Find function $f(x)$ that satisfying differential relation












7












$begingroup$



Suppose the functions $F(x)$ and $G(x)$ satisfying
$$F(x)=f(x)-frac{1}{f(x)}$$
$$G(x)=f(x)+frac{1}{f(x)}$$
such that $F'(x)=(Gcirc G)(x)$, with initial condition $f(frac{pi}{4})=1$ is given. Find $f(x)$.




I have attempted $F(x)+G(x)=2f(x)$, and try to relate functions $F(x)$ and $G(x)$, but stuck in the composite function $(Gcirc G)(x)$. Taking integration for $F'(x)$ and $(Gcirc G)(x)$ on both sides with respect to $x$ or $F(x)$ do not help that much. Any clue?






calculus integration ordinary-differential-equations functional-equations function-and-relation-composition







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Does $G^2(x)=(Gcirc G)(x)?$ Or is $G^2(x)=(G(x))^2?$
    $endgroup$
    – Adrian Keister
    Dec 11 '18 at 15:29






  • 1




    $begingroup$
    It is composite function, as mentioned in the post.
    $endgroup$
    – weilam06
    Dec 11 '18 at 15:30










  • $begingroup$
    And also, I have thought what if it is $[G(x)]^{2}$...
    $endgroup$
    – weilam06
    Dec 11 '18 at 15:31










  • $begingroup$
    Well, they are two very different things. Which is it?
    $endgroup$
    – Adrian Keister
    Dec 11 '18 at 15:32






  • 4




    $begingroup$
    Judging from Adrian Keister's solution, I think the OP is wrong about the notation. Most likely, the original notation $G^2(x)$ does mean $big(G(x)big)^2$. After all, $f(x)=tan(x+c)$ for some constant $c$ is the solution to the differential equation in Adrian's answer. Thus, $F(x)=-2,cot(2x+2c)$ and $G(x)=2,text{csc}(2x+2c)$. With $fleft(dfrac{pi}{4}right)=1$, we conclude that $f(x)=tan(x)$, $F(x)=-2,cot(2x)$, and $G(x)=2,text{csc}(2x)$ for all $xinmathbb{R}setminus pimathbb{Z}$.
    $endgroup$
    – Batominovski
    Dec 11 '18 at 22:45


















7












$begingroup$



Suppose the functions $F(x)$ and $G(x)$ satisfying
$$F(x)=f(x)-frac{1}{f(x)}$$
$$G(x)=f(x)+frac{1}{f(x)}$$
such that $F'(x)=(Gcirc G)(x)$, with initial condition $f(frac{pi}{4})=1$ is given. Find $f(x)$.




I have attempted $F(x)+G(x)=2f(x)$, and try to relate functions $F(x)$ and $G(x)$, but stuck in the composite function $(Gcirc G)(x)$. Taking integration for $F'(x)$ and $(Gcirc G)(x)$ on both sides with respect to $x$ or $F(x)$ do not help that much. Any clue?






calculus integration ordinary-differential-equations functional-equations function-and-relation-composition







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Does $G^2(x)=(Gcirc G)(x)?$ Or is $G^2(x)=(G(x))^2?$
    $endgroup$
    – Adrian Keister
    Dec 11 '18 at 15:29






  • 1




    $begingroup$
    It is composite function, as mentioned in the post.
    $endgroup$
    – weilam06
    Dec 11 '18 at 15:30










  • $begingroup$
    And also, I have thought what if it is $[G(x)]^{2}$...
    $endgroup$
    – weilam06
    Dec 11 '18 at 15:31










  • $begingroup$
    Well, they are two very different things. Which is it?
    $endgroup$
    – Adrian Keister
    Dec 11 '18 at 15:32






  • 4




    $begingroup$
    Judging from Adrian Keister's solution, I think the OP is wrong about the notation. Most likely, the original notation $G^2(x)$ does mean $big(G(x)big)^2$. After all, $f(x)=tan(x+c)$ for some constant $c$ is the solution to the differential equation in Adrian's answer. Thus, $F(x)=-2,cot(2x+2c)$ and $G(x)=2,text{csc}(2x+2c)$. With $fleft(dfrac{pi}{4}right)=1$, we conclude that $f(x)=tan(x)$, $F(x)=-2,cot(2x)$, and $G(x)=2,text{csc}(2x)$ for all $xinmathbb{R}setminus pimathbb{Z}$.
    $endgroup$
    – Batominovski
    Dec 11 '18 at 22:45
















7












7








7


0



$begingroup$



Suppose the functions $F(x)$ and $G(x)$ satisfying
$$F(x)=f(x)-frac{1}{f(x)}$$
$$G(x)=f(x)+frac{1}{f(x)}$$
such that $F'(x)=(Gcirc G)(x)$, with initial condition $f(frac{pi}{4})=1$ is given. Find $f(x)$.




