Mixing
Functional equation involving $f(x^4)+f(x^2)+f(x)$
$begingroup$
Find all increasing functions $f$ from positive reals to positive reals satisfying $f(x^4) + f(x^2) + f(x) = x^4 + x^2 + x$.
It's easy to show that $f(1)=1$, and I was also able to show that
$$f(x)-x = f(x^{(8^k)}) - x^{(8^k)}$$
for all integers $k$.
But where to go from here? Any hints would be much appreciated.
real-analysis functional-equations
edited Jan 24 at 15:40
6005
37k751127
asked Jan 24 at 13:49
PrasiortlePrasiortle
1827
$endgroup$
add a comment |
$begingroup$
Find all increasing functions $f$ from positive reals to positive reals satisfying $f(x^4) + f(x^2) + f(x) = x^4 + x^2 + x$.
It's easy to show that $f(1)=1$, and I was also able to show that
$$f(x)-x = f(x^{(8^k)}) - x^{(8^k)}$$
for all integers $k$.
But where to go from here? Any hints would be much appreciated.
real-analysis functional-equations
edited Jan 24 at 15:40
6005
37k751127
asked Jan 24 at 13:49
PrasiortlePrasiortle
1827
$endgroup$
1
$begingroup$
Well, it's clear that $f(x)=x$ satisfies the equation, but I'm not sure that's the only one.
$endgroup$
– Adrian Keister
Jan 24 at 14:03
1
$begingroup$
Just a thought: If you set $g(x):=f(x)-x$ then the equation above becomes $g(x^4)+g(x^2)+g(x)=0$. However, now it's harder to take into account the fact that $f$ is increasing.
$endgroup$
– Yanko
Jan 24 at 14:24
$begingroup$
@Cesareo, Is it increasing or $(0,infty)$-valued over $(0,infty)$ for any non-trivial choices of $C_1'$ and $C_2'$?
$endgroup$
– Sangchul Lee
Jan 24 at 17:40
$begingroup$
@SangchulLee You got it!. I simply assumed a real function. I didn't see the increasing. My fault.
$endgroup$
– Cesareo
Jan 24 at 19:01
add a comment |
$begingroup$
Find all increasing functions $f$ from positive reals to positive reals satisfying $f(x^4) + f(x^2) + f(x) = x^4 + x^2 + x$.
It's easy to show that $f(1)=1$, and I was also able to show that
$$f(x)-x = f(x^{(8^k)}) - x^{(8^k)}$$
for all integers $k$.
But where to go from here? Any hints would be much appreciated.
real-analysis functional-equations
edited Jan 24 at 15:40
6005
37k751127
asked Jan 24 at 13:49
PrasiortlePrasiortle
1827
$endgroup$
Find all increasing functions $f$ from positive reals to positive reals satisfying $f(x^4) + f(x^2) + f(x) = x^4 + x^2 + x$.
It's easy to show that $f(1)=1$, and I was also able to show that
$$f(x)-x = f(x^{(8^k)}) - x^{(8^k)}$$
for all integers $k$.
But where to go from here? Any hints would be much appreciated.
real-analysis functional-equations
real-analysis functional-equations
edited Jan 24 at 15:40
6005
37k751127
asked Jan 24 at 13:49
PrasiortlePrasiortle
1827
edited Jan 24 at 15:40
6005
37k751127
asked Jan 24 at 13:49
PrasiortlePrasiortle
1827
edited Jan 24 at 15:40
6005
37k751127
edited Jan 24 at 15:40
6005
37k751127
edited Jan 24 at 15:40
6005
37k751127
37k751127
asked Jan 24 at 13:49
PrasiortlePrasiortle
1827
asked Jan 24 at 13:49
PrasiortlePrasiortle
1827
asked Jan 24 at 13:49
PrasiortlePrasiortle
1827
1827
1
$begingroup$
Well, it's clear that $f(x)=x$ satisfies the equation, but I'm not sure that's the only one.
$endgroup$
– Adrian Keister
Jan 24 at 14:03
1
$begingroup$
Just a thought: If you set $g(x):=f(x)-x$ then the equation above becomes $g(x^4)+g(x^2)+g(x)=0$. However, now it's harder to take into account the fact that $f$ is increasing.
$endgroup$
– Yanko
Jan 24 at 14:24
$begingroup$
@Cesareo, Is it increasing or $(0,infty)$-valued over $(0,infty)$ for any non-trivial choices of $C_1'$ and $C_2'$?
$endgroup$
– Sangchul Lee
Jan 24 at 17:40
$begingroup$
@SangchulLee You got it!. I simply assumed a real function. I didn't see the increasing. My fault.
$endgroup$
– Cesareo
Jan 24 at 19:01
add a comment |
1
$begingroup$
Well, it's clear that $f(x)=x$ satisfies the equation, but I'm not sure that's the only one.
$endgroup$
– Adrian Keister
Jan 24 at 14:03
1
$begingroup$
Just a thought: If you set $g(x):=f(x)-x$ then the equation above becomes $g(x^4)+g(x^2)+g(x)=0$. However, now it's harder to take into account the fact that $f$ is increasing.
$endgroup$
– Yanko
Jan 24 at 14:24
$begingroup$
@Cesareo, Is it increasing or $(0,infty)$-valued over $(0,infty)$ for any non-trivial choices of $C_1'$ and $C_2'$?
$endgroup$
– Sangchul Lee
Jan 24 at 17:40
$begingroup$
@SangchulLee You got it!. I simply assumed a real function. I didn't see the increasing. My fault.
$endgroup$
– Cesareo
Jan 24 at 19:01
1
1
$begingroup$
Well, it's clear that $f(x)=x$ satisfies the equation, but I'm not sure that's the only one.
$endgroup$
– Adrian Keister
Jan 24 at 14:03
$begingroup$
Well, it's clear that $f(x)=x$ satisfies the equation, but I'm not sure that's the only one.
$endgroup$
– Adrian Keister
Jan 24 at 14:03
1
1
$begingroup$
Just a thought: If you set $g(x):=f(x)-x$ then the equation above becomes $g(x^4)+g(x^2)+g(x)=0$. However, now it's harder to take into account the fact that $f$ is increasing.
$endgroup$
– Yanko
Jan 24 at 14:24
$begingroup$
Just a thought: If you set $g(x):=f(x)-x$ then the equation above becomes $g(x^4)+g(x^2)+g(x)=0$. However, now it's harder to take into account the fact that $f$ is increasing.
$endgroup$
– Yanko
Jan 24 at 14:24
$begingroup$
@Cesareo, Is it increasing or $(0,infty)$-valued over $(0,infty)$ for any non-trivial choices of $C_1'$ and $C_2'$?
