Disproving a function exists












1












$begingroup$


Prove/Disprove: There exists a function $f: Bbb R toBbb R$ such that
$$
arctan(f(x)) = 2x/(cos^2(x) + 3)
$$



for every x ∈ $Bbb R$



I know that it's not true because arctan is limited between $frac{-0.5}{pi} leq x leq frac{0.5}{pi}$, but how can you fully prove it without just giving an example? seeing that it is an "exists" proof






functions functional-equations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    While you can't prove something with a million examples, you can disprove something with one.
    $endgroup$
    – Melody
    Dec 12 '18 at 14:19










  • $begingroup$
    MathJax hint: if you put a backslash before common functions you get the right font and spacing, so cos x gives $cos x$ compared to cos x giving $cos x$
    $endgroup$
    – Ross Millikan
    Dec 12 '18 at 14:23
















1












$begingroup$


Prove/Disprove: There exists a function $f: Bbb R toBbb R$ such that
$$
arctan(f(x)) = 2x/(cos^2(x) + 3)
$$



for every x ∈ $Bbb R$



I know that it's not true because arctan is limited between $frac{-0.5}{pi} leq x leq frac{0.5}{pi}$, but how can you fully prove it without just giving an example? seeing that it is an "exists" proof






functions functional-equations







share|cite|improve this question











$endgroup$








  • 2




    $begingroup$
    While you can't prove something with a million examples, you can disprove something with one.
    $endgroup$
    – Melody
    Dec 12 '18 at 14:19










  • $begingroup$
    MathJax hint: if you put a backslash before common functions you get the right font and spacing, so cos x gives $cos x$ compared to cos x giving $cos x$
    $endgroup$
    – Ross Millikan
    Dec 12 '18 at 14:23














1












1








1





$begingroup$


Prove/Disprove: There exists a function $f: Bbb R toBbb R$ such that
$$
arctan(f(x)) = 2x/(cos^2(x) + 3)
$$



for every x ∈ $Bbb R$



I know that it's not true because arctan is limited between $frac{-0.5}{pi} leq x leq frac{0.5}{pi}$, but how can you fully prove it without just giving an example? seeing that it is an "exists" proof






functions functional-equations







share|cite|improve this question











$endgroup$




Prove/Disprove: There exists a function $f: Bbb R toBbb R$ such that
$$
arctan(f(x)) = 2x/(cos^2(x) + 3)
$$



for every x ∈ $Bbb R$



I know that it's not true because arctan is limited between $frac{-0.5}{pi} leq x leq frac{0.5}{pi}$, but how can you fully prove it without just giving an example? seeing that it is an "exists" proof






functions functional-equations




functions functional-equations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 12 '18 at 15:41









daw

24.6k1645




24.6k1645










asked Dec 12 '18 at 14:15









Yuki1112Yuki1112

174




174








  • 2




    $begingroup$
    While you can't prove something with a million examples, you can disprove something with one.
    $endgroup$
    – Melody
    Dec 12 '18 at 14:19










  • $begingroup$
    MathJax hint: if you put a backslash before common functions you get the right font and spacing, so cos x gives $cos x$ compared to cos x giving $cos x$
    $endgroup$
    – Ross Millikan
    Dec 12 '18 at 14:23














  • 2




    $begingroup$
    While you can't prove something with a million examples, you can disprove something with one.
    $endgroup$
    – Melody
    Dec 12 '18 at 14:19










  • $begingroup$
    MathJax hint: if you put a backslash before common functions you get the right font and spacing, so cos x gives $cos x$ compared to cos x giving $cos x$
    $endgroup$
    – Ross Millikan
    Dec 12 '18 at 14:23








2




2




$begingroup$
While you can't prove something with a million examples, you can disprove something with one.
$endgroup$
– Melody
Dec 12 '18 at 14:19




$begingroup$
While you can't prove something with a million examples, you can disprove something with one.
$endgroup$
– Melody
Dec 12 '18 at 14:19












$begingroup$
MathJax hint: if you put a backslash before common functions you get the right font and spacing, so cos x gives $cos x$ compared to cos x giving $cos x$
$endgroup$
– Ross Millikan
Dec 12 '18 at 14:23




$begingroup$
MathJax hint: if you put a backslash before common functions you get the right font and spacing, so cos x gives $cos x$ compared to cos x giving $cos x$
$endgroup$
– Ross Millikan
Dec 12 '18 at 14:23










2 Answers
2






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oldest

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0












$begingroup$

I don't understand your request. Take $x>pi$. Then$$frac{2x}{cos^2(x)+3}geqslantfrac{2x}4=frac x2>fracpi2.$$Therefore, for such a $x$ you cannot possible have$$arctanbigl(f(x)bigr)=frac{2x}{cos^2(x)+3},$$since$$(forall yinmathbb{R}):arctan(y)<fracpi2.$$What can possibly be wrong with this proof?






share|cite|improve this answer









$endgroup$













  • $begingroup$
    that was actually my request, to prove something on a general x, not take a specific x (as in, let there be an x)
    $endgroup$
    – Yuki1112
    Dec 13 '18 at 6:32



















1












$begingroup$

Exactly what you did.



