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Equation of function s.t., $f(x^2)+2f(x)=0$
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Are there function $f:mathbbR to mathbbR$ s.t., $fnot =0$ and $f(x^2)+2f(x)=0$?
I tried polynomial. But it is impossible because of its degree. I tried logarithm, but I couldn't.
real-analysis functional-equations
asked Jul 25 at 19:08
B.T.O
295111
 |Â
show 6 more comments
up vote
3
down vote
favorite
Are there function $f:mathbbR to mathbbR$ s.t., $fnot =0$ and $f(x^2)+2f(x)=0$?
I tried polynomial. But it is impossible because of its degree. I tried logarithm, but I couldn't.
real-analysis functional-equations
asked Jul 25 at 19:08
B.T.O
295111
2
There is always the good old $y=0$
– Sorfosh
Jul 25 at 19:11
1
ok nvm you edited the question
– Sorfosh
Jul 25 at 19:11
1
Are you assuming continuity? If so, I think you can quickly show that $f$ must be identically $0$.
– lulu
Jul 25 at 19:14
@lulu care making that an answer?
– Kenny Lau
Jul 25 at 19:15
@KennyLau No...the problem doesn't specify continuity.
– lulu
Jul 25 at 19:15
 |Â
show 6 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Are there function $f:mathbbR to mathbbR$ s.t., $fnot =0$ and $f(x^2)+2f(x)=0$?
I tried polynomial. But it is impossible because of its degree. I tried logarithm, but I couldn't.
real-analysis functional-equations
asked Jul 25 at 19:08
B.T.O
295111
Are there function $f:mathbbR to mathbbR$ s.t., $fnot =0$ and $f(x^2)+2f(x)=0$?
I tried polynomial. But it is impossible because of its degree. I tried logarithm, but I couldn't.
real-analysis functional-equations
asked Jul 25 at 19:08
B.T.O
295111
edited Jul 25 at 19:11
asked Jul 25 at 19:08
B.T.O
295111
asked Jul 25 at 19:08
B.T.O
295111
asked Jul 25 at 19:08
B.T.O
295111
295111
2
There is always the good old $y=0$
– Sorfosh
Jul 25 at 19:11
1
ok nvm you edited the question
– Sorfosh
Jul 25 at 19:11
1
Are you assuming continuity? If so, I think you can quickly show that $f$ must be identically $0$.
– lulu
Jul 25 at 19:14
@lulu care making that an answer?
– Kenny Lau
Jul 25 at 19:15
@KennyLau No...the problem doesn't specify continuity.
– lulu
Jul 25 at 19:15
 |Â
show 6 more comments
2
There is always the good old $y=0$
– Sorfosh
Jul 25 at 19:11
1
ok nvm you edited the question
– Sorfosh
Jul 25 at 19:11
1
Are you assuming continuity? If so, I think you can quickly show that $f$ must be identically $0$.
– lulu
Jul 25 at 19:14
@lulu care making that an answer?
– Kenny Lau
Jul 25 at 19:15
@KennyLau No...the problem doesn't specify continuity.
– lulu
Jul 25 at 19:15
2
2
There is always the good old $y=0$
– Sorfosh
Jul 25 at 19:11
There is always the good old $y=0$
– Sorfosh
Jul 25 at 19:11
1
1
ok nvm you edited the question
– Sorfosh
Jul 25 at 19:11
ok nvm you edited the question
– Sorfosh
Jul 25 at 19:11
1
1
Are you assuming continuity? If so, I think you can quickly show that $f$ must be identically $0$.
– lulu
Jul 25 at 19:14
Are you assuming continuity? If so, I think you can quickly show that $f$ must be identically $0$.
– lulu
Jul 25 at 19:14
@lulu care making that an answer?
– Kenny Lau
Jul 25 at 19:15
@lulu care making that an answer?
– Kenny Lau
Jul 25 at 19:15
@KennyLau No...the problem doesn't specify continuity.
– lulu
Jul 25 at 19:15
@KennyLau No...the problem doesn't specify continuity.