$endgroup$
– Sangchul Lee
Jan 24 at 17:40
$begingroup$
@Cesareo, Is it increasing or $(0,infty)$-valued over $(0,infty)$ for any non-trivial choices of $C_1'$ and $C_2'$?
$endgroup$
– Sangchul Lee
Jan 24 at 17:40
$begingroup$
@SangchulLee You got it!. I simply assumed a real function. I didn't see the increasing. My fault.
$endgroup$
– Cesareo
Jan 24 at 19:01
$begingroup$
@SangchulLee You got it!. I simply assumed a real function. I didn't see the increasing. My fault.
$endgroup$
– Cesareo
Jan 24 at 19:01
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Let $g(x) = f(x)-x$. We obtain $g(x^4)+g(x^2)+g(x)=0$. Define $a_n = g(x^{2^n})$ for every $ninBbb Z$ where $xne 1,x>0$. It follows that
$$
a_{n+2}+a_{n+1}+a_n=0,
$$ and from this 2nd order linear equation we get for some $A,B$,$$
a_n =begin{cases}A,quad nequiv 0 ;text{(mod }3)\ B,quad nequiv 1 ;text{(mod }3)\-A-B,quad nequiv 2 ;text{(mod }3)end{cases}.
$$
It says that $(a_n)$ is a $3$-periodic sequence. Now we claim that $a_n = 0$ for every $n$. Since $f(x)=x+g(x)$ is increasing, it holds for each $k=0,1,2$ and all $nequiv ktext{ (mod} ;3)$, $x>1$ that
$$
x^{2^n}+a_k le x^{2^{n+1}}+a_{k+1}le x^{2^{n+2}}+a_{k+2}.
$$ Let $nequiv ktext{ (mod} ;3)$ tend to $-infty$ to obtain
$$
a_kle a_{k+1}le a_{k+2}
$$ for each $k=0,1,2$. This implies $a_0=a_1=a_2=0$ and thus $a_n = 0, forall n$ follows for $x>1$. The case where $0<x<1$ can be treated similarly by noting that
$$
x^{2^n}+a_k ge x^{2^{n+1}}+a_{k+1}ge x^{2^{n+2}}+a_{k+2}.
$$
This proves $f(x)=x+g(x)=x$ for all $x>0$.
answered Jan 24 at 14:34
SongSong
18.1k21449
$endgroup$
add a comment |
$begingroup$
First we show that $lim_{x to 1} f(x) = f(1) = 1$. Note that the limit of an increasing function from below or from above always exists. Let the limit from below be $a$ and the limit from above be $b$. If we take the limit from below of $f(x) + f(x^2) + f(x^4) = x + x^2 + x^4$, we get $a + a + a = 3$, so $a = 1$ (since as $x$ approaches $1$, $x^2$ and $x^4$ also approach $1$). Similarly from above, $b + b + b = 3$, so $b = 1$. Finally, we already know $f(1) = 1$. So
$$
lim_{x to 1} f(1) = 1.
$$
(In fact, we could show $f$ is continuous for all $x$, using a similar argument.)
To finish, consider your equation $f(x) + f(x^{8^k}) = x + x^{8^{k}}$. Take the limit as $k to -infty$. Regardless of $x$, $x^{8^k}$ approaches $x^{0} = 1$. So, we get
begin{align*}
f(x) + 1 = x + 1 \
implies & boxed{f(x) = x},
end{align*}
which holds for all $x$. So $f(x) = x$ is the only solution.
answered Jan 24 at 15:35
60056005
37k751127
$endgroup$
$begingroup$
Did you mean $x^4$, not $x^3$?
$endgroup$
– Prasiortle
Jan 24 at 15:45
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A minor error: $limlimits_{kto-infty} x^{8^k}=1$.
$endgroup$
– Song
Jan 24 at 15:48
$begingroup$
@Prasiortle Thanks, fixed
$endgroup$
– 6005
Jan 24 at 15:55
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@Song Thanks for the correction. So I needed to do the limit at $x = 1$, not $x = 0$. Fixed now.
$endgroup$
– 6005
Jan 24 at 16:01
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@Song Very nice solution by the way :)
$endgroup$
– 6005
Jan 24 at 16:04
|
show 5 more comments
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If we are interested in functions smooth enough to admit some power series expansion (around $x=0$):
$$f(x) = sum_{k=0}^infty c_kx^k$$
Since $$f(x^n) = sum_{k=0}^infty c_kx^{kn}$$
It will force relations between the coefficients in a systematic way which you can then investigate.
answered Jan 24 at 15:33
mathreadlermathreadler
15.1k72263
$endgroup$
1
$begingroup$
Doing this we immediately derive $f(x) = x$ since all other coefficients must be $0$. But how does it help? Most functions cannot be given as a power series.
$endgroup$
– 6005
Jan 24 at 15:38
$begingroup$
@6005 : Yes the assumption here is if the function is smooth enough to admit a power series around $x = 0$.
$endgroup$
– mathreadler
Jan 24 at 15:42
add a comment |
$begingroup$
Assuming $f(x)$ as a real function, calling $g(x) = f(x)-x$ we have
$$
g(x^4)+g(x^2)+g(x) = 0
$$
now calling $e^y = x$ we have
$$
g(e^{4y})+g(e^{2y})+g(e^y) = 0
$$
or
$$
G(4y)+G(2y)+G(y) = 0
$$
This recurrence equation has as solution
$$
G(y) = C_1 y^{-frac{2 i pi }{3 ln 2}}+C_2 y^{frac{2 i pi }{3 ln 2}}
$$
because
$$
mathcal{G}(log_2 (4y))+mathcal{G}(log_2(2y))+mathcal{G}(log_2 y)=0\
mathcal{G}(log_2 y+2)+mathcal{G}(log_2 y +1)+mathcal{G}(log_2 y)=0
$$
and making $z = log_2 y$ we get
$$
mathcal{G}(z+2)+mathcal{G}(z+1)+mathcal{G}(z)=0
$$
which can be characterized as second order linear difference equation.
Then follows
$$
G(y) = C'_1 cos left(frac{2 pi}{3} log_2 yright)+C'_2sin left(frac{2 pi}{3} log_2 yright) = g(e^y) = f(e^y)-e^y
$$
and finally
$$
f(x) = C'_1cosleft(frac{2pi}{3}log_2(ln x)right)+C'_2sinleft(frac{2pi}{3}log_2(ln x)right)+x
$$
In case of $f(x)$ increasing then the solution is $f(x) = x$ with $C'_1=C'_2=0$
NOTE
Use this MATHEMATICA script to verify the recurrence equation as well as the final solution.
