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Functional equation $3f(-3x) -f(x) = 3x^2$
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I am trying to solve the following functional equation: $f(x)$ is a continuous function, satisfying
(1) $f(f(x))=x$
(2) $ 3f(-3x)-f(x)= 3x^2 $ for $x>0$.
From (1) & (2), I found that $f(x)$ should be decreasing, and $f(x) = f^-1 (x)$.
But, I cannot figure out how to use (2).
How can I use (2) to solve this?
Thanks!
functional-equations inverse-function
asked Jul 17 at 4:30
five manifold
262
add a comment |Â
up vote
5
down vote
favorite
I am trying to solve the following functional equation: $f(x)$ is a continuous function, satisfying
(1) $f(f(x))=x$
(2) $ 3f(-3x)-f(x)= 3x^2 $ for $x>0$.
From (1) & (2), I found that $f(x)$ should be decreasing, and $f(x) = f^-1 (x)$.
But, I cannot figure out how to use (2).
How can I use (2) to solve this?
Thanks!
functional-equations inverse-function
asked Jul 17 at 4:30
five manifold
262
5
"From (1), I found that $f(x)$ should be decreasing" What about $f(x)=x$? It satisfies (1) and is increasing.
– Arthur
Jul 17 at 4:36
1
Yes, only from (1), $f(x)$ can be the solution. But from (2), if $f(x)=x$ for $x>0$, $f(x)$ is not one-to-one.
– five manifold
Jul 17 at 4:39
2
Sure. But it means "From (1), I found that $f(x)$ should be decreasing" is wrong.
– Arthur
Jul 17 at 4:40
For (2) you have condition $x>0$. Is there any condition for (1)?
– Oldboy
Jul 17 at 10:42
(1) is Babbage's equation with prodigious orbits of solutions. (2) has the particular solution $(3/26)x^2 +a/|x|$, but it may not be easy to fit onto one of those orbits....
– Cosmas Zachos
Jul 17 at 15:57
add a comment |Â
up vote
5
down vote
favorite
up vote
5
down vote
favorite
I am trying to solve the following functional equation: $f(x)$ is a continuous function, satisfying
(1) $f(f(x))=x$
(2) $ 3f(-3x)-f(x)= 3x^2 $ for $x>0$.
From (1) & (2), I found that $f(x)$ should be decreasing, and $f(x) = f^-1 (x)$.
But, I cannot figure out how to use (2).
How can I use (2) to solve this?
Thanks!
functional-equations inverse-function
asked Jul 17 at 4:30
five manifold
262
I am trying to solve the following functional equation: $f(x)$ is a continuous function, satisfying
(1) $f(f(x))=x$
(2) $ 3f(-3x)-f(x)= 3x^2 $ for $x>0$.
From (1) & (2), I found that $f(x)$ should be decreasing, and $f(x) = f^-1 (x)$.
But, I cannot figure out how to use (2).
How can I use (2) to solve this?
Thanks!
functional-equations inverse-function
asked Jul 17 at 4:30
five manifold
262
edited Jul 17 at 4:50
asked Jul 17 at 4:30
five manifold
262
asked Jul 17 at 4:30
five manifold
262
asked Jul 17 at 4:30
five manifold
262
262
5
"From (1), I found that $f(x)$ should be decreasing" What about $f(x)=x$? It satisfies (1) and is increasing.
– Arthur
Jul 17 at 4:36
1
Yes, only from (1), $f(x)$ can be the solution. But from (2), if $f(x)=x$ for $x>0$, $f(x)$ is not one-to-one.
– five manifold
Jul 17 at 4:39
2
Sure. But it means "From (1), I found that $f(x)$ should be decreasing" is wrong.
– Arthur
Jul 17 at 4:40
For (2) you have condition $x>0$. Is there any condition for (1)?
– Oldboy
Jul 17 at 10:42
(1) is Babbage's equation with prodigious orbits of solutions. (2) has the particular solution $(3/26)x^2 +a/|x|$, but it may not be easy to fit onto one of those orbits....
– Cosmas Zachos
Jul 17 at 15:57
add a comment |Â
5
"From (1), I found that $f(x)$ should be decreasing" What about $f(x)=x$? It satisfies (1) and is increasing.
– Arthur
Jul 17 at 4:36
1
Yes, only from (1), $f(x)$ can be the solution. But from (2), if $f(x)=x$ for $x>0$, $f(x)$ is not one-to-one.
– five manifold
Jul 17 at 4:39
2
Sure. But it means "From (1), I found that $f(x)$ should be decreasing" is wrong.
– Arthur
Jul 17 at 4:40
For (2) you have condition $x>0$. Is there any condition for (1)?
