Inverse of a function containing $(-1)^n$












1












$begingroup$


Is it possible to find the inverse of a function which contains the term $(-1)^n$?



Specifically: I want to find the inverse of the following function, which gives the nth number of the form $6kpm1$:
$$ f(n) = 3n + frac{3}{2} - frac{(-1)^n}{2} $$



Since it is not possible to take the log to base -1 and I'm not a mathematician, I didn't suceed with my attempts of solving for $f(n)$. (Google search and Wolfram Alpha didn't help either, and based on the fact that $(-1)^n$ "looses" information about n I doubt it's possible in general)






inverse-function







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$endgroup$












  • $begingroup$
    I think you mean to tabulate numbers of the form $6kpm 1$ but your function does not do that.
    $endgroup$
    – Ross Millikan
    Jan 2 at 14:44
















1












$begingroup$


Is it possible to find the inverse of a function which contains the term $(-1)^n$?



Specifically: I want to find the inverse of the following function, which gives the nth number of the form $6kpm1$:
$$ f(n) = 3n + frac{3}{2} - frac{(-1)^n}{2} $$



Since it is not possible to take the log to base -1 and I'm not a mathematician, I didn't suceed with my attempts of solving for $f(n)$. (Google search and Wolfram Alpha didn't help either, and based on the fact that $(-1)^n$ "looses" information about n I doubt it's possible in general)






inverse-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    I think you mean to tabulate numbers of the form $6kpm 1$ but your function does not do that.
    $endgroup$
    – Ross Millikan
    Jan 2 at 14:44














1












1








1





$begingroup$


Is it possible to find the inverse of a function which contains the term $(-1)^n$?



Specifically: I want to find the inverse of the following function, which gives the nth number of the form $6kpm1$:
$$ f(n) = 3n + frac{3}{2} - frac{(-1)^n}{2} $$



Since it is not possible to take the log to base -1 and I'm not a mathematician, I didn't suceed with my attempts of solving for $f(n)$. (Google search and Wolfram Alpha didn't help either, and based on the fact that $(-1)^n$ "looses" information about n I doubt it's possible in general)






inverse-function







share|cite|improve this question











$endgroup$




Is it possible to find the inverse of a function which contains the term $(-1)^n$?



Specifically: I want to find the inverse of the following function, which gives the nth number of the form $6kpm1$:
$$ f(n) = 3n + frac{3}{2} - frac{(-1)^n}{2} $$



Since it is not possible to take the log to base -1 and I'm not a mathematician, I didn't suceed with my attempts of solving for $f(n)$. (Google search and Wolfram Alpha didn't help either, and based on the fact that $(-1)^n$ "looses" information about n I doubt it's possible in general)






inverse-function




inverse-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 2 at 15:22







g3nuine

















asked Jan 2 at 14:38









g3nuineg3nuine

1598




1598












  • $begingroup$
    I think you mean to tabulate numbers of the form $6kpm 1$ but your function does not do that.
    $endgroup$
    – Ross Millikan
    Jan 2 at 14:44


















  • $begingroup$
    I think you mean to tabulate numbers of the form $6kpm 1$ but your function does not do that.
    $endgroup$
    – Ross Millikan
    Jan 2 at 14:44
















$begingroup$
I think you mean to tabulate numbers of the form $6kpm 1$ but your function does not do that.
$endgroup$
– Ross Millikan
Jan 2 at 14:44




$begingroup$
I think you mean to tabulate numbers of the form $6kpm 1$ but your function does not do that.
$endgroup$
– Ross Millikan
Jan 2 at 14:44










2 Answers
2






active

oldest

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3












$begingroup$

For $n$ even we have $$f(n)=3n+1$$ and you can write $$frac{f(n)-1}{3}=n$$
If $n$ is odd then you will get $$f(n)=3n+2$$ so $$n=frac{f(n)-2}{3}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    The problem is that I don't know n's parity im advance, but I guess a single function being the inverse instead of two is not possible?
    $endgroup$
    – g3nuine
    Jan 2 at 15:28










  • $begingroup$
    I think only one term as inverse is not possible.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 15:30



















1












$begingroup$

Consider the equation $f(n)=m$:




  • if $n=2k$, you get the equation $6k+1=m$, whose solution is $k=frac{m-1}6left(iff n=frac{m-1}3right)$;

  • if $n=2k-1$, you get the equation $6k-1=m$, whose solution is $k=frac{m+1}6left(iff n=frac{m+1}3right)$.


