A functional equation defined on the real axis.Easy functional equationFunctional equation: $R(1/x)/x^2 = R(x) $Solving functional equation 1Functional Equation and totally multiplicative functionsFunctional equation extended solutionFunctional equation in $a,x,y$Mean Value theorem functional equationFunctional equation : $f(xf(y))=yf(x)$quadratic functional equation?Functional equation on $mathbbR^+$
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A functional equation defined on the real axis.
Easy functional equationFunctional equation: $R(1/x)/x^2 = R(x) $Solving functional equation 1Functional Equation and totally multiplicative functionsFunctional equation extended solutionFunctional equation in $a,x,y$Mean Value theorem functional equationFunctional equation : $f(xf(y))=yf(x)$quadratic functional equation?Functional equation on $mathbbR^+$
$begingroup$
$$
f: mathbbR to mathbbRqquad fracf(x+y)x+y = fracf(x)-f(y)x-y, qquad forall x,yin mathbbR, left|xright| neq left|yright|
$$
Can I prove anything interesting about this function? I need to find it.
functions functional-equations
asked Mar 6 at 8:20
furfurfurfur
509
$endgroup$
This question has an open bounty worth +50
reputation from furfur ending ending at 2019-03-15 15:41:32Z">in 4 days.
This question has not received enough attention.
add a comment |
$begingroup$
$$
f: mathbbR to mathbbRqquad fracf(x+y)x+y = fracf(x)-f(y)x-y, qquad forall x,yin mathbbR, left|xright| neq left|yright|
$$
Can I prove anything interesting about this function? I need to find it.
functions functional-equations
asked Mar 6 at 8:20
furfurfurfur
509
$endgroup$
This question has an open bounty worth +50
reputation from furfur ending ending at 2019-03-15 15:41:32Z">in 4 days.
This question has not received enough attention.
4
$begingroup$
$f(x)=ax^2$ is also a solution.
$endgroup$
– Kavi Rama Murthy
Mar 6 at 8:26
2
$begingroup$
And thus so is $f(x) = ax^2+bx$. Indeed, any linear combination of solutions is again a solution.
$endgroup$
– Greg Martin
Mar 6 at 8:35
add a comment |
$begingroup$
$$
f: mathbbR to mathbbRqquad fracf(x+y)x+y = fracf(x)-f(y)x-y, qquad forall x,yin mathbbR, left|xright| neq left|yright|
$$
Can I prove anything interesting about this function? I need to find it.
functions functional-equations
asked Mar 6 at 8:20
furfurfurfur
509
$endgroup$
$$
f: mathbbR to mathbbRqquad fracf(x+y)x+y = fracf(x)-f(y)x-y, qquad forall x,yin mathbbR, left|xright| neq left|yright|
$$
Can I prove anything interesting about this function? I need to find it.
functions functional-equations
functions functional-equations
asked Mar 6 at 8:20
furfurfurfur
509
asked Mar 6 at 8:20
furfurfurfur
509
edited Mar 8 at 15:40
furfur
asked Mar 6 at 8:20
furfurfurfur
509
asked Mar 6 at 8:20
furfurfurfur
509
asked Mar 6 at 8:20
furfurfurfur
509
509
This question has an open bounty worth +50
reputation from furfur ending ending at 2019-03-15 15:41:32Z">in 4 days.
This question has not received enough attention.
This question has an open bounty worth +50
reputation from furfur ending ending at 2019-03-15 15:41:32Z">in 4 days.
This question has not received enough attention.
4
$begingroup$
$f(x)=ax^2$ is also a solution.
$endgroup$
– Kavi Rama Murthy
Mar 6 at 8:26
2
$begingroup$
And thus so is $f(x) = ax^2+bx$. Indeed, any linear combination of solutions is again a solution.
$endgroup$
– Greg Martin
Mar 6 at 8:35
add a comment |
4
$begingroup$
$f(x)=ax^2$ is also a solution.
$endgroup$
– Kavi Rama Murthy
Mar 6 at 8:26
2
$begingroup$
And thus so is $f(x) = ax^2+bx$. Indeed, any linear combination of solutions is again a solution.
$endgroup$
– Greg Martin
Mar 6 at 8:35
4
4
$begingroup$
$f(x)=ax^2$ is also a solution.
