Mixing
The recursive systems of equations with the following form, have they ever been considered?
$begingroup$
Suppose a system of 2 equations defined as:
$$begin{cases}
x_{n+1}&=f_x(x_n,y_n) \
y_{n+1}&=f_y(x_n,y_n)
end{cases}$$
where initial conditions for $x_0$ and $y_0$ are defined, and $x_n,y_ninBbb R$, $f_x,f_y:Bbb R^2to Bbb R$.
I found that if $f_x(x,y)=cos(x+y)+sin(x-y)$ and $f_y(x,y)=cos(x-y)+sin(x+y)$. Then the system becomes
$$begin{cases}
x_{n+1}&=cos(x_n+y_n)+sin(x_n-y_n)\
y_{n+1}&=cos(x_n-y_n)+sin(x_n+y_n)
end{cases}$$
which for any starting values for $x_0$ and $y_0$ it will produce the following points(plotted $x$ horizontally and until $x_{1.000.000}$):
which creates a rather interesting structure(I think its called an attractor), and I'd love to hear more about these systems, but couldn't find anything. Any help would be appreciated. Thanks.
EDIT:
If the system is
$$begin{cases}
x_{n+1}&= cos(x_n^2+y_n^2) \
y_{n+1}&= sin(x_n^2-y_n^2)
end{cases}$$
The points would be the following:
Are these coming from some differential equations?
recurrence-relations systems-of-equations
edited Feb 1 at 22:45
flawr
11.8k32546
asked Feb 1 at 20:14
GarmekainGarmekain
1,502720
$endgroup$
add a comment |
$begingroup$
Suppose a system of 2 equations defined as:
$$begin{cases}
x_{n+1}&=f_x(x_n,y_n) \
y_{n+1}&=f_y(x_n,y_n)
end{cases}$$
where initial conditions for $x_0$ and $y_0$ are defined, and $x_n,y_ninBbb R$, $f_x,f_y:Bbb R^2to Bbb R$.
I found that if $f_x(x,y)=cos(x+y)+sin(x-y)$ and $f_y(x,y)=cos(x-y)+sin(x+y)$. Then the system becomes
$$begin{cases}
x_{n+1}&=cos(x_n+y_n)+sin(x_n-y_n)\
y_{n+1}&=cos(x_n-y_n)+sin(x_n+y_n)
end{cases}$$
which for any starting values for $x_0$ and $y_0$ it will produce the following points(plotted $x$ horizontally and until $x_{1.000.000}$):
which creates a rather interesting structure(I think its called an attractor), and I'd love to hear more about these systems, but couldn't find anything. Any help would be appreciated. Thanks.
EDIT:
If the system is
$$begin{cases}
x_{n+1}&= cos(x_n^2+y_n^2) \
y_{n+1}&= sin(x_n^2-y_n^2)
end{cases}$$
The points would be the following:
Are these coming from some differential equations?
recurrence-relations systems-of-equations
edited Feb 1 at 22:45
flawr
11.8k32546
asked Feb 1 at 20:14
GarmekainGarmekain
1,502720
$endgroup$
1
$begingroup$
Maybe using the equation $cos(a+b)=Re(e^{i(a+b)})$ and $sin(a + b) = Im(e^{i(a + b)})$ and combining two levels of the recursion it's possible to find simpler equation for the same system?
$endgroup$
– Esa Pulkkinen
Feb 1 at 20:32
add a comment |
$begingroup$
Suppose a system of 2 equations defined as:
$$begin{cases}
x_{n+1}&=f_x(x_n,y_n) \
y_{n+1}&=f_y(x_n,y_n)
end{cases}$$
where initial conditions for $x_0$ and $y_0$ are defined, and $x_n,y_ninBbb R$, $f_x,f_y:Bbb R^2to Bbb R$.
I found that if $f_x(x,y)=cos(x+y)+sin(x-y)$ and $f_y(x,y)=cos(x-y)+sin(x+y)$. Then the system becomes
$$begin{cases}
x_{n+1}&=cos(x_n+y_n)+sin(x_n-y_n)\
y_{n+1}&=cos(x_n-y_n)+sin(x_n+y_n)
end{cases}$$
which for any starting values for $x_0$ and $y_0$ it will produce the following points(plotted $x$ horizontally and until $x_{1.000.000}$):
which creates a rather interesting structure(I think its called an attractor), and I'd love to hear more about these systems, but couldn't find anything. Any help would be appreciated. Thanks.
EDIT:
If the system is
$$begin{cases}
x_{n+1}&= cos(x_n^2+y_n^2) \
y_{n+1}&= sin(x_n^2-y_n^2)
end{cases}$$
The points would be the following:
Are these coming from some differential equations?
recurrence-relations systems-of-equations
edited Feb 1 at 22:45
flawr
11.8k32546
asked Feb 1 at 20:14
GarmekainGarmekain
1,502720
$endgroup$
Suppose a system of 2 equations defined as:
$$begin{cases}
x_{n+1}&=f_x(x_n,y_n) \
y_{n+1}&=f_y(x_n,y_n)
end{cases}$$
where initial conditions for $x_0$ and $y_0$ are defined, and $x_n,y_ninBbb R$, $f_x,f_y:Bbb R^2to Bbb R$.