Suppose it exists. Then for $x = 100$ (some big number), $arctan f(x) = frac{200}{cos^2 100 + 3} > frac{200}{4} = 50 > max arctan$, contradiction.






share|cite|improve this answer









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    2 Answers
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    2 Answers
    2






    active

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    0












    $begingroup$

    I don't understand your request. Take $x>pi$. Then$$frac{2x}{cos^2(x)+3}geqslantfrac{2x}4=frac x2>fracpi2.$$Therefore, for such a $x$ you cannot possible have$$arctanbigl(f(x)bigr)=frac{2x}{cos^2(x)+3},$$since$$(forall yinmathbb{R}):arctan(y)<fracpi2.$$What can possibly be wrong with this proof?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      that was actually my request, to prove something on a general x, not take a specific x (as in, let there be an x)
      $endgroup$
      – Yuki1112
      Dec 13 '18 at 6:32
















    0












    $begingroup$

    I don't understand your request. Take $x>pi$. Then$$frac{2x}{cos^2(x)+3}geqslantfrac{2x}4=frac x2>fracpi2.$$Therefore, for such a $x$ you cannot possible have$$arctanbigl(f(x)bigr)=frac{2x}{cos^2(x)+3},$$since$$(forall yinmathbb{R}):arctan(y)<fracpi2.$$What can possibly be wrong with this proof?






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      that was actually my request, to prove something on a general x, not take a specific x (as in, let there be an x)
      $endgroup$
      – Yuki1112
      Dec 13 '18 at 6:32














    0












    0








    0





    $begingroup$

    I don't understand your request. Take $x>pi$. Then$$frac{2x}{cos^2(x)+3}geqslantfrac{2x}4=frac x2>fracpi2.$$Therefore, for such a $x$ you cannot possible have$$arctanbigl(f(x)bigr)=frac{2x}{cos^2(x)+3},$$since$$(forall yinmathbb{R}):arctan(y)<fracpi2.$$What can possibly be wrong with this proof?






    share|cite|improve this answer









    $endgroup$



    I don't understand your request. Take $x>pi$. Then$$frac{2x}{cos^2(x)+3}geqslantfrac{2x}4=frac x2>fracpi2.$$Therefore, for such a $x$ you cannot possible have$$arctanbigl(f(x)bigr)=frac{2x}{cos^2(x)+3},$$since$$(forall yinmathbb{R}):arctan(y)<fracpi2.$$What can possibly be wrong with this proof?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Dec 12 '18 at 14:20









    José Carlos SantosJosé Carlos Santos

    164k22131234




    164k22131234












    • $begingroup$
      that was actually my request, to prove something on a general x, not take a specific x (as in, let there be an x)
      $endgroup$
      – Yuki1112
      Dec 13 '18 at 6:32


















    • $begingroup$
      that was actually my request, to prove something on a general x, not take a specific x (as in, let there be an x)
      $endgroup$
      – Yuki1112
      Dec 13 '18 at 6:32
















    $begingroup$
    that was actually my request, to prove something on a general x, not take a specific x (as in, let there be an x)
    $endgroup$
    – Yuki1112
    Dec 13 '18 at 6:32




    $begingroup$
    that was actually my request, to prove something on a general x, not take a specific x (as in, let there be an x)
    $endgroup$
    – Yuki1112
    Dec 13 '18 at 6:32











    1












    $begingroup$

    Exactly what you did.



    Suppose it exists. Then for $x = 100$ (some big number), $arctan f(x) = frac{200}{cos^2 100 + 3} > frac{200}{4} = 50 > max arctan$, contradiction.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Exactly what you did.



      Suppose it exists. Then for $x = 100$ (some big number), $arctan f(x) = frac{200}{cos^2 100 + 3} > frac{200}{4} = 50 > max arctan$, contradiction.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Exactly what you did.



        Suppose it exists. Then for $x = 100$ (some big number), $arctan f(x) = frac{200}{cos^2 100 + 3} > frac{200}{4} = 50 > max arctan$, contradiction.






        share|cite|improve this answer









        $endgroup$



        Exactly what you did.



        Suppose it exists. Then for $x = 100$ (some big number), $arctan f(x) = frac{200}{cos^2 100 + 3} > frac{200}{4} = 50 > max arctan$, contradiction.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 12 '18 at 14:21









        Lucas HenriqueLucas Henrique

        1,026414




        1,026414






























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