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Find all functions that satisfy $f(x+f(y))=f(x)-y$

Recurrence relations on a continuous domainFind all the functions which satisfy a given functional equationWhat is the family of functions that satisfiesFind all real functions $f:mathbb R to mathbb R$ satisfying the equation $f(x^2+y.f(x))=x.f(x+y)$Find all functions $f$ such that $f(x-f(y)) = f(f(x)) - f(y) - 1$The functional equation $ f(x-f(y))=f(f(y))+xf(y)+f(x)-1$Functions $f:mathbb R to mathbb R$ which satisfy $f(x^2+f(y))=y+(f(x))^2$Additive Cauchy functional equation with quotient of functions $fracf(x+y)g(x+y) + B = fracf(x)g(x) + fracf(y)g(y) $Do I Need To Find All Functions Satisfying A Given Equation In Such Cases?Finding all functions that verify a functional equation

6

1

$begingroup$

here is the problem

Here is my solution :

$x=y=0$ gives $f(f(0))=f(0)$

$x=0; y=f(0)$ gives $f(f(f(0)=0=f(0)$ (because $f(f(0))=f(0) iff f(f(f(0)))=f(f(0))=f(0)$)

$x=0$ gives $f(f(y))=-y$

$x=0 ; y=f(y)$ gives $f(-y)=-f(y)iff f(-f(y))=y$

so $y=-f(y)$ gives $f(2x)=2f(x)$

now we can prove by induction that $f(nx)=nf(x)$

it is true for $n=2$

let's suppose that it is true for $n$

we have $f(nx+f(-f(x))=f(nx)+f(x)iff f((n+1)x)=(n+1)x$

so it is true now we have $f(x)=xf(1)$

plugging this in the first equation gives $f(1)^2=-1$ which is impossible

I just want to know if my solution is right

In the official solution they prove that $f$ is bijective (which is true because $f(f(y))=-y$ ;) ) then they use cauchy function , they find that $f(x)=cx$ , plugging this in the first equation gives an impossible result.

functional-equations

share|cite|improve this question

edited yesterday

user600785

asked yesterday

user600785user600785

11310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @vadim123 Because from the first line, $f(f(0)) = f(0)$.
  $endgroup$
  – rogerl
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Looks right; you meant "which is impossible" near the bottom :)
  $endgroup$
  – rogerl
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok updated .....
  $endgroup$
  – user600785
  yesterday
  

  
  
  
  

add a comment | 

6

1

$begingroup$

here is the problem

Here is my solution :

$x=y=0$ gives $f(f(0))=f(0)$

$x=0; y=f(0)$ gives $f(f(f(0)=0=f(0)$ (because $f(f(0))=f(0) iff f(f(f(0)))=f(f(0))=f(0)$)

$x=0$ gives $f(f(y))=-y$

$x=0 ; y=f(y)$ gives $f(-y)=-f(y)iff f(-f(y))=y$

so $y=-f(y)$ gives $f(2x)=2f(x)$

now we can prove by induction that $f(nx)=nf(x)$

it is true for $n=2$

let's suppose that it is true for $n$

we have $f(nx+f(-f(x))=f(nx)+f(x)iff f((n+1)x)=(n+1)x$

so it is true now we have $f(x)=xf(1)$

plugging this in the first equation gives $f(1)^2=-1$ which is impossible

I just want to know if my solution is right

In the official solution they prove that $f$ is bijective (which is true because $f(f(y))=-y$ ;) ) then they use cauchy function , they find that $f(x)=cx$ , plugging this in the first equation gives an impossible result.

functional-equations

share|cite|improve this question

edited yesterday

user600785

asked yesterday

user600785user600785

11310

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @vadim123 Because from the first line, $f(f(0)) = f(0)$.
  $endgroup$
  – rogerl
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Looks right; you meant "which is impossible" near the bottom :)
  $endgroup$
  – rogerl
  yesterday
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok updated .....
  $endgroup$
  – user600785
  yesterday
  

  
  
  
  

add a comment | 

$begingroup$

here is the problem

Here is my solution :

$x=y=0$ gives $f(f(0))=f(0)$

$x=0; y=f(0)$ gives $f(f(f(0)=0=f(0)$ (because $f(f(0))=f(0) iff f(f(f(0)))=f(f(0))=f(0)$)

$x=0$ gives $f(f(y))=-y$

$x=0 ; y=f(y)$ gives $f(-y)=-f(y)iff f(-f(y))=y$

so $y=-f(y)$ gives $f(2x)=2f(x)$

now we can prove by induction that $f(nx)=nf(x)$

it is true for $n=2$

let's suppose that it is true for $n$

we have $f(nx+f(-f(x))=f(nx)+f(x)iff f((n+1)x)=(n+1)x$

so it is true now we have $f(x)=xf(1)$

plugging this in the first equation gives $f(1)^2=-1$ which is impossible

I just want to know if my solution is right

In the official solution they prove that $f$ is bijective (which is true because $f(f(y))=-y$ ;) ) then they use cauchy function , they find that $f(x)=cx$ , plugging this in the first equation gives an impossible result.

functional-equations

share|cite|improve this question

edited yesterday

user600785

asked yesterday

user600785user600785

11310

$endgroup$

share|cite|improve this question

edited yesterday

user600785

asked yesterday

user600785user600785

11310

share|cite|improve this question

edited yesterday

user600785

asked yesterday

user600785user600785

11310

asked yesterday

user600785user600785

11310

asked yesterday

user600785user600785

11310