Clear[G]
G[y_] := Cos[(2 [Pi] Log[2, y])/3] c1 + Sin[(2 [Pi] Log[2, y])/3 ] c2
G[4 y] + G[2 y] + G[y] // FullSimplify
f[x_] := Cos[(2 [Pi] Log[2, Log[x]])/3] c1 + Sin[(2 [Pi] Log[2, Log[x]])/3 ] c2 + x
Assuming[x > 0, f[x^4] + f[x^2] + f[x] - x^4 - x^2 - x // FullSimplify]
answered Jan 24 at 15:26
CesareoCesareo
9,3963517
$endgroup$
$begingroup$
I don't understand your solution. $C_1$ and $C_2$ differ depending on which $y$ you pick.
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– 6005
Jan 24 at 15:37
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@6005 Did you check if $G(y)$ verify the recurrence equation $G(4y)+G(2y)+G(y)=0$ ?
$endgroup$
– Cesareo
Jan 24 at 15:44
$begingroup$
what do you mean? I agree it verifies that recurrence, but the $C_1$ and $C_2$ are different depending on which $y$ you pick. There is actually an infinite-dimensional family of solutions.
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– 6005
Jan 24 at 16:02
$begingroup$
@6005 A recursive equation is quite similar to a differential equation. This is a second order linear recursive equation so we have two free constant which can be determined depending on boundary conditions. So the solution is complete with two arbitrary constant.
$endgroup$
– Cesareo
Jan 24 at 16:07
1
$begingroup$
In general, we may pick for each $y in [1,2)$ a different pair of constants $C_1'$ and $C_2'$. That's infinitely many constants. Then, $G(2y)$, $G(4y)$, $G(8y)$ and so on are determined by the constants $C_1'$ and $C_2'$ for that particular $y in [1,2)$.
$endgroup$
– 6005
Jan 24 at 16:11
|
show 14 more comments
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $g(x) = f(x)-x$. We obtain $g(x^4)+g(x^2)+g(x)=0$. Define $a_n = g(x^{2^n})$ for every $ninBbb Z$ where $xne 1,x>0$. It follows that
$$
a_{n+2}+a_{n+1}+a_n=0,
$$ and from this 2nd order linear equation we get for some $A,B$,$$
a_n =begin{cases}A,quad nequiv 0 ;text{(mod }3)\ B,quad nequiv 1 ;text{(mod }3)\-A-B,quad nequiv 2 ;text{(mod }3)end{cases}.
$$
It says that $(a_n)$ is a $3$-periodic sequence. Now we claim that $a_n = 0$ for every $n$. Since $f(x)=x+g(x)$ is increasing, it holds for each $k=0,1,2$ and all $nequiv ktext{ (mod} ;3)$, $x>1$ that
$$
x^{2^n}+a_k le x^{2^{n+1}}+a_{k+1}le x^{2^{n+2}}+a_{k+2}.
$$ Let $nequiv ktext{ (mod} ;3)$ tend to $-infty$ to obtain
$$
a_kle a_{k+1}le a_{k+2}
$$ for each $k=0,1,2$. This implies $a_0=a_1=a_2=0$ and thus $a_n = 0, forall n$ follows for $x>1$. The case where $0<x<1$ can be treated similarly by noting that
$$
x^{2^n}+a_k ge x^{2^{n+1}}+a_{k+1}ge x^{2^{n+2}}+a_{k+2}.
$$
This proves $f(x)=x+g(x)=x$ for all $x>0$.
answered Jan 24 at 14:34
SongSong
18.1k21449
$endgroup$
add a comment |
$begingroup$
Let $g(x) = f(x)-x$. We obtain $g(x^4)+g(x^2)+g(x)=0$. Define $a_n = g(x^{2^n})$ for every $ninBbb Z$ where $xne 1,x>0$. It follows that
$$
a_{n+2}+a_{n+1}+a_n=0,
$$ and from this 2nd order linear equation we get for some $A,B$,$$
a_n =begin{cases}A,quad nequiv 0 ;text{(mod }3)\ B,quad nequiv 1 ;text{(mod }3)\-A-B,quad nequiv 2 ;text{(mod }3)end{cases}.
$$
It says that $(a_n)$ is a $3$-periodic sequence. Now we claim that $a_n = 0$ for every $n$. Since $f(x)=x+g(x)$ is increasing, it holds for each $k=0,1,2$ and all $nequiv ktext{ (mod} ;3)$, $x>1$ that
$$
x^{2^n}+a_k le x^{2^{n+1}}+a_{k+1}le x^{2^{n+2}}+a_{k+2}.
$$ Let $nequiv ktext{ (mod} ;3)$ tend to $-infty$ to obtain
$$
a_kle a_{k+1}le a_{k+2}
$$ for each $k=0,1,2$. This implies $a_0=a_1=a_2=0$ and thus $a_n = 0, forall n$ follows for $x>1$. The case where $0<x<1$ can be treated similarly by noting that
$$
x^{2^n}+a_k ge x^{2^{n+1}}+a_{k+1}ge x^{2^{n+2}}+a_{k+2}.
$$
This proves $f(x)=x+g(x)=x$ for all $x>0$.
answered Jan 24 at 14:34
SongSong
18.1k21449
$endgroup$
add a comment |
$begingroup$
Let $g(x) = f(x)-x$. We obtain $g(x^4)+g(x^2)+g(x)=0$. Define $a_n = g(x^{2^n})$ for every $ninBbb Z$ where $xne 1,x>0$. It follows that
$$
a_{n+2}+a_{n+1}+a_n=0,
$$ and from this 2nd order linear equation we get for some $A,B$,$$
a_n =begin{cases}A,quad nequiv 0 ;text{(mod }3)\ B,quad nequiv 1 ;text{(mod }3)\-A-B,quad nequiv 2 ;text{(mod }3)end{cases}.
$$
It says that $(a_n)$ is a $3$-periodic sequence. Now we claim that $a_n = 0$ for every $n$. Since $f(x)=x+g(x)$ is increasing, it holds for each $k=0,1,2$ and all $nequiv ktext{ (mod} ;3)$, $x>1$ that
$$
x^{2^n}+a_k le x^{2^{n+1}}+a_{k+1}le x^{2^{n+2}}+a_{k+2}.
$$ Let $nequiv ktext{ (mod} ;3)$ tend to $-infty$ to obtain
$$
a_kle a_{k+1}le a_{k+2}
$$ for each $k=0,1,2$. This implies $a_0=a_1=a_2=0$ and thus $a_n = 0, forall n$ follows for $x>1$. The case where $0<x<1$ can be treated similarly by noting that
$$
x^{2^n}+a_k ge x^{2^{n+1}}+a_{k+1}ge x^{2^{n+2}}+a_{k+2}.