I have attempted $F(x)+G(x)=2f(x)$, and try to relate functions $F(x)$ and $G(x)$, but stuck in the composite function $(Gcirc G)(x)$. Taking integration for $F'(x)$ and $(Gcirc G)(x)$ on both sides with respect to $x$ or $F(x)$ do not help that much. Any clue?






calculus integration ordinary-differential-equations functional-equations function-and-relation-composition







share|cite|improve this question











$endgroup$





Suppose the functions $F(x)$ and $G(x)$ satisfying
$$F(x)=f(x)-frac{1}{f(x)}$$
$$G(x)=f(x)+frac{1}{f(x)}$$
such that $F'(x)=(Gcirc G)(x)$, with initial condition $f(frac{pi}{4})=1$ is given. Find $f(x)$.




I have attempted $F(x)+G(x)=2f(x)$, and try to relate functions $F(x)$ and $G(x)$, but stuck in the composite function $(Gcirc G)(x)$. Taking integration for $F'(x)$ and $(Gcirc G)(x)$ on both sides with respect to $x$ or $F(x)$ do not help that much. Any clue?






calculus integration ordinary-differential-equations functional-equations function-and-relation-composition




calculus integration ordinary-differential-equations functional-equations function-and-relation-composition






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 11 '18 at 22:04









Batominovski

1




1










asked Dec 11 '18 at 15:17









weilam06weilam06

9511




9511








  • 1




    $begingroup$
    Does $G^2(x)=(Gcirc G)(x)?$ Or is $G^2(x)=(G(x))^2?$
    $endgroup$
    – Adrian Keister
    Dec 11 '18 at 15:29






  • 1




    $begingroup$
    It is composite function, as mentioned in the post.
    $endgroup$
    – weilam06
    Dec 11 '18 at 15:30










  • $begingroup$
    And also, I have thought what if it is $[G(x)]^{2}$...
    $endgroup$
    – weilam06
    Dec 11 '18 at 15:31










  • $begingroup$
    Well, they are two very different things. Which is it?
    $endgroup$
    – Adrian Keister
    Dec 11 '18 at 15:32






  • 4




    $begingroup$
    Judging from Adrian Keister's solution, I think the OP is wrong about the notation. Most likely, the original notation $G^2(x)$ does mean $big(G(x)big)^2$. After all, $f(x)=tan(x+c)$ for some constant $c$ is the solution to the differential equation in Adrian's answer. Thus, $F(x)=-2,cot(2x+2c)$ and $G(x)=2,text{csc}(2x+2c)$. With $fleft(dfrac{pi}{4}right)=1$, we conclude that $f(x)=tan(x)$, $F(x)=-2,cot(2x)$, and $G(x)=2,text{csc}(2x)$ for all $xinmathbb{R}setminus pimathbb{Z}$.
    $endgroup$
    – Batominovski
    Dec 11 '18 at 22:45
















  • 1




    $begingroup$
    Does $G^2(x)=(Gcirc G)(x)?$ Or is $G^2(x)=(G(x))^2?$
    $endgroup$
    – Adrian Keister
    Dec 11 '18 at 15:29






  • 1




    $begingroup$
    It is composite function, as mentioned in the post.
    $endgroup$
    – weilam06
    Dec 11 '18 at 15:30










  • $begingroup$
    And also, I have thought what if it is $[G(x)]^{2}$...
    $endgroup$
    – weilam06
    Dec 11 '18 at 15:31










  • $begingroup$
    Well, they are two very different things. Which is it?
    $endgroup$
    – Adrian Keister
    Dec 11 '18 at 15:32