– Oldboy
Jul 17 at 10:42
(1) is Babbage's equation with prodigious orbits of solutions. (2) has the particular solution $(3/26)x^2 +a/|x|$, but it may not be easy to fit onto one of those orbits....
– Cosmas Zachos
Jul 17 at 15:57
5
5
"From (1), I found that $f(x)$ should be decreasing" What about $f(x)=x$? It satisfies (1) and is increasing.
– Arthur
Jul 17 at 4:36
"From (1), I found that $f(x)$ should be decreasing" What about $f(x)=x$? It satisfies (1) and is increasing.
– Arthur
Jul 17 at 4:36
1
1
Yes, only from (1), $f(x)$ can be the solution. But from (2), if $f(x)=x$ for $x>0$, $f(x)$ is not one-to-one.
– five manifold
Jul 17 at 4:39
Yes, only from (1), $f(x)$ can be the solution. But from (2), if $f(x)=x$ for $x>0$, $f(x)$ is not one-to-one.
– five manifold
Jul 17 at 4:39
2
2
Sure. But it means "From (1), I found that $f(x)$ should be decreasing" is wrong.
– Arthur
Jul 17 at 4:40
Sure. But it means "From (1), I found that $f(x)$ should be decreasing" is wrong.
– Arthur
Jul 17 at 4:40
For (2) you have condition $x>0$. Is there any condition for (1)?
– Oldboy
Jul 17 at 10:42
For (2) you have condition $x>0$. Is there any condition for (1)?
– Oldboy
Jul 17 at 10:42
(1) is Babbage's equation with prodigious orbits of solutions. (2) has the particular solution $(3/26)x^2 +a/|x|$, but it may not be easy to fit onto one of those orbits....
– Cosmas Zachos
Jul 17 at 15:57
(1) is Babbage's equation with prodigious orbits of solutions. (2) has the particular solution $(3/26)x^2 +a/|x|$, but it may not be easy to fit onto one of those orbits....
– Cosmas Zachos
Jul 17 at 15:57
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
There are no solutions.
(The question already notes $f$ is decreasing; I include a proof for completeness).
By (1), $f$ is bijective.
Since $f$ is assumed continuous, this implies $f$ is strictly monotone by this result. That is, $f$ is either strictly increasing or strictly decreasing.
Taking the limit as $xto0^+$ in (2) gives $f(0)=0$. Setting $x=1$ gives
$$
3f(-3)-f(1)=3.tag3
$$
If $f$ is increasing, then $f(-3)<0$ and $f(1)>0$, contradicting (3). Thus $f$ is decreasing. In particular $f(-3)>0$, so (3) gives
$$
f(1)=3f(-3)-3>-3.
$$
Thus $1=f(f(1))<f(-3)$. Now (3) gives
$$
f(1)=3f(-3)-3>0=f(0),
$$
a contradiction.
answered Jul 18 at 5:55
stewbasic
5,6081826
(2) is for x>0 , but since f continuous (2) true for x = 0.
– Anton Sorokovskiy
Jul 18 at 6:22
@AntonSorokovskiy Good point, thanks.
– stewbasic
Jul 18 at 21:51
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
There are no solutions.
(The question already notes $f$ is decreasing; I include a proof for completeness).
By (1), $f$ is bijective.
Since $f$ is assumed continuous, this implies $f$ is strictly monotone by this result. That is, $f$ is either strictly increasing or strictly decreasing.
Taking the limit as $xto0^+$ in (2) gives $f(0)=0$. Setting $x=1$ gives
$$
3f(-3)-f(1)=3.tag3
$$
If $f$ is increasing, then $f(-3)<0$ and $f(1)>0$, contradicting (3). Thus $f$ is decreasing. In particular $f(-3)>0$, so (3) gives
$$
f(1)=3f(-3)-3>-3.
$$
Thus $1=f(f(1))<f(-3)$. Now (3) gives
$$
f(1)=3f(-3)-3>0=f(0),
$$
a contradiction.
answered Jul 18 at 5:55
stewbasic
5,6081826
(2) is for x>0 , but since f continuous (2) true for x = 0.
– Anton Sorokovskiy
Jul 18 at 6:22
@AntonSorokovskiy Good point, thanks.
– stewbasic
Jul 18 at 21:51
add a comment |Â
up vote
1
down vote
There are no solutions.
(The question already notes $f$ is decreasing; I include a proof for completeness).
By (1), $f$ is bijective.
Since $f$ is assumed continuous, this implies $f$ is strictly monotone by this result. That is, $f$ is either strictly increasing or strictly decreasing.