$$
This proves $f(x)=x+g(x)=x$ for all $x>0$.
answered Jan 24 at 14:34
SongSong
18.1k21449
$endgroup$
Let $g(x) = f(x)-x$. We obtain $g(x^4)+g(x^2)+g(x)=0$. Define $a_n = g(x^{2^n})$ for every $ninBbb Z$ where $xne 1,x>0$. It follows that
$$
a_{n+2}+a_{n+1}+a_n=0,
$$ and from this 2nd order linear equation we get for some $A,B$,$$
a_n =begin{cases}A,quad nequiv 0 ;text{(mod }3)\ B,quad nequiv 1 ;text{(mod }3)\-A-B,quad nequiv 2 ;text{(mod }3)end{cases}.
$$
It says that $(a_n)$ is a $3$-periodic sequence. Now we claim that $a_n = 0$ for every $n$. Since $f(x)=x+g(x)$ is increasing, it holds for each $k=0,1,2$ and all $nequiv ktext{ (mod} ;3)$, $x>1$ that
$$
x^{2^n}+a_k le x^{2^{n+1}}+a_{k+1}le x^{2^{n+2}}+a_{k+2}.
$$ Let $nequiv ktext{ (mod} ;3)$ tend to $-infty$ to obtain
$$
a_kle a_{k+1}le a_{k+2}
$$ for each $k=0,1,2$. This implies $a_0=a_1=a_2=0$ and thus $a_n = 0, forall n$ follows for $x>1$. The case where $0<x<1$ can be treated similarly by noting that
$$
x^{2^n}+a_k ge x^{2^{n+1}}+a_{k+1}ge x^{2^{n+2}}+a_{k+2}.
$$
This proves $f(x)=x+g(x)=x$ for all $x>0$.
answered Jan 24 at 14:34
SongSong
18.1k21449
edited Jan 24 at 14:51
answered Jan 24 at 14:34
SongSong
18.1k21449
answered Jan 24 at 14:34
SongSong
18.1k21449
answered Jan 24 at 14:34
SongSong
18.1k21449
18.1k21449
add a comment |
add a comment |
$begingroup$
First we show that $lim_{x to 1} f(x) = f(1) = 1$. Note that the limit of an increasing function from below or from above always exists. Let the limit from below be $a$ and the limit from above be $b$. If we take the limit from below of $f(x) + f(x^2) + f(x^4) = x + x^2 + x^4$, we get $a + a + a = 3$, so $a = 1$ (since as $x$ approaches $1$, $x^2$ and $x^4$ also approach $1$). Similarly from above, $b + b + b = 3$, so $b = 1$. Finally, we already know $f(1) = 1$. So
$$
lim_{x to 1} f(1) = 1.
$$
(In fact, we could show $f$ is continuous for all $x$, using a similar argument.)
To finish, consider your equation $f(x) + f(x^{8^k}) = x + x^{8^{k}}$. Take the limit as $k to -infty$. Regardless of $x$, $x^{8^k}$ approaches $x^{0} = 1$. So, we get
begin{align*}
f(x) + 1 = x + 1 \
implies & boxed{f(x) = x},
end{align*}
which holds for all $x$. So $f(x) = x$ is the only solution.
answered Jan 24 at 15:35
60056005
37k751127
$endgroup$
$begingroup$
Did you mean $x^4$, not $x^3$?
$endgroup$
– Prasiortle
Jan 24 at 15:45
$begingroup$
A minor error: $limlimits_{kto-infty} x^{8^k}=1$.
$endgroup$
– Song
Jan 24 at 15:48
$begingroup$
@Prasiortle Thanks, fixed
$endgroup$
– 6005
Jan 24 at 15:55
$begingroup$
@Song Thanks for the correction. So I needed to do the limit at $x = 1$, not $x = 0$. Fixed now.
$endgroup$
– 6005
Jan 24 at 16:01
$begingroup$
@Song Very nice solution by the way :)
$endgroup$
– 6005
Jan 24 at 16:04
|
show 5 more comments
$begingroup$
First we show that $lim_{x to 1} f(x) = f(1) = 1$. Note that the limit of an increasing function from below or from above always exists. Let the limit from below be $a$ and the limit from above be $b$. If we take the limit from below of $f(x) + f(x^2) + f(x^4) = x + x^2 + x^4$, we get $a + a + a = 3$, so $a = 1$ (since as $x$ approaches $1$, $x^2$ and $x^4$ also approach $1$). Similarly from above, $b + b + b = 3$, so $b = 1$. Finally, we already know $f(1) = 1$. So
$$
lim_{x to 1} f(1) = 1.
$$
(In fact, we could show $f$ is continuous for all $x$, using a similar argument.)
To finish, consider your equation $f(x) + f(x^{8^k}) = x + x^{8^{k}}$. Take the limit as $k to -infty$. Regardless of $x$, $x^{8^k}$ approaches $x^{0} = 1$. So, we get
begin{align*}
f(x) + 1 = x + 1 \
implies & boxed{f(x) = x},
end{align*}
which holds for all $x$. So $f(x) = x$ is the only solution.
answered Jan 24 at 15:35
60056005
37k751127
$endgroup$
$begingroup$
Did you mean $x^4$, not $x^3$?
$endgroup$
– Prasiortle
Jan 24 at 15:45
$begingroup$
A minor error: $limlimits_{kto-infty} x^{8^k}=1$.
$endgroup$
– Song
Jan 24 at 15:48
$begingroup$
@Prasiortle Thanks, fixed
$endgroup$
– 6005
Jan 24 at 15:55
$begingroup$
@Song Thanks for the correction. So I needed to do the limit at $x = 1$, not $x = 0$. Fixed now.
$endgroup$
– 6005
Jan 24 at 16:01
$begingroup$
@Song Very nice solution by the way :)
$endgroup$
– 6005
Jan 24 at 16:04
|
show 5 more comments
$begingroup$
First we show that $lim_{x to 1} f(x) = f(1) = 1$. Note that the limit of an increasing function from below or from above always exists. Let the limit from below be $a$ and the limit from above be $b$. If we take the limit from below of $f(x) + f(x^2) + f(x^4) = x + x^2 + x^4$, we get $a + a + a = 3$, so $a = 1$ (since as $x$ approaches $1$, $x^2$ and $x^4$ also approach $1$). Similarly from above, $b + b + b = 3$, so $b = 1$. Finally, we already know $f(1) = 1$. So
$$
lim_{x to 1} f(1) = 1.
$$
(In fact, we could show $f$ is continuous for all $x$, using a similar argument.)