  • 4




    $begingroup$
    Judging from Adrian Keister's solution, I think the OP is wrong about the notation. Most likely, the original notation $G^2(x)$ does mean $big(G(x)big)^2$. After all, $f(x)=tan(x+c)$ for some constant $c$ is the solution to the differential equation in Adrian's answer. Thus, $F(x)=-2,cot(2x+2c)$ and $G(x)=2,text{csc}(2x+2c)$. With $fleft(dfrac{pi}{4}right)=1$, we conclude that $f(x)=tan(x)$, $F(x)=-2,cot(2x)$, and $G(x)=2,text{csc}(2x)$ for all $xinmathbb{R}setminus pimathbb{Z}$.
    $endgroup$
    – Batominovski
    Dec 11 '18 at 22:45










1




1




$begingroup$
Does $G^2(x)=(Gcirc G)(x)?$ Or is $G^2(x)=(G(x))^2?$
$endgroup$
– Adrian Keister
Dec 11 '18 at 15:29




$begingroup$
Does $G^2(x)=(Gcirc G)(x)?$ Or is $G^2(x)=(G(x))^2?$
$endgroup$
– Adrian Keister
Dec 11 '18 at 15:29




1




1




$begingroup$
It is composite function, as mentioned in the post.
$endgroup$
– weilam06
Dec 11 '18 at 15:30




$begingroup$
It is composite function, as mentioned in the post.
$endgroup$
– weilam06
Dec 11 '18 at 15:30












$begingroup$
And also, I have thought what if it is $[G(x)]^{2}$...
$endgroup$
– weilam06
Dec 11 '18 at 15:31




$begingroup$
And also, I have thought what if it is $[G(x)]^{2}$...
$endgroup$
– weilam06
Dec 11 '18 at 15:31












$begingroup$
Well, they are two very different things. Which is it?
$endgroup$
– Adrian Keister
Dec 11 '18 at 15:32




$begingroup$
Well, they are two very different things. Which is it?
$endgroup$
– Adrian Keister
Dec 11 '18 at 15:32




4




4




$begingroup$
Judging from Adrian Keister's solution, I think the OP is wrong about the notation. Most likely, the original notation $G^2(x)$ does mean $big(G(x)big)^2$. After all, $f(x)=tan(x+c)$ for some constant $c$ is the solution to the differential equation in Adrian's answer. Thus, $F(x)=-2,cot(2x+2c)$ and $G(x)=2,text{csc}(2x+2c)$. With $fleft(dfrac{pi}{4}right)=1$, we conclude that $f(x)=tan(x)$, $F(x)=-2,cot(2x)$, and $G(x)=2,text{csc}(2x)$ for all $xinmathbb{R}setminus pimathbb{Z}$.
$endgroup$
– Batominovski
Dec 11 '18 at 22:45






$begingroup$
Judging from Adrian Keister's solution, I think the OP is wrong about the notation. Most likely, the original notation $G^2(x)$ does mean $big(G(x)big)^2$. After all, $f(x)=tan(x+c)$ for some constant $c$ is the solution to the differential equation in Adrian's answer. Thus, $F(x)=-2,cot(2x+2c)$ and $G(x)=2,text{csc}(2x+2c)$. With $fleft(dfrac{pi}{4}right)=1$, we conclude that $f(x)=tan(x)$, $F(x)=-2,cot(2x)$, and $G(x)=2,text{csc}(2x)$ for all $xinmathbb{R}setminus pimathbb{Z}$.
$endgroup$
– Batominovski
Dec 11 '18 at 22:45












2 Answers
2






active

oldest

votes


















4












$begingroup$

Assuming $G^2(x)=(G(x))^2,$ that is, ordinary multiplication, which was the original question's notation, we have that
begin{align*}
F'&=f'-frac{-f'}{f^2}=frac{f'(1+f^2)}{f^2}, ; text{and} \
G^2&=f^2+2+frac{1}{f^2}.
end{align*}
Then
$$frac{f'(1+f^2)}{f^2}=f^2+2+frac{1}{f^2},$$
or
$$f'(1+f^2)=f^4+2f^2+1=(1+f^2)^2;$$
and therefore
$$f'=1+f^2,$$
which is separable.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Nice! I can't vote it up right now because I used up my votes for the day. However, I have a suggestion. Since the question wants $Gcirc G$, not $Gcdot G$ (but I know that the OP is also curious about using $Gcdot G$), there may be people who will downvote your answer. So, maybe you can put a remark in your answer that you are answering the OP's curiosity that is not directly in the question.
    $endgroup$
    – Batominovski
    Dec 11 '18 at 22:02












