If $f:\mathbb{R}-\{-1\}\rightarrow \mathbb{R}$ and $f$ is a differentiable function that satisfies $$f(x+f(y)+xf(y)) = y+f(x)+yf(x)\forall x,y \in \mathbb{R}-\{-1\}\;,$$ Then value of $\displaystyle 2016(1+f(2015)) = $

$\bf{My\; Try::}$ Using partial Differentiation, Differentiate w r to $x$ and $y$ as a constant

$$f'(x+f(y)+xf(y)) \cdot (1+f(y)) = f'(x)+yf'(x)$$

Similarly Differentiate w r to $y$ and $x$ as a constant

$$f'(x+f(y)+xf(y))\cdot (f'(y)+xf'(y)) = 1+f(x)$$

Now Divide these two equation, We get $$\frac{1+f(y)}{(1+x)f'(y)} = \frac{(1+y)f'(x)}{1+f(x)}$$

Now How can i solve it after that, Help required, Thanks

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    $\begingroup$ $f$ is the identity map. $\endgroup$ – Mattos Nov 6 '16 at 8:33
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    $\begingroup$ @Mattos … or the constant map $-1$, and I haven't proved that they are the only possible maps. $\endgroup$ – Patrick Stevens Nov 6 '16 at 8:33
  • $\begingroup$ @PatrickStevens Then $\frac{1}{(1+x) f'(y)}$ is undefined. $\endgroup$ – Mattos Nov 6 '16 at 8:34
  • $\begingroup$ @Mattos But the question doesn't talk about fractions. Only the supplied working talks about fractions. $\endgroup$ – Patrick Stevens Nov 6 '16 at 8:35
  • $\begingroup$ @PatrickStevens Sorry, I didn't actually read the full post. I just realised that the second last line is clearly satisfied by $f = -1$. $\endgroup$ – Mattos Nov 6 '16 at 8:35

Your last equation tells us that $$ \frac{(1+x)f'(x)}{1+f(x)} $$ has to be constant, and that this constant has to be equal to its reciprocal, i.e., it is $\pm1$.

Integration gives $$ \ln|1+f(x)|=\pm \ln|1+x|+C $$ and thus $$ f(x)=C(1+x)-1\text{ or } f(x)=\frac{C}{1+x}-1 $$

This now needs to be tested against the original functional equation.

The equation can be rewritten as $1+f((1+x)(1+f(y))-1)=(1+y)(1+f(x))$.

\begin{array}{rl|l} f(x)&=C(1+x)-1&=\dfrac{C}{1+x}-1\\ \hline (1+x)(1+f(y))&=C(1+x)(1+y) & =C\dfrac{1+x}{1+y}\\ 1+f((1+x)(1+f(y))-1)&=C^2(1+x)(1+y) & =\dfrac{1+y}{1+x}\\ (1+y)(1+f(x))&=C(1+y)(1+x) & = C\dfrac{1+y}{1+x} \end{array}

which gives $f(x)=-1$, $f(x)=x$ and $f(x)=\dfrac{1}{1+x}-1=-\dfrac{x}{1+x}$ as valid solutions.

Thus the possible values for $(1+x)(1+f(x))$, for $x=2015$ or elsewhere, are $0$, $(1+x)^2$ or $-1$.

PS: As discussed with Patrick Stevens and egreg, for the function $f\equiv -1$ the functional equation is nowhere defined, as the argument of $f$ on the left would be $x(1+f(y))+f(y)=-1$ and thus outside the domain of $f$.

If one wants to simplify the functional equation first, set $g(x)=1+f(x-1)$, $u=1+x$, $v=1+x$ then the reduced form is $$g(ug(v))=vg(u).$$


Summing up the comments; both the identity map and the constant map $-1$ are differentiable functions that satisfy the functional equation. Hence the value of $2016(1+f(2015))$ is not uniquely determined by the given data.

  • $\begingroup$ Actually I realised that the constant $-1$ does not satisfy the functional equation. Let $x=y=0$ to obtain $f(-1) = -1$, which contradicts that $f$ is undefined at $-1$. $\endgroup$ – Patrick Stevens Nov 6 '16 at 10:27
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    $\begingroup$ @PatrickStevens: That it is undefined in $-1$ does not imply that the function can not be extended to that point in a sensible manner. $\endgroup$ – LutzL Nov 6 '16 at 10:35
  • $\begingroup$ @LutzL It's a direct contradiction. "Suppose $f$ is a function from $\mathbb{R} \setminus \{1\}$ to $\mathbb{R}$ satisfying…. Then we show that in fact $f$ must be defined on $\{1\}$. This is a contradiction." We aren't asked to find a function $f: \mathbb{R} \to \mathbb{R}$ which satisfies the condition; we're asked to find a function $\mathbb{R} \setminus \{1\} \to \mathbb{R}$, and there isn't one which is always $-1$. If there were one which were always $-1$, we could deduce that it was also $-1$ on the input $-1$, which is a contradiction. $\endgroup$ – Patrick Stevens Nov 6 '16 at 10:36
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    $\begingroup$ @PatrickStevens: No, it is not a contradiction. It is called extension of a function outside its (original) domain. $1/x$ is not defined on $x=0$. However, one can define a function $f(x)=\begin{cases}0&x=0\\1/x&x\ne 0\end{cases}$ which is defined everywhere. $\endgroup$ – LutzL Nov 6 '16 at 10:39
  • $\begingroup$ @LutzL But the function $1/x$ is not defined in such a way that it is required to satisfy a certain recurrence for all $x, y \not = 0$. $\endgroup$ – Patrick Stevens Nov 6 '16 at 10:40

(Work in progress; I was unexpectedly called away. The aim of this post is to do it without assuming differentiability.)