To finish, consider your equation $f(x) + f(x^{8^k}) = x + x^{8^{k}}$. Take the limit as $k to -infty$. Regardless of $x$, $x^{8^k}$ approaches $x^{0} = 1$. So, we get
begin{align*}
f(x) + 1 = x + 1 \
implies & boxed{f(x) = x},
end{align*}
which holds for all $x$. So $f(x) = x$ is the only solution.
answered Jan 24 at 15:35
60056005
37k751127
$endgroup$
First we show that $lim_{x to 1} f(x) = f(1) = 1$. Note that the limit of an increasing function from below or from above always exists. Let the limit from below be $a$ and the limit from above be $b$. If we take the limit from below of $f(x) + f(x^2) + f(x^4) = x + x^2 + x^4$, we get $a + a + a = 3$, so $a = 1$ (since as $x$ approaches $1$, $x^2$ and $x^4$ also approach $1$). Similarly from above, $b + b + b = 3$, so $b = 1$. Finally, we already know $f(1) = 1$. So
$$
lim_{x to 1} f(1) = 1.
$$
(In fact, we could show $f$ is continuous for all $x$, using a similar argument.)
To finish, consider your equation $f(x) + f(x^{8^k}) = x + x^{8^{k}}$. Take the limit as $k to -infty$. Regardless of $x$, $x^{8^k}$ approaches $x^{0} = 1$. So, we get
begin{align*}
f(x) + 1 = x + 1 \
implies & boxed{f(x) = x},
end{align*}
which holds for all $x$. So $f(x) = x$ is the only solution.
answered Jan 24 at 15:35
60056005
37k751127
edited Jan 24 at 18:18
answered Jan 24 at 15:35
60056005
37k751127
answered Jan 24 at 15:35
60056005
37k751127
answered Jan 24 at 15:35
60056005
37k751127
37k751127
$begingroup$
Did you mean $x^4$, not $x^3$?
$endgroup$
– Prasiortle
Jan 24 at 15:45
$begingroup$
A minor error: $limlimits_{kto-infty} x^{8^k}=1$.
$endgroup$
– Song
Jan 24 at 15:48
$begingroup$
@Prasiortle Thanks, fixed
$endgroup$
– 6005
Jan 24 at 15:55
$begingroup$
@Song Thanks for the correction. So I needed to do the limit at $x = 1$, not $x = 0$. Fixed now.
$endgroup$
– 6005
Jan 24 at 16:01
$begingroup$
@Song Very nice solution by the way :)
$endgroup$
– 6005
Jan 24 at 16:04
|
show 5 more comments
$begingroup$
Did you mean $x^4$, not $x^3$?
$endgroup$
– Prasiortle
Jan 24 at 15:45
$begingroup$
A minor error: $limlimits_{kto-infty} x^{8^k}=1$.
$endgroup$
– Song
Jan 24 at 15:48
$begingroup$
@Prasiortle Thanks, fixed
$endgroup$
– 6005
Jan 24 at 15:55
$begingroup$
@Song Thanks for the correction. So I needed to do the limit at $x = 1$, not $x = 0$. Fixed now.
$endgroup$
– 6005
Jan 24 at 16:01
$begingroup$
@Song Very nice solution by the way :)
$endgroup$
– 6005
Jan 24 at 16:04
$begingroup$
Did you mean $x^4$, not $x^3$?
$endgroup$
– Prasiortle
Jan 24 at 15:45
$begingroup$
Did you mean $x^4$, not $x^3$?
$endgroup$
– Prasiortle
Jan 24 at 15:45
$begingroup$
A minor error: $limlimits_{kto-infty} x^{8^k}=1$.
$endgroup$
– Song
Jan 24 at 15:48
$begingroup$
A minor error: $limlimits_{kto-infty} x^{8^k}=1$.
$endgroup$
– Song
Jan 24 at 15:48
$begingroup$
@Prasiortle Thanks, fixed
$endgroup$
– 6005
Jan 24 at 15:55
$begingroup$
@Prasiortle Thanks, fixed
$endgroup$
– 6005
Jan 24 at 15:55
$begingroup$
@Song Thanks for the correction. So I needed to do the limit at $x = 1$, not $x = 0$. Fixed now.
$endgroup$
– 6005
Jan 24 at 16:01
$begingroup$
@Song Thanks for the correction. So I needed to do the limit at $x = 1$, not $x = 0$. Fixed now.
$endgroup$
– 6005
Jan 24 at 16:01
$begingroup$
@Song Very nice solution by the way :)
$endgroup$
– 6005
Jan 24 at 16:04
$begingroup$
@Song Very nice solution by the way :)
$endgroup$
– 6005
Jan 24 at 16:04
|
show 5 more comments
$begingroup$
If we are interested in functions smooth enough to admit some power series expansion (around $x=0$):
$$f(x) = sum_{k=0}^infty c_kx^k$$
Since $$f(x^n) = sum_{k=0}^infty c_kx^{kn}$$
It will force relations between the coefficients in a systematic way which you can then investigate.
answered Jan 24 at 15:33
mathreadlermathreadler
15.1k72263
$endgroup$
1
$begingroup$
Doing this we immediately derive $f(x) = x$ since all other coefficients must be $0$. But how does it help? Most functions cannot be given as a power series.
$endgroup$
– 6005
Jan 24 at 15:38
$begingroup$
@6005 : Yes the assumption here is if the function is smooth enough to admit a power series around $x = 0$.
$endgroup$
– mathreadler
Jan 24 at 15:42
add a comment |
$begingroup$
If we are interested in functions smooth enough to admit some power series expansion (around $x=0$):
$$f(x) = sum_{k=0}^infty c_kx^k$$
Since $$f(x^n) = sum_{k=0}^infty c_kx^{kn}$$
It will force relations between the coefficients in a systematic way which you can then investigate.
answered Jan 24 at 15:33
mathreadlermathreadler
15.1k72263
$endgroup$
1
$begingroup$
Doing this we immediately derive $f(x) = x$ since all other coefficients must be $0$. But how does it help? Most functions cannot be given as a power series.
$endgroup$
– 6005
Jan 24 at 15:38
$begingroup$
@6005 : Yes the assumption here is if the function is smooth enough to admit a power series around $x = 0$.
