Which continuous function $f$ satisfies $f(a+b)={f(a)+f(b)over 1−f(a)f(b)}$? [closed]












0














If $f:(-pi/2,pi/2) tomathbb{R} $ and $$f(a+b)=frac{f(a)+f(b)}{1-f(a)f(b)}$$ and $f$ is continuous at $x=0$, then show that $f$ is continuous on its domain.



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closed as off-topic by Martin R, Travis, Nosrati, Saad, Paul Frost Nov 25 '18 at 15:51


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Travis, Nosrati, Saad, Paul Frost

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  • 5




    A good question without questioner attempts!
    – Nosrati
    Nov 25 '18 at 9:33










  • Check whether $tan$ is continous on $(-pi/2,pi/2)$
    – Yadati Kiran
    Nov 25 '18 at 9:49












  • Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    – Martin R
    Nov 25 '18 at 13:54
















0














If $f:(-pi/2,pi/2) tomathbb{R} $ and $$f(a+b)=frac{f(a)+f(b)}{1-f(a)f(b)}$$ and $f$ is continuous at $x=0$, then show that $f$ is continuous on its domain.



Where to start?










share|cite|improve this question















closed as off-topic by Martin R, Travis, Nosrati, Saad, Paul Frost Nov 25 '18 at 15:51


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Travis, Nosrati, Saad, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.









  • 5




    A good question without questioner attempts!
    – Nosrati
    Nov 25 '18 at 9:33










  • Check whether $tan$ is continous on $(-pi/2,pi/2)$
    – Yadati Kiran
    Nov 25 '18 at 9:49












  • Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    – Martin R
    Nov 25 '18 at 13:54














0












0








0







If $f:(-pi/2,pi/2) tomathbb{R} $ and $$f(a+b)=frac{f(a)+f(b)}{1-f(a)f(b)}$$ and $f$ is continuous at $x=0$, then show that $f$ is continuous on its domain.



Where to start?










share|cite|improve this question















If $f:(-pi/2,pi/2) tomathbb{R} $ and $$f(a+b)=frac{f(a)+f(b)}{1-f(a)f(b)}$$ and $f$ is continuous at $x=0$, then show that $f$ is continuous on its domain.



Where to start?







calculus analysis functions continuity functional-equations






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edited Nov 25 '18 at 13:32









Yiorgos S. Smyrlis

62.5k1383162




62.5k1383162










asked Nov 25 '18 at 9:19









Misaqe Nedi

183




183




closed as off-topic by Martin R, Travis, Nosrati, Saad, Paul Frost Nov 25 '18 at 15:51


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Travis, Nosrati, Saad, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Martin R, Travis, Nosrati, Saad, Paul Frost Nov 25 '18 at 15:51


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Martin R, Travis, Nosrati, Saad, Paul Frost

If this question can be reworded to fit the rules in the help center, please edit the question.








  • 5




    A good question without questioner attempts!
    – Nosrati
    Nov 25 '18 at 9:33










  • Check whether $tan$ is continous on $(-pi/2,pi/2)$
    – Yadati Kiran
    Nov 25 '18 at 9:49












  • Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    – Martin R
    Nov 25 '18 at 13:54














  • 5




    A good question without questioner attempts!
    – Nosrati
    Nov 25 '18 at 9:33










  • Check whether $tan$ is continous on $(-pi/2,pi/2)$
    – Yadati Kiran
    Nov 25 '18 at 9:49












  • Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
    – Martin R
    Nov 25 '18 at 13:54








5




5




A good question without questioner attempts!
– Nosrati
Nov 25 '18 at 9:33




A good question without questioner attempts!
– Nosrati
Nov 25 '18 at 9:33












Check whether $tan$ is continous on $(-pi/2,pi/2)$
– Yadati Kiran
Nov 25 '18 at 9:49






Check whether $tan$ is continous on $(-pi/2,pi/2)$
– Yadati Kiran
Nov 25 '18 at 9:49














Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– Martin R
Nov 25 '18 at 13:54




Your question is phrased as an isolated problem, without any further information or context. This does not match many users' quality standards, so it may attract downvotes, or be put on hold. To prevent that, please edit the question. This will help you recognise and resolve the issues. Concretely: please provide context, and include your work and thoughts on the problem. These changes can help in formulating more appropriate answers.
– Martin R
Nov 25 '18 at 13:54










2 Answers
2






active

oldest

votes


















3














Hints. Write $I:=left(-dfrac{pi}{2},+dfrac{pi}{2}right)$. Presumably, the functional equation is this. Let $f:Itomathbb{R}$ be such that $f(a),f(b)neq 1$ and
$$f(a+b)=frac{f(a)+f(b)}{1-f(a),f(b)},,$$
for all $a,bin I$ such that $a+bin I$.



Let $g:Ito I$ be defined by $$g(x):=arctanbig(f(x)big)$$ for all $xin I$. Then, show that $$tanbig(g(x+y)big)=tanbig(g(x)+g(y)big)$$ for all $x,yin I$ such that $x+yin I$. If $f$ is continuous at $0$, then $g$ is continuous at $0$, which then shows that $g$ is continuous on the whole $I$. It follows also that there exists $kinmathbb{R}$ with $|k| leq 1$ such that $f(x)=tan(kx)$ for all $xin I$. (Note that $g$ does not exactly satisfy Cauchy's functional equation. So, the proof must take this into consideration.)






share|cite|improve this answer























  • How to prove the uniqueness of the solution?
    – Tianlalu
    Nov 25 '18 at 9:50










  • The uniqueness follows from continuity. You must first show that $g(x)=kx$ for $x$ in a small open neighborhood of $0$.
    – Batominovski
    Nov 25 '18 at 9:53



















2














Hint. The functional relation is written also as
$$
f(a+b)big(1-f(a)f(b)big)=f(a)+f(b).
$$

So, for $a=b=0$, we have that
$$
f(0)big(1-f^2(0)big)=2f(0)
$$

or
$$
f(0)big(,f^2(0)+1)=0,
$$

and hence $f(0)=0$.



Next, for $b=h$, we have
$$
lim_{hto 0}f(a+h)=lim_{hto 0}frac{f(a)+f(h)}{1-f(a)f(h)}=frac{f(a)+f(0)}{1-f(a)f(0)}=f(a),
$$

hence $f$ continuous everywhere.






share|cite|improve this answer






























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    Hints. Write $I:=left(-dfrac{pi}{2},+dfrac{pi}{2}right)$. Presumably, the functional equation is this. Let $f:Itomathbb{R}$ be such that $f(a),f(b)neq 1$ and
    $$f(a+b)=frac{f(a)+f(b)}{1-f(a),f(b)},,$$
    for all $a,bin I$ such that $a+bin I$.



    Let $g:Ito I$ be defined by $$g(x):=arctanbig(f(x)big)$$ for all $xin I$. Then, show that $$tanbig(g(x+y)big)=tanbig(g(x)+g(y)big)$$ for all $x,yin I$ such that $x+yin I$. If $f$ is continuous at $0$, then $g$ is continuous at $0$, which then shows that $g$ is continuous on the whole $I$. It follows also that there exists $kinmathbb{R}$ with $|k| leq 1$ such that $f(x)=tan(kx)$ for all $xin I$. (Note that $g$ does not exactly satisfy Cauchy's functional equation. So, the proof must take this into consideration.)






    share|cite|improve this answer























    • How to prove the uniqueness of the solution?
      – Tianlalu
      Nov 25 '18 at 9:50










    • The uniqueness follows from continuity. You must first show that $g(x)=kx$ for $x$ in a small open neighborhood of $0$.
      – Batominovski
      Nov 25 '18 at 9:53
















    3














    Hints. Write $I:=left(-dfrac{pi}{2},+dfrac{pi}{2}right)$. Presumably, the functional equation is this. Let $f:Itomathbb{R}$ be such that $f(a),f(b)neq 1$ and
    $$f(a+b)=frac{f(a)+f(b)}{1-f(a),f(b)},,$$
    for all $a,bin I$ such that $a+bin I$.