I found that if $f_x(x,y)=cos(x+y)+sin(x-y)$ and $f_y(x,y)=cos(x-y)+sin(x+y)$. Then the system becomes
$$begin{cases}
x_{n+1}&=cos(x_n+y_n)+sin(x_n-y_n)\
y_{n+1}&=cos(x_n-y_n)+sin(x_n+y_n)
end{cases}$$
which for any starting values for $x_0$ and $y_0$ it will produce the following points(plotted $x$ horizontally and until $x_{1.000.000}$):
which creates a rather interesting structure(I think its called an attractor), and I'd love to hear more about these systems, but couldn't find anything. Any help would be appreciated. Thanks.
EDIT:
If the system is
$$begin{cases}
x_{n+1}&= cos(x_n^2+y_n^2) \
y_{n+1}&= sin(x_n^2-y_n^2)
end{cases}$$
The points would be the following:
Are these coming from some differential equations?
recurrence-relations systems-of-equations
recurrence-relations systems-of-equations
edited Feb 1 at 22:45
flawr
11.8k32546
asked Feb 1 at 20:14
GarmekainGarmekain
1,502720
edited Feb 1 at 22:45
flawr
11.8k32546
asked Feb 1 at 20:14
GarmekainGarmekain
1,502720
edited Feb 1 at 22:45
flawr
11.8k32546
edited Feb 1 at 22:45
flawr
11.8k32546
edited Feb 1 at 22:45
flawr
11.8k32546
11.8k32546
asked Feb 1 at 20:14
GarmekainGarmekain
1,502720
asked Feb 1 at 20:14
GarmekainGarmekain
1,502720
asked Feb 1 at 20:14
GarmekainGarmekain
1,502720
1,502720
1
$begingroup$
Maybe using the equation $cos(a+b)=Re(e^{i(a+b)})$ and $sin(a + b) = Im(e^{i(a + b)})$ and combining two levels of the recursion it's possible to find simpler equation for the same system?
$endgroup$
– Esa Pulkkinen
Feb 1 at 20:32
add a comment |
1
$begingroup$
Maybe using the equation $cos(a+b)=Re(e^{i(a+b)})$ and $sin(a + b) = Im(e^{i(a + b)})$ and combining two levels of the recursion it's possible to find simpler equation for the same system?
$endgroup$
– Esa Pulkkinen
Feb 1 at 20:32
1
1
$begingroup$
Maybe using the equation $cos(a+b)=Re(e^{i(a+b)})$ and $sin(a + b) = Im(e^{i(a + b)})$ and combining two levels of the recursion it's possible to find simpler equation for the same system?
$endgroup$
– Esa Pulkkinen
Feb 1 at 20:32
$begingroup$
Maybe using the equation $cos(a+b)=Re(e^{i(a+b)})$ and $sin(a + b) = Im(e^{i(a + b)})$ and combining two levels of the recursion it's possible to find simpler equation for the same system?
$endgroup$
– Esa Pulkkinen
Feb 1 at 20:32
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, such recurrences have been studied extensively and are still subject to resear in the field of Dynamical Systems, discrete time dynamical systems in particular. In this field the behaviour of repeated applications of a function (in your case $mathbb R^2 to mathbb R^2$) to some inputs is studied.
answered Feb 1 at 22:35
flawrflawr
11.8k32546
$endgroup$
add a comment |
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$begingroup$
Yes, such recurrences have been studied extensively and are still subject to resear in the field of Dynamical Systems, discrete time dynamical systems in particular. In this field the behaviour of repeated applications of a function (in your case $mathbb R^2 to mathbb R^2$) to some inputs is studied.
answered Feb 1 at 22:35
flawrflawr
11.8k32546
$endgroup$
add a comment |
$begingroup$
Yes, such recurrences have been studied extensively and are still subject to resear in the field of Dynamical Systems, discrete time dynamical systems in particular. In this field the behaviour of repeated applications of a function (in your case $mathbb R^2 to mathbb R^2$) to some inputs is studied.
answered Feb 1 at 22:35
flawrflawr
11.8k32546
$endgroup$
add a comment |
$begingroup$
Yes, such recurrences have been studied extensively and are still subject to resear in the field of Dynamical Systems, discrete time dynamical systems in particular. In this field the behaviour of repeated applications of a function (in your case $mathbb R^2 to mathbb R^2$) to some inputs is studied.
answered Feb 1 at 22:35
flawrflawr
11.8k32546
$endgroup$
Yes, such recurrences have been studied extensively and are still subject to resear in the field of Dynamical Systems, discrete time dynamical systems in particular. In this field the behaviour of repeated applications of a function (in your case $mathbb R^2 to mathbb R^2$) to some inputs is studied.
answered Feb 1 at 22:35
flawrflawr
11.8k32546
answered Feb 1 at 22:35
flawrflawr
11.8k32546
answered Feb 1 at 22:35
flawrflawr
11.8k32546
answered Feb 1 at 22:35
flawrflawr
11.8k32546
11.8k32546
add a comment |
add a comment |
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$begingroup$
Maybe using the equation $cos(a+b)=Re(e^{i(a+b)})$ and $sin(a + b) = Im(e^{i(a + b)})$ and combining two levels of the recursion it's possible to find simpler equation for the same system?
$endgroup$
– Esa Pulkkinen
Feb 1 at 20:32