– lulu
Jul 25 at 19:15
 |Â
show 6 more comments
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
First, $f(0)=0$ after we plug in $0$ for $x$. Then $f(x)=f(-x)$, so it is enough to find a function $f:J=(0,infty)toBbb R$ that satisfies the given equation. Any $x$ in the interval $J$ can be written uniquely as $x=e^t$, $tinBbb R$. So it is enough to search for a function $g:Bbb RtoBbb R$ so that $f(e^t)=g(t)$. The functional equation for $f$ transposes now in an equation for $g$, which is
$g(2t)=-2g(t)$.
Of course, we look at the action of $Bbb Z$ on $Bbb R$ given by letting $1$ act as $tto 2t$. Then $0$ is fixed, and any other element in $R$ can be brought by this action to an element in either $[1,2)$, or $[-2,-1)$.
We can immediately forget about this action, but this is the reason for the following.
We have immediately $g(0)=0$ as a condition.
Fix now $g$ arbitrarily on $[1,2)sqcup [-2,1)$, and extend (uniquely) it so that the above $g(2y)=-2g(y)$ is satisfied.
(So there are infinitely many solutions.)
answered Jul 25 at 19:25
dan_fulea
4,1421211
@KennyLau An action cannot be wrong. And maybe my action applies to a different function. And your answer has no action. And i inserted this only to give the idea. And your answer was edited too many times in the last seconds....
– dan_fulea
Jul 25 at 19:39
add a comment |Â
up vote
1
down vote
Observations:
$f(0) = f(1) = 0$: substitute $x=0$ and $x=1$ respectively.
The function must be even: $f(x) = -frac12f(x^2) = -frac12f((-x)^2) = f(-x)$, so we focus on $x ge 0$.
Assume $x>0$, $x ne 1$. If we know $f(x)$, then we know $f(y)$ for all $y = x^2^n$ where $n in Bbb Z$. Under the right group action, $ x^2^n mid n in Bbb Z $ can be interpreted as the orbit of $x$. Specifically, $f(x^2^n) = (-2)^n f(x)$. Moreover, these values are the only values that $f(x)$ can control.
Therefore, we partition $(0,1) cup (1, infty)$ by using $ x^2^n mid n in Bbb Z $, which gives us (under ZFC) continuumly many equivalence classes. Each class can have continummly many possible values, and we conclude that there are continuum-powerset ($2^mathfrak c$) many possible such functions.
With more work, one can actually prove that there is a unique representative of $ x^2^n mid n in Bbb Z $ in $[2,4)$, and then it follows that such functions are in bijection to functions $f_0 : [2,4) to Bbb R$, i.e. any such function $f_0 : [2,4) to Bbb R$ can be extended to a unique function $f : Bbb R to Bbb R$ satisfying your equation, and (more trivially) any function $f : Bbb R to Bbb R$ can be restricted to a function $f_0 : [2,4) to Bbb R$.
$[2,4)$ can be generalized to any $[y, y^2)$ with $y > 1$ or $(y^2, y]$ with $0 < y < 1$.
answered Jul 25 at 19:32
Kenny Lau
18.5k2157
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
First, $f(0)=0$ after we plug in $0$ for $x$. Then $f(x)=f(-x)$, so it is enough to find a function $f:J=(0,infty)toBbb R$ that satisfies the given equation. Any $x$ in the interval $J$ can be written uniquely as $x=e^t$, $tinBbb R$. So it is enough to search for a function $g:Bbb RtoBbb R$ so that $f(e^t)=g(t)$. The functional equation for $f$ transposes now in an equation for $g$, which is
$g(2t)=-2g(t)$.
Of course, we look at the action of $Bbb Z$ on $Bbb R$ given by letting $1$ act as $tto 2t$. Then $0$ is fixed, and any other element in $R$ can be brought by this action to an element in either $[1,2)$, or $[-2,-1)$.
We can immediately forget about this action, but this is the reason for the following.
We have immediately $g(0)=0$ as a condition.
Fix now $g$ arbitrarily on $[1,2)sqcup [-2,1)$, and extend (uniquely) it so that the above $g(2y)=-2g(y)$ is satisfied.