To find the roots of $f$: suppose $f(y) = 0$. Then $f(x) = y+f(x)+y f(x)$ and so $(1+f(x))y = 0$; therefore $f(x) = -1$ for all $x$, or $y=0$. So we have two possibilities: $f(x) = -1$ for all $x \not = -1$, or $f(0) = 0$.

But if $f(x) = -1$ for all $x \not = -1$, then letting $x=0$ we obtain $$f(-1) = -1$$ which is a contradiction to the undefinedness of $f(-1)$. (There's a slightly simpler contradiction if we simply suppose there is any $y$ with $f(y) = -1$, as egreg points out.)

Therefore $f$ has exactly one root, and it is at $0$. (Strictly, we should go back through this, and verify that we never tried to give $f$ the argument $-1$. The easiest way to do this is to let $x$ be some real such that $f(x) \not = -1$, and then just go through the same proof again.)

Using the root

Letting $x=0$, we see that $f(f(y)) = y$ for all $y$.

Letting $y=f(x)$, we obtain $$f(2x+x^2) = 2f(x)+f(x)^2$$ so, if $x=-2$, we get $f(0) = 2 f(-2) + f(-2)^2$; that is, $f(-2) = 0$ or $f(-2) = -2$. We already know $f$ has exactly one root, so $f(-2) = -2$.

Letting $x = f(y)$, we obtain $$f(2f(y)+f(y)^2) = 2y+y^2$$ so if $f(y) = -2$, we obtain $y = 0$ (contradiction) or $y=-2$.

Therefore $f$ hits $-2$ at exactly one input: namely $-2$.

Letting $y=-2$, then, we obtain $f(-x-2) = -2+f(x)-2f(x)$, so $$f(-x-2) = -f(x)-2$$ and so the behaviour of the function is determined precisely by its behaviour on $x>-1$.

Actually, under the assumption that there is $x$ such that $f(x) = 1$, we obtain $f(x + f(y) + x f(y)) = 1 + 2 y$ and hence (by applying $f$ to both sides) $$x + f(y) + x f(y) = f(1 + 2 y)$$ whence (substituting $z=1+2y$) we get $$f(z) = f\left(\frac{z-1}{2}\right) + x f\left(\frac{z-1}{2}\right) + x$$ so any open interval $(-1, r)$ determines the behaviour of $f$ completely.

TODO: complete this. As LutzL points out, there is more than one function satisfying the recurrence :(

  • $\begingroup$ For the last point see math.stackexchange.com/questions/878517/…. $\endgroup$ – LutzL Nov 6 '16 at 9:47
  • $\begingroup$ @LutzL Most of those functions do not satisfy the recurrence, though. $f(f(y))$ is a fact about $f$, but does not come anywhere near to characterising $f$. $\endgroup$ – Patrick Stevens Nov 6 '16 at 9:57
  • $\begingroup$ This was just to show that $f(x)=x$ may not be the only solution. $\endgroup$ – LutzL Nov 6 '16 at 10:27
  • $\begingroup$ @LutzL I'm pretty sure $f(x) = x$ is the only solution; the only thing that remains is the work of verifying that this is so. $\endgroup$ – Patrick Stevens Nov 6 '16 at 10:28
  • $\begingroup$ Per my answer, $f(x)=-\frac{x}{1+x}$ is also a solution And indeed, $f(f(x))=-\frac{-x}{(1+x)-x}=x$. $\endgroup$ – LutzL Nov 6 '16 at 10:44

Let $x=0$ then $$f(f(y))=y\,\big(\,f(0)+1\big)+f(0)\tag{1}$$ and from this we get $f(f(0))=f(0)$ and $f(f(1))=2f(0)+1$. Let $y=0$ and $x=1$ in original equation then $$f\big(1+2f(0)\big)=f(1)\tag{2}$$ applying $\,f$ we get $$f(f(1+2f(0)))=f(f(1))=2f(0)+1$$ Substituting $y=1+2f(0)$ in $(1)$ we obtain $$f(f(1+2f(0)))=(1+2f(0))(f(0)+1)+f(0)=1+3f(0)+2(f(0))^2,$$ and from the last two equations we get $$f(0)\big(1+f(0)\big)=0.$$ If $f(0)=-1$, then $f(f(0))=f(-1)$ which is not defined but we know that $f(f(0))=f(0)$. Thus $f(0)=0$ and from $(1)$ $\:f(f(x))=x$. For the next step see This post


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