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Solve the functional equation $fracf(x)f(y)=fleft( fracx-yf(y) right)$

The functional equation $f(y) + fleft(frac1yright) = 0$D'alembert functional equationSolve the functional equation $ q , fracf(x+1)f(x)=frach(x+1)h(x). $How to solve the functional equation $f(x+a)=f(x)+a$The function $f (x) = f left (frac x2 right ) + f left (frac x2 + frac 12right)$Solve the functional Equation:- $f(x+y)=3^xf(y)+9^yf(x)forall x,yin mathbbR$Solving functional equation $1=e^−frac2kf(ax+bx^2)+2f(x)$How to solve functional equation?Solve Functional Equation $3f(x)=fleft(fracx3right)+fleft(frac1+x3right)$Suggestion for a functional equation $f'left(fracaxright)=fracxf(x)$ where $f:(0,infty)to(0,infty)$ is differentiable

8

2

$begingroup$

Solve the functional equation
$$
fracf(x)f(y)=fleft( fracx-yf(y) right),
$$
here $f: mathbbR to mathbbR$ and $f$ is differentiable at $x=0.$

By set $x=y$ we get $f(0)=1$.

Differentiate
$$
fracf'(x)f(y)=f'left( fracx-yf(y) right) cdot frac1f(y)
$$
and set $x=0$ get
$$
f'(0)=f'left( frac-yf(y) right).
$$
So we reduce the problem to the problem to describe all function $g$
$$
gleft( -fracyg(y) right)=const.
$$

I have no more ideas.

functional-equations

share|cite|improve this question

edited 9 hours ago

Martin Sleziak

45k10122277

asked Jun 6 '15 at 11:15

LeoxLeox

5,3481424

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Leox if $f$ is only differentiable at $x=0$ I don't think you can differentiate it like that.
  $endgroup$
  – pointer
  Jun 6 '15 at 11:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the right differentiation?
  $endgroup$
  – Leox
  Jun 6 '15 at 11:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Define a variable $h$ , where $y=x$ and $x=x+h$ and look for derivative as $hto 0$ at $x=0$
  $endgroup$
  – Mann
  Jun 6 '15 at 11:46
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Mann Do you mean I have to find this limit at $h to 0$ $$ frac1hleft(fleft(fracx+h-yf(y)right) - fleft(fracx-yf(y)right) right)? $$
  $endgroup$
  – Leox
  Jun 6 '15 at 12:21
  
  

  
  
  
  

add a comment | 

8

2

$begingroup$

Solve the functional equation
$$
fracf(x)f(y)=fleft( fracx-yf(y) right),
$$
here $f: mathbbR to mathbbR$ and $f$ is differentiable at $x=0.$

By set $x=y$ we get $f(0)=1$.

Differentiate
$$
fracf'(x)f(y)=f'left( fracx-yf(y) right) cdot frac1f(y)
$$
and set $x=0$ get
$$
f'(0)=f'left( frac-yf(y) right).
$$
So we reduce the problem to the problem to describe all function $g$
$$
gleft( -fracyg(y) right)=const.
$$

I have no more ideas.

functional-equations

share|cite|improve this question

edited 9 hours ago

Martin Sleziak

45k10122277

asked Jun 6 '15 at 11:15

LeoxLeox

5,3481424

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Leox if $f$ is only differentiable at $x=0$ I don't think you can differentiate it like that.
  $endgroup$
  – pointer
  Jun 6 '15 at 11:42
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is the right differentiation?
  $endgroup$
  – Leox
  Jun 6 '15 at 11:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Define a variable $h$ , where $y=x$ and $x=x+h$ and look for derivative as $hto 0$ at $x=0$
  $endgroup$
  – Mann
  Jun 6 '15 at 11:46
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Mann Do you mean I have to find this limit at $h to 0$ $$ frac1hleft(fleft(fracx+h-yf(y)right) - fleft(fracx-yf(y)right) right)? $$
  $endgroup$
  – Leox
  Jun 6 '15 at 12:21
  
  

  
  
  
  

add a comment | 

$begingroup$

Solve the functional equation
$$
fracf(x)f(y)=fleft( fracx-yf(y) right),
$$
here $f: mathbbR to mathbbR$ and $f$ is differentiable at $x=0.$

By set $x=y$ we get $f(0)=1$.

