Help in solving a simple functional equation












6














I need to find all continuous functions satisfying:



$$3f(2x+1)=f(x) + 5x$$



The functional equation looks simple but I am unable to solve it. I tried to convert it into a Cauchy type equation but I wasn't able to do so.






functional-equations







share|cite|improve this question


















  • 1




    Is there any restriction given for $x$? It is rather simple to determine the value of $f(-1)$ but I am not sure whether this is helpful or not.
    – mrtaurho
    Dec 15 at 10:53










  • If the domain of $f$ is all real numbers then Song's answer is correct. If the domain is allowed to be $x>-1$ then Cesareo's answer is correct and includes Song's answer (where $C_0=-1/2$). So the domain of your function $f$ is relevant. What is the domain?
    – Rory Daulton
    Dec 15 at 12:52


















6














I need to find all continuous functions satisfying:



$$3f(2x+1)=f(x) + 5x$$



The functional equation looks simple but I am unable to solve it. I tried to convert it into a Cauchy type equation but I wasn't able to do so.






functional-equations







share|cite|improve this question


















  • 1




    Is there any restriction given for $x$? It is rather simple to determine the value of $f(-1)$ but I am not sure whether this is helpful or not.
    – mrtaurho
    Dec 15 at 10:53










  • If the domain of $f$ is all real numbers then Song's answer is correct. If the domain is allowed to be $x>-1$ then Cesareo's answer is correct and includes Song's answer (where $C_0=-1/2$). So the domain of your function $f$ is relevant. What is the domain?
    – Rory Daulton
    Dec 15 at 12:52
















6












6








6


2





I need to find all continuous functions satisfying:



$$3f(2x+1)=f(x) + 5x$$



The functional equation looks simple but I am unable to solve it. I tried to convert it into a Cauchy type equation but I wasn't able to do so.






functional-equations







share|cite|improve this question













I need to find all continuous functions satisfying:



$$3f(2x+1)=f(x) + 5x$$



The functional equation looks simple but I am unable to solve it. I tried to convert it into a Cauchy type equation but I wasn't able to do so.






functional-equations




functional-equations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 15 at 10:43









saisanjeev

891212




891212








  • 1




    Is there any restriction given for $x$? It is rather simple to determine the value of $f(-1)$ but I am not sure whether this is helpful or not.
    – mrtaurho
    Dec 15 at 10:53










  • If the domain of $f$ is all real numbers then Song's answer is correct. If the domain is allowed to be $x>-1$ then Cesareo's answer is correct and includes Song's answer (where $C_0=-1/2$). So the domain of your function $f$ is relevant. What is the domain?
    – Rory Daulton
    Dec 15 at 12:52
















  • 1




    Is there any restriction given for $x$? It is rather simple to determine the value of $f(-1)$ but I am not sure whether this is helpful or not.
    – mrtaurho
    Dec 15 at 10:53










  • If the domain of $f$ is all real numbers then Song's answer is correct. If the domain is allowed to be $x>-1$ then Cesareo's answer is correct and includes Song's answer (where $C_0=-1/2$). So the domain of your function $f$ is relevant. What is the domain?
    – Rory Daulton
    Dec 15 at 12:52










1




1




Is there any restriction given for $x$? It is rather simple to determine the value of $f(-1)$ but I am not sure whether this is helpful or not.
– mrtaurho
Dec 15 at 10:53




Is there any restriction given for $x$? It is rather simple to determine the value of $f(-1)$ but I am not sure whether this is helpful or not.
– mrtaurho
Dec 15 at 10:53












If the domain of $f$ is all real numbers then Song's answer is correct. If the domain is allowed to be $x>-1$ then Cesareo's answer is correct and includes Song's answer (where $C_0=-1/2$). So the domain of your function $f$ is relevant. What is the domain?
– Rory Daulton
Dec 15 at 12:52






If the domain of $f$ is all real numbers then Song's answer is correct. If the domain is allowed to be $x>-1$ then Cesareo's answer is correct and includes Song's answer (where $C_0=-1/2$). So the domain of your function $f$ is relevant. What is the domain?
– Rory Daulton
Dec 15 at 12:52