Taking the limit as $xto0^+$ in (2) gives $f(0)=0$. Setting $x=1$ gives
$$
3f(-3)-f(1)=3.tag3
$$
If $f$ is increasing, then $f(-3)<0$ and $f(1)>0$, contradicting (3). Thus $f$ is decreasing. In particular $f(-3)>0$, so (3) gives
$$
f(1)=3f(-3)-3>-3.
$$
Thus $1=f(f(1))<f(-3)$. Now (3) gives
$$
f(1)=3f(-3)-3>0=f(0),
$$
a contradiction.
answered Jul 18 at 5:55
stewbasic
5,6081826
(2) is for x>0 , but since f continuous (2) true for x = 0.
– Anton Sorokovskiy
Jul 18 at 6:22
@AntonSorokovskiy Good point, thanks.
– stewbasic
Jul 18 at 21:51
add a comment |Â
up vote
1
down vote
up vote
1
down vote
There are no solutions.
(The question already notes $f$ is decreasing; I include a proof for completeness).
By (1), $f$ is bijective.
Since $f$ is assumed continuous, this implies $f$ is strictly monotone by this result. That is, $f$ is either strictly increasing or strictly decreasing.
Taking the limit as $xto0^+$ in (2) gives $f(0)=0$. Setting $x=1$ gives
$$
3f(-3)-f(1)=3.tag3
$$
If $f$ is increasing, then $f(-3)<0$ and $f(1)>0$, contradicting (3). Thus $f$ is decreasing. In particular $f(-3)>0$, so (3) gives
$$
f(1)=3f(-3)-3>-3.
$$
Thus $1=f(f(1))<f(-3)$. Now (3) gives
$$
f(1)=3f(-3)-3>0=f(0),
$$
a contradiction.
answered Jul 18 at 5:55
stewbasic
5,6081826
There are no solutions.
(The question already notes $f$ is decreasing; I include a proof for completeness).
By (1), $f$ is bijective.
Since $f$ is assumed continuous, this implies $f$ is strictly monotone by this result. That is, $f$ is either strictly increasing or strictly decreasing.
Taking the limit as $xto0^+$ in (2) gives $f(0)=0$. Setting $x=1$ gives
$$
3f(-3)-f(1)=3.tag3
$$
If $f$ is increasing, then $f(-3)<0$ and $f(1)>0$, contradicting (3). Thus $f$ is decreasing. In particular $f(-3)>0$, so (3) gives
$$
f(1)=3f(-3)-3>-3.
$$
Thus $1=f(f(1))<f(-3)$. Now (3) gives
$$
f(1)=3f(-3)-3>0=f(0),
$$
a contradiction.
answered Jul 18 at 5:55
stewbasic
5,6081826
edited Jul 18 at 21:50
answered Jul 18 at 5:55
stewbasic
5,6081826
answered Jul 18 at 5:55
stewbasic
5,6081826
answered Jul 18 at 5:55
stewbasic
5,6081826
5,6081826
(2) is for x>0 , but since f continuous (2) true for x = 0.
– Anton Sorokovskiy
Jul 18 at 6:22
@AntonSorokovskiy Good point, thanks.
– stewbasic
Jul 18 at 21:51
add a comment |Â
(2) is for x>0 , but since f continuous (2) true for x = 0.
– Anton Sorokovskiy
Jul 18 at 6:22
@AntonSorokovskiy Good point, thanks.
– stewbasic
Jul 18 at 21:51
(2) is for x>0 , but since f continuous (2) true for x = 0.
– Anton Sorokovskiy
Jul 18 at 6:22
(2) is for x>0 , but since f continuous (2) true for x = 0.
– Anton Sorokovskiy
Jul 18 at 6:22
@AntonSorokovskiy Good point, thanks.
– stewbasic
Jul 18 at 21:51
@AntonSorokovskiy Good point, thanks.
– stewbasic
Jul 18 at 21:51
add a comment |Â
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5
"From (1), I found that $f(x)$ should be decreasing" What about $f(x)=x$? It satisfies (1) and is increasing.
– Arthur
Jul 17 at 4:36
1
Yes, only from (1), $f(x)$ can be the solution. But from (2), if $f(x)=x$ for $x>0$, $f(x)$ is not one-to-one.
– five manifold
Jul 17 at 4:39
2
Sure. But it means "From (1), I found that $f(x)$ should be decreasing" is wrong.
– Arthur
Jul 17 at 4:40
For (2) you have condition $x>0$. Is there any condition for (1)?
– Oldboy
Jul 17 at 10:42
(1) is Babbage's equation with prodigious orbits of solutions. (2) has the particular solution $(3/26)x^2 +a/|x|$, but it may not be easy to fit onto one of those orbits....
– Cosmas Zachos
Jul 17 at 15:57