RSolve not reducing for a certain recurrence relation











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I'm trying to use RSolve as follows to solve a recurrence relation:



RSolve[{a[0] == 1, a[2 n + 1] == a[2 n]*2, a[2 n] == a[2 n - 1] + 2}, a, n]


I think the meaning of the relation is clear: for odd terms in the sequence, multiply the previous term by two, and for even terms, add two to the previous term.



However, when evaluating this, Mathematica simply echos the input:



output of evaluation



rather than attempting to solve the recurrence.



I can't find anything in the RSolve documentation which talks about cases where RSolve will do nothing, without any error message.



Have I made a syntax error, or does this mean that Mathematica is not able to solve this type of recurrence relation? How can I change my input so that Mathematica does solve the recurrence, assuming it is possible?










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  • Try replacing the next-to-last term in RSolve (a) with a[n].
    – yosimitsu kodanuri
    2 days ago










  • It still simply echoes the input: RSolve[{a[0] == 1, a[1 + 2 n] == 2 a[2 n], a[2 n] == 2 + a[-1 + 2 n]}, a[n], n]
    – konsolas
    2 days ago






  • 1




    Welcome to Mathematica.SE! Interesting question(+1). I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
    – bbgodfrey
    2 days ago

















up vote
4
down vote

favorite
1












I'm trying to use RSolve as follows to solve a recurrence relation:



RSolve[{a[0] == 1, a[2 n + 1] == a[2 n]*2, a[2 n] == a[2 n - 1] + 2}, a, n]


I think the meaning of the relation is clear: for odd terms in the sequence, multiply the previous term by two, and for even terms, add two to the previous term.



However, when evaluating this, Mathematica simply echos the input:



output of evaluation



rather than attempting to solve the recurrence.



I can't find anything in the RSolve documentation which talks about cases where RSolve will do nothing, without any error message.



Have I made a syntax error, or does this mean that Mathematica is not able to solve this type of recurrence relation? How can I change my input so that Mathematica does solve the recurrence, assuming it is possible?










share|improve this question









New contributor




konsolas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • Try replacing the next-to-last term in RSolve (a) with a[n].
    – yosimitsu kodanuri
    2 days ago










  • It still simply echoes the input: RSolve[{a[0] == 1, a[1 + 2 n] == 2 a[2 n], a[2 n] == 2 + a[-1 + 2 n]}, a[n], n]
    – konsolas
    2 days ago






  • 1




    Welcome to Mathematica.SE! Interesting question(+1). I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
    – bbgodfrey
    2 days ago















up vote
4
down vote

favorite
1









up vote
4
down vote

favorite
1






1





I'm trying to use RSolve as follows to solve a recurrence relation:



RSolve[{a[0] == 1, a[2 n + 1] == a[2 n]*2, a[2 n] == a[2 n - 1] + 2}, a, n]


I think the meaning of the relation is clear: for odd terms in the sequence, multiply the previous term by two, and for even terms, add two to the previous term.



However, when evaluating this, Mathematica simply echos the input:



output of evaluation



rather than attempting to solve the recurrence.



I can't find anything in the RSolve documentation which talks about cases where RSolve will do nothing, without any error message.



Have I made a syntax error, or does this mean that Mathematica is not able to solve this type of recurrence relation? How can I change my input so that Mathematica does solve the recurrence, assuming it is possible?










share|improve this question









New contributor




konsolas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











I'm trying to use RSolve as follows to solve a recurrence relation:



RSolve[{a[0] == 1, a[2 n + 1] == a[2 n]*2, a[2 n] == a[2 n - 1] + 2}, a, n]


I think the meaning of the relation is clear: for odd terms in the sequence, multiply the previous term by two, and for even terms, add two to the previous term.



However, when evaluating this, Mathematica simply echos the input:



output of evaluation



rather than attempting to solve the recurrence.



I can't find anything in the RSolve documentation which talks about cases where RSolve will do nothing, without any error message.



