zigzag rising from the point $(1, 0)$. Because of this, the ray from the origin in the direction $(1,1)$ (shown in magenta in the image below) intersects it multiple times.
Hence in the second parameterization, you'd need multiple values for $r(pi/4)$, which is impossible if $r$ is a function.
On the other hand, if you're willing to work with "at most one $y$-value for each $x$-value," then you can parameterize all such paths with the following:
begin{align}
x(t) = (1-t)a + t b & text{where $a > 0, b < 0$}\
y(t) = h(t) * sin(pi t)
end{align}
where $h: [0, 1] to Bbb R^{+}$ is continuous.
This parameterization has the charm that for any continuous curve from the point $a$ (on the positive $x$-axis) to $b$ (on the negative $x$-axis) that meets each $x$-coordinate at most once, there is a unique function $h$, and vice versa.
[I'm assuming that by "upper half plane" you mean the open upper half plane, i.e., points with $y > 0$. If you want to allow the closed upper half plane ($y ge 0$), then the target for $h$ should be the nonnegative reals rather than the positive reals. ]
answered Jan 25 at 14:25
John HughesJohn Hughes
64.6k24191
$endgroup$
$begingroup$
Excellent! Is there anywhere I can read more about that last parametrization? Does it have a name?
$endgroup$
– pureundergrad
Jan 25 at 14:28
$begingroup$
Well...the $x$-part is just "linear interpolation from $a$ to $b$." The $y$-part was inspired by your sine-and-cosine parameterization. So...to be honest, it was just invented. But roughly speaking, it's almost exactly the graph of $h$, except for the sine factor that makes it touch the $x$-axis at the ends. There's nothing really clever or special about it, and it surely doesn't have a name.
$endgroup$
– John Hughes
Jan 25 at 14:39
add a comment |
$begingroup$
The parameterizations are not equivalent. Consider
$$
g(t) = 1
$$
and
$$
f(t) = 1 + sin(100 t)
$$
For $t$ near $0$, this looks like a rapid left-right zigzag rising from the point $(1, 0)$. Because of this, the ray from the origin in the direction $(1,1)$ (shown in magenta in the image below) intersects it multiple times.
Hence in the second parameterization, you'd need multiple values for $r(pi/4)$, which is impossible if $r$ is a function.
On the other hand, if you're willing to work with "at most one $y$-value for each $x$-value," then you can parameterize all such paths with the following:
begin{align}
x(t) = (1-t)a + t b & text{where $a > 0, b < 0$}\
y(t) = h(t) * sin(pi t)
end{align}
where $h: [0, 1] to Bbb R^{+}$ is continuous.
This parameterization has the charm that for any continuous curve from the point $a$ (on the positive $x$-axis) to $b$ (on the negative $x$-axis) that meets each $x$-coordinate at most once, there is a unique function $h$, and vice versa.
[I'm assuming that by "upper half plane" you mean the open upper half plane, i.e., points with $y > 0$. If you want to allow the closed upper half plane ($y ge 0$), then the target for $h$ should be the nonnegative reals rather than the positive reals. ]
answered Jan 25 at 14:25
John HughesJohn Hughes
64.6k24191
$endgroup$
The parameterizations are not equivalent. Consider
$$
g(t) = 1
$$
and
$$
f(t) = 1 + sin(100 t)
$$
For $t$ near $0$, this looks like a rapid left-right zigzag rising from the point $(1, 0)$. Because of this, the ray from the origin in the direction $(1,1)$ (shown in magenta in the image below) intersects it multiple times.
Hence in the second parameterization, you'd need multiple values for $r(pi/4)$, which is impossible if $r$ is a function.
On the other hand, if you're willing to work with "at most one $y$-value for each $x$-value," then you can parameterize all such paths with the following:
begin{align}
x(t) = (1-t)a + t b & text{where $a > 0, b < 0$}\
y(t) = h(t) * sin(pi t)
end{align}
where $h: [0, 1] to Bbb R^{+}$ is continuous.
