2 variable Functional Equations

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Suppose a function $f : R->R$ satisfies the following conditions



  1. $f(4xy) = 2y[f(x+y)+f(x-y)]$


  2. $f(5)=3$


What is the value of $f(2015)$?



I am currently stuck after $x=y$ which gives out:



$f(4y^2) = 2y(f(2y))$



After getting this, I notice that



$4y^2 = (2y)^2$. So maybe the function also has this property



$f(xy) = xf(y)$. So maybe $403 cdot 3 = 1209$ is the answer? It is not proven though. I need help with what to substitute in to the functional equation. Thanks







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  • "Functional Analysis" is something else.
    – uniquesolution
    Aug 15 at 13:12










  • My mistake, it was called functional equations, sorry
    – SuperMage1
    Aug 15 at 13:19














up vote
1
down vote

favorite












Suppose a function $f : R->R$ satisfies the following conditions



  1. $f(4xy) = 2y[f(x+y)+f(x-y)]$


  2. $f(5)=3$


What is the value of $f(2015)$?



I am currently stuck after $x=y$ which gives out:



$f(4y^2) = 2y(f(2y))$



After getting this, I notice that



$4y^2 = (2y)^2$. So maybe the function also has this property



$f(xy) = xf(y)$. So maybe $403 cdot 3 = 1209$ is the answer? It is not proven though. I need help with what to substitute in to the functional equation. Thanks







share|cite|improve this question






















  • "Functional Analysis" is something else.
    – uniquesolution
    Aug 15 at 13:12










  • My mistake, it was called functional equations, sorry
    – SuperMage1
    Aug 15 at 13:19












up vote
1
down vote

favorite









up vote
1
down vote

favorite











Suppose a function $f : R->R$ satisfies the following conditions



  1. $f(4xy) = 2y[f(x+y)+f(x-y)]$


  2. $f(5)=3$


What is the value of $f(2015)$?



I am currently stuck after $x=y$ which gives out:



$f(4y^2) = 2y(f(2y))$



After getting this, I notice that



$4y^2 = (2y)^2$. So maybe the function also has this property



$f(xy) = xf(y)$. So maybe $403 cdot 3 = 1209$ is the answer? It is not proven though. I need help with what to substitute in to the functional equation. Thanks







share|cite|improve this question














Suppose a function $f : R->R$ satisfies the following conditions



  1. $f(4xy) = 2y[f(x+y)+f(x-y)]$


  2. $f(5)=3$


What is the value of $f(2015)$?



I am currently stuck after $x=y$ which gives out:



$f(4y^2) = 2y(f(2y))$



After getting this, I notice that



$4y^2 = (2y)^2$. So maybe the function also has this property



$f(xy) = xf(y)$. So maybe $403 cdot 3 = 1209$ is the answer? It is not proven though. I need help with what to substitute in to the functional equation. Thanks









share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Aug 15 at 13:18

























asked Aug 15 at 12:43









SuperMage1

708210




708210











  • "Functional Analysis" is something else.
    – uniquesolution
    Aug 15 at 13:12










  • My mistake, it was called functional equations, sorry
    – SuperMage1
    Aug 15 at 13:19
















  • "Functional Analysis" is something else.
    – uniquesolution
    Aug 15 at 13:12










  • My mistake, it was called functional equations, sorry
    – SuperMage1
    Aug 15 at 13:19















"Functional Analysis" is something else.
– uniquesolution
Aug 15 at 13:12




"Functional Analysis" is something else.
– uniquesolution
Aug 15 at 13:12












My mistake, it was called functional equations, sorry
– SuperMage1
Aug 15 at 13:19




My mistake, it was called functional equations, sorry
– SuperMage1
Aug 15 at 13:19










1 Answer
1






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If you look at the function $f(x) = frac35x$, you can see that



  1. $2y(f(x+y) + f(x-y)) = 2y(frac35(x+y) + frac35(x-y)) = frac35(2y((x + y) + (x - y))) = frac35(2y cdot 2x) = f(4xy)$

  2. $f(5) = frac35 cdot 5 = 3$

For this function, $f(2015) = 1209$ indeed.



It remains to show that this is the only function that satisfies the above conditions.
It is easy to show that $f(0) = 0$ and $f(-x) = -f(x)$. For the latter, just set $x = 0$ in the first condition. So maybe that's a help.






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    1 Answer
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    1 Answer
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    up vote
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    down vote













    If you look at the function $f(x) = frac35x$, you can see that



    1. $2y(f(x+y) + f(x-y)) = 2y(frac35(x+y) + frac35(x-y)) = frac35(2y((x + y) + (x - y))) = frac35(2y cdot 2x) = f(4xy)$

    2. $f(5) = frac35 cdot 5 = 3$

    For this function, $f(2015) = 1209$ indeed.



    It remains to show that this is the only function that satisfies the above conditions.
    It is easy to show that $f(0) = 0$ and $f(-x) = -f(x)$. For the latter, just set $x = 0$ in the first condition. So maybe that's a help.






    share|cite|improve this answer
























      up vote
      1
      down vote













      If you look at the function $f(x) = frac35x$, you can see that



      1. $2y(f(x+y) + f(x-y)) = 2y(frac35(x+y) + frac35(x-y)) = frac35(2y((x + y) + (x - y))) = frac35(2y cdot 2x) = f(4xy)$

      2. $f(5) = frac35 cdot 5 = 3$

      For this function, $f(2015) = 1209$ indeed.



      It remains to show that this is the only function that satisfies the above conditions.
      It is easy to show that $f(0) = 0$ and $f(-x) = -f(x)$. For the latter, just set $x = 0$ in the first condition. So maybe that's a help.






      share|cite|improve this answer






















        up vote
        1
        down vote










        up vote
        1
        down vote









        If you look at the function $f(x) = frac35x$, you can see that



        1. $2y(f(x+y) + f(x-y)) = 2y(frac35(x+y) + frac35(x-y)) = frac35(2y((x + y) + (x - y))) = frac35(2y cdot 2x) = f(4xy)$

        2. $f(5) = frac35 cdot 5 = 3$

        For this function, $f(2015) = 1209$ indeed.



        It remains to show that this is the only function that satisfies the above conditions.
        It is easy to show that $f(0) = 0$ and $f(-x) = -f(x)$. For the latter, just set $x = 0$ in the first condition. So maybe that's a help.






        share|cite|improve this answer












        If you look at the function $f(x) = frac35x$, you can see that



        1. $2y(f(x+y) + f(x-y)) = 2y(frac35(x+y) + frac35(x-y)) = frac35(2y((x + y) + (x - y))) = frac35(2y cdot 2x) = f(4xy)$

        2. $f(5) = frac35 cdot 5 = 3$

        For this function, $f(2015) = 1209$ indeed.



        It remains to show that this is the only function that satisfies the above conditions.
        It is easy to show that $f(0) = 0$ and $f(-x) = -f(x)$. For the latter, just set $x = 0$ in the first condition. So maybe that's a help.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 15 at 15:05









        Ronald

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