(So there are infinitely many solutions.)
answered Jul 25 at 19:25
dan_fulea
4,1421211
@KennyLau An action cannot be wrong. And maybe my action applies to a different function. And your answer has no action. And i inserted this only to give the idea. And your answer was edited too many times in the last seconds....
– dan_fulea
Jul 25 at 19:39
add a comment |Â
up vote
2
down vote
accepted
First, $f(0)=0$ after we plug in $0$ for $x$. Then $f(x)=f(-x)$, so it is enough to find a function $f:J=(0,infty)toBbb R$ that satisfies the given equation. Any $x$ in the interval $J$ can be written uniquely as $x=e^t$, $tinBbb R$. So it is enough to search for a function $g:Bbb RtoBbb R$ so that $f(e^t)=g(t)$. The functional equation for $f$ transposes now in an equation for $g$, which is
$g(2t)=-2g(t)$.
Of course, we look at the action of $Bbb Z$ on $Bbb R$ given by letting $1$ act as $tto 2t$. Then $0$ is fixed, and any other element in $R$ can be brought by this action to an element in either $[1,2)$, or $[-2,-1)$.
We can immediately forget about this action, but this is the reason for the following.
We have immediately $g(0)=0$ as a condition.
Fix now $g$ arbitrarily on $[1,2)sqcup [-2,1)$, and extend (uniquely) it so that the above $g(2y)=-2g(y)$ is satisfied.
(So there are infinitely many solutions.)
answered Jul 25 at 19:25
dan_fulea
4,1421211
@KennyLau An action cannot be wrong. And maybe my action applies to a different function. And your answer has no action. And i inserted this only to give the idea. And your answer was edited too many times in the last seconds....
– dan_fulea
Jul 25 at 19:39
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
First, $f(0)=0$ after we plug in $0$ for $x$. Then $f(x)=f(-x)$, so it is enough to find a function $f:J=(0,infty)toBbb R$ that satisfies the given equation. Any $x$ in the interval $J$ can be written uniquely as $x=e^t$, $tinBbb R$. So it is enough to search for a function $g:Bbb RtoBbb R$ so that $f(e^t)=g(t)$. The functional equation for $f$ transposes now in an equation for $g$, which is
$g(2t)=-2g(t)$.
Of course, we look at the action of $Bbb Z$ on $Bbb R$ given by letting $1$ act as $tto 2t$. Then $0$ is fixed, and any other element in $R$ can be brought by this action to an element in either $[1,2)$, or $[-2,-1)$.
We can immediately forget about this action, but this is the reason for the following.
We have immediately $g(0)=0$ as a condition.
Fix now $g$ arbitrarily on $[1,2)sqcup [-2,1)$, and extend (uniquely) it so that the above $g(2y)=-2g(y)$ is satisfied.
(So there are infinitely many solutions.)
answered Jul 25 at 19:25
dan_fulea
4,1421211
First, $f(0)=0$ after we plug in $0$ for $x$. Then $f(x)=f(-x)$, so it is enough to find a function $f:J=(0,infty)toBbb R$ that satisfies the given equation. Any $x$ in the interval $J$ can be written uniquely as $x=e^t$, $tinBbb R$. So it is enough to search for a function $g:Bbb RtoBbb R$ so that $f(e^t)=g(t)$. The functional equation for $f$ transposes now in an equation for $g$, which is
$g(2t)=-2g(t)$.
Of course, we look at the action of $Bbb Z$ on $Bbb R$ given by letting $1$ act as $tto 2t$. Then $0$ is fixed, and any other element in $R$ can be brought by this action to an element in either $[1,2)$, or $[-2,-1)$.
We can immediately forget about this action, but this is the reason for the following.
We have immediately $g(0)=0$ as a condition.
Fix now $g$ arbitrarily on $[1,2)sqcup [-2,1)$, and extend (uniquely) it so that the above $g(2y)=-2g(y)$ is satisfied.
(So there are infinitely many solutions.)
answered Jul 25 at 19:25
dan_fulea
4,1421211
answered Jul 25 at 19:25
dan_fulea
4,1421211
answered Jul 25 at 19:25
dan_fulea
4,1421211
answered Jul 25 at 19:25
dan_fulea
4,1421211
4,1421211
@KennyLau An action cannot be wrong. And maybe my action applies to a different function. And your answer has no action. And i inserted this only to give the idea. And your answer was edited too many times in the last seconds....
