Functional Equation Solved Using Differentiation












0














Let $f$ be a function with domain $R$ that satisfies the conditions: $$f(x+y)=f(x)f(y), forall x,y $$ and $$f(0) neq 0$$



(a) Show that $f(0)=1$



(b) Prove that $f(x) neq 0$, for all $xin R$



(c) Assuming that $f'(x)$ exists for all $x in R$, use the definition of the derivative to show that $f(x)$ satisfies the equation $f'(x)=kf(x)$, where $k=f'(0)$



I've tried solving part (a) and (b) by substituting $x=y=0$ and $y=-x$ respectively, but I can't seem to solve part (c) as I can't avoid dividing by 0 when dealing with the limit. Does anyone know how to work around this?






calculus differential-equations derivatives functional-equations







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  • You could also simply use $f'(x+y)=frac{d}{dy}f(x+y)=f(x)f'(y)$ and then set $y=0$.
    – LutzL
    Jan 5 at 23:53






  • 1




    You don't need to assume anything more that the continuity of $f$ at any single point. The functional equation combined with continuity at a single point leads us to the exponential function. For details see this answer : math.stackexchange.com/a/1885860/72031
    – Paramanand Singh
    Jan 6 at 17:00
















0














Let $f$ be a function with domain $R$ that satisfies the conditions: $$f(x+y)=f(x)f(y), forall x,y $$ and $$f(0) neq 0$$



(a) Show that $f(0)=1$



(b) Prove that $f(x) neq 0$, for all $xin R$



(c) Assuming that $f'(x)$ exists for all $x in R$, use the definition of the derivative to show that $f(x)$ satisfies the equation $f'(x)=kf(x)$, where $k=f'(0)$



I've tried solving part (a) and (b) by substituting $x=y=0$ and $y=-x$ respectively, but I can't seem to solve part (c) as I can't avoid dividing by 0 when dealing with the limit. Does anyone know how to work around this?






calculus differential-equations derivatives functional-equations







share|cite|improve this question
























  • You could also simply use $f'(x+y)=frac{d}{dy}f(x+y)=f(x)f'(y)$ and then set $y=0$.
    – LutzL
    Jan 5 at 23:53






  • 1




    You don't need to assume anything more that the continuity of $f$ at any single point. The functional equation combined with continuity at a single point leads us to the exponential function. For details see this answer : math.stackexchange.com/a/1885860/72031
    – Paramanand Singh
    Jan 6 at 17:00














0












0








0







Let $f$ be a function with domain $R$ that satisfies the conditions: $$f(x+y)=f(x)f(y), forall x,y $$ and $$f(0) neq 0$$



(a) Show that $f(0)=1$



(b) Prove that $f(x) neq 0$, for all $xin R$



(c) Assuming that $f'(x)$ exists for all $x in R$, use the definition of the derivative to show that $f(x)$ satisfies the equation $f'(x)=kf(x)$, where $k=f'(0)$



I've tried solving part (a) and (b) by substituting $x=y=0$ and $y=-x$ respectively, but I can't seem to solve part (c) as I can't avoid dividing by 0 when dealing with the limit. Does anyone know how to work around this?






calculus differential-equations derivatives functional-equations







share|cite|improve this question















Let $f$ be a function with domain $R$ that satisfies the conditions: $$f(x+y)=f(x)f(y), forall x,y $$ and $$f(0) neq 0$$



(a) Show that $f(0)=1$



(b) Prove that $f(x) neq 0$, for all $xin R$



(c) Assuming that $f'(x)$ exists for all $x in R$, use the definition of the derivative to show that $f(x)$ satisfies the equation $f'(x)=kf(x)$, where $k=f'(0)$



I've tried solving part (a) and (b) by substituting $x=y=0$ and $y=-x$ respectively, but I can't seem to solve part (c) as I can't avoid dividing by 0 when dealing with the limit. Does anyone know how to work around this?






calculus differential-equations derivatives functional-equations




calculus differential-equations derivatives functional-equations






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share|cite|improve this question













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edited Jan 5 at 23:00









Yiorgos S. Smyrlis

62.8k1383163




62.8k1383163










asked Jan 4 at 18:08









Matthew TanMatthew Tan

555




555












  • You could also simply use $f'(x+y)=frac{d}{dy}f(x+y)=f(x)f'(y)$ and then set $y=0$.
    – LutzL
    Jan 5 at 23:53






  • 1




    You don't need to assume anything more that the continuity of $f$ at any single point. The functional equation combined with continuity at a single point leads us to the exponential function. For details see this answer : math.stackexchange.com/a/1885860/72031
    – Paramanand Singh
    Jan 6 at 17:00


