Have I made a syntax error, or does this mean that Mathematica is not able to solve this type of recurrence relation? How can I change my input so that Mathematica does solve the recurrence, assuming it is possible?







equation-solving symbolic syntax recursion difference-equations






share|improve this question









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share|improve this question









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edited 2 days ago









bbgodfrey

43.7k857107




43.7k857107






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asked 2 days ago









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konsolas is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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  • Try replacing the next-to-last term in RSolve (a) with a[n].
    – yosimitsu kodanuri
    2 days ago










  • It still simply echoes the input: RSolve[{a[0] == 1, a[1 + 2 n] == 2 a[2 n], a[2 n] == 2 + a[-1 + 2 n]}, a[n], n]
    – konsolas
    2 days ago






  • 1




    Welcome to Mathematica.SE! Interesting question(+1). I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
    – bbgodfrey
    2 days ago




















  • Try replacing the next-to-last term in RSolve (a) with a[n].
    – yosimitsu kodanuri
    2 days ago










  • It still simply echoes the input: RSolve[{a[0] == 1, a[1 + 2 n] == 2 a[2 n], a[2 n] == 2 + a[-1 + 2 n]}, a[n], n]
    – konsolas
    2 days ago






  • 1




    Welcome to Mathematica.SE! Interesting question(+1). I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
    – bbgodfrey
    2 days ago


















Try replacing the next-to-last term in RSolve (a) with a[n].
– yosimitsu kodanuri
2 days ago




Try replacing the next-to-last term in RSolve (a) with a[n].
– yosimitsu kodanuri
2 days ago












It still simply echoes the input: RSolve[{a[0] == 1, a[1 + 2 n] == 2 a[2 n], a[2 n] == 2 + a[-1 + 2 n]}, a[n], n]
– konsolas
2 days ago




It still simply echoes the input: RSolve[{a[0] == 1, a[1 + 2 n] == 2 a[2 n], a[2 n] == 2 + a[-1 + 2 n]}, a[n], n]
– konsolas
2 days ago




1




1




Welcome to Mathematica.SE! Interesting question(+1). I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
– bbgodfrey
2 days ago






Welcome to Mathematica.SE! Interesting question(+1). I hope you will become a regular contributor. To get started, 1) take the introductory tour now, 2) when you see good questions and answers, vote them up by clicking the gray triangles, because the credibility of the system is based on the reputation gained by users sharing their knowledge, 3) remember to accept the answer, if any, that solves your problem, by clicking the checkmark sign, and 4) give help too, by answering questions in your areas of expertise.
– bbgodfrey
2 days ago












2 Answers
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active

oldest

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up vote
7
down vote



accepted










It appears that RSolve cannot solve difference equations in which two different equations describe the same variable. If so, a work-around is to represent a[k] with k an even index as c[k], and with k an odd indix as b[k], so that there is only one equation per variable.



FullSimplify[RSolveValue[{c[0] == 1, b[k] == c[k - 1]*2, c[k + 2] == b[k + 1] + 2}, 
{c[k], b[k]}, k] /. C[1] -> 0]
(* {-2 + 3 2^(-1 + k/2) (1 + (-1)^k), -4 - 3 2^(1/2 (-1 + k)) (-1 + (-1)^k)} *)


Then, construct the desired a[k] as even-index terms of c[k] and odd-index terms of b[k].



sol[k_] := If[EvenQ[k], -2 + 3 2^(k/2) , -4 + 3 2^((1 + k)/2) ]
Table[sol[k], {k, 0, 10}]
(* {1, 2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)


as desired.






share|improve this answer






























    up vote
    4
    down vote













    When Mathematica returns the input without any error message, it is unable to evaluate the input.



