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Correctly defining the return of a procedure
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 16 hours ago
JacquesLeenJacquesLeen
161
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 16 hours ago
JacquesLeenJacquesLeen
161
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
15 hours ago
1
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
14 hours ago
$begingroup$
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
13 hours ago
$begingroup$
You might also look intoNestandNestListas an alternative, built-in way to iterate this map.
$endgroup$
– Chris K
13 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 16 hours ago
JacquesLeenJacquesLeen
161
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 16 hours ago
JacquesLeenJacquesLeen
161
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 16 hours ago
JacquesLeenJacquesLeen
161
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 16 hours ago
JacquesLeenJacquesLeen
161
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 16 hours ago
JacquesLeenJacquesLeen
161
asked 16 hours ago
JacquesLeenJacquesLeen
161
161
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
15 hours ago
1
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
14 hours ago
$begingroup$
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
13 hours ago
$begingroup$
You might also look intoNestandNestListas an alternative, built-in way to iterate this map.
$endgroup$
– Chris K
13 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
15 hours ago
1
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
14 hours ago
$begingroup$
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
13 hours ago
$begingroup$
You might also look intoNestandNestListas an alternative, built-in way to iterate this map.
$endgroup$
– Chris K
13 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
15 hours ago
$begingroup$
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
15 hours ago
1
1
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
14 hours ago
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
14 hours ago
$begingroup$
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
13 hours ago
$begingroup$
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
13 hours ago
$begingroup$
You might also look into
Nest and NestList as an alternative, built-in way to iterate this map.$endgroup$
– Chris K
13 hours ago
$begingroup$
You might also look into
Nest and NestList as an alternative, built-in way to iterate this map.$endgroup$
– Chris K
13 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 13 hours ago
MarcoB
38.6k557115
answered 16 hours ago
Innerw0lfInnerw0lf
564
$endgroup$
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
12 hours ago
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 12 hours ago
RomanRoman
4,65511128
$endgroup$
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 13 hours ago
MarcoB
38.6k557115
answered 16 hours ago
Innerw0lfInnerw0lf
564
$endgroup$
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
12 hours ago
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 13 hours ago
MarcoB
38.6k557115
answered 16 hours ago
Innerw0lfInnerw0lf
564
$endgroup$
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
12 hours ago
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 13 hours ago
MarcoB
38.6k557115
answered 16 hours ago
Innerw0lfInnerw0lf
564
$endgroup$
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 13 hours ago
MarcoB
38.6k557115
answered 16 hours ago
Innerw0lfInnerw0lf
564
edited 13 hours ago
MarcoB
38.6k557115
edited 13 hours ago
MarcoB
38.6k557115
edited 13 hours ago
MarcoB
38.6k557115
38.6k557115
answered 16 hours ago
Innerw0lfInnerw0lf
564
answered 16 hours ago
Innerw0lfInnerw0lf
564
answered 16 hours ago
Innerw0lfInnerw0lf
564
564
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
12 hours ago
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
12 hours ago
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
12 hours ago
$begingroup$
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
12 hours ago
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 12 hours ago
RomanRoman
4,65511128
$endgroup$
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 12 hours ago
RomanRoman
4,65511128
$endgroup$
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 12 hours ago
RomanRoman
4,65511128
$endgroup$
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 12 hours ago
RomanRoman
4,65511128
edited 11 hours ago
answered 12 hours ago
RomanRoman
4,65511128
answered 12 hours ago
RomanRoman
4,65511128
answered 12 hours ago
RomanRoman
4,65511128
4,65511128
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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
15 hours ago
1
$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
14 hours ago
$begingroup$
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
13 hours ago
$begingroup$
You might also look into
NestandNestListas an alternative, built-in way to iterate this map.$endgroup$
– Chris K
13 hours ago