$endgroup$
– Kavi Rama Murthy
Mar 6 at 8:26
$begingroup$
$f(x)=ax^2$ is also a solution.
$endgroup$
– Kavi Rama Murthy
Mar 6 at 8:26
2
2
$begingroup$
And thus so is $f(x) = ax^2+bx$. Indeed, any linear combination of solutions is again a solution.
$endgroup$
– Greg Martin
Mar 6 at 8:35
$begingroup$
And thus so is $f(x) = ax^2+bx$. Indeed, any linear combination of solutions is again a solution.
$endgroup$
– Greg Martin
Mar 6 at 8:35
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
See Aczél or Schwaiger, where there is a misprint: $h(x_1,ldots,x_n)$ should read as $$h(x_1+ldots+x_n).$$
In the first paper it is shown that $f$ is a solution of the functions equation in question iff $f(x)=ax^2+bx$ for all $x$ with arbitrary constants $a,b$. The second reference gives certain generalisations.
Edit The results mentioned concerns different equations.
The somehow simpler one $fracf(x)-f(y)x-y=g(x+y)$ for $xnot=y$ and certain generalisations.
answered Mar 6 at 11:26
Jens SchwaigerJens Schwaiger
1,631138
$endgroup$
$begingroup$
As it stands this is a link-only answer, and the second article seems to be behind a paywall. Also, both seem to cover a broader class of functions. Including at least a summary of the relevant results seems appropriate.
$endgroup$
– Servaes
Mar 6 at 12:05
$begingroup$
This doesn’t solve my problem though.. in that link answer he makes x=-y to prove function h is odd. I can’t do x=-y in my problem, as the denominator cannot be 0.
$endgroup$
– furfur
Mar 6 at 13:18
$begingroup$
First you should rewrite the equation by multiplying both sides with $(x+y)(x-y)$. The resulting equation holds true also if $x=y$ or $x=-y$.
$endgroup$
– Jens Schwaiger
Mar 6 at 17:35
$begingroup$
Yes. That’s true. But in my case the function “h” (the one in the solution by Aczel) is equal to f(x+y)/(x+y) which is not defined for x=-y. So I can’t have an h(0). This is means h is defined on R*. So I can’t prove h is odd.
$endgroup$
– furfur
Mar 6 at 18:20
2
$begingroup$
@JensSchwaiger you should delete this "answer"
$endgroup$
– mathworker21
2 days ago
|
show 1 more comment
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
See Aczél or Schwaiger, where there is a misprint: $h(x_1,ldots,x_n)$ should read as $$h(x_1+ldots+x_n).$$
In the first paper it is shown that $f$ is a solution of the functions equation in question iff $f(x)=ax^2+bx$ for all $x$ with arbitrary constants $a,b$. The second reference gives certain generalisations.
Edit The results mentioned concerns different equations.
The somehow simpler one $fracf(x)-f(y)x-y=g(x+y)$ for $xnot=y$ and certain generalisations.
answered Mar 6 at 11:26
Jens SchwaigerJens Schwaiger
1,631138
$endgroup$
$begingroup$
As it stands this is a link-only answer, and the second article seems to be behind a paywall. Also, both seem to cover a broader class of functions. Including at least a summary of the relevant results seems appropriate.
$endgroup$
– Servaes
Mar 6 at 12:05
$begingroup$
This doesn’t solve my problem though.. in that link answer he makes x=-y to prove function h is odd. I can’t do x=-y in my problem, as the denominator cannot be 0.
$endgroup$
– furfur
Mar 6 at 13:18
$begingroup$
First you should rewrite the equation by multiplying both sides with $(x+y)(x-y)$. The resulting equation holds true also if $x=y$ or $x=-y$.
$endgroup$
– Jens Schwaiger
Mar 6 at 17:35
$begingroup$
Yes. That’s true. But in my case the function “h” (the one in the solution by Aczel) is equal to f(x+y)/(x+y) which is not defined for x=-y. So I can’t have an h(0). This is means h is defined on R*. So I can’t prove h is odd.