    With this particular recursion there is another approach. The recursion can be defined by



    Clear[a, ar]

    ar[0] = 1; ar[n_?OddQ] := ar[n - 1]*2; ar[n_?EvenQ] := ar[n - 1] + 2;


    Generating a sequence from this recursion



    seq = ar /@ Range[10]

    (* {2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)


    Use FindSequenceFunction to find the closed form of the recursion



    a[n_] = FindSequenceFunction[seq, n] // FullSimplify

    (* -3 + (-1)^n + 3 2^(-1 + n/2) (1 + Sqrt[2] + (-1)^(1 + n) (-1 + Sqrt[2])) *)


    Checking equivalence outside of the range of seq



    And @@ Table[a[n] == ar[n], {n, 0, 100}]

    (* True *)


    EDIT: Verifying,



    Simplify[{a[2 n + 1] == a[2 n]*2, 
    a[2 n] == a[2 n - 1] + 2}, {Element[n, Integers], n >= 0}]

    (* {True, True} *)





    share|improve this answer























    • Hi, thanks for the alternative solution. Do you know why Mathematica was unable to evaluate the input given?
      – konsolas
      2 days ago










    • Don’t know. Presumably the algorithms used are not sufficiently robust to cover this case.
      – Bob Hanlon
      2 days ago











    Your Answer





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    2 Answers
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    2 Answers
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    active

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    active

    oldest

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    up vote
    7
    down vote



    accepted










    It appears that RSolve cannot solve difference equations in which two different equations describe the same variable. If so, a work-around is to represent a[k] with k an even index as c[k], and with k an odd indix as b[k], so that there is only one equation per variable.



    FullSimplify[RSolveValue[{c[0] == 1, b[k] == c[k - 1]*2, c[k + 2] == b[k + 1] + 2}, 
    {c[k], b[k]}, k] /. C[1] -> 0]
    (* {-2 + 3 2^(-1 + k/2) (1 + (-1)^k), -4 - 3 2^(1/2 (-1 + k)) (-1 + (-1)^k)} *)


    Then, construct the desired a[k] as even-index terms of c[k] and odd-index terms of b[k].



    sol[k_] := If[EvenQ[k], -2 + 3 2^(k/2) , -4 + 3 2^((1 + k)/2) ]
    Table[sol[k], {k, 0, 10}]
    (* {1, 2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)


    as desired.






    share|improve this answer



























      up vote
      7
      down vote



      accepted










      It appears that RSolve cannot solve difference equations in which two different equations describe the same variable. If so, a work-around is to represent a[k] with k an even index as c[k], and with k an odd indix as b[k], so that there is only one equation per variable.



      FullSimplify[RSolveValue[{c[0] == 1, b[k] == c[k - 1]*2, c[k + 2] == b[k + 1] + 2}, 
      {c[k], b[k]}, k] /. C[1] -> 0]
      (* {-2 + 3 2^(-1 + k/2) (1 + (-1)^k), -4 - 3 2^(1/2 (-1 + k)) (-1 + (-1)^k)} *)


      Then, construct the desired a[k] as even-index terms of c[k] and odd-index terms of b[k].



      sol[k_] := If[EvenQ[k], -2 + 3 2^(k/2) , -4 + 3 2^((1 + k)/2) ]
      Table[sol[k], {k, 0, 10}]
      (* {1, 2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)


      as desired.






      share|improve this answer

























        up vote
        7
        down vote



        accepted







        up vote
        7
        down vote



        accepted






        It appears that RSolve cannot solve difference equations in which two different equations describe the same variable. If so, a work-around is to represent a[k] with k an even index as c[k], and with k an odd indix as b[k], so that there is only one equation per variable.