So, the inverse that you're looking for is$$nmapstofrac{n-(-1)^n}3.$$






share|cite|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









    3












    $begingroup$

    For $n$ even we have $$f(n)=3n+1$$ and you can write $$frac{f(n)-1}{3}=n$$
    If $n$ is odd then you will get $$f(n)=3n+2$$ so $$n=frac{f(n)-2}{3}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The problem is that I don't know n's parity im advance, but I guess a single function being the inverse instead of two is not possible?
      $endgroup$
      – g3nuine
      Jan 2 at 15:28










    • $begingroup$
      I think only one term as inverse is not possible.
      $endgroup$
      – Dr. Sonnhard Graubner
      Jan 2 at 15:30
















    3












    $begingroup$

    For $n$ even we have $$f(n)=3n+1$$ and you can write $$frac{f(n)-1}{3}=n$$
    If $n$ is odd then you will get $$f(n)=3n+2$$ so $$n=frac{f(n)-2}{3}$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      The problem is that I don't know n's parity im advance, but I guess a single function being the inverse instead of two is not possible?
      $endgroup$
      – g3nuine
      Jan 2 at 15:28










    • $begingroup$
      I think only one term as inverse is not possible.
      $endgroup$
      – Dr. Sonnhard Graubner
      Jan 2 at 15:30














    3












    3








    3





    $begingroup$

    For $n$ even we have $$f(n)=3n+1$$ and you can write $$frac{f(n)-1}{3}=n$$
    If $n$ is odd then you will get $$f(n)=3n+2$$ so $$n=frac{f(n)-2}{3}$$






    share|cite|improve this answer









    $endgroup$



    For $n$ even we have $$f(n)=3n+1$$ and you can write $$frac{f(n)-1}{3}=n$$
    If $n$ is odd then you will get $$f(n)=3n+2$$ so $$n=frac{f(n)-2}{3}$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 2 at 14:46









    Dr. Sonnhard GraubnerDr. Sonnhard Graubner

    78.4k42867




    78.4k42867












    • $begingroup$
      The problem is that I don't know n's parity im advance, but I guess a single function being the inverse instead of two is not possible?
      $endgroup$
      – g3nuine
      Jan 2 at 15:28










    • $begingroup$
      I think only one term as inverse is not possible.
      $endgroup$
      – Dr. Sonnhard Graubner
      Jan 2 at 15:30


















    • $begingroup$
      The problem is that I don't know n's parity im advance, but I guess a single function being the inverse instead of two is not possible?
      $endgroup$
      – g3nuine
      Jan 2 at 15:28










    • $begingroup$
      I think only one term as inverse is not possible.
      $endgroup$
      – Dr. Sonnhard Graubner
      Jan 2 at 15:30
















    $begingroup$
    The problem is that I don't know n's parity im advance, but I guess a single function being the inverse instead of two is not possible?
    $endgroup$
    – g3nuine
    Jan 2 at 15:28




    $begingroup$
    The problem is that I don't know n's parity im advance, but I guess a single function being the inverse instead of two is not possible?
    $endgroup$
    – g3nuine
    Jan 2 at 15:28












    $begingroup$
    I think only one term as inverse is not possible.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 15:30




    $begingroup$
    I think only one term as inverse is not possible.
    $endgroup$
    – Dr. Sonnhard Graubner
    Jan 2 at 15:30











    1












    $begingroup$

    Consider the equation $f(n)=m$:




    • if $n=2k$, you get the equation $6k+1=m$, whose solution is $k=frac{m-1}6left(iff n=frac{m-1}3right)$;

    • if $n=2k-1$, you get the equation $6k-1=m$, whose solution is $k=frac{m+1}6left(iff n=frac{m+1}3right)$.


    So, the inverse that you're looking for is$$nmapstofrac{n-(-1)^n}3.$$






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Consider the equation $f(n)=m$:




      • if $n=2k$, you get the equation $6k+1=m$, whose solution is $k=frac{m-1}6left(iff n=frac{m-1}3right)$;

      • if $n=2k-1$, you get the equation $6k-1=m$, whose solution is $k=frac{m+1}6left(iff n=frac{m+1}3right)$.


      So, the inverse that you're looking for is$$nmapstofrac{n-(-1)^n}3.$$






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Consider the equation $f(n)=m$:




        • if $n=2k$, you get the equation $6k+1=m$, whose solution is $k=frac{m-1}6left(iff n=frac{m-1}3right)$;

        • if $n=2k-1$, you get the equation $6k-1=m$, whose solution is $k=frac{m+1}6left(iff n=frac{m+1}3right)$.


        So, the inverse that you're looking for is$$nmapstofrac{n-(-1)^n}3.$$






        share|cite|improve this answer









        $endgroup$



        Consider the equation $f(n)=m$:




        • if $n=2k$, you get the equation $6k+1=m$, whose solution is $k=frac{m-1}6left(iff n=frac{m-1}3right)$;

        • if $n=2k-1$, you get the equation $6k-1=m$, whose solution is $k=frac{m+1}6left(iff n=frac{m+1}3right)$.


        So, the inverse that you're looking for is$$nmapstofrac{n-(-1)^n}3.$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 2 at 14:45









        José Carlos SantosJosé Carlos Santos

        171k23132240




        171k23132240






























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