2 Answers
2






active

oldest

votes


















11














Let $g(x) = f(x-1)$. Then we have
$$
3g(2x+2) = g(x+1) + 5x,
$$ or equivalently
$$
g(x)-frac{1}{3}g(frac{x}{2}) = frac{5x-10}{6}=: phi(x).$$
Note that $g(0) = -2.5$. Hence we have
$$begin{eqnarray}
g(x) = g(x) -lim_{jtoinfty}3^{-j}g(2^{-j}x) &=& sum_{j=0}^infty left(3^{-j} g(2^{-j}x) -3^{-j-1}g(2^{-j-1}x)right)\ &=& sum_{j=0}^infty 3^{-j}phi(2^{-j}x)\
&=&frac{1}{6}sum_{j=0}^infty 3^{-j}(5cdot2^{-j}x-10)\
&=&frac{6x-15}{6} = x -frac{5}{2}.
end{eqnarray}$$ This establishes $f(x) = x -frac{3}{2}$.



$textbf{EDIT:}$ I implicitly assumed that the domain of definition of $g$ is $mathbb{R}$. If the domain of $g$ contains $0$, then the unique continuous solution is given by $g(x) = x-frac{5}{2} $ as we can see from the above argument. Otherwise, the argument collapses, and one can see that $$g(x) = ( x-frac{5}{2}) + h(x)$$ is a solution of $g(x)-frac{1}{3}g(frac{x}{2}) = phi(x)$ whenever it holds that $$
h(x) = frac{1}{3}h(frac{x}{2})quadcdots(*).
$$ Note that any continuous function $k : [1,2]tomathbb{R}$ with $k(2) = frac{1}{3}k(1)$ can be extended uniquely to continuous $overline{k} :(0,infty)tomathbb{R}$ satisfying $(*)$. This shows that there are as many solutions $$g:xin(0,infty)mapsto ( x-frac{5}{2}) + overline{k}(x)$$ as there are $k:[1,2] tomathbb{R}$ with $k(2) = frac{1}{3}k(1)$. And the same is also true for $g(x)$ on $(-infty,0)$.






share|cite|improve this answer































    4














    This is a linear difference equation that can be easily solved as



    $$
    f(x) = f_h(x)+f_p(x)
    $$



    the homogeneous solution gives



    $$
    f_h(x) = C_0 3^{1-log_2(x+1)}
    $$



    The complete solution is



    $$
    f(x) = frac{1}{2} left(left(2 C_0+1right) 3^{1-log_2 (x+1)}+2 x-3right)
    $$






    share|cite|improve this answer





















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      2 Answers
      2






      active

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      2 Answers
      2






      active

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      active

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      active

      oldest

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      11














      Let $g(x) = f(x-1)$. Then we have
      $$
      3g(2x+2) = g(x+1) + 5x,
      $$       or equivalently
      $$
      g(x)-frac{1}{3}g(frac{x}{2}) = frac{5x-10}{6}=: phi(x).$$
      Note that $g(0) = -2.5$. Hence we have
      $$begin{eqnarray}
      g(x) = g(x) -lim_{jtoinfty}3^{-j}g(2^{-j}x) &=& sum_{j=0}^infty left(3^{-j} g(2^{-j}x) -3^{-j-1}g(2^{-j-1}x)right)\ &=& sum_{j=0}^infty 3^{-j}phi(2^{-j}x)\
      &=&frac{1}{6}sum_{j=0}^infty 3^{-j}(5cdot2^{-j}x-10)\
      &=&frac{6x-15}{6} = x -frac{5}{2}.
      end{eqnarray}$$       This establishes $f(x) = x -frac{3}{2}$.



      $textbf{EDIT:}$ I implicitly assumed that the domain of definition of $g$ is $mathbb{R}$. If the domain of $g$ contains $0$, then the unique continuous solution is given by $g(x) = x-frac{5}{2} $ as we can see from the above argument. Otherwise, the argument collapses, and one can see that $$g(x) = ( x-frac{5}{2}) + h(x)$$ is a solution of $g(x)-frac{1}{3}g(frac{x}{2}) = phi(x)$ whenever it holds that $$
      h(x) = frac{1}{3}h(frac{x}{2})quadcdots(*).
      $$       Note that any continuous function $k : [1,2]tomathbb{R}$ with $k(2) = frac{1}{3}k(1)$ can be extended uniquely to continuous $overline{k} :(0,infty)tomathbb{R}$ satisfying $(*)$. This shows that there are as many solutions $$g:xin(0,infty)mapsto ( x-frac{5}{2}) + overline{k}(x)$$ as there are $k:[1,2] tomathbb{R}$ with $k(2) = frac{1}{3}k(1)$. And the same is also true for $g(x)$ on $(-infty,0)$.






