Mixing
What is this curve?
Clash Royale CLAN TAG#URR8PPP
up vote
16
down vote
favorite
Lines (same angle space between) radiating outward from a point and intersecting a line:
This is the density distribution of the points on the line:
I used a Python script to calculate this. The angular interval is 0.01 degrees. X = tan(deg) rounded to the nearest tenth, so many points will have the same x. Y is the number of points for its X. You can see on the plot that ~5,700 points were between -0.05 and 0.05 on the line.
What is this graph curve called? What's the function equation?
geometry graphing-functions
asked Sep 9 at 20:36
clickbait
1868
New contributor
clickbait is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |Â
up vote
16
down vote
favorite
Lines (same angle space between) radiating outward from a point and intersecting a line:
This is the density distribution of the points on the line:
I used a Python script to calculate this. The angular interval is 0.01 degrees. X = tan(deg) rounded to the nearest tenth, so many points will have the same x. Y is the number of points for its X. You can see on the plot that ~5,700 points were between -0.05 and 0.05 on the line.
What is this graph curve called? What's the function equation?
geometry graphing-functions
asked Sep 9 at 20:36
clickbait
1868
New contributor
clickbait is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What are the numbers in your graph supposed to represent?
– Aretino
Sep 9 at 20:42
There are a few functions you could make look roughly like this, but I think its going to be of the form $amathrme^-bx^2$
– aidangallagher4
Sep 9 at 20:52
5
en.wikipedia.org/wiki/Cauchy_distribution
– Aretino
Sep 9 at 20:54
1
mathworld.wolfram.com/LorentzianFunction.html
– Aretino
Sep 9 at 20:58
add a comment |Â
up vote
16
down vote
favorite
up vote
16
down vote
favorite
Lines (same angle space between) radiating outward from a point and intersecting a line:
This is the density distribution of the points on the line:
I used a Python script to calculate this. The angular interval is 0.01 degrees. X = tan(deg) rounded to the nearest tenth, so many points will have the same x. Y is the number of points for its X. You can see on the plot that ~5,700 points were between -0.05 and 0.05 on the line.
What is this graph curve called? What's the function equation?
geometry graphing-functions
asked Sep 9 at 20:36
clickbait
1868
New contributor
clickbait is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Lines (same angle space between) radiating outward from a point and intersecting a line:
This is the density distribution of the points on the line:
I used a Python script to calculate this. The angular interval is 0.01 degrees. X = tan(deg) rounded to the nearest tenth, so many points will have the same x. Y is the number of points for its X. You can see on the plot that ~5,700 points were between -0.05 and 0.05 on the line.
What is this graph curve called? What's the function equation?
geometry graphing-functions
geometry graphing-functions
asked Sep 9 at 20:36
clickbait
1868
New contributor
clickbait is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Sep 9 at 20:36
clickbait
1868
New contributor
clickbait is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
asked Sep 9 at 20:36
clickbait
1868
New contributor
clickbait is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Sep 9 at 20:36
clickbait
1868
asked Sep 9 at 20:36
clickbait
1868
1868
New contributor
clickbait is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
clickbait is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
clickbait is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What are the numbers in your graph supposed to represent?
– Aretino
Sep 9 at 20:42
There are a few functions you could make look roughly like this, but I think its going to be of the form $amathrme^-bx^2$
– aidangallagher4
Sep 9 at 20:52
5
en.wikipedia.org/wiki/Cauchy_distribution
– Aretino
Sep 9 at 20:54
1
mathworld.wolfram.com/LorentzianFunction.html
– Aretino
Sep 9 at 20:58
add a comment |Â
What are the numbers in your graph supposed to represent?
– Aretino
Sep 9 at 20:42
There are a few functions you could make look roughly like this, but I think its going to be of the form $amathrme^-bx^2$
– aidangallagher4
Sep 9 at 20:52
5
en.wikipedia.org/wiki/Cauchy_distribution
– Aretino
Sep 9 at 20:54
1
mathworld.wolfram.com/LorentzianFunction.html
– Aretino
Sep 9 at 20:58
What are the numbers in your graph supposed to represent?
