How do we write $f(x+2)$ in terms of $f(x)$?












0












$begingroup$


$$f : R^+ rightarrow R^+$$



$$f(x) = dfrac{x}{x+1} $$



How do we write $f(x+2)$ in terms of $f(x)$? This is a general question that I wondered how to algebraically write it.



Regards






functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Literally divide $frac{f(x+2)}{f(x)}$ and you will get what you want.
    $endgroup$
    – Lucas Henrique
    Dec 10 '18 at 5:45






  • 3




    $begingroup$
    One way to do it is $f(x+2) = frac{x+2}{x+3} times frac{x+1}{x} times f(x)$. It's not really clear what you're looking for.
    $endgroup$
    – platty
    Dec 10 '18 at 5:45










  • $begingroup$
    If I get you right, you want a product/sum independent of x as a factor but only as an argument, because, as in my first comment, it would be obvious.
    $endgroup$
    – Lucas Henrique
    Dec 10 '18 at 5:47










  • $begingroup$
    Something like $f(x+2) = f(f^{-1}(f(x))+2) = frac{f(x)-2}{2f(x)-3}$???
    $endgroup$
    – achille hui
    Dec 10 '18 at 5:50
















0












$begingroup$


$$f : R^+ rightarrow R^+$$



$$f(x) = dfrac{x}{x+1} $$



How do we write $f(x+2)$ in terms of $f(x)$? This is a general question that I wondered how to algebraically write it.



Regards






functions







share|cite|improve this question









$endgroup$












  • $begingroup$
    Literally divide $frac{f(x+2)}{f(x)}$ and you will get what you want.
    $endgroup$
    – Lucas Henrique
    Dec 10 '18 at 5:45






  • 3




    $begingroup$
    One way to do it is $f(x+2) = frac{x+2}{x+3} times frac{x+1}{x} times f(x)$. It's not really clear what you're looking for.
    $endgroup$
    – platty
    Dec 10 '18 at 5:45










  • $begingroup$
    If I get you right, you want a product/sum independent of x as a factor but only as an argument, because, as in my first comment, it would be obvious.
    $endgroup$
    – Lucas Henrique
    Dec 10 '18 at 5:47










  • $begingroup$
    Something like $f(x+2) = f(f^{-1}(f(x))+2) = frac{f(x)-2}{2f(x)-3}$???
    $endgroup$
    – achille hui
    Dec 10 '18 at 5:50














0












0








0





$begingroup$


$$f : R^+ rightarrow R^+$$



$$f(x) = dfrac{x}{x+1} $$



How do we write $f(x+2)$ in terms of $f(x)$? This is a general question that I wondered how to algebraically write it.



Regards






functions







share|cite|improve this question









$endgroup$




$$f : R^+ rightarrow R^+$$



$$f(x) = dfrac{x}{x+1} $$



How do we write $f(x+2)$ in terms of $f(x)$? This is a general question that I wondered how to algebraically write it.



Regards






functions




functions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 10 '18 at 5:42









HamiltonHamilton

1778




1778












  • $begingroup$
    Literally divide $frac{f(x+2)}{f(x)}$ and you will get what you want.
    $endgroup$
    – Lucas Henrique
    Dec 10 '18 at 5:45






  • 3




    $begingroup$
    One way to do it is $f(x+2) = frac{x+2}{x+3} times frac{x+1}{x} times f(x)$. It's not really clear what you're looking for.
    $endgroup$
    – platty
    Dec 10 '18 at 5:45










  • $begingroup$
    If I get you right, you want a product/sum independent of x as a factor but only as an argument, because, as in my first comment, it would be obvious.
    $endgroup$
    – Lucas Henrique
    Dec 10 '18 at 5:47










  • $begingroup$
    Something like $f(x+2) = f(f^{-1}(f(x))+2) = frac{f(x)-2}{2f(x)-3}$???
    $endgroup$
    – achille hui
    Dec 10 '18 at 5:50


















  • $begingroup$
    Literally divide $frac{f(x+2)}{f(x)}$ and you will get what you want.
    $endgroup$
    – Lucas Henrique
    Dec 10 '18 at 5:45