$endgroup$
– furfur
Mar 6 at 18:20
2
$begingroup$
@JensSchwaiger you should delete this "answer"
$endgroup$
– mathworker21
2 days ago
|
show 1 more comment
$begingroup$
See Aczél or Schwaiger, where there is a misprint: $h(x_1,ldots,x_n)$ should read as $$h(x_1+ldots+x_n).$$
In the first paper it is shown that $f$ is a solution of the functions equation in question iff $f(x)=ax^2+bx$ for all $x$ with arbitrary constants $a,b$. The second reference gives certain generalisations.
Edit The results mentioned concerns different equations.
The somehow simpler one $fracf(x)-f(y)x-y=g(x+y)$ for $xnot=y$ and certain generalisations.
answered Mar 6 at 11:26
Jens SchwaigerJens Schwaiger
1,631138
$endgroup$
$begingroup$
As it stands this is a link-only answer, and the second article seems to be behind a paywall. Also, both seem to cover a broader class of functions. Including at least a summary of the relevant results seems appropriate.
$endgroup$
– Servaes
Mar 6 at 12:05
$begingroup$
This doesn’t solve my problem though.. in that link answer he makes x=-y to prove function h is odd. I can’t do x=-y in my problem, as the denominator cannot be 0.
$endgroup$
– furfur
Mar 6 at 13:18
$begingroup$
First you should rewrite the equation by multiplying both sides with $(x+y)(x-y)$. The resulting equation holds true also if $x=y$ or $x=-y$.
$endgroup$
– Jens Schwaiger
Mar 6 at 17:35
$begingroup$
Yes. That’s true. But in my case the function “h” (the one in the solution by Aczel) is equal to f(x+y)/(x+y) which is not defined for x=-y. So I can’t have an h(0). This is means h is defined on R*. So I can’t prove h is odd.
$endgroup$
– furfur
Mar 6 at 18:20
2
$begingroup$
@JensSchwaiger you should delete this "answer"
$endgroup$
– mathworker21
2 days ago
|
show 1 more comment
$begingroup$
See Aczél or Schwaiger, where there is a misprint: $h(x_1,ldots,x_n)$ should read as $$h(x_1+ldots+x_n).$$
In the first paper it is shown that $f$ is a solution of the functions equation in question iff $f(x)=ax^2+bx$ for all $x$ with arbitrary constants $a,b$. The second reference gives certain generalisations.
Edit The results mentioned concerns different equations.
The somehow simpler one $fracf(x)-f(y)x-y=g(x+y)$ for $xnot=y$ and certain generalisations.
answered Mar 6 at 11:26
Jens SchwaigerJens Schwaiger
1,631138
$endgroup$
See Aczél or Schwaiger, where there is a misprint: $h(x_1,ldots,x_n)$ should read as $$h(x_1+ldots+x_n).$$
In the first paper it is shown that $f$ is a solution of the functions equation in question iff $f(x)=ax^2+bx$ for all $x$ with arbitrary constants $a,b$. The second reference gives certain generalisations.
Edit The results mentioned concerns different equations.
The somehow simpler one $fracf(x)-f(y)x-y=g(x+y)$ for $xnot=y$ and certain generalisations.
answered Mar 6 at 11:26
Jens SchwaigerJens Schwaiger
1,631138
edited 2 days ago
answered Mar 6 at 11:26
Jens SchwaigerJens Schwaiger
1,631138
answered Mar 6 at 11:26
Jens SchwaigerJens Schwaiger
1,631138
answered Mar 6 at 11:26
Jens SchwaigerJens Schwaiger
1,631138
1,631138
$begingroup$
As it stands this is a link-only answer, and the second article seems to be behind a paywall. Also, both seem to cover a broader class of functions. Including at least a summary of the relevant results seems appropriate.
$endgroup$
– Servaes
Mar 6 at 12:05
$begingroup$
This doesn’t solve my problem though.. in that link answer he makes x=-y to prove function h is odd. I can’t do x=-y in my problem, as the denominator cannot be 0.
$endgroup$
– furfur
Mar 6 at 13:18
$begingroup$
First you should rewrite the equation by multiplying both sides with $(x+y)(x-y)$. The resulting equation holds true also if $x=y$ or $x=-y$.