– Aretino
Sep 9 at 20:42
What are the numbers in your graph supposed to represent?
– Aretino
Sep 9 at 20:42
There are a few functions you could make look roughly like this, but I think its going to be of the form $amathrme^-bx^2$
– aidangallagher4
Sep 9 at 20:52
There are a few functions you could make look roughly like this, but I think its going to be of the form $amathrme^-bx^2$
– aidangallagher4
Sep 9 at 20:52
5
5
en.wikipedia.org/wiki/Cauchy_distribution
– Aretino
Sep 9 at 20:54
en.wikipedia.org/wiki/Cauchy_distribution
– Aretino
Sep 9 at 20:54
1
1
mathworld.wolfram.com/LorentzianFunction.html
– Aretino
Sep 9 at 20:58
mathworld.wolfram.com/LorentzianFunction.html
– Aretino
Sep 9 at 20:58
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
33
down vote
accepted
The curve is a density function. The idea is the following. From your first picture, assume that the angles of the rays are spaced evenly, the angle between two rays being $alpha$. I.e. the nth ray has angle $n cdot alpha$. The nth ray's intersection point $x$ with the line then follows $tan (n cdot alpha) = x/h$ where h is the distance of the line to the origin. So from 0 to $x$, n rays cross the line.
Now you are interested in the density $p(x)$, i.e. how many rays intersect the line at $x$, per line interval $Delta x$. In the limit of small $alpha$, you have $int_0^x p(x') dx' =c n = fraccalphaarctan (x/h)$ and correspondingly, $p(x) = fracddxfraccalphaarctan (x/h) = fracc halpha (x^2+h^2)$. The constant $c$ is determined since the integral over the density function must be $1$ (in probability sense), hence
$p(x)= frachpi (x^2+h^2)$.
This curve is called a Cauchy distribution.
Obviously, $p(x)$ can be multiplied with a constant $K$ to give an expectation value distribution $E(x) = K p(x)$ over $x$, instead of a probability distribution. This explains the large value of $E(0) = 5700$ or so in your picture. The value $h$ is also called a scale parameter, it specifies the half-width at half-maximum (HWHM) of the curve and can be roughly read off to be $1$. If we are really "counting rays", then with angle spacing $alpha$, in total $pi/alpha$ many rays intersect the line and hence we must have
$$
pi/alpha = int_-infty^inftyE(x) dx = int_-infty^inftyK p(x) dx = K
$$
So the expectation value distribution of the number of rays intersecting one unit of the line at position $x$ is
$$
E(x) = fracpialpha p(x)= frachalpha (x^2+h^2)
$$
as we already had with the constant $c=1$. Reading off approximately $E(0) = 5700$ and $h=1$ gives $
E(x) = frac5700x^2+1
$ and $alpha = 1/5700$ (in rads), or in other words, $pi/alpha simeq 17900$ rays (lower half plane) intersect the line.
answered Sep 9 at 21:17
Andreas
6,9051036
add a comment |Â
up vote
6
down vote
It can be shown assuming that the quantity from the ray is proportional to the angular interval that the denisty distribuiton function is in the form
$$y=fracdH+x^2$$
where $d$ is related to the intensity and $H$ is the distance of the source point from the line.
Here is the plot for the case $d=H=1$
answered Sep 9 at 20:56
gimusi
71.4k73786
add a comment |Â
up vote
1
down vote
This is also known, especially among we physicists as a Lorentz distribution.
We know that every point on the line has the same vertical distance from the source point. Let's call this $l_0$. We also know the ratio of the horizontal component of the distance to the vertical component is $tan(theta)$, where theta goes from $-piover2$ to $piover2$ in the increments given. So horizontal distance=$x=l_0*tan(theta)$.
To find the densities, first take the derivative of both sides, getting $dx=l_0*sec^2(theta)*dtheta$. From trig, we know $sec^2(theta)=1+tan^2(theta)$, and from above, we know that $tan(theta)$ is $xoverl_0$. So we can isolate $theta$ to get
$$dx over l_0* left( 1+left(xoverl_0 right)^2 right) =dtheta$$
We know theta is uniformly distributed, so we can divide both sides by $pi$. Now the probability of a given $theta$ falling between $theta$ and $theta+d theta$ is equal to the function of $x$ on the left.
edited 2 days ago
FundThmCalculus
2,286720
answered 2 days ago
R. Romero
313
3
To clear up any possible confusion, the Lorentz distribution is an alternative name for the Cauchy distribution.
