Correctly defining the return of a procedure
$begingroup$
I am sort of new to mathematica and I am struggling with how to correctly set the return of a specific function which is a procedure.
I am give the following task:
Given the map $x_{n+1} = r x_n (1-x_n)$:
1- set $x_0 = 0.5$
2- write a function $f(r)$ which evaluates the first 1000 terms fo the sequence, takes the last 100 and then selects distinct elements.
This is what I came up with so far
x[0] = 0.5
f[r_]:={
l = Table[0,1000]; (*init table of 1000 elems*)
l[[1]] = x[0]; (*set x_0*)
For[n=1,n<1000,n++,
{
x[n] = r * x[n-1] *(1-x[n-1]); (*evaluate x_n*)
l[[n+1]] = x[n]; (*set nth elem of list*)
}
];
l= Union[Take[l, -100]] (*modify list*)
Return[l] (*return list*)
}
but this does not work at all. thanks tho everyone who is keen to partecipate and give some suggestion :)
list-manipulation
asked yesterday
JacquesLeenJacquesLeen
303
New contributor
JacquesLeen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
I am sort of new to mathematica and I am struggling with how to correctly set the return of a specific function which is a procedure.
I am give the following task:
Given the map $x_{n+1} = r x_n (1-x_n)$:
1- set $x_0 = 0.5$
2- write a function $f(r)$ which evaluates the first 1000 terms fo the sequence, takes the last 100 and then selects distinct elements.
This is what I came up with so far
x[0] = 0.5
f[r_]:={
l = Table[0,1000]; (*init table of 1000 elems*)
l[[1]] = x[0]; (*set x_0*)
For[n=1,n<1000,n++,
{
x[n] = r * x[n-1] *(1-x[n-1]); (*evaluate x_n*)
l[[n+1]] = x[n]; (*set nth elem of list*)
}
];
l= Union[Take[l, -100]] (*modify list*)
Return[l] (*return list*)
}
but this does not work at all. thanks tho everyone who is keen to partecipate and give some suggestion :)
list-manipulation
asked yesterday
JacquesLeenJacquesLeen
303
New contributor
JacquesLeen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Curly braces are for lists, not for code blocks. UseModule,Block, orWithfor code blocks. And try not to useFor.
$endgroup$
– Roman
yesterday
3
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
yesterday
1
$begingroup$
Returndoesn't do what you think it does. Don't use it until you understand it. Once you understand it, you'll probably never use it.
$endgroup$
– John Doty
yesterday
$begingroup$
You might also look intoNestandNestListas an alternative, built-in way to iterate this map.
$endgroup$
– Chris K
yesterday
add a comment |
$begingroup$
I am sort of new to mathematica and I am struggling with how to correctly set the return of a specific function which is a procedure.
I am give the following task:
Given the map $x_{n+1} = r x_n (1-x_n)$:
1- set $x_0 = 0.5$
2- write a function $f(r)$ which evaluates the first 1000 terms fo the sequence, takes the last 100 and then selects distinct elements.
This is what I came up with so far
x[0] = 0.5
f[r_]:={
l = Table[0,1000]; (*init table of 1000 elems*)
l[[1]] = x[0]; (*set x_0*)
For[n=1,n<1000,n++,
{
x[n] = r * x[n-1] *(1-x[n-1]); (*evaluate x_n*)
l[[n+1]] = x[n]; (*set nth elem of list*)
}
];
l= Union[Take[l, -100]] (*modify list*)
Return[l] (*return list*)
}
but this does not work at all. thanks tho everyone who is keen to partecipate and give some suggestion :)
list-manipulation
asked yesterday
JacquesLeenJacquesLeen
303
New contributor
JacquesLeen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am sort of new to mathematica and I am struggling with how to correctly set the return of a specific function which is a procedure.
I am give the following task:
Given the map $x_{n+1} = r x_n (1-x_n)$:
1- set $x_0 = 0.5$
2- write a function $f(r)$ which evaluates the first 1000 terms fo the sequence, takes the last 100 and then selects distinct elements.