    Let $g:Ito I$ be defined by $$g(x):=arctanbig(f(x)big)$$ for all $xin I$. Then, show that $$tanbig(g(x+y)big)=tanbig(g(x)+g(y)big)$$ for all $x,yin I$ such that $x+yin I$. If $f$ is continuous at $0$, then $g$ is continuous at $0$, which then shows that $g$ is continuous on the whole $I$. It follows also that there exists $kinmathbb{R}$ with $|k| leq 1$ such that $f(x)=tan(kx)$ for all $xin I$. (Note that $g$ does not exactly satisfy Cauchy's functional equation. So, the proof must take this into consideration.)






    share|cite|improve this answer























    • How to prove the uniqueness of the solution?
      – Tianlalu
      Nov 25 '18 at 9:50










    • The uniqueness follows from continuity. You must first show that $g(x)=kx$ for $x$ in a small open neighborhood of $0$.
      – Batominovski
      Nov 25 '18 at 9:53














    3












    3








    3






    Hints. Write $I:=left(-dfrac{pi}{2},+dfrac{pi}{2}right)$. Presumably, the functional equation is this. Let $f:Itomathbb{R}$ be such that $f(a),f(b)neq 1$ and
    $$f(a+b)=frac{f(a)+f(b)}{1-f(a),f(b)},,$$
    for all $a,bin I$ such that $a+bin I$.



    Let $g:Ito I$ be defined by $$g(x):=arctanbig(f(x)big)$$ for all $xin I$. Then, show that $$tanbig(g(x+y)big)=tanbig(g(x)+g(y)big)$$ for all $x,yin I$ such that $x+yin I$. If $f$ is continuous at $0$, then $g$ is continuous at $0$, which then shows that $g$ is continuous on the whole $I$. It follows also that there exists $kinmathbb{R}$ with $|k| leq 1$ such that $f(x)=tan(kx)$ for all $xin I$. (Note that $g$ does not exactly satisfy Cauchy's functional equation. So, the proof must take this into consideration.)






    share|cite|improve this answer














    Hints. Write $I:=left(-dfrac{pi}{2},+dfrac{pi}{2}right)$. Presumably, the functional equation is this. Let $f:Itomathbb{R}$ be such that $f(a),f(b)neq 1$ and
    $$f(a+b)=frac{f(a)+f(b)}{1-f(a),f(b)},,$$
    for all $a,bin I$ such that $a+bin I$.



    Let $g:Ito I$ be defined by $$g(x):=arctanbig(f(x)big)$$ for all $xin I$. Then, show that $$tanbig(g(x+y)big)=tanbig(g(x)+g(y)big)$$ for all $x,yin I$ such that $x+yin I$. If $f$ is continuous at $0$, then $g$ is continuous at $0$, which then shows that $g$ is continuous on the whole $I$. It follows also that there exists $kinmathbb{R}$ with $|k| leq 1$ such that $f(x)=tan(kx)$ for all $xin I$. (Note that $g$ does not exactly satisfy Cauchy's functional equation. So, the proof must take this into consideration.)