  • 3




    $begingroup$
    One way to do it is $f(x+2) = frac{x+2}{x+3} times frac{x+1}{x} times f(x)$. It's not really clear what you're looking for.
    $endgroup$
    – platty
    Dec 10 '18 at 5:45










  • $begingroup$
    If I get you right, you want a product/sum independent of x as a factor but only as an argument, because, as in my first comment, it would be obvious.
    $endgroup$
    – Lucas Henrique
    Dec 10 '18 at 5:47










  • $begingroup$
    Something like $f(x+2) = f(f^{-1}(f(x))+2) = frac{f(x)-2}{2f(x)-3}$???
    $endgroup$
    – achille hui
    Dec 10 '18 at 5:50
















$begingroup$
Literally divide $frac{f(x+2)}{f(x)}$ and you will get what you want.
$endgroup$
– Lucas Henrique
Dec 10 '18 at 5:45




$begingroup$
Literally divide $frac{f(x+2)}{f(x)}$ and you will get what you want.
$endgroup$
– Lucas Henrique
Dec 10 '18 at 5:45




3




3




$begingroup$
One way to do it is $f(x+2) = frac{x+2}{x+3} times frac{x+1}{x} times f(x)$. It's not really clear what you're looking for.
$endgroup$
– platty
Dec 10 '18 at 5:45




$begingroup$
One way to do it is $f(x+2) = frac{x+2}{x+3} times frac{x+1}{x} times f(x)$. It's not really clear what you're looking for.
$endgroup$
– platty
Dec 10 '18 at 5:45












$begingroup$
If I get you right, you want a product/sum independent of x as a factor but only as an argument, because, as in my first comment, it would be obvious.
$endgroup$
– Lucas Henrique
Dec 10 '18 at 5:47




$begingroup$
If I get you right, you want a product/sum independent of x as a factor but only as an argument, because, as in my first comment, it would be obvious.
$endgroup$
– Lucas Henrique
Dec 10 '18 at 5:47












$begingroup$
Something like $f(x+2) = f(f^{-1}(f(x))+2) = frac{f(x)-2}{2f(x)-3}$???
$endgroup$
– achille hui
Dec 10 '18 at 5:50




$begingroup$
Something like $f(x+2) = f(f^{-1}(f(x))+2) = frac{f(x)-2}{2f(x)-3}$???
$endgroup$
– achille hui
Dec 10 '18 at 5:50










3 Answers
3






active

oldest

votes


















1












$begingroup$

I can't quite tell, but I think you're looking for a function $g$ so that



$$g(f(x))=f(x+2).$$



To do this, we find $f^{-1}(y)$ first. If $f(x)=y$, then



$$frac{x}{x+1}=yimplies x=y(x+1)implies y=x(1-y)implies x=frac{y}{1-y}.$$



So, if $f(x)=y$, then



$$f(x+2)=frac{x+2}{x+3}=frac{frac{y}{1-y}+2}{frac{y}{1-y}+3}=frac{y+2(1-y)}{y+3(1-y)}=frac{2-y}{3-2y}.$$



So, your answer is



$$f(x+2)=frac{2-f(x)}{3-2f(x)}.$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Oh...I was too slow typing the answer :)
    $endgroup$
    – tonychow0929
    Dec 10 '18 at 5:53





















1












$begingroup$

A more general method:



$$f(x) = frac{x}{x+1}$$
$$f(x)(x+1) = x$$
$$(f(x)-1)x=-f(x)$$
$$x=frac{f(x)}{1-f(x)}$$
Hence,
$$f(x+2)=frac{x+2}{x+3}=frac{frac{f(x)}{1-f(x)}+2}{frac{f(x)}{1-f(x)}+3}=frac{f(x)+2(1-f(x))}{f(x)+3(1-f(x))}=frac{2-f(x)}{3-2f(x)}$$



Note: I usually think that this can easily go wrong, so you may want to do some checking: $f(1+2)=frac{3}{4}$ by the original expression, and indeed $f(1+2)=frac{2-f(1)}{3-2f(1)}=frac{2-frac{1}{2}}{3-2(frac{1}{2})}=frac{frac{3}{2}}{2}=frac{3}{4}$.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    If I get your question correctly, then you can try this:



    $$f(x+2) = frac{x+2} {x+3}$$
    $$f(x) = frac{x} {x+1}$$



    $$ x = x.f(x) + f(x)$$
    $$implies x(1-f(x)) = f(x)$$



    $$implies x = frac {f(x)}{1-f(x)}$$



    Now substitute this $x$ in $f(x+2)$ and get the desired answer.