        FullSimplify[RSolveValue[{c[0] == 1, b[k] == c[k - 1]*2, c[k + 2] == b[k + 1] + 2}, 
        {c[k], b[k]}, k] /. C[1] -> 0]
        (* {-2 + 3 2^(-1 + k/2) (1 + (-1)^k), -4 - 3 2^(1/2 (-1 + k)) (-1 + (-1)^k)} *)


        Then, construct the desired a[k] as even-index terms of c[k] and odd-index terms of b[k].



        sol[k_] := If[EvenQ[k], -2 + 3 2^(k/2) , -4 + 3 2^((1 + k)/2) ]
        Table[sol[k], {k, 0, 10}]
        (* {1, 2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)


        as desired.






        share|improve this answer














        It appears that RSolve cannot solve difference equations in which two different equations describe the same variable. If so, a work-around is to represent a[k] with k an even index as c[k], and with k an odd indix as b[k], so that there is only one equation per variable.



        FullSimplify[RSolveValue[{c[0] == 1, b[k] == c[k - 1]*2, c[k + 2] == b[k + 1] + 2}, 
        {c[k], b[k]}, k] /. C[1] -> 0]
        (* {-2 + 3 2^(-1 + k/2) (1 + (-1)^k), -4 - 3 2^(1/2 (-1 + k)) (-1 + (-1)^k)} *)


        Then, construct the desired a[k] as even-index terms of c[k] and odd-index terms of b[k].



        sol[k_] := If[EvenQ[k], -2 + 3 2^(k/2) , -4 + 3 2^((1 + k)/2) ]
        Table[sol[k], {k, 0, 10}]
        (* {1, 2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)


        as desired.







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited yesterday

























        answered 2 days ago









        bbgodfrey

        43.7k857107




        43.7k857107






















            up vote
            4
            down vote













            When Mathematica returns the input without any error message, it is unable to evaluate the input.



            With this particular recursion there is another approach. The recursion can be defined by



            Clear[a, ar]

            ar[0] = 1; ar[n_?OddQ] := ar[n - 1]*2; ar[n_?EvenQ] := ar[n - 1] + 2;


            Generating a sequence from this recursion



            seq = ar /@ Range[10]

            (* {2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)


            Use FindSequenceFunction to find the closed form of the recursion



            a[n_] = FindSequenceFunction[seq, n] // FullSimplify

            (* -3 + (-1)^n + 3 2^(-1 + n/2) (1 + Sqrt[2] + (-1)^(1 + n) (-1 + Sqrt[2])) *)


            Checking equivalence outside of the range of seq



            And @@ Table[a[n] == ar[n], {n, 0, 100}]

            (* True *)


            EDIT: Verifying,



            Simplify[{a[2 n + 1] == a[2 n]*2, 
            a[2 n] == a[2 n - 1] + 2}, {Element[n, Integers], n >= 0}]

            (* {True, True} *)





            share|improve this answer























            • Hi, thanks for the alternative solution. Do you know why Mathematica was unable to evaluate the input given?
              – konsolas
              2 days ago










            • Don’t know. Presumably the algorithms used are not sufficiently robust to cover this case.
              – Bob Hanlon
              2 days ago















            up vote
            4
            down vote













            When Mathematica returns the input without any error message, it is unable to evaluate the input.



            With this particular recursion there is another approach. The recursion can be defined by



            Clear[a, ar]

            ar[0] = 1; ar[n_?OddQ] := ar[n - 1]*2; ar[n_?EvenQ] := ar[n - 1] + 2;


            Generating a sequence from this recursion



            seq = ar /@ Range[10]

            (* {2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)


            Use FindSequenceFunction to find the closed form of the recursion



            a[n_] = FindSequenceFunction[seq, n] // FullSimplify

            (* -3 + (-1)^n + 3 2^(-1 + n/2) (1 + Sqrt[2] + (-1)^(1 + n) (-1 + Sqrt[2])) *)


            Checking equivalence outside of the range of seq



            And @@ Table[a[n] == ar[n], {n, 0, 100}]

            (* True *)


            EDIT: Verifying,



            Simplify[{a[2 n + 1] == a[2 n]*2, 
            a[2 n] == a[2 n - 1] + 2}, {Element[n, Integers], n >= 0}]

            (* {True, True} *)





            share|improve this answer























            • Hi, thanks for the alternative solution. Do you know why Mathematica was unable to evaluate the input given?
              – konsolas
              2 days ago










            • Don’t know. Presumably the algorithms used are not sufficiently robust to cover this case.
              – Bob Hanlon
              2 days ago













            up vote
            4
            down vote










            up vote
            4
            down vote









            When Mathematica returns the input without any error message, it is unable to evaluate the input.