      share|cite|improve this answer




























        11














        Let $g(x) = f(x-1)$. Then we have
        $$
        3g(2x+2) = g(x+1) + 5x,
        $$         or equivalently
        $$
        g(x)-frac{1}{3}g(frac{x}{2}) = frac{5x-10}{6}=: phi(x).$$
        Note that $g(0) = -2.5$. Hence we have
        $$begin{eqnarray}
        g(x) = g(x) -lim_{jtoinfty}3^{-j}g(2^{-j}x) &=& sum_{j=0}^infty left(3^{-j} g(2^{-j}x) -3^{-j-1}g(2^{-j-1}x)right)\ &=& sum_{j=0}^infty 3^{-j}phi(2^{-j}x)\
        &=&frac{1}{6}sum_{j=0}^infty 3^{-j}(5cdot2^{-j}x-10)\
        &=&frac{6x-15}{6} = x -frac{5}{2}.
        end{eqnarray}$$         This establishes $f(x) = x -frac{3}{2}$.



        $textbf{EDIT:}$ I implicitly assumed that the domain of definition of $g$ is $mathbb{R}$. If the domain of $g$ contains $0$, then the unique continuous solution is given by $g(x) = x-frac{5}{2} $ as we can see from the above argument. Otherwise, the argument collapses, and one can see that $$g(x) = ( x-frac{5}{2}) + h(x)$$ is a solution of $g(x)-frac{1}{3}g(frac{x}{2}) = phi(x)$ whenever it holds that $$
        h(x) = frac{1}{3}h(frac{x}{2})quadcdots(*).
        $$         Note that any continuous function $k : [1,2]tomathbb{R}$ with $k(2) = frac{1}{3}k(1)$ can be extended uniquely to continuous $overline{k} :(0,infty)tomathbb{R}$ satisfying $(*)$. This shows that there are as many solutions $$g:xin(0,infty)mapsto ( x-frac{5}{2}) + overline{k}(x)$$ as there are $k:[1,2] tomathbb{R}$ with $k(2) = frac{1}{3}k(1)$. And the same is also true for $g(x)$ on $(-infty,0)$.






        share|cite|improve this answer


























          11












          11








          11






          Let $g(x) = f(x-1)$. Then we have
          $$
          3g(2x+2) = g(x+1) + 5x,
          $$           or equivalently
          $$
          g(x)-frac{1}{3}g(frac{x}{2}) = frac{5x-10}{6}=: phi(x).$$
          Note that $g(0) = -2.5$. Hence we have
          $$begin{eqnarray}
          g(x) = g(x) -lim_{jtoinfty}3^{-j}g(2^{-j}x) &=& sum_{j=0}^infty left(3^{-j} g(2^{-j}x) -3^{-j-1}g(2^{-j-1}x)right)\ &=& sum_{j=0}^infty 3^{-j}phi(2^{-j}x)\
          &=&frac{1}{6}sum_{j=0}^infty 3^{-j}(5cdot2^{-j}x-10)\
          &=&frac{6x-15}{6} = x -frac{5}{2}.
          end{eqnarray}$$           This establishes $f(x) = x -frac{3}{2}$.



          $textbf{EDIT:}$ I implicitly assumed that the domain of definition of $g$ is $mathbb{R}$. If the domain of $g$ contains $0$, then the unique continuous solution is given by $g(x) = x-frac{5}{2} $ as we can see from the above argument. Otherwise, the argument collapses, and one can see that $$g(x) = ( x-frac{5}{2}) + h(x)$$ is a solution of $g(x)-frac{1}{3}g(frac{x}{2}) = phi(x)$ whenever it holds that $$
          h(x) = frac{1}{3}h(frac{x}{2})quadcdots(*).
          $$           Note that any continuous function $k : [1,2]tomathbb{R}$ with $k(2) = frac{1}{3}k(1)$ can be extended uniquely to continuous $overline{k} :(0,infty)tomathbb{R}$ satisfying $(*)$. This shows that there are as many solutions $$g:xin(0,infty)mapsto ( x-frac{5}{2}) + overline{k}(x)$$ as there are $k:[1,2] tomathbb{R}$ with $k(2) = frac{1}{3}k(1)$. And the same is also true for $g(x)$ on $(-infty,0)$.