This is what I came up with so far
x[0] = 0.5
f[r_]:={
l = Table[0,1000]; (*init table of 1000 elems*)
l[[1]] = x[0]; (*set x_0*)
For[n=1,n<1000,n++,
{
x[n] = r * x[n-1] *(1-x[n-1]); (*evaluate x_n*)
l[[n+1]] = x[n]; (*set nth elem of list*)
}
];
l= Union[Take[l, -100]] (*modify list*)
Return[l] (*return list*)
}
but this does not work at all. thanks tho everyone who is keen to partecipate and give some suggestion :)
list-manipulation
list-manipulation
asked yesterday
JacquesLeenJacquesLeen
303
New contributor
JacquesLeen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
JacquesLeenJacquesLeen
303
New contributor
JacquesLeen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
JacquesLeenJacquesLeen
303
New contributor
JacquesLeen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
JacquesLeenJacquesLeen
303
asked yesterday
JacquesLeenJacquesLeen
303
303
New contributor
JacquesLeen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
JacquesLeen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
JacquesLeen is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Curly braces are for lists, not for code blocks. UseModule,Block, orWithfor code blocks. And try not to useFor.
$endgroup$
– Roman
yesterday
3
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
yesterday
1
$begingroup$
Returndoesn't do what you think it does. Don't use it until you understand it. Once you understand it, you'll probably never use it.
$endgroup$
– John Doty
yesterday
$begingroup$
You might also look intoNestandNestListas an alternative, built-in way to iterate this map.
$endgroup$
– Chris K
yesterday
add a comment |
1
$begingroup$
Curly braces are for lists, not for code blocks. UseModule,Block, orWithfor code blocks. And try not to useFor.
$endgroup$
– Roman
yesterday
3
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
yesterday
1
$begingroup$
Returndoesn't do what you think it does. Don't use it until you understand it. Once you understand it, you'll probably never use it.
$endgroup$
– John Doty
yesterday
$begingroup$
You might also look intoNestandNestListas an alternative, built-in way to iterate this map.
$endgroup$
– Chris K
yesterday
1
1
$begingroup$
Curly braces are for lists, not for code blocks. Use
Module, Block, or With for code blocks. And try not to use For.$endgroup$
– Roman
yesterday
$begingroup$
Curly braces are for lists, not for code blocks. Use
Module, Block, or With for code blocks. And try not to use For.$endgroup$
– Roman
yesterday
3
3
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
yesterday
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
yesterday
1
1
$begingroup$
Return doesn't do what you think it does. Don't use it until you understand it. Once you understand it, you'll probably never use it.$endgroup$
– John Doty
yesterday
$begingroup$
Return doesn't do what you think it does. Don't use it until you understand it. Once you understand it, you'll probably never use it.$endgroup$
– John Doty
yesterday
$begingroup$
You might also look into
Nest and NestList as an alternative, built-in way to iterate this map.$endgroup$
– Chris K
yesterday
$begingroup$
You might also look into
Nest and NestList as an alternative, built-in way to iterate this map.$endgroup$
– Chris K
yesterday
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Try this:
f[r_] := Union[
Take[
RecurrenceTable[
{a[n + 1] == r*(1 - a[n])*a[n], a[1] == .5},
a, {n, 1, 1000}
],
-100]
]
edited yesterday
MarcoB
38.6k557115
answered yesterday
Innerw0lfInnerw0lf
814
$endgroup$
2
$begingroup$
Or, a little more efficiently,f[r_] := Union[ RecurrenceTable[{a[n + 1] == r*(1 - a[n])*a[n], a[1] == 0.5}, a, {n, 901, 1000}]]
$endgroup$
– Bob Hanlon
yesterday
add a comment |
$begingroup$
For a specific value of $r$ you can do
With[{r = 3.7},
NestList[r*#*(1-#) &, 0.5, 1000][[-100 ;;]]]
{0.783499, 0.627626, 0.864733, 0.432788, 0.908285, 0.308222,
0.788918, 0.616148, 0.875086, 0.404449, 0.891219, 0.358706, 0.851134,
0.468809, 0.9214, 0.26796, 0.725783, 0.736382, 0.718257, 0.748746,
0.696065, 0.782767, 0.629159, 0.863277, 0.436711, 0.910179, 0.302485,
0.780655, 0.63356, 0.858998, 0.448145, 0.915051, 0.287611, 0.758096,
0.67853, 0.80707, 0.576119, 0.903562, 0.32241, 0.808309, 0.573299,
0.905121, 0.317745, 0.802098, 0.587326, 0.896784, 0.34248, 0.833193,
0.514234, 0.92425, 0.259043, 0.710177, 0.761555, 0.67188, 0.815692,
0.556253, 0.913292, 0.293003, 0.766463, 0.662291, 0.827548, 0.528036,
0.922092, 0.265802, 0.722061, 0.74255, 0.707328, 0.765957, 0.663288,
0.826347, 0.530942, 0.921458, 0.267782, 0.725477, 0.736893, 0.717362,
0.750189, 0.6934, 0.786607, 0.621069, 0.870766, 0.41637, 0.899122,
0.335595, 0.824992, 0.534206, 0.920671, 0.270233, 0.729667, 0.729837,
0.729548, 0.730039, 0.729203, 0.730623, 0.728207, 0.732309, 0.72532,
0.737154, 0.716905, 0.750923}
I don't think there are any duplicates in this list ($r$ is in the chaotic region). For other values of $r$ there are indeed duplicates:
With[{r = 3.5},
NestList[r*#*(1 - #) &, 0.5, 1000][[-100 ;;]]] // DeleteDuplicates // Sort
{0.38282, 0.500884, 0.826941, 0.874997}
// Union does the same thing as // DeleteDuplicates // Sort if you prefer.
