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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 $mathbbR^+$

2

1

$begingroup$

$$
f: mathbbR to mathbbRqquad fracf(x+y)x+y = fracf(x)-f(y)x-y, qquad forall x,yin mathbbR, left|xright| neq left|yright|
$$

Can I prove anything interesting about this function? I need to find it.

functions functional-equations

share|cite|improve this question

edited Mar 8 at 15:40

furfur

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 | 

2

1

$begingroup$

$$
f: mathbbR to mathbbRqquad fracf(x+y)x+y = fracf(x)-f(y)x-y, qquad forall x,yin mathbbR, left|xright| neq left|yright|
$$

Can I prove anything interesting about this function? I need to find it.

functions functional-equations

share|cite|improve this question

edited Mar 8 at 15:40

furfur

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

share|cite|improve this question

edited Mar 8 at 15:40

furfur

asked Mar 6 at 8:20

furfurfurfur

509

$endgroup$

share|cite|improve this question

edited Mar 8 at 15:40

furfur

asked Mar 6 at 8:20

furfurfurfur

509

share|cite|improve this question

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

1 Answer
1

active

oldest

votes

0

$begingroup$

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.

share|cite|improve this answer

edited 2 days ago

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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0

$begingroup$

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.

share|cite|improve this answer

edited 2 days ago

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

0

$begingroup$

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.

share|cite|improve this answer

edited 2 days ago

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.

share|cite|improve this answer

edited 2 days ago

answered Mar 6 at 11:26

Jens SchwaigerJens Schwaiger

1,631138

$endgroup$

share|cite|improve this answer

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