  • You could also simply use $f'(x+y)=frac{d}{dy}f(x+y)=f(x)f'(y)$ and then set $y=0$.
    – LutzL
    Jan 5 at 23:53






  • 1




    You don't need to assume anything more that the continuity of $f$ at any single point. The functional equation combined with continuity at a single point leads us to the exponential function. For details see this answer : math.stackexchange.com/a/1885860/72031
    – Paramanand Singh
    Jan 6 at 17:00
















You could also simply use $f'(x+y)=frac{d}{dy}f(x+y)=f(x)f'(y)$ and then set $y=0$.
– LutzL
Jan 5 at 23:53




You could also simply use $f'(x+y)=frac{d}{dy}f(x+y)=f(x)f'(y)$ and then set $y=0$.
– LutzL
Jan 5 at 23:53




1




1




You don't need to assume anything more that the continuity of $f$ at any single point. The functional equation combined with continuity at a single point leads us to the exponential function. For details see this answer : math.stackexchange.com/a/1885860/72031
– Paramanand Singh
Jan 6 at 17:00




You don't need to assume anything more that the continuity of $f$ at any single point. The functional equation combined with continuity at a single point leads us to the exponential function. For details see this answer : math.stackexchange.com/a/1885860/72031
– Paramanand Singh
Jan 6 at 17:00










3 Answers
3






active

oldest

votes


















4














Note that, as $hto 0$, then
$$
f'(x)leftarrowfrac{f(x+h)-f(x)}{h}=frac{f(x)f(h)-f(x)f(0)}{h}=f(x)frac{f(h)-f(0)}{h}to f(x),f'(0)
$$
and hence
$$
f'(x)=f'(0),f(x),
$$
which means that $f'(x)=k,f(x)$, where $k=f'(0)$.



Note. Clearly, as $f$ satisfies the ODE, $y'=ky$, then it is of the form $f(x)=ce^{kx}$, and since $f(0)=1$, then
$$
f(x)=e^{kx}=e^{f'(0)x}.
$$






share|cite|improve this answer































    0














    You can guess the function to be:



    $$f(x) = a^x$$



    since



    $a^{p+q} =a^pa^q $






    share|cite|improve this answer





















    • But is there any way to solve the questions using differentiation?
      – Matthew Tan
      Jan 4 at 18:13



















    0














    $$f(0)=1$$
    is apparent because when $x=y=0$, we have $f(0)=f^2(0)$ which can only happen at $0$ or $1$, and $0$ is ruled out by the question.



    For $b)$, notice that $f(x)=0implies f(x+y)=0f(y)=0$ for any $y$. So in effect, if one value of $x$ yields $0$, all of them will. This is contradicted by part $a)$.






    share|cite|improve this answer























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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      4














      Note that, as $hto 0$, then
      $$
      f'(x)leftarrowfrac{f(x+h)-f(x)}{h}=frac{f(x)f(h)-f(x)f(0)}{h}=f(x)frac{f(h)-f(0)}{h}to f(x),f'(0)
      $$
      and hence
      $$
      f'(x)=f'(0),f(x),
      $$
      which means that $f'(x)=k,f(x)$, where $k=f'(0)$.



      Note. Clearly, as $f$ satisfies the ODE, $y'=ky$, then it is of the form $f(x)=ce^{kx}$, and since $f(0)=1$, then
      $$
      f(x)=e^{kx}=e^{f'(0)x}.
      $$






      share|cite|improve this answer




























        4














        Note that, as $hto 0$, then
        $$
        f'(x)leftarrowfrac{f(x+h)-f(x)}{h}=frac{f(x)f(h)-f(x)f(0)}{h}=f(x)frac{f(h)-f(0)}{h}to f(x),f'(0)
        $$
        and hence
        $$
        f'(x)=f'(0),f(x),
        $$
        which means that $f'(x)=k,f(x)$, where $k=f'(0)$.



        Note. Clearly, as $f$ satisfies the ODE, $y'=ky$, then it is of the form $f(x)=ce^{kx}$, and since $f(0)=1$, then
        $$
        f(x)=e^{kx}=e^{f'(0)x}.
        $$






        share|cite|improve this answer


























          4












          4








          4






          Note that, as $hto 0$, then
          $$
          f'(x)leftarrowfrac{f(x+h)-f(x)}{h}=frac{f(x)f(h)-f(x)f(0)}{h}=f(x)frac{f(h)-f(0)}{h}to f(x),f'(0)
          $$
          and hence
          $$
          f'(x)=f'(0),f(x),
          $$
          which means that $f'(x)=k,f(x)$, where $k=f'(0)$.