$endgroup$
– mathreadler
Jan 24 at 15:42
add a comment |
$begingroup$
If we are interested in functions smooth enough to admit some power series expansion (around $x=0$):
$$f(x) = sum_{k=0}^infty c_kx^k$$
Since $$f(x^n) = sum_{k=0}^infty c_kx^{kn}$$
It will force relations between the coefficients in a systematic way which you can then investigate.
answered Jan 24 at 15:33
mathreadlermathreadler
15.1k72263
$endgroup$
If we are interested in functions smooth enough to admit some power series expansion (around $x=0$):
$$f(x) = sum_{k=0}^infty c_kx^k$$
Since $$f(x^n) = sum_{k=0}^infty c_kx^{kn}$$
It will force relations between the coefficients in a systematic way which you can then investigate.
answered Jan 24 at 15:33
mathreadlermathreadler
15.1k72263
edited Jan 24 at 15:43
answered Jan 24 at 15:33
mathreadlermathreadler
15.1k72263
answered Jan 24 at 15:33
mathreadlermathreadler
15.1k72263
answered Jan 24 at 15:33
mathreadlermathreadler
15.1k72263
15.1k72263
1
$begingroup$
Doing this we immediately derive $f(x) = x$ since all other coefficients must be $0$. But how does it help? Most functions cannot be given as a power series.
$endgroup$
– 6005
Jan 24 at 15:38
$begingroup$
@6005 : Yes the assumption here is if the function is smooth enough to admit a power series around $x = 0$.
$endgroup$
– mathreadler
Jan 24 at 15:42
add a comment |
1
$begingroup$
Doing this we immediately derive $f(x) = x$ since all other coefficients must be $0$. But how does it help? Most functions cannot be given as a power series.
$endgroup$
– 6005
Jan 24 at 15:38
$begingroup$
@6005 : Yes the assumption here is if the function is smooth enough to admit a power series around $x = 0$.
$endgroup$
– mathreadler
Jan 24 at 15:42
1
1
$begingroup$
Doing this we immediately derive $f(x) = x$ since all other coefficients must be $0$. But how does it help? Most functions cannot be given as a power series.
$endgroup$
– 6005
Jan 24 at 15:38
$begingroup$
Doing this we immediately derive $f(x) = x$ since all other coefficients must be $0$. But how does it help? Most functions cannot be given as a power series.
$endgroup$
– 6005
Jan 24 at 15:38
$begingroup$
@6005 : Yes the assumption here is if the function is smooth enough to admit a power series around $x = 0$.
$endgroup$
– mathreadler
Jan 24 at 15:42
$begingroup$
@6005 : Yes the assumption here is if the function is smooth enough to admit a power series around $x = 0$.
$endgroup$
– mathreadler
Jan 24 at 15:42
add a comment |
$begingroup$
Assuming $f(x)$ as a real function, calling $g(x) = f(x)-x$ we have
$$
g(x^4)+g(x^2)+g(x) = 0
$$
now calling $e^y = x$ we have
$$
g(e^{4y})+g(e^{2y})+g(e^y) = 0
$$
or
$$
G(4y)+G(2y)+G(y) = 0
$$
This recurrence equation has as solution
$$
G(y) = C_1 y^{-frac{2 i pi }{3 ln 2}}+C_2 y^{frac{2 i pi }{3 ln 2}}
$$
because
$$
mathcal{G}(log_2 (4y))+mathcal{G}(log_2(2y))+mathcal{G}(log_2 y)=0\
mathcal{G}(log_2 y+2)+mathcal{G}(log_2 y +1)+mathcal{G}(log_2 y)=0
$$
and making $z = log_2 y$ we get
$$
mathcal{G}(z+2)+mathcal{G}(z+1)+mathcal{G}(z)=0
$$
which can be characterized as second order linear difference equation.
Then follows
$$
G(y) = C'_1 cos left(frac{2 pi}{3} log_2 yright)+C'_2sin left(frac{2 pi}{3} log_2 yright) = g(e^y) = f(e^y)-e^y
$$
and finally
$$
f(x) = C'_1cosleft(frac{2pi}{3}log_2(ln x)right)+C'_2sinleft(frac{2pi}{3}log_2(ln x)right)+x
$$
In case of $f(x)$ increasing then the solution is $f(x) = x$ with $C'_1=C'_2=0$
NOTE
Use this MATHEMATICA script to verify the recurrence equation as well as the final solution.
Clear[G]
G[y_] := Cos[(2 [Pi] Log[2, y])/3] c1 + Sin[(2 [Pi] Log[2, y])/3 ] c2
G[4 y] + G[2 y] + G[y] // FullSimplify
f[x_] := Cos[(2 [Pi] Log[2, Log[x]])/3] c1 + Sin[(2 [Pi] Log[2, Log[x]])/3 ] c2 + x
Assuming[x > 0, f[x^4] + f[x^2] + f[x] - x^4 - x^2 - x // FullSimplify]
answered Jan 24 at 15:26
CesareoCesareo
9,3963517
$endgroup$
$begingroup$
I don't understand your solution. $C_1$ and $C_2$ differ depending on which $y$ you pick.
$endgroup$
– 6005
Jan 24 at 15:37
$begingroup$
@6005 Did you check if $G(y)$ verify the recurrence equation $G(4y)+G(2y)+G(y)=0$ ?
$endgroup$
– Cesareo
Jan 24 at 15:44
$begingroup$
what do you mean? I agree it verifies that recurrence, but the $C_1$ and $C_2$ are different depending on which $y$ you pick. There is actually an infinite-dimensional family of solutions.
$endgroup$
– 6005
Jan 24 at 16:02
$begingroup$
@6005 A recursive equation is quite similar to a differential equation. This is a second order linear recursive equation so we have two free constant which can be determined depending on boundary conditions. So the solution is complete with two arbitrary constant.
$endgroup$
– Cesareo
Jan 24 at 16:07
1
$begingroup$
In general, we may pick for each $y in [1,2)$ a different pair of constants $C_1'$ and $C_2'$. That's infinitely many constants. Then, $G(2y)$, $G(4y)$, $G(8y)$ and so on are determined by the constants $C_1'$ and $C_2'$ for that particular $y in [1,2)$.
$endgroup$
– 6005
Jan 24 at 16:11
|
show 14 more comments
$begingroup$
Assuming $f(x)$ as a real function, calling $g(x) = f(x)-x$ we have
$$
g(x^4)+g(x^2)+g(x) = 0
$$
now calling $e^y = x$ we have
$$
g(e^{4y})+g(e^{2y})+g(e^y) = 0
$$
or
$$
G(4y)+G(2y)+G(y) = 0
$$
This recurrence equation has as solution
$$
G(y) = C_1 y^{-frac{2 i pi }{3 ln 2}}+C_2 y^{frac{2 i pi }{3 ln 2}}
$$
because
$$
mathcal{G}(log_2 (4y))+mathcal{G}(log_2(2y))+mathcal{G}(log_2 y)=0\
mathcal{G}(log_2 y+2)+mathcal{G}(log_2 y +1)+mathcal{G}(log_2 y)=0
$$
and making $z = log_2 y$ we get
$$
mathcal{G}(z+2)+mathcal{G}(z+1)+mathcal{G}(z)=0
$$
which can be characterized as second order linear difference equation.