– Ingolifs
2 days ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
33
down vote
accepted
The curve is a density function. The idea is the following. From your first picture, assume that the angles of the rays are spaced evenly, the angle between two rays being $alpha$. I.e. the nth ray has angle $n cdot alpha$. The nth ray's intersection point $x$ with the line then follows $tan (n cdot alpha) = x/h$ where h is the distance of the line to the origin. So from 0 to $x$, n rays cross the line.
Now you are interested in the density $p(x)$, i.e. how many rays intersect the line at $x$, per line interval $Delta x$. In the limit of small $alpha$, you have $int_0^x p(x') dx' =c n = fraccalphaarctan (x/h)$ and correspondingly, $p(x) = fracddxfraccalphaarctan (x/h) = fracc halpha (x^2+h^2)$. The constant $c$ is determined since the integral over the density function must be $1$ (in probability sense), hence
$p(x)= frachpi (x^2+h^2)$.
This curve is called a Cauchy distribution.
Obviously, $p(x)$ can be multiplied with a constant $K$ to give an expectation value distribution $E(x) = K p(x)$ over $x$, instead of a probability distribution. This explains the large value of $E(0) = 5700$ or so in your picture. The value $h$ is also called a scale parameter, it specifies the half-width at half-maximum (HWHM) of the curve and can be roughly read off to be $1$. If we are really "counting rays", then with angle spacing $alpha$, in total $pi/alpha$ many rays intersect the line and hence we must have
$$
pi/alpha = int_-infty^inftyE(x) dx = int_-infty^inftyK p(x) dx = K
$$
So the expectation value distribution of the number of rays intersecting one unit of the line at position $x$ is
$$
E(x) = fracpialpha p(x)= frachalpha (x^2+h^2)
$$
as we already had with the constant $c=1$. Reading off approximately $E(0) = 5700$ and $h=1$ gives $
E(x) = frac5700x^2+1
$ and $alpha = 1/5700$ (in rads), or in other words, $pi/alpha simeq 17900$ rays (lower half plane) intersect the line.
answered Sep 9 at 21:17
Andreas
6,9051036
add a comment |Â
up vote
33
down vote
accepted
The curve is a density function. The idea is the following. From your first picture, assume that the angles of the rays are spaced evenly, the angle between two rays being $alpha$. I.e. the nth ray has angle $n cdot alpha$. The nth ray's intersection point $x$ with the line then follows $tan (n cdot alpha) = x/h$ where h is the distance of the line to the origin. So from 0 to $x$, n rays cross the line.
Now you are interested in the density $p(x)$, i.e. how many rays intersect the line at $x$, per line interval $Delta x$. In the limit of small $alpha$, you have $int_0^x p(x') dx' =c n = fraccalphaarctan (x/h)$ and correspondingly, $p(x) = fracddxfraccalphaarctan (x/h) = fracc halpha (x^2+h^2)$. The constant $c$ is determined since the integral over the density function must be $1$ (in probability sense), hence
$p(x)= frachpi (x^2+h^2)$.
This curve is called a Cauchy distribution.
Obviously, $p(x)$ can be multiplied with a constant $K$ to give an expectation value distribution $E(x) = K p(x)$ over $x$, instead of a probability distribution. This explains the large value of $E(0) = 5700$ or so in your picture. The value $h$ is also called a scale parameter, it specifies the half-width at half-maximum (HWHM) of the curve and can be roughly read off to be $1$. If we are really "counting rays", then with angle spacing $alpha$, in total $pi/alpha$ many rays intersect the line and hence we must have
$$
pi/alpha = int_-infty^inftyE(x) dx = int_-infty^inftyK p(x) dx = K
$$
So the expectation value distribution of the number of rays intersecting one unit of the line at position $x$ is
$$
E(x) = fracpialpha p(x)= frachalpha (x^2+h^2)
$$
as we already had with the constant $c=1$. Reading off approximately $E(0) = 5700$ and $h=1$ gives $
E(x) = frac5700x^2+1
$ and $alpha = 1/5700$ (in rads), or in other words, $pi/alpha simeq 17900$ rays (lower half plane) intersect the line.