    share|cite|improve this answer











    $endgroup$









    • 2




      $begingroup$
      Your sign of $x$ is wrong. That should be $1-f(x)$ instead of $f(x)-1$.
      $endgroup$
      – tonychow0929
      Dec 10 '18 at 5:56






    • 1




      $begingroup$
      @tonychow0929 I am sorry, thank you for that :)
      $endgroup$
      – PradyumanDixit
      Dec 10 '18 at 5:56













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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    I can't quite tell, but I think you're looking for a function $g$ so that



    $$g(f(x))=f(x+2).$$



    To do this, we find $f^{-1}(y)$ first. If $f(x)=y$, then



    $$frac{x}{x+1}=yimplies x=y(x+1)implies y=x(1-y)implies x=frac{y}{1-y}.$$



    So, if $f(x)=y$, then



    $$f(x+2)=frac{x+2}{x+3}=frac{frac{y}{1-y}+2}{frac{y}{1-y}+3}=frac{y+2(1-y)}{y+3(1-y)}=frac{2-y}{3-2y}.$$



    So, your answer is



    $$f(x+2)=frac{2-f(x)}{3-2f(x)}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Oh...I was too slow typing the answer :)
      $endgroup$
      – tonychow0929
      Dec 10 '18 at 5:53


















    1












    $begingroup$

    I can't quite tell, but I think you're looking for a function $g$ so that



    $$g(f(x))=f(x+2).$$



    To do this, we find $f^{-1}(y)$ first. If $f(x)=y$, then



    $$frac{x}{x+1}=yimplies x=y(x+1)implies y=x(1-y)implies x=frac{y}{1-y}.$$



    So, if $f(x)=y$, then



    $$f(x+2)=frac{x+2}{x+3}=frac{frac{y}{1-y}+2}{frac{y}{1-y}+3}=frac{y+2(1-y)}{y+3(1-y)}=frac{2-y}{3-2y}.$$



    So, your answer is



    $$f(x+2)=frac{2-f(x)}{3-2f(x)}.$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Oh...I was too slow typing the answer :)
      $endgroup$
      – tonychow0929
      Dec 10 '18 at 5:53
















    1












    1








    1





    $begingroup$

    I can't quite tell, but I think you're looking for a function $g$ so that



    $$g(f(x))=f(x+2).$$



    To do this, we find $f^{-1}(y)$ first. If $f(x)=y$, then



    $$frac{x}{x+1}=yimplies x=y(x+1)implies y=x(1-y)implies x=frac{y}{1-y}.$$



    So, if $f(x)=y$, then



    $$f(x+2)=frac{x+2}{x+3}=frac{frac{y}{1-y}+2}{frac{y}{1-y}+3}=frac{y+2(1-y)}{y+3(1-y)}=frac{2-y}{3-2y}.$$



    So, your answer is



    $$f(x+2)=frac{2-f(x)}{3-2f(x)}.$$






    share|cite|improve this answer









    $endgroup$



    I can't quite tell, but I think you're looking for a function $g$ so that



    $$g(f(x))=f(x+2).$$



    To do this, we find $f^{-1}(y)$ first. If $f(x)=y$, then



    $$frac{x}{x+1}=yimplies x=y(x+1)implies y=x(1-y)implies x=frac{y}{1-y}.$$



    So, if $f(x)=y$, then



    $$f(x+2)=frac{x+2}{x+3}=frac{frac{y}{1-y}+2}{frac{y}{1-y}+3}=frac{y+2(1-y)}{y+3(1-y)}=frac{2-y}{3-2y}.$$



    So, your answer is



    $$f(x+2)=frac{2-f(x)}{3-2f(x)}.$$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 10 '18 at 5:51









    Carl SchildkrautCarl Schildkraut

    11.7k11443




    11.7k11443












    • $begingroup$
      Oh...I was too slow typing the answer :)
      $endgroup$
      – tonychow0929
      Dec 10 '18 at 5:53




