$endgroup$
– Jens Schwaiger
Mar 6 at 17:35
$begingroup$
Yes. That’s true. But in my case the function “h” (the one in the solution by Aczel) is equal to f(x+y)/(x+y) which is not defined for x=-y. So I can’t have an h(0). This is means h is defined on R*. So I can’t prove h is odd.
$endgroup$
– furfur
Mar 6 at 18:20
2
$begingroup$
@JensSchwaiger you should delete this "answer"
$endgroup$
– mathworker21
2 days ago
|
show 1 more comment
$begingroup$
As it stands this is a link-only answer, and the second article seems to be behind a paywall. Also, both seem to cover a broader class of functions. Including at least a summary of the relevant results seems appropriate.
$endgroup$
– Servaes
Mar 6 at 12:05
$begingroup$
This doesn’t solve my problem though.. in that link answer he makes x=-y to prove function h is odd. I can’t do x=-y in my problem, as the denominator cannot be 0.
$endgroup$
– furfur
Mar 6 at 13:18
$begingroup$
First you should rewrite the equation by multiplying both sides with $(x+y)(x-y)$. The resulting equation holds true also if $x=y$ or $x=-y$.
$endgroup$
– Jens Schwaiger
Mar 6 at 17:35
$begingroup$
Yes. That’s true. But in my case the function “h” (the one in the solution by Aczel) is equal to f(x+y)/(x+y) which is not defined for x=-y. So I can’t have an h(0). This is means h is defined on R*. So I can’t prove h is odd.
$endgroup$
– furfur
Mar 6 at 18:20
2
$begingroup$
@JensSchwaiger you should delete this "answer"
$endgroup$
– mathworker21
2 days ago
$begingroup$
As it stands this is a link-only answer, and the second article seems to be behind a paywall. Also, both seem to cover a broader class of functions. Including at least a summary of the relevant results seems appropriate.
$endgroup$
– Servaes
Mar 6 at 12:05
$begingroup$
As it stands this is a link-only answer, and the second article seems to be behind a paywall. Also, both seem to cover a broader class of functions. Including at least a summary of the relevant results seems appropriate.
$endgroup$
– Servaes
Mar 6 at 12:05
$begingroup$
This doesn’t solve my problem though.. in that link answer he makes x=-y to prove function h is odd. I can’t do x=-y in my problem, as the denominator cannot be 0.
$endgroup$
– furfur
Mar 6 at 13:18
$begingroup$
This doesn’t solve my problem though.. in that link answer he makes x=-y to prove function h is odd. I can’t do x=-y in my problem, as the denominator cannot be 0.
$endgroup$
– furfur
Mar 6 at 13:18
$begingroup$
First you should rewrite the equation by multiplying both sides with $(x+y)(x-y)$. The resulting equation holds true also if $x=y$ or $x=-y$.
$endgroup$
– Jens Schwaiger
Mar 6 at 17:35
$begingroup$
First you should rewrite the equation by multiplying both sides with $(x+y)(x-y)$. The resulting equation holds true also if $x=y$ or $x=-y$.
$endgroup$
– Jens Schwaiger
Mar 6 at 17:35
$begingroup$
Yes. That’s true. But in my case the function “h” (the one in the solution by Aczel) is equal to f(x+y)/(x+y) which is not defined for x=-y. So I can’t have an h(0). This is means h is defined on R*. So I can’t prove h is odd.
$endgroup$
– furfur
Mar 6 at 18:20
$begingroup$
Yes. That’s true. But in my case the function “h” (the one in the solution by Aczel) is equal to f(x+y)/(x+y) which is not defined for x=-y. So I can’t have an h(0). This is means h is defined on R*. So I can’t prove h is odd.
$endgroup$
– furfur
Mar 6 at 18:20
2
2
$begingroup$
@JensSchwaiger you should delete this "answer"
$endgroup$
– mathworker21
2 days ago
$begingroup$
@JensSchwaiger you should delete this "answer"
$endgroup$
– mathworker21
2 days ago
|
show 1 more comment
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4
$begingroup$
$f(x)=ax^2$ is also a solution.
$endgroup$
– Kavi Rama Murthy
Mar 6 at 8:26
2
$begingroup$
And thus so is $f(x) = ax^2+bx$. Indeed, any linear combination of solutions is again a solution.
$endgroup$
– Greg Martin
Mar 6 at 8:35