  • $begingroup$
    @Batominovski: Thanks for the suggestion! I've put an explanation in the answer.
    $endgroup$
    – Adrian Keister
    Dec 11 '18 at 22:11










  • $begingroup$
    Note that $G^2 = F^2+4$, so you can very easily solve for $F$ first. I think OP was wrong and this is the correct interpretation.
    $endgroup$
    – Dylan
    Dec 13 '18 at 6:18





















3












$begingroup$

Here is a very unfinished attempt, but the last differential equation in $G$ may (or may not) be helpful. Here, I put my trust on the OP's interpretation that $G^2(x)$ means $(Gcirc G)(x)$.



Note that
$$(Gcirc G)(x)=F'(x)=left(1+frac{1}{big(f(x)big)^2}right),f'(x)=G(x),frac{f'(x)}{f(x)},.$$
We then have
$$frac{f'(x)}{f(x)}=frac{(Gcirc G)(x)}{G(x)},.$$
Similarly, observe that
$$G'(x)=left(1-frac{1}{big(f(x)big)^2}right),f'(x)=F(x),frac{f'(x)}{f(x)}=F(x),left(frac{(Gcirc G)(x)}{G(x)}right),.$$
Therefore,
$$F(x)=frac{G(x),G'(x)}{(Gcirc G)(x)},.$$
Taking derivative, we get
$$(Gcirc G)(x)=F'(x)=frac{big(G'(x)big)^2+G(x),G''(x)}{(Gcirc G)(x)}-frac{G(x),G'(x)}{big((Gcirc G)(x)big)^2},G'big(G(x)big),G'(x),.$$
Consequently,
$$(Gcirc G)(x),big(G'(x)big)^2+(Gcirc G)(x),G(x),G''(x)=G(x),big(G'(x)big)^2,G'big(G(x)big)+big((Gcirc G)(x)big)^3,,$$
with $Gleft(dfrac{pi}{4}right)=2$ and $G'left(dfrac{pi}{4}right)=0$.
Well, this looks hopeless.






share|cite|improve this answer











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    2 Answers
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    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    Assuming $G^2(x)=(G(x))^2,$ that is, ordinary multiplication, which was the original question's notation, we have that
    begin{align*}
    F'&=f'-frac{-f'}{f^2}=frac{f'(1+f^2)}{f^2}, ; text{and} \
    G^2&=f^2+2+frac{1}{f^2}.
    end{align*}
    Then
    $$frac{f'(1+f^2)}{f^2}=f^2+2+frac{1}{f^2},$$
    or
    $$f'(1+f^2)=f^4+2f^2+1=(1+f^2)^2;$$
    and therefore
    $$f'=1+f^2,$$
    which is separable.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Nice! I can't vote it up right now because I used up my votes for the day. However, I have a suggestion. Since the question wants $Gcirc G$, not $Gcdot G$ (but I know that the OP is also curious about using $Gcdot G$), there may be people who will downvote your answer. So, maybe you can put a remark in your answer that you are answering the OP's curiosity that is not directly in the question.
      $endgroup$
      – Batominovski
      Dec 11 '18 at 22:02












    • $begingroup$
      @Batominovski: Thanks for the suggestion! I've put an explanation in the answer.
      $endgroup$
      – Adrian Keister
      Dec 11 '18 at 22:11










    • $begingroup$
      Note that $G^2 = F^2+4$, so you can very easily solve for $F$ first. I think OP was wrong and this is the correct interpretation.
      $endgroup$
      – Dylan
      Dec 13 '18 at 6:18


















    4












    $begingroup$

    Assuming $G^2(x)=(G(x))^2,$ that is, ordinary multiplication, which was the original question's notation, we have that
    begin{align*}
    F'&=f'-frac{-f'}{f^2}=frac{f'(1+f^2)}{f^2}, ; text{and} \
    G^2&=f^2+2+frac{1}{f^2}.
    end{align*}
    Then
    $$frac{f'(1+f^2)}{f^2}=f^2+2+frac{1}{f^2},$$
    or
    $$f'(1+f^2)=f^4+2f^2+1=(1+f^2)^2;$$
    and therefore
    $$f'=1+f^2,$$
    which is separable.