Then follows
$$
G(y) = C'_1 cos left(frac{2 pi}{3} log_2 yright)+C'_2sin left(frac{2 pi}{3} log_2 yright) = g(e^y) = f(e^y)-e^y
$$
and finally
$$
f(x) = C'_1cosleft(frac{2pi}{3}log_2(ln x)right)+C'_2sinleft(frac{2pi}{3}log_2(ln x)right)+x
$$
In case of $f(x)$ increasing then the solution is $f(x) = x$ with $C'_1=C'_2=0$
NOTE
Use this MATHEMATICA script to verify the recurrence equation as well as the final solution.
Clear[G]
G[y_] := Cos[(2 [Pi] Log[2, y])/3] c1 + Sin[(2 [Pi] Log[2, y])/3 ] c2
G[4 y] + G[2 y] + G[y] // FullSimplify
f[x_] := Cos[(2 [Pi] Log[2, Log[x]])/3] c1 + Sin[(2 [Pi] Log[2, Log[x]])/3 ] c2 + x
Assuming[x > 0, f[x^4] + f[x^2] + f[x] - x^4 - x^2 - x // FullSimplify]
answered Jan 24 at 15:26
CesareoCesareo
9,3963517
$endgroup$
$begingroup$
I don't understand your solution. $C_1$ and $C_2$ differ depending on which $y$ you pick.
$endgroup$
– 6005
Jan 24 at 15:37
$begingroup$
@6005 Did you check if $G(y)$ verify the recurrence equation $G(4y)+G(2y)+G(y)=0$ ?
$endgroup$
– Cesareo
Jan 24 at 15:44
$begingroup$
what do you mean? I agree it verifies that recurrence, but the $C_1$ and $C_2$ are different depending on which $y$ you pick. There is actually an infinite-dimensional family of solutions.
$endgroup$
– 6005
Jan 24 at 16:02
$begingroup$
@6005 A recursive equation is quite similar to a differential equation. This is a second order linear recursive equation so we have two free constant which can be determined depending on boundary conditions. So the solution is complete with two arbitrary constant.
$endgroup$
– Cesareo
Jan 24 at 16:07
1
$begingroup$
In general, we may pick for each $y in [1,2)$ a different pair of constants $C_1'$ and $C_2'$. That's infinitely many constants. Then, $G(2y)$, $G(4y)$, $G(8y)$ and so on are determined by the constants $C_1'$ and $C_2'$ for that particular $y in [1,2)$.
$endgroup$
– 6005
Jan 24 at 16:11
|
show 14 more comments
$begingroup$
Assuming $f(x)$ as a real function, calling $g(x) = f(x)-x$ we have
$$
g(x^4)+g(x^2)+g(x) = 0
$$
now calling $e^y = x$ we have
$$
g(e^{4y})+g(e^{2y})+g(e^y) = 0
$$
or
$$
G(4y)+G(2y)+G(y) = 0
$$
This recurrence equation has as solution
$$
G(y) = C_1 y^{-frac{2 i pi }{3 ln 2}}+C_2 y^{frac{2 i pi }{3 ln 2}}
$$
because
$$
mathcal{G}(log_2 (4y))+mathcal{G}(log_2(2y))+mathcal{G}(log_2 y)=0\
mathcal{G}(log_2 y+2)+mathcal{G}(log_2 y +1)+mathcal{G}(log_2 y)=0
$$
and making $z = log_2 y$ we get
$$
mathcal{G}(z+2)+mathcal{G}(z+1)+mathcal{G}(z)=0
$$
which can be characterized as second order linear difference equation.
Then follows
$$
G(y) = C'_1 cos left(frac{2 pi}{3} log_2 yright)+C'_2sin left(frac{2 pi}{3} log_2 yright) = g(e^y) = f(e^y)-e^y
$$
and finally
$$
f(x) = C'_1cosleft(frac{2pi}{3}log_2(ln x)right)+C'_2sinleft(frac{2pi}{3}log_2(ln x)right)+x
$$
In case of $f(x)$ increasing then the solution is $f(x) = x$ with $C'_1=C'_2=0$
NOTE
Use this MATHEMATICA script to verify the recurrence equation as well as the final solution.
Clear[G]
G[y_] := Cos[(2 [Pi] Log[2, y])/3] c1 + Sin[(2 [Pi] Log[2, y])/3 ] c2
G[4 y] + G[2 y] + G[y] // FullSimplify
f[x_] := Cos[(2 [Pi] Log[2, Log[x]])/3] c1 + Sin[(2 [Pi] Log[2, Log[x]])/3 ] c2 + x
Assuming[x > 0, f[x^4] + f[x^2] + f[x] - x^4 - x^2 - x // FullSimplify]
answered Jan 24 at 15:26
CesareoCesareo
9,3963517
$endgroup$
Assuming $f(x)$ as a real function, calling $g(x) = f(x)-x$ we have
$$
g(x^4)+g(x^2)+g(x) = 0
$$
now calling $e^y = x$ we have
$$
g(e^{4y})+g(e^{2y})+g(e^y) = 0
$$
or
$$
G(4y)+G(2y)+G(y) = 0
$$
This recurrence equation has as solution
$$
G(y) = C_1 y^{-frac{2 i pi }{3 ln 2}}+C_2 y^{frac{2 i pi }{3 ln 2}}
$$
because
$$
mathcal{G}(log_2 (4y))+mathcal{G}(log_2(2y))+mathcal{G}(log_2 y)=0\
mathcal{G}(log_2 y+2)+mathcal{G}(log_2 y +1)+mathcal{G}(log_2 y)=0
$$
and making $z = log_2 y$ we get
$$
mathcal{G}(z+2)+mathcal{G}(z+1)+mathcal{G}(z)=0
$$
which can be characterized as second order linear difference equation.
Then follows
$$
G(y) = C'_1 cos left(frac{2 pi}{3} log_2 yright)+C'_2sin left(frac{2 pi}{3} log_2 yright) = g(e^y) = f(e^y)-e^y
$$
and finally
$$
f(x) = C'_1cosleft(frac{2pi}{3}log_2(ln x)right)+C'_2sinleft(frac{2pi}{3}log_2(ln x)right)+x
$$
In case of $f(x)$ increasing then the solution is $f(x) = x$ with $C'_1=C'_2=0$
NOTE
Use this MATHEMATICA script to verify the recurrence equation as well as the final solution.