answered Sep 9 at 21:17
Andreas
6,9051036
add a comment |Â
up vote
33
down vote
accepted
up vote
33
down vote
accepted
The curve is a density function. The idea is the following. From your first picture, assume that the angles of the rays are spaced evenly, the angle between two rays being $alpha$. I.e. the nth ray has angle $n cdot alpha$. The nth ray's intersection point $x$ with the line then follows $tan (n cdot alpha) = x/h$ where h is the distance of the line to the origin. So from 0 to $x$, n rays cross the line.
Now you are interested in the density $p(x)$, i.e. how many rays intersect the line at $x$, per line interval $Delta x$. In the limit of small $alpha$, you have $int_0^x p(x') dx' =c n = fraccalphaarctan (x/h)$ and correspondingly, $p(x) = fracddxfraccalphaarctan (x/h) = fracc halpha (x^2+h^2)$. The constant $c$ is determined since the integral over the density function must be $1$ (in probability sense), hence
$p(x)= frachpi (x^2+h^2)$.
This curve is called a Cauchy distribution.
Obviously, $p(x)$ can be multiplied with a constant $K$ to give an expectation value distribution $E(x) = K p(x)$ over $x$, instead of a probability distribution. This explains the large value of $E(0) = 5700$ or so in your picture. The value $h$ is also called a scale parameter, it specifies the half-width at half-maximum (HWHM) of the curve and can be roughly read off to be $1$. If we are really "counting rays", then with angle spacing $alpha$, in total $pi/alpha$ many rays intersect the line and hence we must have
$$
pi/alpha = int_-infty^inftyE(x) dx = int_-infty^inftyK p(x) dx = K
$$
So the expectation value distribution of the number of rays intersecting one unit of the line at position $x$ is
$$
E(x) = fracpialpha p(x)= frachalpha (x^2+h^2)
$$
as we already had with the constant $c=1$. Reading off approximately $E(0) = 5700$ and $h=1$ gives $
E(x) = frac5700x^2+1
$ and $alpha = 1/5700$ (in rads), or in other words, $pi/alpha simeq 17900$ rays (lower half plane) intersect the line.
answered Sep 9 at 21:17
Andreas
6,9051036
The curve is a density function. The idea is the following. From your first picture, assume that the angles of the rays are spaced evenly, the angle between two rays being $alpha$. I.e. the nth ray has angle $n cdot alpha$. The nth ray's intersection point $x$ with the line then follows $tan (n cdot alpha) = x/h$ where h is the distance of the line to the origin. So from 0 to $x$, n rays cross the line.
Now you are interested in the density $p(x)$, i.e. how many rays intersect the line at $x$, per line interval $Delta x$. In the limit of small $alpha$, you have $int_0^x p(x') dx' =c n = fraccalphaarctan (x/h)$ and correspondingly, $p(x) = fracddxfraccalphaarctan (x/h) = fracc halpha (x^2+h^2)$. The constant $c$ is determined since the integral over the density function must be $1$ (in probability sense), hence
$p(x)= frachpi (x^2+h^2)$.
This curve is called a Cauchy distribution.
Obviously, $p(x)$ can be multiplied with a constant $K$ to give an expectation value distribution $E(x) = K p(x)$ over $x$, instead of a probability distribution. This explains the large value of $E(0) = 5700$ or so in your picture. The value $h$ is also called a scale parameter, it specifies the half-width at half-maximum (HWHM) of the curve and can be roughly read off to be $1$. If we are really "counting rays", then with angle spacing $alpha$, in total $pi/alpha$ many rays intersect the line and hence we must have
$$
pi/alpha = int_-infty^inftyE(x) dx = int_-infty^inftyK p(x) dx = K
$$
So the expectation value distribution of the number of rays intersecting one unit of the line at position $x$ is
$$
E(x) = fracpialpha p(x)= frachalpha (x^2+h^2)
$$
as we already had with the constant $c=1$. Reading off approximately $E(0) = 5700$ and $h=1$ gives $
E(x) = frac5700x^2+1
$ and $alpha = 1/5700$ (in rads), or in other words, $pi/alpha simeq 17900$ rays (lower half plane) intersect the line.