    share|cite|improve this answer











    $endgroup$









    • 1




      $begingroup$
      Nice! I can't vote it up right now because I used up my votes for the day. However, I have a suggestion. Since the question wants $Gcirc G$, not $Gcdot G$ (but I know that the OP is also curious about using $Gcdot G$), there may be people who will downvote your answer. So, maybe you can put a remark in your answer that you are answering the OP's curiosity that is not directly in the question.
      $endgroup$
      – Batominovski
      Dec 11 '18 at 22:02












    • $begingroup$
      @Batominovski: Thanks for the suggestion! I've put an explanation in the answer.
      $endgroup$
      – Adrian Keister
      Dec 11 '18 at 22:11










    • $begingroup$
      Note that $G^2 = F^2+4$, so you can very easily solve for $F$ first. I think OP was wrong and this is the correct interpretation.
      $endgroup$
      – Dylan
      Dec 13 '18 at 6:18
















    4












    4








    4





    $begingroup$

    Assuming $G^2(x)=(G(x))^2,$ that is, ordinary multiplication, which was the original question's notation, we have that
    begin{align*}
    F'&=f'-frac{-f'}{f^2}=frac{f'(1+f^2)}{f^2}, ; text{and} \
    G^2&=f^2+2+frac{1}{f^2}.
    end{align*}
    Then
    $$frac{f'(1+f^2)}{f^2}=f^2+2+frac{1}{f^2},$$
    or
    $$f'(1+f^2)=f^4+2f^2+1=(1+f^2)^2;$$
    and therefore
    $$f'=1+f^2,$$
    which is separable.






    share|cite|improve this answer











    $endgroup$



    Assuming $G^2(x)=(G(x))^2,$ that is, ordinary multiplication, which was the original question's notation, we have that
    begin{align*}
    F'&=f'-frac{-f'}{f^2}=frac{f'(1+f^2)}{f^2}, ; text{and} \
    G^2&=f^2+2+frac{1}{f^2}.
    end{align*}
    Then
    $$frac{f'(1+f^2)}{f^2}=f^2+2+frac{1}{f^2},$$
    or
    $$f'(1+f^2)=f^4+2f^2+1=(1+f^2)^2;$$
    and therefore
    $$f'=1+f^2,$$
    which is separable.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 11 '18 at 22:10

























    answered Dec 11 '18 at 15:44









    Adrian KeisterAdrian Keister

    5,19571933




    5,19571933








    • 1




      $begingroup$
      Nice! I can't vote it up right now because I used up my votes for the day. However, I have a suggestion. Since the question wants $Gcirc G$, not $Gcdot G$ (but I know that the OP is also curious about using $Gcdot G$), there may be people who will downvote your answer. So, maybe you can put a remark in your answer that you are answering the OP's curiosity that is not directly in the question.
      $endgroup$
      – Batominovski
      Dec 11 '18 at 22:02












    • $begingroup$
      @Batominovski: Thanks for the suggestion! I've put an explanation in the answer.
      $endgroup$
      – Adrian Keister
      Dec 11 '18 at 22:11










    • $begingroup$
      Note that $G^2 = F^2+4$, so you can very easily solve for $F$ first. I think OP was wrong and this is the correct interpretation.
      $endgroup$
      – Dylan
      Dec 13 '18 at 6:18
















    • 1




      $begingroup$
      Nice! I can't vote it up right now because I used up my votes for the day. However, I have a suggestion. Since the question wants $Gcirc G$, not $Gcdot G$ (but I know that the OP is also curious about using $Gcdot G$), there may be people who will downvote your answer. So, maybe you can put a remark in your answer that you are answering the OP's curiosity that is not directly in the question.
      $endgroup$
      – Batominovski
      Dec 11 '18 at 22:02












    • $begingroup$
      @Batominovski: Thanks for the suggestion! I've put an explanation in the answer.
      $endgroup$
      – Adrian Keister
      Dec 11 '18 at 22:11










    • $begingroup$
      Note that $G^2 = F^2+4$, so you can very easily solve for $F$ first. I think OP was wrong and this is the correct interpretation.
      $endgroup$
      – Dylan
      Dec 13 '18 at 6:18