answered yesterday
RomanRoman
4,89011130
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Try this:
f[r_] := Union[
Take[
RecurrenceTable[
{a[n + 1] == r*(1 - a[n])*a[n], a[1] == .5},
a, {n, 1, 1000}
],
-100]
]
edited yesterday
MarcoB
38.6k557115
answered yesterday
Innerw0lfInnerw0lf
814
$endgroup$
2
$begingroup$
Or, a little more efficiently,f[r_] := Union[ RecurrenceTable[{a[n + 1] == r*(1 - a[n])*a[n], a[1] == 0.5}, a, {n, 901, 1000}]]
$endgroup$
– Bob Hanlon
yesterday
add a comment |
$begingroup$
Try this:
f[r_] := Union[
Take[
RecurrenceTable[
{a[n + 1] == r*(1 - a[n])*a[n], a[1] == .5},
a, {n, 1, 1000}
],
-100]
]
edited yesterday
MarcoB
38.6k557115
answered yesterday
Innerw0lfInnerw0lf
814
$endgroup$
2
$begingroup$
Or, a little more efficiently,f[r_] := Union[ RecurrenceTable[{a[n + 1] == r*(1 - a[n])*a[n], a[1] == 0.5}, a, {n, 901, 1000}]]
$endgroup$
– Bob Hanlon
yesterday
add a comment |
$begingroup$
Try this:
f[r_] := Union[
Take[
RecurrenceTable[
{a[n + 1] == r*(1 - a[n])*a[n], a[1] == .5},
a, {n, 1, 1000}
],
-100]
]
edited yesterday
MarcoB
38.6k557115
answered yesterday
Innerw0lfInnerw0lf
814
$endgroup$
Try this:
f[r_] := Union[
Take[
RecurrenceTable[
{a[n + 1] == r*(1 - a[n])*a[n], a[1] == .5},
a, {n, 1, 1000}
],
-100]
]
edited yesterday
MarcoB
38.6k557115
answered yesterday
Innerw0lfInnerw0lf
814
edited yesterday
MarcoB
38.6k557115
edited yesterday
MarcoB
38.6k557115
edited yesterday
MarcoB
38.6k557115
38.6k557115
answered yesterday
Innerw0lfInnerw0lf
814
answered yesterday
Innerw0lfInnerw0lf
814
answered yesterday
Innerw0lfInnerw0lf
814
814
2
$begingroup$
Or, a little more efficiently,f[r_] := Union[ RecurrenceTable[{a[n + 1] == r*(1 - a[n])*a[n], a[1] == 0.5}, a, {n, 901, 1000}]]
$endgroup$
– Bob Hanlon
yesterday
add a comment |
2
$begingroup$
Or, a little more efficiently,f[r_] := Union[ RecurrenceTable[{a[n + 1] == r*(1 - a[n])*a[n], a[1] == 0.5}, a, {n, 901, 1000}]]
$endgroup$
– Bob Hanlon
yesterday
2
2
$begingroup$
Or, a little more efficiently,
f[r_] := Union[ RecurrenceTable[{a[n + 1] == r*(1 - a[n])*a[n], a[1] == 0.5}, a, {n, 901, 1000}]]$endgroup$
– Bob Hanlon
yesterday
$begingroup$
Or, a little more efficiently,
f[r_] := Union[ RecurrenceTable[{a[n + 1] == r*(1 - a[n])*a[n], a[1] == 0.5}, a, {n, 901, 1000}]]$endgroup$
– Bob Hanlon
yesterday
add a comment |
$begingroup$
For a specific value of $r$ you can do
With[{r = 3.7},
NestList[r*#*(1-#) &, 0.5, 1000][[-100 ;;]]]
{0.783499, 0.627626, 0.864733, 0.432788, 0.908285, 0.308222,
0.788918, 0.616148, 0.875086, 0.404449, 0.891219, 0.358706, 0.851134,
0.468809, 0.9214, 0.26796, 0.725783, 0.736382, 0.718257, 0.748746,
