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Find all solutions to the functional equation $f(x+y)-f(y)=\cfrac{x}{y(x+y)}$

I've tried the substitution technique but I didn't really get something useful.

For $y=1$ I have

$F(x+1)-F(1)=\cfrac{x}{x+1}$

A pattern I've found in this example is that if I iterate $n$ times the function $g(x)=\cfrac{x}{x+1}$ I have that $g^n(x)=\cfrac{x}{nx +1}$ ,which may be a clue about the general behaviour of the function ( ?) .

I am really kinda of clueless ,it seems like the problem is calling some slick way of solving it.

Can you guys give me a hint ?

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  • $\begingroup$ What can $x,y$ be? Anything as long as $y(x+y) \neq 0$? $\endgroup$ – wythagoras Dec 27 '15 at 17:37
  • $\begingroup$ As the question is stated in my book,it doesnt make any restrictions . $\endgroup$ – Mr. Y Dec 27 '15 at 17:39
  • $\begingroup$ Then we should assume that $x,y$ are real and $y\neq0$, $x+y \neq 0$. $\endgroup$ – wythagoras Dec 27 '15 at 17:41
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Hint: Let $x=1-y$, hence $x+y=1$.

Then we get $$f(1)-f(y)=\frac{1-y}{y}$$

I will include the full solution in a spoiler, since you only asked for a hint.

This gives $f(0)=c$ and $f(1)=d$, then $f(y)=\frac{y-1}{y}+d$ for all other $y$.
This simplifies to $f(0)=c$ and $f(y)=\frac{y-1}{y}+d$ for all other $y$. We have to check whether every such function satisfies. Note that neither $y$ nor $x+y$ can ever be zero, so $f(0)$ can indeed be anything. Now we have to check the other values: Does

$$\left(\frac{x+y-1}{x+y} + d\right) - \left(\frac{y-1}{y} +d \right) = \frac{y}{y(x+y)} ?$$ $$\frac{y(x+y-1)-(y-1)(x+y)}{y(x+y)} = \frac{y}{y(x+y)} ?$$
It turns out that the functions do satisfy the functional equation.

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  • $\begingroup$ Give me some time to fully comprehend the meaning of it. $\endgroup$ – Mr. Y Dec 27 '15 at 17:45
  • $\begingroup$ @Mr.Y Sorry I mistyped it. Now the equation is the correct one and you should now be able to solve it. There are a few tricky parts left. $\endgroup$ – wythagoras Dec 27 '15 at 17:48
  • $\begingroup$ If I rearrange your hint I have that $f(1)-f(y)=\cfrac{1}{y}-1$ from which I see ,matching terms,that the requested function is $f(y)=-\cfrac{1}{y}$ since $y$ can be any value.Is this a bad way to extrapolate the solution ? $\endgroup$ – Mr. Y Dec 27 '15 at 18:08
  • $\begingroup$ @wythagoras: Why are you entitled to assume that $x +y = 1$? $\endgroup$ – Geoff Robinson Dec 27 '15 at 18:14
  • $\begingroup$ @GeoffRobinson We may assume that $x=1-y$ since $x,y$ can be anything but those satisfying $x+y=0$ and $y=0$. Rearranging $x=1-y$ gives $x+y=1$. $\endgroup$ – wythagoras Dec 27 '15 at 18:57
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$f(x+y)-f(y) =\cfrac{x}{y(x+y)} =\frac1{y}-\frac1{x+y} $ so $f(x+y)+\frac1{x+y} =f(y)+\frac1{y} $.

Therefore, $f(x)+\frac1{x}$ is constant, so $f(x) =d-\frac1{x} $ for some $d$.

Substituting this, the $d$s cancel out, so any $d$ works, and the solution is $f(x) =d-\frac1{x} $ for any $d$.

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  • $\begingroup$ This is also really neat.Thank you marty cohen.I love when I get many beautiful answers on one question.I wish I could accept all of your answers. :) Thanks again ! $\endgroup$ – Mr. Y Dec 27 '15 at 21:13
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Another approach: I assume that $f$ is a function of a single real variable.Write the defining equation as $\frac{f(y+x) - f(y)}{(y+x)- y} = \frac{1}{y(x+y)}$ (for $x,y,(x+y) \neq 0).$ Take the limit as $x \to 0$. On the one hand this limit is $\frac{1}{y^{2}}.$ On the other hand, the limit is, by definition, the derivative $f^{\prime}(y)$, (and we have proved that this exists for $y \neq 0$). An antiderivative of $\frac{1}{y^{2}}$ is $\frac{-1}{y}$. Hence $f(y) = \frac{-1}{y} + C$ for a constant $C$, as long as $y \neq 0$. (Strictly speaking this proves that any function $f$ which satisfies the equation has to be of the described form. It should be checked that such functions satisfy the equation- but they do).

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  • $\begingroup$ This is really a total different way to look at the problem,really neat !!Thank you Geoff Robinson! $\endgroup$ – Mr. Y Dec 27 '15 at 19:51
  • $\begingroup$ The answer by Marty Cohen is more direct though! $\endgroup$ – Geoff Robinson Dec 28 '15 at 1:53

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