Differentiate
$$
fracf'(x)f(y)=f'left( fracx-yf(y) right) cdot frac1f(y)
$$
and set $x=0$ get
$$
f'(0)=f'left( frac-yf(y) right).
$$
So we reduce the problem to the problem to describe all function $g$
$$
gleft( -fracyg(y) right)=const.
$$

I have no more ideas.

functional-equations

share|cite|improve this question

edited 9 hours ago

Martin Sleziak

45k10122277

asked Jun 6 '15 at 11:15

LeoxLeox

5,3481424

$endgroup$

share|cite|improve this question

edited 9 hours ago

Martin Sleziak

45k10122277

asked Jun 6 '15 at 11:15

LeoxLeox

5,3481424

share|cite|improve this question

edited 9 hours ago

Martin Sleziak

45k10122277

asked Jun 6 '15 at 11:15

LeoxLeox

5,3481424

edited 9 hours ago

Martin Sleziak

45k10122277

edited 9 hours ago

Martin Sleziak

45k10122277

asked Jun 6 '15 at 11:15

LeoxLeox

5,3481424

asked Jun 6 '15 at 11:15

LeoxLeox

5,3481424

2 Answers
2

active

oldest

votes

10

$begingroup$

Let us define our variables as $x=x+h$ and $y=x$, for convenience.

Then our functional equation becomes,

$$fracf(x+h)f(x)=fleft( frachf(x) right)$$

With $h=0$ and $x=0$, we can get that $f(0)=1$. (If $f$ exists and is finite as well for a single $x_0$).

Using $f(x+h)=f(x)times fleft( frachf(x) right) $ and the definition of derivative,

$$f'(x)=lim_limitsh to 0fracf(x+h)-f(x)h=lim_limitsh to 0;f(x) times fracfleft( frachf(x) right)-1h=lim_limitsh to 0fracfleft( frachf(x) right)-1frachf(x)$$

Then, $lim_limitsx to x_0 f'(x)$ exists for arbitrary $x_0$ and is equal to $f'(0)$. Hence, $f'(x)$ is continuous 'almost everywhere' or everywhere and is equal to $f'(0)$. We can integrate $f'(x)$ to obtain a linear straight line, because Null sets have measure $0$. The null set only consists the point where $f(x) $ doesn't exist.

Under the assumption that $f(x)$ is finite at the point of inspection $x$ and also the fact that $f(0)=1$. We see that this limit is simply $f'(0)$ which exists. We get $$f'(x)=f'(0)$$

Which on integrating gives general solution, $$f(x)=f'(0)x+c=ax+c$$

also $f(0)=1implies c=1$ which gives $$f(x)=ax+1$$ except possibly on a Null set which contains the points $x$ where $f(x)$ doesn't exist or $f(x)=0$

share|cite|improve this answer

edited 9 hours ago

Martin Sleziak

45k10122277

answered Jun 6 '15 at 12:24

MannMann

2,3501727

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Very beautiful solution!
  $endgroup$
  – pointer
  Jun 6 '15 at 12:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, the only linear functions?
  $endgroup$
  – Leox
  Jun 6 '15 at 12:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Only Linear, yes.
  $endgroup$
  – Mann
  Mar 29 at 23:37
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $f(x)=x+1$. It is obviously a solution.

share|cite|improve this answer

answered Jun 6 '15 at 11:17

Alex FokAlex Fok

3,866818

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Moreover, $ax+1$ is a solution for any $a$ and there is no other linear solution.
  $endgroup$
  – pointer
  Jun 6 '15 at 11:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what about nonlinear solutions?
  $endgroup$
  – Leox
  Jun 6 '15 at 11:37
  

  
  
  
  

add a comment | 

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10

$begingroup$

Let us define our variables as $x=x+h$ and $y=x$, for convenience.