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Nov 25 '18 at 13:27

























    answered Nov 25 '18 at 9:48









    Batominovski

    33.7k33292




    33.7k33292












    • How to prove the uniqueness of the solution?
      – Tianlalu
      Nov 25 '18 at 9:50










    • The uniqueness follows from continuity. You must first show that $g(x)=kx$ for $x$ in a small open neighborhood of $0$.
      – Batominovski
      Nov 25 '18 at 9:53


















    • How to prove the uniqueness of the solution?
      – Tianlalu
      Nov 25 '18 at 9:50










    • The uniqueness follows from continuity. You must first show that $g(x)=kx$ for $x$ in a small open neighborhood of $0$.
      – Batominovski
      Nov 25 '18 at 9:53
















    How to prove the uniqueness of the solution?
    – Tianlalu
    Nov 25 '18 at 9:50




    How to prove the uniqueness of the solution?
    – Tianlalu
    Nov 25 '18 at 9:50












    The uniqueness follows from continuity. You must first show that $g(x)=kx$ for $x$ in a small open neighborhood of $0$.
    – Batominovski
    Nov 25 '18 at 9:53




    The uniqueness follows from continuity. You must first show that $g(x)=kx$ for $x$ in a small open neighborhood of $0$.
    – Batominovski
    Nov 25 '18 at 9:53











    2














    Hint. The functional relation is written also as
    $$
    f(a+b)big(1-f(a)f(b)big)=f(a)+f(b).
    $$

    So, for $a=b=0$, we have that
    $$
    f(0)big(1-f^2(0)big)=2f(0)
    $$

    or
    $$
    f(0)big(,f^2(0)+1)=0,
    $$

    and hence $f(0)=0$.



    Next, for $b=h$, we have
    $$
    lim_{hto 0}f(a+h)=lim_{hto 0}frac{f(a)+f(h)}{1-f(a)f(h)}=frac{f(a)+f(0)}{1-f(a)f(0)}=f(a),
    $$

    hence $f$ continuous everywhere.






    share|cite|improve this answer




























      2














      Hint. The functional relation is written also as
      $$
      f(a+b)big(1-f(a)f(b)big)=f(a)+f(b).
      $$

      So, for $a=b=0$, we have that
      $$
      f(0)big(1-f^2(0)big)=2f(0)
      $$

      or
      $$
      f(0)big(,f^2(0)+1)=0,
      $$

      and hence $f(0)=0$.



      Next, for $b=h$, we have
      $$
      lim_{hto 0}f(a+h)=lim_{hto 0}frac{f(a)+f(h)}{1-f(a)f(h)}=frac{f(a)+f(0)}{1-f(a)f(0)}=f(a),
      $$

      hence $f$ continuous everywhere.






      share|cite|improve this answer


























        2












        2








        2






        Hint. The functional relation is written also as
        $$
        f(a+b)big(1-f(a)f(b)big)=f(a)+f(b).
        $$

        So, for $a=b=0$, we have that
        $$
        f(0)big(1-f^2(0)big)=2f(0)
        $$

        or
        $$
        f(0)big(,f^2(0)+1)=0,
        $$

        and hence $f(0)=0$.



        Next, for $b=h$, we have
        $$
        lim_{hto 0}f(a+h)=lim_{hto 0}frac{f(a)+f(h)}{1-f(a)f(h)}=frac{f(a)+f(0)}{1-f(a)f(0)}=f(a),
        $$

        hence $f$ continuous everywhere.






        share|cite|improve this answer














        Hint. The functional relation is written also as
        $$
        f(a+b)big(1-f(a)f(b)big)=f(a)+f(b).
        $$

        So, for $a=b=0$, we have that
        $$
        f(0)big(1-f^2(0)big)=2f(0)
        $$

        or
        $$
        f(0)big(,f^2(0)+1)=0,
        $$

        and hence $f(0)=0$.



        Next, for $b=h$, we have
        $$
        lim_{hto 0}f(a+h)=lim_{hto 0}frac{f(a)+f(h)}{1-f(a)f(h)}=frac{f(a)+f(0)}{1-f(a)f(0)}=f(a),
        $$

        hence $f$ continuous everywhere.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 25 '18 at 13:38

























        answered Nov 25 '18 at 13:33









        Yiorgos S. Smyrlis

        62.5k1383162




        62.5k1383162















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