Correctly defining the return of a procedure The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why should I avoid the For loop in Mathematica?Defining a string based sort functionDefining functions in a loopReturn value of Reap when using tagsHow to define even permutations correctly?How to exclude some indices in defining Table?How to make a repetive procedure with LinearModelFitTable with List iterator return unpacked listreverse procedure of KroneckerProductDefine the substitution procedure as a functionRebuild a vector defining the sign of the elements
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Correctly defining the return of a procedure
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Why should I avoid the For loop in Mathematica?Defining a string based sort functionDefining functions in a loopReturn value of Reap when using tagsHow to define even permutations correctly?How to exclude some indices in defining Table?How to make a repetive procedure with LinearModelFitTable with List iterator return unpacked listreverse procedure of KroneckerProductDefine the substitution procedure as a functionRebuild a vector defining the sign of the elements
$begingroup$
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 Apr 8 at 9:54
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 Apr 8 at 9:54
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
Apr 8 at 10:45
3
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
Apr 8 at 11:34
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
Apr 8 at 13:17
$begingroup$
You might also look intoNestandNestListas an alternative, built-in way to iterate this map.
$endgroup$
– Chris K
Apr 8 at 13:20
$begingroup$
Don't just guess at the syntax, read a tutorial. Plenty can be found with a simple websearch.
$endgroup$
– Szabolcs
17 hours ago
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 Apr 8 at 9:54
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 Apr 8 at 9:54
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 Apr 8 at 9:54
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 Apr 8 at 9:54
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 Apr 8 at 9:54
JacquesLeenJacquesLeen
303
asked Apr 8 at 9:54
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
Apr 8 at 10:45
3
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
Apr 8 at 11:34
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
Apr 8 at 13:17
$begingroup$
You might also look intoNestandNestListas an alternative, built-in way to iterate this map.
$endgroup$
– Chris K
Apr 8 at 13:20
$begingroup$
Don't just guess at the syntax, read a tutorial. Plenty can be found with a simple websearch.
$endgroup$
– Szabolcs
17 hours ago
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
Apr 8 at 10:45
3
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
Apr 8 at 11:34
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
Apr 8 at 13:17
$begingroup$
You might also look intoNestandNestListas an alternative, built-in way to iterate this map.
$endgroup$
– Chris K
Apr 8 at 13:20
$begingroup$
Don't just guess at the syntax, read a tutorial. Plenty can be found with a simple websearch.
$endgroup$
– Szabolcs
17 hours ago
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
Apr 8 at 10:45
$begingroup$
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
Apr 8 at 10:45
3
3
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
Apr 8 at 11:34
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
Apr 8 at 11:34
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
Apr 8 at 13:17
$begingroup$
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
Apr 8 at 13:17
$begingroup$
You might also look into
Nest and NestList as an alternative, built-in way to iterate this map.$endgroup$
– Chris K
Apr 8 at 13:20
$begingroup$
You might also look into
Nest and NestList as an alternative, built-in way to iterate this map.$endgroup$
– Chris K
Apr 8 at 13:20
$begingroup$
Don't just guess at the syntax, read a tutorial. Plenty can be found with a simple websearch.
$endgroup$
– Szabolcs
17 hours ago
$begingroup$
Don't just guess at the syntax, read a tutorial. Plenty can be found with a simple websearch.
$endgroup$
– Szabolcs
17 hours ago
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 Apr 8 at 13:08
MarcoB
38.7k557116
answered Apr 8 at 10:29
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
Apr 8 at 13:50
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 Apr 8 at 14:01
RomanRoman
5,20011131
$endgroup$
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 Apr 8 at 13:08
MarcoB
38.7k557116
answered Apr 8 at 10:29
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
Apr 8 at 13:50
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 Apr 8 at 13:08
MarcoB
38.7k557116
answered Apr 8 at 10:29
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
Apr 8 at 13:50
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 Apr 8 at 13:08
MarcoB
38.7k557116
answered Apr 8 at 10:29
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 Apr 8 at 13:08
MarcoB
38.7k557116
answered Apr 8 at 10:29
Innerw0lfInnerw0lf
814
edited Apr 8 at 13:08
MarcoB
38.7k557116
edited Apr 8 at 13:08
MarcoB
38.7k557116
edited Apr 8 at 13:08
MarcoB
38.7k557116
38.7k557116
answered Apr 8 at 10:29
Innerw0lfInnerw0lf
814
answered Apr 8 at 10:29
Innerw0lfInnerw0lf
814
answered Apr 8 at 10:29
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
Apr 8 at 13:50
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
Apr 8 at 13:50
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
Apr 8 at 13:50
$begingroup$
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
Apr 8 at 13:50
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 Apr 8 at 14:01
RomanRoman
5,20011131
$endgroup$
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 Apr 8 at 14:01
RomanRoman
5,20011131
$endgroup$
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 Apr 8 at 14:01
RomanRoman
5,20011131
$endgroup$
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 Apr 8 at 14:01
RomanRoman
5,20011131
edited Apr 8 at 15:04
answered Apr 8 at 14:01
RomanRoman
5,20011131
answered Apr 8 at 14:01
RomanRoman
5,20011131
answered Apr 8 at 14:01
RomanRoman
5,20011131
5,20011131
add a comment |
add a comment |
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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
Apr 8 at 10:45
3
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
Apr 8 at 11:34
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
Apr 8 at 13:17
$begingroup$
You might also look into
NestandNestListas an alternative, built-in way to iterate this map.$endgroup$
– Chris K
Apr 8 at 13:20
$begingroup$
Don't just guess at the syntax, read a tutorial. Plenty can be found with a simple websearch.
$endgroup$
– Szabolcs
17 hours ago