– dan_fulea
Jul 25 at 19:39
add a comment |Â
@KennyLau An action cannot be wrong. And maybe my action applies to a different function. And your answer has no action. And i inserted this only to give the idea. And your answer was edited too many times in the last seconds....
– dan_fulea
Jul 25 at 19:39
@KennyLau An action cannot be wrong. And maybe my action applies to a different function. And your answer has no action. And i inserted this only to give the idea. And your answer was edited too many times in the last seconds....
– dan_fulea
Jul 25 at 19:39
@KennyLau An action cannot be wrong. And maybe my action applies to a different function. And your answer has no action. And i inserted this only to give the idea. And your answer was edited too many times in the last seconds....
– dan_fulea
Jul 25 at 19:39
add a comment |Â
up vote
1
down vote
Observations:
$f(0) = f(1) = 0$: substitute $x=0$ and $x=1$ respectively.
The function must be even: $f(x) = -frac12f(x^2) = -frac12f((-x)^2) = f(-x)$, so we focus on $x ge 0$.
Assume $x>0$, $x ne 1$. If we know $f(x)$, then we know $f(y)$ for all $y = x^2^n$ where $n in Bbb Z$. Under the right group action, $ x^2^n mid n in Bbb Z $ can be interpreted as the orbit of $x$. Specifically, $f(x^2^n) = (-2)^n f(x)$. Moreover, these values are the only values that $f(x)$ can control.
Therefore, we partition $(0,1) cup (1, infty)$ by using $ x^2^n mid n in Bbb Z $, which gives us (under ZFC) continuumly many equivalence classes. Each class can have continummly many possible values, and we conclude that there are continuum-powerset ($2^mathfrak c$) many possible such functions.
With more work, one can actually prove that there is a unique representative of $ x^2^n mid n in Bbb Z $ in $[2,4)$, and then it follows that such functions are in bijection to functions $f_0 : [2,4) to Bbb R$, i.e. any such function $f_0 : [2,4) to Bbb R$ can be extended to a unique function $f : Bbb R to Bbb R$ satisfying your equation, and (more trivially) any function $f : Bbb R to Bbb R$ can be restricted to a function $f_0 : [2,4) to Bbb R$.
$[2,4)$ can be generalized to any $[y, y^2)$ with $y > 1$ or $(y^2, y]$ with $0 < y < 1$.
answered Jul 25 at 19:32
Kenny Lau
18.5k2157
add a comment |Â
up vote
1
down vote
Observations:
$f(0) = f(1) = 0$: substitute $x=0$ and $x=1$ respectively.
The function must be even: $f(x) = -frac12f(x^2) = -frac12f((-x)^2) = f(-x)$, so we focus on $x ge 0$.
Assume $x>0$, $x ne 1$. If we know $f(x)$, then we know $f(y)$ for all $y = x^2^n$ where $n in Bbb Z$. Under the right group action, $ x^2^n mid n in Bbb Z $ can be interpreted as the orbit of $x$. Specifically, $f(x^2^n) = (-2)^n f(x)$. Moreover, these values are the only values that $f(x)$ can control.
Therefore, we partition $(0,1) cup (1, infty)$ by using $ x^2^n mid n in Bbb Z $, which gives us (under ZFC) continuumly many equivalence classes. Each class can have continummly many possible values, and we conclude that there are continuum-powerset ($2^mathfrak c$) many possible such functions.
With more work, one can actually prove that there is a unique representative of $ x^2^n mid n in Bbb Z $ in $[2,4)$, and then it follows that such functions are in bijection to functions $f_0 : [2,4) to Bbb R$, i.e. any such function $f_0 : [2,4) to Bbb R$ can be extended to a unique function $f : Bbb R to Bbb R$ satisfying your equation, and (more trivially) any function $f : Bbb R to Bbb R$ can be restricted to a function $f_0 : [2,4) to Bbb R$.