answered Sep 9 at 21:17
Andreas
6,9051036
edited 2 days ago
answered Sep 9 at 21:17
Andreas
6,9051036
answered Sep 9 at 21:17
Andreas
6,9051036
answered Sep 9 at 21:17
Andreas
6,9051036
6,9051036
add a comment |Â
add a comment |Â
up vote
6
down vote
It can be shown assuming that the quantity from the ray is proportional to the angular interval that the denisty distribuiton function is in the form
$$y=fracdH+x^2$$
where $d$ is related to the intensity and $H$ is the distance of the source point from the line.
Here is the plot for the case $d=H=1$
answered Sep 9 at 20:56
gimusi
71.4k73786
add a comment |Â
up vote
6
down vote
It can be shown assuming that the quantity from the ray is proportional to the angular interval that the denisty distribuiton function is in the form
$$y=fracdH+x^2$$
where $d$ is related to the intensity and $H$ is the distance of the source point from the line.
Here is the plot for the case $d=H=1$
answered Sep 9 at 20:56
gimusi
71.4k73786
add a comment |Â
up vote
6
down vote
up vote
6
down vote
It can be shown assuming that the quantity from the ray is proportional to the angular interval that the denisty distribuiton function is in the form
$$y=fracdH+x^2$$
where $d$ is related to the intensity and $H$ is the distance of the source point from the line.
Here is the plot for the case $d=H=1$
answered Sep 9 at 20:56
gimusi
71.4k73786
It can be shown assuming that the quantity from the ray is proportional to the angular interval that the denisty distribuiton function is in the form
$$y=fracdH+x^2$$
where $d$ is related to the intensity and $H$ is the distance of the source point from the line.
Here is the plot for the case $d=H=1$
answered Sep 9 at 20:56
gimusi
71.4k73786
answered Sep 9 at 20:56
gimusi
71.4k73786
answered Sep 9 at 20:56
gimusi
71.4k73786
answered Sep 9 at 20:56
gimusi
71.4k73786
71.4k73786
add a comment |Â
add a comment |Â
up vote
1
down vote
This is also known, especially among we physicists as a Lorentz distribution.
We know that every point on the line has the same vertical distance from the source point. Let's call this $l_0$. We also know the ratio of the horizontal component of the distance to the vertical component is $tan(theta)$, where theta goes from $-piover2$ to $piover2$ in the increments given. So horizontal distance=$x=l_0*tan(theta)$.
To find the densities, first take the derivative of both sides, getting $dx=l_0*sec^2(theta)*dtheta$. From trig, we know $sec^2(theta)=1+tan^2(theta)$, and from above, we know that $tan(theta)$ is $xoverl_0$. So we can isolate $theta$ to get
$$dx over l_0* left( 1+left(xoverl_0 right)^2 right) =dtheta$$
We know theta is uniformly distributed, so we can divide both sides by $pi$. Now the probability of a given $theta$ falling between $theta$ and $theta+d theta$ is equal to the function of $x$ on the left.
edited 2 days ago
FundThmCalculus
2,286720
answered 2 days ago
R. Romero
313
3
To clear up any possible confusion, the Lorentz distribution is an alternative name for the Cauchy distribution.
– Ingolifs
2 days ago
add a comment |Â
up vote
1
down vote
This is also known, especially among we physicists as a Lorentz distribution.
We know that every point on the line has the same vertical distance from the source point. Let's call this $l_0$. We also know the ratio of the horizontal component of the distance to the vertical component is $tan(theta)$, where theta goes from $-piover2$ to $piover2$ in the increments given. So horizontal distance=$x=l_0*tan(theta)$.