    1




    1




    $begingroup$
    Nice! I can't vote it up right now because I used up my votes for the day. However, I have a suggestion. Since the question wants $Gcirc G$, not $Gcdot G$ (but I know that the OP is also curious about using $Gcdot G$), there may be people who will downvote your answer. So, maybe you can put a remark in your answer that you are answering the OP's curiosity that is not directly in the question.
    $endgroup$
    – Batominovski
    Dec 11 '18 at 22:02






    $begingroup$
    Nice! I can't vote it up right now because I used up my votes for the day. However, I have a suggestion. Since the question wants $Gcirc G$, not $Gcdot G$ (but I know that the OP is also curious about using $Gcdot G$), there may be people who will downvote your answer. So, maybe you can put a remark in your answer that you are answering the OP's curiosity that is not directly in the question.
    $endgroup$
    – Batominovski
    Dec 11 '18 at 22:02














    $begingroup$
    @Batominovski: Thanks for the suggestion! I've put an explanation in the answer.
    $endgroup$
    – Adrian Keister
    Dec 11 '18 at 22:11




    $begingroup$
    @Batominovski: Thanks for the suggestion! I've put an explanation in the answer.
    $endgroup$
    – Adrian Keister
    Dec 11 '18 at 22:11












    $begingroup$
    Note that $G^2 = F^2+4$, so you can very easily solve for $F$ first. I think OP was wrong and this is the correct interpretation.
    $endgroup$
    – Dylan
    Dec 13 '18 at 6:18






    $begingroup$
    Note that $G^2 = F^2+4$, so you can very easily solve for $F$ first. I think OP was wrong and this is the correct interpretation.
    $endgroup$
    – Dylan
    Dec 13 '18 at 6:18













    3












    $begingroup$

    Here is a very unfinished attempt, but the last differential equation in $G$ may (or may not) be helpful. Here, I put my trust on the OP's interpretation that $G^2(x)$ means $(Gcirc G)(x)$.



    Note that
    $$(Gcirc G)(x)=F'(x)=left(1+frac{1}{big(f(x)big)^2}right),f'(x)=G(x),frac{f'(x)}{f(x)},.$$
    We then have
    $$frac{f'(x)}{f(x)}=frac{(Gcirc G)(x)}{G(x)},.$$
    Similarly, observe that
    $$G'(x)=left(1-frac{1}{big(f(x)big)^2}right),f'(x)=F(x),frac{f'(x)}{f(x)}=F(x),left(frac{(Gcirc G)(x)}{G(x)}right),.$$
    Therefore,
    $$F(x)=frac{G(x),G'(x)}{(Gcirc G)(x)},.$$
    Taking derivative, we get
    $$(Gcirc G)(x)=F'(x)=frac{big(G'(x)big)^2+G(x),G''(x)}{(Gcirc G)(x)}-frac{G(x),G'(x)}{big((Gcirc G)(x)big)^2},G'big(G(x)big),G'(x),.$$
    Consequently,
    $$(Gcirc G)(x),big(G'(x)big)^2+(Gcirc G)(x),G(x),G''(x)=G(x),big(G'(x)big)^2,G'big(G(x)big)+big((Gcirc G)(x)big)^3,,$$
    with $Gleft(dfrac{pi}{4}right)=2$ and $G'left(dfrac{pi}{4}right)=0$.
    Well, this looks hopeless.






    share|cite|improve this answer











    $endgroup$


















      3












      $begingroup$

      Here is a very unfinished attempt, but the last differential equation in $G$ may (or may not) be helpful. Here, I put my trust on the OP's interpretation that $G^2(x)$ means $(Gcirc G)(x)$.