Then our functional equation becomes,

$$fracf(x+h)f(x)=fleft( frachf(x) right)$$

With $h=0$ and $x=0$, we can get that $f(0)=1$. (If $f$ exists and is finite as well for a single $x_0$).

Using $f(x+h)=f(x)times fleft( frachf(x) right) $ and the definition of derivative,

$$f'(x)=lim_limitsh to 0fracf(x+h)-f(x)h=lim_limitsh to 0;f(x) times fracfleft( frachf(x) right)-1h=lim_limitsh to 0fracfleft( frachf(x) right)-1frachf(x)$$

Then, $lim_limitsx to x_0 f'(x)$ exists for arbitrary $x_0$ and is equal to $f'(0)$. Hence, $f'(x)$ is continuous 'almost everywhere' or everywhere and is equal to $f'(0)$. We can integrate $f'(x)$ to obtain a linear straight line, because Null sets have measure $0$. The null set only consists the point where $f(x) $ doesn't exist.

Under the assumption that $f(x)$ is finite at the point of inspection $x$ and also the fact that $f(0)=1$. We see that this limit is simply $f'(0)$ which exists. We get $$f'(x)=f'(0)$$

Which on integrating gives general solution, $$f(x)=f'(0)x+c=ax+c$$

also $f(0)=1implies c=1$ which gives $$f(x)=ax+1$$ except possibly on a Null set which contains the points $x$ where $f(x)$ doesn't exist or $f(x)=0$

share|cite|improve this answer

edited 9 hours ago

Martin Sleziak

45k10122277

answered Jun 6 '15 at 12:24

MannMann

2,3501727

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Very beautiful solution!
  $endgroup$
  – pointer
  Jun 6 '15 at 12:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, the only linear functions?
  $endgroup$
  – Leox
  Jun 6 '15 at 12:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Only Linear, yes.
  $endgroup$
  – Mann
  Mar 29 at 23:37
  

  
  
  
  

add a comment | 

10

$begingroup$

Let us define our variables as $x=x+h$ and $y=x$, for convenience.

Then our functional equation becomes,

$$fracf(x+h)f(x)=fleft( frachf(x) right)$$

With $h=0$ and $x=0$, we can get that $f(0)=1$. (If $f$ exists and is finite as well for a single $x_0$).

Using $f(x+h)=f(x)times fleft( frachf(x) right) $ and the definition of derivative,

$$f'(x)=lim_limitsh to 0fracf(x+h)-f(x)h=lim_limitsh to 0;f(x) times fracfleft( frachf(x) right)-1h=lim_limitsh to 0fracfleft( frachf(x) right)-1frachf(x)$$

Then, $lim_limitsx to x_0 f'(x)$ exists for arbitrary $x_0$ and is equal to $f'(0)$. Hence, $f'(x)$ is continuous 'almost everywhere' or everywhere and is equal to $f'(0)$. We can integrate $f'(x)$ to obtain a linear straight line, because Null sets have measure $0$. The null set only consists the point where $f(x) $ doesn't exist.

Under the assumption that $f(x)$ is finite at the point of inspection $x$ and also the fact that $f(0)=1$. We see that this limit is simply $f'(0)$ which exists. We get $$f'(x)=f'(0)$$

Which on integrating gives general solution, $$f(x)=f'(0)x+c=ax+c$$

also $f(0)=1implies c=1$ which gives $$f(x)=ax+1$$ except possibly on a Null set which contains the points $x$ where $f(x)$ doesn't exist or $f(x)=0$

share|cite|improve this answer

edited 9 hours ago

Martin Sleziak

45k10122277

answered Jun 6 '15 at 12:24

MannMann

2,3501727

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  Very beautiful solution!
  $endgroup$
  – pointer
  Jun 6 '15 at 12:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  So, the only linear functions?
  $endgroup$
  – Leox
  Jun 6 '15 at 12:38
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Only Linear, yes.
  $endgroup$
  – Mann
  Mar 29 at 23:37
  

  
  
  
  

add a comment | 

$begingroup$

Let us define our variables as $x=x+h$ and $y=x$, for convenience.