          share|cite|improve this answer














          Let $g(x) = f(x-1)$. Then we have
          $$
          3g(2x+2) = g(x+1) + 5x,
          $$           or equivalently
          $$
          g(x)-frac{1}{3}g(frac{x}{2}) = frac{5x-10}{6}=: phi(x).$$
          Note that $g(0) = -2.5$. Hence we have
          $$begin{eqnarray}
          g(x) = g(x) -lim_{jtoinfty}3^{-j}g(2^{-j}x) &=& sum_{j=0}^infty left(3^{-j} g(2^{-j}x) -3^{-j-1}g(2^{-j-1}x)right)\ &=& sum_{j=0}^infty 3^{-j}phi(2^{-j}x)\
          &=&frac{1}{6}sum_{j=0}^infty 3^{-j}(5cdot2^{-j}x-10)\
          &=&frac{6x-15}{6} = x -frac{5}{2}.
          end{eqnarray}$$           This establishes $f(x) = x -frac{3}{2}$.



          $textbf{EDIT:}$ I implicitly assumed that the domain of definition of $g$ is $mathbb{R}$. If the domain of $g$ contains $0$, then the unique continuous solution is given by $g(x) = x-frac{5}{2} $ as we can see from the above argument. Otherwise, the argument collapses, and one can see that $$g(x) = ( x-frac{5}{2}) + h(x)$$ is a solution of $g(x)-frac{1}{3}g(frac{x}{2}) = phi(x)$ whenever it holds that $$
          h(x) = frac{1}{3}h(frac{x}{2})quadcdots(*).
          $$           Note that any continuous function $k : [1,2]tomathbb{R}$ with $k(2) = frac{1}{3}k(1)$ can be extended uniquely to continuous $overline{k} :(0,infty)tomathbb{R}$ satisfying $(*)$. This shows that there are as many solutions $$g:xin(0,infty)mapsto ( x-frac{5}{2}) + overline{k}(x)$$ as there are $k:[1,2] tomathbb{R}$ with $k(2) = frac{1}{3}k(1)$. And the same is also true for $g(x)$ on $(-infty,0)$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 15 at 15:05

























          answered Dec 15 at 11:06









          Song

          4,240316




          4,240316























              4














              This is a linear difference equation that can be easily solved as



              $$
              f(x) = f_h(x)+f_p(x)
              $$



              the homogeneous solution gives



              $$
              f_h(x) = C_0 3^{1-log_2(x+1)}
              $$



              The complete solution is



              $$
              f(x) = frac{1}{2} left(left(2 C_0+1right) 3^{1-log_2 (x+1)}+2 x-3right)
              $$






              share|cite|improve this answer


























                4














                This is a linear difference equation that can be easily solved as



                $$
                f(x) = f_h(x)+f_p(x)
                $$



                the homogeneous solution gives



                $$
                f_h(x) = C_0 3^{1-log_2(x+1)}
                $$



                The complete solution is



                $$
                f(x) = frac{1}{2} left(left(2 C_0+1right) 3^{1-log_2 (x+1)}+2 x-3right)
                $$






                share|cite|improve this answer
























                  4












                  4








                  4






                  This is a linear difference equation that can be easily solved as



                  $$
                  f(x) = f_h(x)+f_p(x)
                  $$



                  the homogeneous solution gives



                  $$
                  f_h(x) = C_0 3^{1-log_2(x+1)}
                  $$



                  The complete solution is



                  $$
                  f(x) = frac{1}{2} left(left(2 C_0+1right) 3^{1-log_2 (x+1)}+2 x-3right)
                  $$






                  share|cite|improve this answer












                  This is a linear difference equation that can be easily solved as



                  $$
                  f(x) = f_h(x)+f_p(x)
                  $$



                  the homogeneous solution gives



                  $$
                  f_h(x) = C_0 3^{1-log_2(x+1)}
                  $$



                  The complete solution is



                  $$
                  f(x) = frac{1}{2} left(left(2 C_0+1right) 3^{1-log_2 (x+1)}+2 x-3right)
                  $$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 15 at 11:52









                  Cesareo

                  8,1333516




                  8,1333516






























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