0.696065, 0.782767, 0.629159, 0.863277, 0.436711, 0.910179, 0.302485,
0.780655, 0.63356, 0.858998, 0.448145, 0.915051, 0.287611, 0.758096,
0.67853, 0.80707, 0.576119, 0.903562, 0.32241, 0.808309, 0.573299,
0.905121, 0.317745, 0.802098, 0.587326, 0.896784, 0.34248, 0.833193,
0.514234, 0.92425, 0.259043, 0.710177, 0.761555, 0.67188, 0.815692,
0.556253, 0.913292, 0.293003, 0.766463, 0.662291, 0.827548, 0.528036,
0.922092, 0.265802, 0.722061, 0.74255, 0.707328, 0.765957, 0.663288,
0.826347, 0.530942, 0.921458, 0.267782, 0.725477, 0.736893, 0.717362,
0.750189, 0.6934, 0.786607, 0.621069, 0.870766, 0.41637, 0.899122,
0.335595, 0.824992, 0.534206, 0.920671, 0.270233, 0.729667, 0.729837,
0.729548, 0.730039, 0.729203, 0.730623, 0.728207, 0.732309, 0.72532,
0.737154, 0.716905, 0.750923}
I don't think there are any duplicates in this list ($r$ is in the chaotic region). For other values of $r$ there are indeed duplicates:
With[{r = 3.5},
NestList[r*#*(1 - #) &, 0.5, 1000][[-100 ;;]]] // DeleteDuplicates // Sort
{0.38282, 0.500884, 0.826941, 0.874997}
// Union does the same thing as // DeleteDuplicates // Sort if you prefer.
answered yesterday
RomanRoman
4,89011130
$endgroup$
add a comment |
$begingroup$
For a specific value of $r$ you can do
With[{r = 3.7},
NestList[r*#*(1-#) &, 0.5, 1000][[-100 ;;]]]
{0.783499, 0.627626, 0.864733, 0.432788, 0.908285, 0.308222,
0.788918, 0.616148, 0.875086, 0.404449, 0.891219, 0.358706, 0.851134,
0.468809, 0.9214, 0.26796, 0.725783, 0.736382, 0.718257, 0.748746,
0.696065, 0.782767, 0.629159, 0.863277, 0.436711, 0.910179, 0.302485,
0.780655, 0.63356, 0.858998, 0.448145, 0.915051, 0.287611, 0.758096,
0.67853, 0.80707, 0.576119, 0.903562, 0.32241, 0.808309, 0.573299,
0.905121, 0.317745, 0.802098, 0.587326, 0.896784, 0.34248, 0.833193,
0.514234, 0.92425, 0.259043, 0.710177, 0.761555, 0.67188, 0.815692,
0.556253, 0.913292, 0.293003, 0.766463, 0.662291, 0.827548, 0.528036,
0.922092, 0.265802, 0.722061, 0.74255, 0.707328, 0.765957, 0.663288,
0.826347, 0.530942, 0.921458, 0.267782, 0.725477, 0.736893, 0.717362,
0.750189, 0.6934, 0.786607, 0.621069, 0.870766, 0.41637, 0.899122,
0.335595, 0.824992, 0.534206, 0.920671, 0.270233, 0.729667, 0.729837,
0.729548, 0.730039, 0.729203, 0.730623, 0.728207, 0.732309, 0.72532,
0.737154, 0.716905, 0.750923}
I don't think there are any duplicates in this list ($r$ is in the chaotic region). For other values of $r$ there are indeed duplicates:
With[{r = 3.5},
NestList[r*#*(1 - #) &, 0.5, 1000][[-100 ;;]]] // DeleteDuplicates // Sort
{0.38282, 0.500884, 0.826941, 0.874997}
// Union does the same thing as // DeleteDuplicates // Sort if you prefer.