      Note that
      $$(Gcirc G)(x)=F'(x)=left(1+frac{1}{big(f(x)big)^2}right),f'(x)=G(x),frac{f'(x)}{f(x)},.$$
      We then have
      $$frac{f'(x)}{f(x)}=frac{(Gcirc G)(x)}{G(x)},.$$
      Similarly, observe that
      $$G'(x)=left(1-frac{1}{big(f(x)big)^2}right),f'(x)=F(x),frac{f'(x)}{f(x)}=F(x),left(frac{(Gcirc G)(x)}{G(x)}right),.$$
      Therefore,
      $$F(x)=frac{G(x),G'(x)}{(Gcirc G)(x)},.$$
      Taking derivative, we get
      $$(Gcirc G)(x)=F'(x)=frac{big(G'(x)big)^2+G(x),G''(x)}{(Gcirc G)(x)}-frac{G(x),G'(x)}{big((Gcirc G)(x)big)^2},G'big(G(x)big),G'(x),.$$
      Consequently,
      $$(Gcirc G)(x),big(G'(x)big)^2+(Gcirc G)(x),G(x),G''(x)=G(x),big(G'(x)big)^2,G'big(G(x)big)+big((Gcirc G)(x)big)^3,,$$
      with $Gleft(dfrac{pi}{4}right)=2$ and $G'left(dfrac{pi}{4}right)=0$.
      Well, this looks hopeless.






      share|cite|improve this answer











      $endgroup$
















        3












        3








        3





        $begingroup$

        Here is a very unfinished attempt, but the last differential equation in $G$ may (or may not) be helpful. Here, I put my trust on the OP's interpretation that $G^2(x)$ means $(Gcirc G)(x)$.



        Note that
        $$(Gcirc G)(x)=F'(x)=left(1+frac{1}{big(f(x)big)^2}right),f'(x)=G(x),frac{f'(x)}{f(x)},.$$
        We then have
        $$frac{f'(x)}{f(x)}=frac{(Gcirc G)(x)}{G(x)},.$$
        Similarly, observe that
        $$G'(x)=left(1-frac{1}{big(f(x)big)^2}right),f'(x)=F(x),frac{f'(x)}{f(x)}=F(x),left(frac{(Gcirc G)(x)}{G(x)}right),.$$
        Therefore,
        $$F(x)=frac{G(x),G'(x)}{(Gcirc G)(x)},.$$
        Taking derivative, we get
        $$(Gcirc G)(x)=F'(x)=frac{big(G'(x)big)^2+G(x),G''(x)}{(Gcirc G)(x)}-frac{G(x),G'(x)}{big((Gcirc G)(x)big)^2},G'big(G(x)big),G'(x),.$$
        Consequently,
        $$(Gcirc G)(x),big(G'(x)big)^2+(Gcirc G)(x),G(x),G''(x)=G(x),big(G'(x)big)^2,G'big(G(x)big)+big((Gcirc G)(x)big)^3,,$$
        with $Gleft(dfrac{pi}{4}right)=2$ and $G'left(dfrac{pi}{4}right)=0$.
        Well, this looks hopeless.






        share|cite|improve this answer











        $endgroup$



        Here is a very unfinished attempt, but the last differential equation in $G$ may (or may not) be helpful. Here, I put my trust on the OP's interpretation that $G^2(x)$ means $(Gcirc G)(x)$.



        Note that
        $$(Gcirc G)(x)=F'(x)=left(1+frac{1}{big(f(x)big)^2}right),f'(x)=G(x),frac{f'(x)}{f(x)},.$$
        We then have
        $$frac{f'(x)}{f(x)}=frac{(Gcirc G)(x)}{G(x)},.$$
        Similarly, observe that
        $$G'(x)=left(1-frac{1}{big(f(x)big)^2}right),f'(x)=F(x),frac{f'(x)}{f(x)}=F(x),left(frac{(Gcirc G)(x)}{G(x)}right),.$$
        Therefore,
        $$F(x)=frac{G(x),G'(x)}{(Gcirc G)(x)},.$$
        Taking derivative, we get
        $$(Gcirc G)(x)=F'(x)=frac{big(G'(x)big)^2+G(x),G''(x)}{(Gcirc G)(x)}-frac{G(x),G'(x)}{big((Gcirc G)(x)big)^2},G'big(G(x)big),G'(x),.$$
        Consequently,
        $$(Gcirc G)(x),big(G'(x)big)^2+(Gcirc G)(x),G(x),G''(x)=G(x),big(G'(x)big)^2,G'big(G(x)big)+big((Gcirc G)(x)big)^3,,$$
        with $Gleft(dfrac{pi}{4}right)=2$ and $G'left(dfrac{pi}{4}right)=0$.
        Well, this looks hopeless.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 11 '18 at 22:42

























        answered Dec 11 '18 at 22:26









        BatominovskiBatominovski

        1




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