This parameterization has the charm that for any continuous curve from the point $a$ (on the positive $x$-axis) to $b$ (on the negative $x$-axis) that meets each $x$-coordinate at most once, there is a unique function $h$, and vice versa.
[I'm assuming that by "upper half plane" you mean the open upper half plane, i.e., points with $y > 0$. If you want to allow the closed upper half plane ($y ge 0$), then the target for $h$ should be the nonnegative reals rather than the positive reals. ]
answered Jan 25 at 14:25
John HughesJohn Hughes
64.6k24191
edited Jan 25 at 14:39
answered Jan 25 at 14:25
John HughesJohn Hughes
64.6k24191
answered Jan 25 at 14:25
John HughesJohn Hughes
64.6k24191
answered Jan 25 at 14:25
John HughesJohn Hughes
64.6k24191
64.6k24191
$begingroup$
Excellent! Is there anywhere I can read more about that last parametrization? Does it have a name?
$endgroup$
– pureundergrad
Jan 25 at 14:28
$begingroup$
Well...the $x$-part is just "linear interpolation from $a$ to $b$." The $y$-part was inspired by your sine-and-cosine parameterization. So...to be honest, it was just invented. But roughly speaking, it's almost exactly the graph of $h$, except for the sine factor that makes it touch the $x$-axis at the ends. There's nothing really clever or special about it, and it surely doesn't have a name.
$endgroup$
– John Hughes
Jan 25 at 14:39
add a comment |
$begingroup$
Excellent! Is there anywhere I can read more about that last parametrization? Does it have a name?
$endgroup$
– pureundergrad
Jan 25 at 14:28
$begingroup$
Well...the $x$-part is just "linear interpolation from $a$ to $b$." The $y$-part was inspired by your sine-and-cosine parameterization. So...to be honest, it was just invented. But roughly speaking, it's almost exactly the graph of $h$, except for the sine factor that makes it touch the $x$-axis at the ends. There's nothing really clever or special about it, and it surely doesn't have a name.
$endgroup$
– John Hughes
Jan 25 at 14:39
$begingroup$
Excellent! Is there anywhere I can read more about that last parametrization? Does it have a name?
$endgroup$
– pureundergrad
Jan 25 at 14:28
$begingroup$
Excellent! Is there anywhere I can read more about that last parametrization? Does it have a name?
$endgroup$
– pureundergrad
Jan 25 at 14:28
$begingroup$
Well...the $x$-part is just "linear interpolation from $a$ to $b$." The $y$-part was inspired by your sine-and-cosine parameterization. So...to be honest, it was just invented. But roughly speaking, it's almost exactly the graph of $h$, except for the sine factor that makes it touch the $x$-axis at the ends. There's nothing really clever or special about it, and it surely doesn't have a name.
$endgroup$
– John Hughes
Jan 25 at 14:39
$begingroup$
Well...the $x$-part is just "linear interpolation from $a$ to $b$." The $y$-part was inspired by your sine-and-cosine parameterization. So...to be honest, it was just invented. But roughly speaking, it's almost exactly the graph of $h$, except for the sine factor that makes it touch the $x$-axis at the ends. There's nothing really clever or special about it, and it surely doesn't have a name.
$endgroup$
– John Hughes
Jan 25 at 14:39
add a comment |
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$begingroup$
Am I correct in believing that you only care about paths that take on each $x$-value at most once? Otherwise, both these approaches miss out on almost all paths. Think, for example, of a random walk in the open upper half-plane, but starting at, say $(1,0)$ and ending at $(-1, 0)$. It's likely to cross the $y$-axis a great many times, but in both your first and second parameterizations, there's only one point with $t = pi/2$.
$endgroup$
– John Hughes
Jan 25 at 13:56
$begingroup$
Good point, I would need to think about that, but for now lets rule that possibility out
$endgroup$
– pureundergrad
Jan 25 at 14:02
$begingroup$
I was mistaken about the first parameterization --- it DOES allow some kinds of backtracking, although not across the $y$-axis. More details in my answer below.
$endgroup$
– John Hughes
Jan 25 at 14:11