          Note. Clearly, as $f$ satisfies the ODE, $y'=ky$, then it is of the form $f(x)=ce^{kx}$, and since $f(0)=1$, then
          $$
          f(x)=e^{kx}=e^{f'(0)x}.
          $$






          share|cite|improve this answer














          Note that, as $hto 0$, then
          $$
          f'(x)leftarrowfrac{f(x+h)-f(x)}{h}=frac{f(x)f(h)-f(x)f(0)}{h}=f(x)frac{f(h)-f(0)}{h}to f(x),f'(0)
          $$
          and hence
          $$
          f'(x)=f'(0),f(x),
          $$
          which means that $f'(x)=k,f(x)$, where $k=f'(0)$.



          Note. Clearly, as $f$ satisfies the ODE, $y'=ky$, then it is of the form $f(x)=ce^{kx}$, and since $f(0)=1$, then
          $$
          f(x)=e^{kx}=e^{f'(0)x}.
          $$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited 2 days ago

























          answered Jan 4 at 18:13









          Yiorgos S. SmyrlisYiorgos S. Smyrlis

          62.8k1383163




          62.8k1383163























              0














              You can guess the function to be:



              $$f(x) = a^x$$



              since



              $a^{p+q} =a^pa^q $






              share|cite|improve this answer





















              • But is there any way to solve the questions using differentiation?
                – Matthew Tan
                Jan 4 at 18:13
















              0














              You can guess the function to be:



              $$f(x) = a^x$$



              since



              $a^{p+q} =a^pa^q $






              share|cite|improve this answer





















              • But is there any way to solve the questions using differentiation?
                – Matthew Tan
                Jan 4 at 18:13














              0












              0








              0






              You can guess the function to be:



              $$f(x) = a^x$$



              since



              $a^{p+q} =a^pa^q $






              share|cite|improve this answer












              You can guess the function to be:



              $$f(x) = a^x$$



              since



              $a^{p+q} =a^pa^q $







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Jan 4 at 18:10









              AbcdAbcd

              3,02321135




              3,02321135












              • But is there any way to solve the questions using differentiation?
                – Matthew Tan
                Jan 4 at 18:13


















              • But is there any way to solve the questions using differentiation?
                – Matthew Tan
                Jan 4 at 18:13
















              But is there any way to solve the questions using differentiation?
              – Matthew Tan
              Jan 4 at 18:13




              But is there any way to solve the questions using differentiation?
              – Matthew Tan
              Jan 4 at 18:13











              0














              $$f(0)=1$$
              is apparent because when $x=y=0$, we have $f(0)=f^2(0)$ which can only happen at $0$ or $1$, and $0$ is ruled out by the question.



              For $b)$, notice that $f(x)=0implies f(x+y)=0f(y)=0$ for any $y$. So in effect, if one value of $x$ yields $0$, all of them will. This is contradicted by part $a)$.






              share|cite|improve this answer




























                0














                $$f(0)=1$$
                is apparent because when $x=y=0$, we have $f(0)=f^2(0)$ which can only happen at $0$ or $1$, and $0$ is ruled out by the question.



                For $b)$, notice that $f(x)=0implies f(x+y)=0f(y)=0$ for any $y$. So in effect, if one value of $x$ yields $0$, all of them will. This is contradicted by part $a)$.






                share|cite|improve this answer


























                  0












                  0








                  0






                  $$f(0)=1$$
                  is apparent because when $x=y=0$, we have $f(0)=f^2(0)$ which can only happen at $0$ or $1$, and $0$ is ruled out by the question.



                  For $b)$, notice that $f(x)=0implies f(x+y)=0f(y)=0$ for any $y$. So in effect, if one value of $x$ yields $0$, all of them will. This is contradicted by part $a)$.






                  share|cite|improve this answer














                  $$f(0)=1$$
                  is apparent because when $x=y=0$, we have $f(0)=f^2(0)$ which can only happen at $0$ or $1$, and $0$ is ruled out by the question.



                  For $b)$, notice that $f(x)=0implies f(x+y)=0f(y)=0$ for any $y$. So in effect, if one value of $x$ yields $0$, all of them will. This is contradicted by part $a)$.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 4 at 18:23

























                  answered Jan 4 at 18:17









                  Rhys HughesRhys Hughes

                  5,0991427




                  5,0991427






























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