Find function $f(x)$ that satisfying differential relation
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
add a comment |
$begingroup$
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
$endgroup$
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
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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
add a comment |
$begingroup$
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.
$endgroup$
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.
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
add a comment |
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
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
edited Dec 11 '18 at 22:42
answered Dec 11 '18 at 22:26
BatominovskiBatominovski
1
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$begingroup$
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