    • $begingroup$
      Oh...I was too slow typing the answer :)
      $endgroup$
      – tonychow0929
      Dec 10 '18 at 5:53


















    $begingroup$
    Oh...I was too slow typing the answer :)
    $endgroup$
    – tonychow0929
    Dec 10 '18 at 5:53






    $begingroup$
    Oh...I was too slow typing the answer :)
    $endgroup$
    – tonychow0929
    Dec 10 '18 at 5:53













    1












    $begingroup$

    A more general method:



    $$f(x) = frac{x}{x+1}$$
    $$f(x)(x+1) = x$$
    $$(f(x)-1)x=-f(x)$$
    $$x=frac{f(x)}{1-f(x)}$$
    Hence,
    $$f(x+2)=frac{x+2}{x+3}=frac{frac{f(x)}{1-f(x)}+2}{frac{f(x)}{1-f(x)}+3}=frac{f(x)+2(1-f(x))}{f(x)+3(1-f(x))}=frac{2-f(x)}{3-2f(x)}$$



    Note: I usually think that this can easily go wrong, so you may want to do some checking: $f(1+2)=frac{3}{4}$ by the original expression, and indeed $f(1+2)=frac{2-f(1)}{3-2f(1)}=frac{2-frac{1}{2}}{3-2(frac{1}{2})}=frac{frac{3}{2}}{2}=frac{3}{4}$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      A more general method:



      $$f(x) = frac{x}{x+1}$$
      $$f(x)(x+1) = x$$
      $$(f(x)-1)x=-f(x)$$
      $$x=frac{f(x)}{1-f(x)}$$
      Hence,
      $$f(x+2)=frac{x+2}{x+3}=frac{frac{f(x)}{1-f(x)}+2}{frac{f(x)}{1-f(x)}+3}=frac{f(x)+2(1-f(x))}{f(x)+3(1-f(x))}=frac{2-f(x)}{3-2f(x)}$$



      Note: I usually think that this can easily go wrong, so you may want to do some checking: $f(1+2)=frac{3}{4}$ by the original expression, and indeed $f(1+2)=frac{2-f(1)}{3-2f(1)}=frac{2-frac{1}{2}}{3-2(frac{1}{2})}=frac{frac{3}{2}}{2}=frac{3}{4}$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        A more general method:



        $$f(x) = frac{x}{x+1}$$
        $$f(x)(x+1) = x$$
        $$(f(x)-1)x=-f(x)$$
        $$x=frac{f(x)}{1-f(x)}$$
        Hence,
        $$f(x+2)=frac{x+2}{x+3}=frac{frac{f(x)}{1-f(x)}+2}{frac{f(x)}{1-f(x)}+3}=frac{f(x)+2(1-f(x))}{f(x)+3(1-f(x))}=frac{2-f(x)}{3-2f(x)}$$



        Note: I usually think that this can easily go wrong, so you may want to do some checking: $f(1+2)=frac{3}{4}$ by the original expression, and indeed $f(1+2)=frac{2-f(1)}{3-2f(1)}=frac{2-frac{1}{2}}{3-2(frac{1}{2})}=frac{frac{3}{2}}{2}=frac{3}{4}$.






        share|cite|improve this answer









        $endgroup$



        A more general method:



        $$f(x) = frac{x}{x+1}$$
        $$f(x)(x+1) = x$$
        $$(f(x)-1)x=-f(x)$$
        $$x=frac{f(x)}{1-f(x)}$$
        Hence,
        $$f(x+2)=frac{x+2}{x+3}=frac{frac{f(x)}{1-f(x)}+2}{frac{f(x)}{1-f(x)}+3}=frac{f(x)+2(1-f(x))}{f(x)+3(1-f(x))}=frac{2-f(x)}{3-2f(x)}$$



        Note: I usually think that this can easily go wrong, so you may want to do some checking: $f(1+2)=frac{3}{4}$ by the original expression, and indeed $f(1+2)=frac{2-f(1)}{3-2f(1)}=frac{2-frac{1}{2}}{3-2(frac{1}{2})}=frac{frac{3}{2}}{2}=frac{3}{4}$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 5:53









        tonychow0929tonychow0929

        28825




        28825























            1












            $begingroup$

            If I get your question correctly, then you can try this:



            $$f(x+2) = frac{x+2} {x+3}$$
            $$f(x) = frac{x} {x+1}$$



            $$ x = x.f(x) + f(x)$$
            $$implies x(1-f(x)) = f(x)$$



            $$implies x = frac {f(x)}{1-f(x)}$$



            Now substitute this $x$ in $f(x+2)$ and get the desired answer.