To find the densities, first take the derivative of both sides, getting $dx=l_0*sec^2(theta)*dtheta$. From trig, we know $sec^2(theta)=1+tan^2(theta)$, and from above, we know that $tan(theta)$ is $xoverl_0$. So we can isolate $theta$ to get
$$dx over l_0* left( 1+left(xoverl_0 right)^2 right) =dtheta$$
We know theta is uniformly distributed, so we can divide both sides by $pi$. Now the probability of a given $theta$ falling between $theta$ and $theta+d theta$ is equal to the function of $x$ on the left.
edited 2 days ago
FundThmCalculus
2,286720
answered 2 days ago
R. Romero
313
3
To clear up any possible confusion, the Lorentz distribution is an alternative name for the Cauchy distribution.
– Ingolifs
2 days ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
This is also known, especially among we physicists as a Lorentz distribution.
We know that every point on the line has the same vertical distance from the source point. Let's call this $l_0$. We also know the ratio of the horizontal component of the distance to the vertical component is $tan(theta)$, where theta goes from $-piover2$ to $piover2$ in the increments given. So horizontal distance=$x=l_0*tan(theta)$.
To find the densities, first take the derivative of both sides, getting $dx=l_0*sec^2(theta)*dtheta$. From trig, we know $sec^2(theta)=1+tan^2(theta)$, and from above, we know that $tan(theta)$ is $xoverl_0$. So we can isolate $theta$ to get
$$dx over l_0* left( 1+left(xoverl_0 right)^2 right) =dtheta$$
We know theta is uniformly distributed, so we can divide both sides by $pi$. Now the probability of a given $theta$ falling between $theta$ and $theta+d theta$ is equal to the function of $x$ on the left.
edited 2 days ago
FundThmCalculus
2,286720
answered 2 days ago
R. Romero
313
This is also known, especially among we physicists as a Lorentz distribution.
We know that every point on the line has the same vertical distance from the source point. Let's call this $l_0$. We also know the ratio of the horizontal component of the distance to the vertical component is $tan(theta)$, where theta goes from $-piover2$ to $piover2$ in the increments given. So horizontal distance=$x=l_0*tan(theta)$.
To find the densities, first take the derivative of both sides, getting $dx=l_0*sec^2(theta)*dtheta$. From trig, we know $sec^2(theta)=1+tan^2(theta)$, and from above, we know that $tan(theta)$ is $xoverl_0$. So we can isolate $theta$ to get
$$dx over l_0* left( 1+left(xoverl_0 right)^2 right) =dtheta$$
We know theta is uniformly distributed, so we can divide both sides by $pi$. Now the probability of a given $theta$ falling between $theta$ and $theta+d theta$ is equal to the function of $x$ on the left.
edited 2 days ago
FundThmCalculus
2,286720
answered 2 days ago
R. Romero
313
edited 2 days ago
FundThmCalculus
2,286720
edited 2 days ago
FundThmCalculus
2,286720
edited 2 days ago
FundThmCalculus
2,286720
2,286720
answered 2 days ago
R. Romero
313
answered 2 days ago
R. Romero
313
answered 2 days ago
R. Romero
313
313
3
To clear up any possible confusion, the Lorentz distribution is an alternative name for the Cauchy distribution.
– Ingolifs
2 days ago
add a comment |Â
3
To clear up any possible confusion, the Lorentz distribution is an alternative name for the Cauchy distribution.
– Ingolifs
2 days ago
3
3
To clear up any possible confusion, the Lorentz distribution is an alternative name for the Cauchy distribution.
– Ingolifs
2 days ago
To clear up any possible confusion, the Lorentz distribution is an alternative name for the Cauchy distribution.
– Ingolifs
2 days ago
add a comment |Â
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What are the numbers in your graph supposed to represent?
– Aretino
Sep 9 at 20:42
There are a few functions you could make look roughly like this, but I think its going to be of the form $amathrme^-bx^2$
– aidangallagher4
Sep 9 at 20:52
5
en.wikipedia.org/wiki/Cauchy_distribution
– Aretino
Sep 9 at 20:54
1
mathworld.wolfram.com/LorentzianFunction.html
– Aretino
Sep 9 at 20:58