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Correctly defining the return of a procedure



The 2019 Stack Overflow Developer Survey Results Are InWhy 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










3












$begingroup$


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 :)










share|improve this question







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. Use Module, Block, or With for code blocks. And try not to use For.
    $endgroup$
    – Roman
    2 days ago






  • 3




    $begingroup$
    Why should I avoid the For loop in Mathematica?
    $endgroup$
    – corey979
    2 days ago






  • 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
    2 days ago










  • $begingroup$
    You might also look into Nest and NestList as an alternative, built-in way to iterate this map.
    $endgroup$
    – Chris K
    2 days ago















3












$begingroup$


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 :)










share|improve this question







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. Use Module, Block, or With for code blocks. And try not to use For.
    $endgroup$
    – Roman
    2 days ago






  • 3




    $begingroup$
    Why should I avoid the For loop in Mathematica?
    $endgroup$
    – corey979
    2 days ago






  • 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
    2 days ago










  • $begingroup$
    You might also look into Nest and NestList as an alternative, built-in way to iterate this map.
    $endgroup$
    – Chris K
    2 days ago













3












3








3


0



$begingroup$


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 :)










share|improve this question







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






share|improve this question







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.











share|improve this question







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.









share|improve this question




share|improve this question






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 2 days ago









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. Use Module, Block, or With for code blocks. And try not to use For.
    $endgroup$
    – Roman
    2 days ago






  • 3




    $begingroup$
    Why should I avoid the For loop in Mathematica?
    $endgroup$
    – corey979
    2 days ago






  • 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
    2 days ago










  • $begingroup$
    You might also look into Nest and NestList as an alternative, built-in way to iterate this map.
    $endgroup$
    – Chris K
    2 days ago












  • 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
    2 days ago






  • 3




    $begingroup$
    Why should I avoid the For loop in Mathematica?
    $endgroup$
    – corey979
    2 days ago






  • 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
    2 days ago










  • $begingroup$
    You might also look into Nest and NestList as an alternative, built-in way to iterate this map.
    $endgroup$
    – Chris K
    2 days 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
2 days 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
2 days ago




3




3




$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
2 days ago




$begingroup$
Why should I avoid the For loop in Mathematica?
$endgroup$
– corey979
2 days ago




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
2 days 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
2 days ago












$begingroup$
You might also look into Nest and NestList as an alternative, built-in way to iterate this map.
$endgroup$
– Chris K
2 days ago




$begingroup$
You might also look into Nest and NestList as an alternative, built-in way to iterate this map.
$endgroup$
– Chris K
2 days ago










2 Answers
2






active

oldest

votes


















6












$begingroup$

Try this:



f[r_] := Union[
Take[
RecurrenceTable[
a[n + 1] == r*(1 - a[n])*a[n], a[1] == .5,
a, n, 1, 1000
],
-100]
]





share|improve this answer











$endgroup$








  • 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
    2 days ago


















3












$begingroup$

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.






share|improve this answer











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    Try this:



    f[r_] := Union[
    Take[
    RecurrenceTable[
    a[n + 1] == r*(1 - a[n])*a[n], a[1] == .5,
    a, n, 1, 1000
    ],
    -100]
    ]





    share|improve this answer











    $endgroup$








    • 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
      2 days ago















    6












    $begingroup$

    Try this:



    f[r_] := Union[
    Take[
    RecurrenceTable[
    a[n + 1] == r*(1 - a[n])*a[n], a[1] == .5,
    a, n, 1, 1000
    ],
    -100]
    ]





    share|improve this answer











    $endgroup$








    • 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
      2 days ago













    6












    6








    6





    $begingroup$

    Try this:



    f[r_] := Union[
    Take[
    RecurrenceTable[
    a[n + 1] == r*(1 - a[n])*a[n], a[1] == .5,
    a, n, 1, 1000
    ],
    -100]
    ]





    share|improve this answer











    $endgroup$



    Try this:



    f[r_] := Union[
    Take[
    RecurrenceTable[
    a[n + 1] == r*(1 - a[n])*a[n], a[1] == .5,
    a, n, 1, 1000
    ],
    -100]
    ]






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 2 days ago









    MarcoB

    38.6k557115




    38.6k557115










    answered 2 days ago









    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
      2 days ago












    • 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
      2 days 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
    2 days 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
    2 days ago











    3












    $begingroup$

    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.






    share|improve this answer











    $endgroup$

















      3












      $begingroup$

      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.






      share|improve this answer











      $endgroup$















        3












        3








        3





        $begingroup$

        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.






        share|improve this answer











        $endgroup$



        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.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 2 days ago

























        answered 2 days ago









        RomanRoman

        4,95511130




        4,95511130




















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