            With this particular recursion there is another approach. The recursion can be defined by



            Clear[a, ar]

            ar[0] = 1; ar[n_?OddQ] := ar[n - 1]*2; ar[n_?EvenQ] := ar[n - 1] + 2;


            Generating a sequence from this recursion



            seq = ar /@ Range[10]

            (* {2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)


            Use FindSequenceFunction to find the closed form of the recursion



            a[n_] = FindSequenceFunction[seq, n] // FullSimplify

            (* -3 + (-1)^n + 3 2^(-1 + n/2) (1 + Sqrt[2] + (-1)^(1 + n) (-1 + Sqrt[2])) *)


            Checking equivalence outside of the range of seq



            And @@ Table[a[n] == ar[n], {n, 0, 100}]

            (* True *)


            EDIT: Verifying,



            Simplify[{a[2 n + 1] == a[2 n]*2, 
            a[2 n] == a[2 n - 1] + 2}, {Element[n, Integers], n >= 0}]

            (* {True, True} *)





            share|improve this answer














            When Mathematica returns the input without any error message, it is unable to evaluate the input.



            With this particular recursion there is another approach. The recursion can be defined by



            Clear[a, ar]

            ar[0] = 1; ar[n_?OddQ] := ar[n - 1]*2; ar[n_?EvenQ] := ar[n - 1] + 2;


            Generating a sequence from this recursion



            seq = ar /@ Range[10]

            (* {2, 4, 8, 10, 20, 22, 44, 46, 92, 94} *)


            Use FindSequenceFunction to find the closed form of the recursion



            a[n_] = FindSequenceFunction[seq, n] // FullSimplify

            (* -3 + (-1)^n + 3 2^(-1 + n/2) (1 + Sqrt[2] + (-1)^(1 + n) (-1 + Sqrt[2])) *)


            Checking equivalence outside of the range of seq



            And @@ Table[a[n] == ar[n], {n, 0, 100}]

            (* True *)


            EDIT: Verifying,



            Simplify[{a[2 n + 1] == a[2 n]*2, 
            a[2 n] == a[2 n - 1] + 2}, {Element[n, Integers], n >= 0}]

            (* {True, True} *)






            share|improve this answer














            share|improve this answer



            share|improve this answer








            edited yesterday

























            answered 2 days ago









            Bob Hanlon

            57.9k23593




            57.9k23593












            • Hi, thanks for the alternative solution. Do you know why Mathematica was unable to evaluate the input given?
              – konsolas
              2 days ago










            • Don’t know. Presumably the algorithms used are not sufficiently robust to cover this case.
              – Bob Hanlon
              2 days ago


















            • Hi, thanks for the alternative solution. Do you know why Mathematica was unable to evaluate the input given?
              – konsolas
              2 days ago










            • Don’t know. Presumably the algorithms used are not sufficiently robust to cover this case.
              – Bob Hanlon
              2 days ago
















            Hi, thanks for the alternative solution. Do you know why Mathematica was unable to evaluate the input given?
            – konsolas
            2 days ago




            Hi, thanks for the alternative solution. Do you know why Mathematica was unable to evaluate the input given?
            – konsolas
            2 days ago












            Don’t know. Presumably the algorithms used are not sufficiently robust to cover this case.
            – Bob Hanlon
            2 days ago




            Don’t know. Presumably the algorithms used are not sufficiently robust to cover this case.
            – Bob Hanlon
            2 days ago










            konsolas is a new contributor. Be nice, and check out our Code of Conduct.










             

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