answered yesterday
RomanRoman
4,89011130
$endgroup$
add a comment |
$begingroup$
For a specific value of $r$ you can do
With[{r = 3.7},
NestList[r*#*(1-#) &, 0.5, 1000][[-100 ;;]]]
{0.783499, 0.627626, 0.864733, 0.432788, 0.908285, 0.308222,
0.788918, 0.616148, 0.875086, 0.404449, 0.891219, 0.358706, 0.851134,
0.468809, 0.9214, 0.26796, 0.725783, 0.736382, 0.718257, 0.748746,
0.696065, 0.782767, 0.629159, 0.863277, 0.436711, 0.910179, 0.302485,
0.780655, 0.63356, 0.858998, 0.448145, 0.915051, 0.287611, 0.758096,
0.67853, 0.80707, 0.576119, 0.903562, 0.32241, 0.808309, 0.573299,
0.905121, 0.317745, 0.802098, 0.587326, 0.896784, 0.34248, 0.833193,
0.514234, 0.92425, 0.259043, 0.710177, 0.761555, 0.67188, 0.815692,
0.556253, 0.913292, 0.293003, 0.766463, 0.662291, 0.827548, 0.528036,
0.922092, 0.265802, 0.722061, 0.74255, 0.707328, 0.765957, 0.663288,
0.826347, 0.530942, 0.921458, 0.267782, 0.725477, 0.736893, 0.717362,
0.750189, 0.6934, 0.786607, 0.621069, 0.870766, 0.41637, 0.899122,
0.335595, 0.824992, 0.534206, 0.920671, 0.270233, 0.729667, 0.729837,
0.729548, 0.730039, 0.729203, 0.730623, 0.728207, 0.732309, 0.72532,
0.737154, 0.716905, 0.750923}
I don't think there are any duplicates in this list ($r$ is in the chaotic region). For other values of $r$ there are indeed duplicates:
With[{r = 3.5},
NestList[r*#*(1 - #) &, 0.5, 1000][[-100 ;;]]] // DeleteDuplicates // Sort
{0.38282, 0.500884, 0.826941, 0.874997}
// Union does the same thing as // DeleteDuplicates // Sort if you prefer.
answered yesterday
RomanRoman
4,89011130
$endgroup$
For a specific value of $r$ you can do
With[{r = 3.7},
NestList[r*#*(1-#) &, 0.5, 1000][[-100 ;;]]]
{0.783499, 0.627626, 0.864733, 0.432788, 0.908285, 0.308222,
0.788918, 0.616148, 0.875086, 0.404449, 0.891219, 0.358706, 0.851134,
0.468809, 0.9214, 0.26796, 0.725783, 0.736382, 0.718257, 0.748746,
0.696065, 0.782767, 0.629159, 0.863277, 0.436711, 0.910179, 0.302485,
0.780655, 0.63356, 0.858998, 0.448145, 0.915051, 0.287611, 0.758096,
0.67853, 0.80707, 0.576119, 0.903562, 0.32241, 0.808309, 0.573299,
0.905121, 0.317745, 0.802098, 0.587326, 0.896784, 0.34248, 0.833193,
0.514234, 0.92425, 0.259043, 0.710177, 0.761555, 0.67188, 0.815692,
0.556253, 0.913292, 0.293003, 0.766463, 0.662291, 0.827548, 0.528036,
0.922092, 0.265802, 0.722061, 0.74255, 0.707328, 0.765957, 0.663288,
0.826347, 0.530942, 0.921458, 0.267782, 0.725477, 0.736893, 0.717362,
0.750189, 0.6934, 0.786607, 0.621069, 0.870766, 0.41637, 0.899122,
0.335595, 0.824992, 0.534206, 0.920671, 0.270233, 0.729667, 0.729837,
0.729548, 0.730039, 0.729203, 0.730623, 0.728207, 0.732309, 0.72532,
0.737154, 0.716905, 0.750923}
I don't think there are any duplicates in this list ($r$ is in the chaotic region). For other values of $r$ there are indeed duplicates:
With[{r = 3.5},
NestList[r*#*(1 - #) &, 0.5, 1000][[-100 ;;]]] // DeleteDuplicates // Sort
{0.38282, 0.500884, 0.826941, 0.874997}
// Union does the same thing as // DeleteDuplicates // Sort if you prefer.
answered yesterday
RomanRoman
4,89011130
edited yesterday
answered yesterday
RomanRoman
4,89011130
answered yesterday
RomanRoman
4,89011130
answered yesterday
RomanRoman
4,89011130
4,89011130
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1
$begingroup$
Curly braces are for lists, not for code blocks. Use
Module,Block, orWithfor code blocks. And try not to useFor.$endgroup$
– Roman
yesterday
3
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
yesterday
1
$begingroup$
Returndoesn't do what you think it does. Don't use it until you understand it. Once you understand it, you'll probably never use it.$endgroup$
– John Doty
yesterday
$begingroup$
You might also look into
NestandNestListas an alternative, built-in way to iterate this map.$endgroup$
– Chris K
yesterday