Then our functional equation becomes,

$$fracf(x+h)f(x)=fleft( frachf(x) right)$$

With $h=0$ and $x=0$, we can get that $f(0)=1$. (If $f$ exists and is finite as well for a single $x_0$).

Using $f(x+h)=f(x)times fleft( frachf(x) right) $ and the definition of derivative,

$$f'(x)=lim_limitsh to 0fracf(x+h)-f(x)h=lim_limitsh to 0;f(x) times fracfleft( frachf(x) right)-1h=lim_limitsh to 0fracfleft( frachf(x) right)-1frachf(x)$$

Then, $lim_limitsx to x_0 f'(x)$ exists for arbitrary $x_0$ and is equal to $f'(0)$. Hence, $f'(x)$ is continuous 'almost everywhere' or everywhere and is equal to $f'(0)$. We can integrate $f'(x)$ to obtain a linear straight line, because Null sets have measure $0$. The null set only consists the point where $f(x) $ doesn't exist.

Under the assumption that $f(x)$ is finite at the point of inspection $x$ and also the fact that $f(0)=1$. We see that this limit is simply $f'(0)$ which exists. We get $$f'(x)=f'(0)$$

Which on integrating gives general solution, $$f(x)=f'(0)x+c=ax+c$$

also $f(0)=1implies c=1$ which gives $$f(x)=ax+1$$ except possibly on a Null set which contains the points $x$ where $f(x)$ doesn't exist or $f(x)=0$

share|cite|improve this answer

edited 9 hours ago

Martin Sleziak

45k10122277

answered Jun 6 '15 at 12:24

MannMann

2,3501727

$endgroup$

share|cite|improve this answer

edited 9 hours ago

Martin Sleziak

45k10122277

answered Jun 6 '15 at 12:24

MannMann

2,3501727

edited 9 hours ago

Martin Sleziak

45k10122277

edited 9 hours ago

Martin Sleziak

45k10122277

answered Jun 6 '15 at 12:24

MannMann

2,3501727

answered Jun 6 '15 at 12:24

MannMann

2,3501727

2

$begingroup$

Let $f(x)=x+1$. It is obviously a solution.

share|cite|improve this answer

answered Jun 6 '15 at 11:17

Alex FokAlex Fok

3,866818

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Moreover, $ax+1$ is a solution for any $a$ and there is no other linear solution.
  $endgroup$
  – pointer
  Jun 6 '15 at 11:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what about nonlinear solutions?
  $endgroup$
  – Leox
  Jun 6 '15 at 11:37
  

  
  
  
  

add a comment | 

2

$begingroup$

Let $f(x)=x+1$. It is obviously a solution.

share|cite|improve this answer

answered Jun 6 '15 at 11:17

Alex FokAlex Fok

3,866818

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Moreover, $ax+1$ is a solution for any $a$ and there is no other linear solution.
  $endgroup$
  – pointer
  Jun 6 '15 at 11:34
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  what about nonlinear solutions?
  $endgroup$
  – Leox
  Jun 6 '15 at 11:37
  

  
  
  
  

add a comment | 

$begingroup$

Let $f(x)=x+1$. It is obviously a solution.

share|cite|improve this answer

answered Jun 6 '15 at 11:17

Alex FokAlex Fok

3,866818

$endgroup$

share|cite|improve this answer

answered Jun 6 '15 at 11:17

Alex FokAlex Fok

3,866818

answered Jun 6 '15 at 11:17

Alex FokAlex Fok

3,866818

answered Jun 6 '15 at 11:17

Alex FokAlex Fok

3,866818