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              Your sign of $x$ is wrong. That should be $1-f(x)$ instead of $f(x)-1$.
              $endgroup$
              – tonychow0929
              Dec 10 '18 at 5:56






            • 1




              $begingroup$
              @tonychow0929 I am sorry, thank you for that :)
              $endgroup$
              – PradyumanDixit
              Dec 10 '18 at 5:56


















            1












            $begingroup$

            If I get your question correctly, then you can try this:



            $$f(x+2) = frac{x+2} {x+3}$$
            $$f(x) = frac{x} {x+1}$$



            $$ x = x.f(x) + f(x)$$
            $$implies x(1-f(x)) = f(x)$$



            $$implies x = frac {f(x)}{1-f(x)}$$



            Now substitute this $x$ in $f(x+2)$ and get the desired answer.






            share|cite|improve this answer











            $endgroup$









            • 2




              $begingroup$
              Your sign of $x$ is wrong. That should be $1-f(x)$ instead of $f(x)-1$.
              $endgroup$
              – tonychow0929
              Dec 10 '18 at 5:56






            • 1




              $begingroup$
              @tonychow0929 I am sorry, thank you for that :)
              $endgroup$
              – PradyumanDixit
              Dec 10 '18 at 5:56
















            1












            1








            1





            $begingroup$

            If I get your question correctly, then you can try this:



            $$f(x+2) = frac{x+2} {x+3}$$
            $$f(x) = frac{x} {x+1}$$



            $$ x = x.f(x) + f(x)$$
            $$implies x(1-f(x)) = f(x)$$



            $$implies x = frac {f(x)}{1-f(x)}$$



            Now substitute this $x$ in $f(x+2)$ and get the desired answer.






            share|cite|improve this answer











            $endgroup$



            If I get your question correctly, then you can try this:



            $$f(x+2) = frac{x+2} {x+3}$$
            $$f(x) = frac{x} {x+1}$$



            $$ x = x.f(x) + f(x)$$
            $$implies x(1-f(x)) = f(x)$$



            $$implies x = frac {f(x)}{1-f(x)}$$



            Now substitute this $x$ in $f(x+2)$ and get the desired answer.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 10 '18 at 5:57

























            answered Dec 10 '18 at 5:53









            PradyumanDixitPradyumanDixit

            847214




            847214








            • 2




              $begingroup$
              Your sign of $x$ is wrong. That should be $1-f(x)$ instead of $f(x)-1$.
              $endgroup$
              – tonychow0929
              Dec 10 '18 at 5:56






            • 1




              $begingroup$
              @tonychow0929 I am sorry, thank you for that :)
              $endgroup$
              – PradyumanDixit
              Dec 10 '18 at 5:56
















            • 2




              $begingroup$
              Your sign of $x$ is wrong. That should be $1-f(x)$ instead of $f(x)-1$.
              $endgroup$
              – tonychow0929
              Dec 10 '18 at 5:56






            • 1




              $begingroup$
              @tonychow0929 I am sorry, thank you for that :)
              $endgroup$
              – PradyumanDixit
              Dec 10 '18 at 5:56










            2




            2




            $begingroup$
            Your sign of $x$ is wrong. That should be $1-f(x)$ instead of $f(x)-1$.
            $endgroup$
            – tonychow0929
            Dec 10 '18 at 5:56




            $begingroup$
            Your sign of $x$ is wrong. That should be $1-f(x)$ instead of $f(x)-1$.
            $endgroup$
            – tonychow0929
            Dec 10 '18 at 5:56




            1




            1




            $begingroup$
            @tonychow0929 I am sorry, thank you for that :)
            $endgroup$
            – PradyumanDixit
            Dec 10 '18 at 5:56






            $begingroup$
            @tonychow0929 I am sorry, thank you for that :)
            $endgroup$
            – PradyumanDixit
            Dec 10 '18 at 5:56




















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