$[2,4)$ can be generalized to any $[y, y^2)$ with $y > 1$ or $(y^2, y]$ with $0 < y < 1$.
answered Jul 25 at 19:32
Kenny Lau
18.5k2157
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Observations:
$f(0) = f(1) = 0$: substitute $x=0$ and $x=1$ respectively.
The function must be even: $f(x) = -frac12f(x^2) = -frac12f((-x)^2) = f(-x)$, so we focus on $x ge 0$.
Assume $x>0$, $x ne 1$. If we know $f(x)$, then we know $f(y)$ for all $y = x^2^n$ where $n in Bbb Z$. Under the right group action, $ x^2^n mid n in Bbb Z $ can be interpreted as the orbit of $x$. Specifically, $f(x^2^n) = (-2)^n f(x)$. Moreover, these values are the only values that $f(x)$ can control.
Therefore, we partition $(0,1) cup (1, infty)$ by using $ x^2^n mid n in Bbb Z $, which gives us (under ZFC) continuumly many equivalence classes. Each class can have continummly many possible values, and we conclude that there are continuum-powerset ($2^mathfrak c$) many possible such functions.
With more work, one can actually prove that there is a unique representative of $ x^2^n mid n in Bbb Z $ in $[2,4)$, and then it follows that such functions are in bijection to functions $f_0 : [2,4) to Bbb R$, i.e. any such function $f_0 : [2,4) to Bbb R$ can be extended to a unique function $f : Bbb R to Bbb R$ satisfying your equation, and (more trivially) any function $f : Bbb R to Bbb R$ can be restricted to a function $f_0 : [2,4) to Bbb R$.
$[2,4)$ can be generalized to any $[y, y^2)$ with $y > 1$ or $(y^2, y]$ with $0 < y < 1$.
answered Jul 25 at 19:32
Kenny Lau
18.5k2157
Observations:
$f(0) = f(1) = 0$: substitute $x=0$ and $x=1$ respectively.
The function must be even: $f(x) = -frac12f(x^2) = -frac12f((-x)^2) = f(-x)$, so we focus on $x ge 0$.
Assume $x>0$, $x ne 1$. If we know $f(x)$, then we know $f(y)$ for all $y = x^2^n$ where $n in Bbb Z$. Under the right group action, $ x^2^n mid n in Bbb Z $ can be interpreted as the orbit of $x$. Specifically, $f(x^2^n) = (-2)^n f(x)$. Moreover, these values are the only values that $f(x)$ can control.
Therefore, we partition $(0,1) cup (1, infty)$ by using $ x^2^n mid n in Bbb Z $, which gives us (under ZFC) continuumly many equivalence classes. Each class can have continummly many possible values, and we conclude that there are continuum-powerset ($2^mathfrak c$) many possible such functions.
With more work, one can actually prove that there is a unique representative of $ x^2^n mid n in Bbb Z $ in $[2,4)$, and then it follows that such functions are in bijection to functions $f_0 : [2,4) to Bbb R$, i.e. any such function $f_0 : [2,4) to Bbb R$ can be extended to a unique function $f : Bbb R to Bbb R$ satisfying your equation, and (more trivially) any function $f : Bbb R to Bbb R$ can be restricted to a function $f_0 : [2,4) to Bbb R$.
$[2,4)$ can be generalized to any $[y, y^2)$ with $y > 1$ or $(y^2, y]$ with $0 < y < 1$.
answered Jul 25 at 19:32
Kenny Lau
18.5k2157
answered Jul 25 at 19:32
Kenny Lau
18.5k2157
answered Jul 25 at 19:32
Kenny Lau
18.5k2157
answered Jul 25 at 19:32
Kenny Lau
18.5k2157
18.5k2157
add a comment |Â
add a comment |Â
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2
There is always the good old $y=0$
– Sorfosh
Jul 25 at 19:11
1
ok nvm you edited the question
– Sorfosh
Jul 25 at 19:11
1
Are you assuming continuity? If so, I think you can quickly show that $f$ must be identically $0$.
– lulu
Jul 25 at 19:14
@lulu care making that an answer?
– Kenny Lau
Jul 25 at 19:15
@KennyLau No...the problem doesn't specify continuity.
– lulu
Jul 25 at 19:15