Clear[G]
G[y_] := Cos[(2 [Pi] Log[2, y])/3] c1 + Sin[(2 [Pi] Log[2, y])/3 ] c2
G[4 y] + G[2 y] + G[y] // FullSimplify
f[x_] := Cos[(2 [Pi] Log[2, Log[x]])/3] c1 + Sin[(2 [Pi] Log[2, Log[x]])/3 ] c2 + x
Assuming[x > 0, f[x^4] + f[x^2] + f[x] - x^4 - x^2 - x // FullSimplify]
answered Jan 24 at 15:26
CesareoCesareo
9,3963517
edited Jan 28 at 8:46
answered Jan 24 at 15:26
CesareoCesareo
9,3963517
answered Jan 24 at 15:26
CesareoCesareo
9,3963517
answered Jan 24 at 15:26
CesareoCesareo
9,3963517
9,3963517
$begingroup$
I don't understand your solution. $C_1$ and $C_2$ differ depending on which $y$ you pick.
$endgroup$
– 6005
Jan 24 at 15:37
$begingroup$
@6005 Did you check if $G(y)$ verify the recurrence equation $G(4y)+G(2y)+G(y)=0$ ?
$endgroup$
– Cesareo
Jan 24 at 15:44
$begingroup$
what do you mean? I agree it verifies that recurrence, but the $C_1$ and $C_2$ are different depending on which $y$ you pick. There is actually an infinite-dimensional family of solutions.
$endgroup$
– 6005
Jan 24 at 16:02
$begingroup$
@6005 A recursive equation is quite similar to a differential equation. This is a second order linear recursive equation so we have two free constant which can be determined depending on boundary conditions. So the solution is complete with two arbitrary constant.
$endgroup$
– Cesareo
Jan 24 at 16:07
1
$begingroup$
In general, we may pick for each $y in [1,2)$ a different pair of constants $C_1'$ and $C_2'$. That's infinitely many constants. Then, $G(2y)$, $G(4y)$, $G(8y)$ and so on are determined by the constants $C_1'$ and $C_2'$ for that particular $y in [1,2)$.
$endgroup$
– 6005
Jan 24 at 16:11
|
show 14 more comments
$begingroup$
I don't understand your solution. $C_1$ and $C_2$ differ depending on which $y$ you pick.
$endgroup$
– 6005
Jan 24 at 15:37
$begingroup$
@6005 Did you check if $G(y)$ verify the recurrence equation $G(4y)+G(2y)+G(y)=0$ ?
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– Cesareo
Jan 24 at 15:44
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what do you mean? I agree it verifies that recurrence, but the $C_1$ and $C_2$ are different depending on which $y$ you pick. There is actually an infinite-dimensional family of solutions.
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– 6005
Jan 24 at 16:02
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@6005 A recursive equation is quite similar to a differential equation. This is a second order linear recursive equation so we have two free constant which can be determined depending on boundary conditions. So the solution is complete with two arbitrary constant.
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– Cesareo
Jan 24 at 16:07
1
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In general, we may pick for each $y in [1,2)$ a different pair of constants $C_1'$ and $C_2'$. That's infinitely many constants. Then, $G(2y)$, $G(4y)$, $G(8y)$ and so on are determined by the constants $C_1'$ and $C_2'$ for that particular $y in [1,2)$.
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– 6005
Jan 24 at 16:11
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I don't understand your solution. $C_1$ and $C_2$ differ depending on which $y$ you pick.
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– 6005
Jan 24 at 15:37
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I don't understand your solution. $C_1$ and $C_2$ differ depending on which $y$ you pick.
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– 6005
Jan 24 at 15:37
$begingroup$
@6005 Did you check if $G(y)$ verify the recurrence equation $G(4y)+G(2y)+G(y)=0$ ?
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– Cesareo
Jan 24 at 15:44
$begingroup$
@6005 Did you check if $G(y)$ verify the recurrence equation $G(4y)+G(2y)+G(y)=0$ ?
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– Cesareo
Jan 24 at 15:44
$begingroup$
what do you mean? I agree it verifies that recurrence, but the $C_1$ and $C_2$ are different depending on which $y$ you pick. There is actually an infinite-dimensional family of solutions.
$endgroup$
– 6005
Jan 24 at 16:02
$begingroup$
what do you mean? I agree it verifies that recurrence, but the $C_1$ and $C_2$ are different depending on which $y$ you pick. There is actually an infinite-dimensional family of solutions.
$endgroup$
– 6005
Jan 24 at 16:02
$begingroup$
@6005 A recursive equation is quite similar to a differential equation. This is a second order linear recursive equation so we have two free constant which can be determined depending on boundary conditions. So the solution is complete with two arbitrary constant.
$endgroup$
– Cesareo
Jan 24 at 16:07
$begingroup$
@6005 A recursive equation is quite similar to a differential equation. This is a second order linear recursive equation so we have two free constant which can be determined depending on boundary conditions. So the solution is complete with two arbitrary constant.
$endgroup$
– Cesareo
Jan 24 at 16:07
1
1
$begingroup$
In general, we may pick for each $y in [1,2)$ a different pair of constants $C_1'$ and $C_2'$. That's infinitely many constants. Then, $G(2y)$, $G(4y)$, $G(8y)$ and so on are determined by the constants $C_1'$ and $C_2'$ for that particular $y in [1,2)$.
$endgroup$
– 6005
Jan 24 at 16:11
$begingroup$
In general, we may pick for each $y in [1,2)$ a different pair of constants $C_1'$ and $C_2'$. That's infinitely many constants. Then, $G(2y)$, $G(4y)$, $G(8y)$ and so on are determined by the constants $C_1'$ and $C_2'$ for that particular $y in [1,2)$.
$endgroup$
– 6005
Jan 24 at 16:11
|
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Well, it's clear that $f(x)=x$ satisfies the equation, but I'm not sure that's the only one.
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– Adrian Keister
Jan 24 at 14:03
1
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Just a thought: If you set $g(x):=f(x)-x$ then the equation above becomes $g(x^4)+g(x^2)+g(x)=0$. However, now it's harder to take into account the fact that $f$ is increasing.
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– Yanko
Jan 24 at 14:24
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@Cesareo, Is it increasing or $(0,infty)$-valued over $(0,infty)$ for any non-trivial choices of $C_1'$ and $C_2'$?
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– Sangchul Lee
Jan 24 at 17:40
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@SangchulLee You got it!. I simply assumed a real function. I didn't see the increasing. My fault.
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– Cesareo
Jan 24 at 19:01