Find all functions $ f : \mathbb R \rightarrow \mathbb R$ such that $ (x + y)(f(x) - f(y)) = (x -y)f(x + y)$ for all $ x, y\in \mathbb R$
Problem
Source: Singapore IMO TST 2008, Problem 5
Tags: function, algebra proposed, algebra
07.09.2008 15:06
mr.danh wrote: Find all functions $ f : \mathbb R \rightarrow \mathbb R$ so that $ (x + y)(f(x) - f(y)) = (x - y)f(x + y)$ for all $ x, y\in \mathbb R$ we have $ (x-y)(f(x)-f(-y))=(x+y)(f(x-y)$ then $ (f(x)-f(y))$ we have $ (x-y)(f(x)-f(-y))=(x+y)(f(x-y)$ then $ (f(x)-f(y))(f(x)-(f-y)=f(x+y)(f(x-y)$ let $ g(x-y)=\frac {f(x)-f(y)}{f(x-y)}$ then$ g(x-y)g(x+y)=1$ let $ x-y=u,x+y=v$ then $ g(x)g(y)=1$ for all $ x,y,\in R$ then $ g(x)=1$ then $ f(x)-f(y)=f(x-y)$ next step,we only let $ h(x-y)=\frac{f(x-y)}{x-y}$ and we will find $ h(x)=h(y)$ for all $ x,y\in R$ then $ f(x)=kx$we have done Am I right ?
07.09.2008 20:46
Allnames wrote: let $ g(x - y) = \frac {f(x) - f(y)}{f(x - y)}$ This step is erroneous (even disregarding the possibility that $ f(x-y)=0$). $ g$ is not a well-defined function unless the right hand side is genuinely a function of $ x - y$, and we don't know a priori whether or not it is (turns out it's not). By assuming that it is, you've effectively assumed that $ f$ satisfies $ f(x) - f(y) = f(x - y)$, and so the linear solution you came up with is essentially circular reasoning. There are nonlinear solutions.
17.09.2008 20:20
By plugging in : $ x=x, y=1; x=x+1, y=1; x=x, y=2$ we get a system of 3 equations and 3 unknowns and we get the result,, smthing like mr.danh wrote
14.09.2015 18:11
You are right.
15.09.2015 03:57
Observe that the set of solutions forms an $\mathbb R$-vector space, and that $f(x) \equiv x$, $f(x) \equiv x^2$ are solutions. Consider the linear subspace of solutions $f$ satisfying $f(1) = f(-1) = 0$; we show that in this case $f(x) \equiv 0$. First, $(x,y)=(1,-1)$ gives $f(0) = 0$. Now note that \begin{align*} (x,y) = (a,1) \implies (a+1)f(a) = (a-1)f(a+1) \\ (x,y) = (a+1,-1) \implies af(a+1) = (a+2)f(a) \end{align*}View this as a linear system in $f(a)$ and $f(a+1)$. Now note that \[ \det \left( \begin{array}{cc} a+1 & -(a-1) \\ a+2 & -a \end{array} \right) = -2 \]thus it follows that $f(a) = f(a+1) = 0$ for all $a$ as needed. From this we deduce that the set of solutions to the original functional equation are the span of $x$ and $x^2$, id est those functions of the form $f(x) = ax^2 + bx$ for $a, b \in \mathbb R$.
27.04.2019 20:43
Letting $y=1$, we get that $$(x-1)f(x+1)=(x+1)(f(x)-f(1)).$$We shall use this to compute $f(x+2)$. Note that $xf(x+2)=(x+2)(f(x+1)-f(1))$. It follows $$x(x-1)f(x+2)=(x+2)((x-1)f(x+1)-(x-1)f(1))$$$$=(x+2)((x+1)(f(x)-f(1))-(x-1)f(1))=(x+2)((x+1)f(x)-2xf(1))$$Taking $y=2$ now gives $(x-2)f(x+2)=(x+2)(f(x)-f(2))$. Multiplying the above relation by $(x-2)$ and using the previous result, $$(x-2)(x+2)((x+1)f(x)-2xf(1))=x(x-1)(x-2)f(x+2)=x(x-1)(x+2)(f(x)-f(2))$$Assuming $x=\neq -2$, we may simplify to get $$2f(x)=(-2x^2+4x)f(1)+(x^2-x)f(2)\implies f(x)=ax^2+bx\forall x\neq -2.$$But plugging $(x,y)=(-3,1)$ and using the above result shows $f(-2)=4a-2b$, so we have the solution $\boxed{f(x)=ax^2+bx}$.
02.08.2019 16:04
I don't understand how to solve this kind of problem Someone could teach me, I would appreciate it very much I'm from Peru, but I know how to speak English
21.12.2020 07:47
i am getting something different (x+y)(f(x)-f(y))=(x-y)(f(x+y)) so on assuming f(x) is some polynomial and on using (x-y)|f(x)-f(y) we get the obvious answer as a linear function but others are getting something different
29.12.2023 14:35
If $f, g$ are solutions, then $f + g, f - g$ are also solutions, so we can assume $f(1) = f(2) = 0$. Then plugging $(x, y) = (x, 1)$ gives $f(x+1) = f(x)\frac{x+1}{x-1}$ for all $x \neq 1$. Plugging $(x, y) = (x, 2)$ gives $f(x+2) = f(x)\frac{x+2}{x-2}$ for all $x \neq 2$. Note that for all $x \neq 0, 1, 2$ we have $\frac{x+2}{x-2}f(x) = f(x+2) = \frac{x+2}{x}f(x+1) = \frac{(x+2)(x+1)}{x(x-1)}f(x)$, or equivalently $\frac{f(x)}{x-2} = \frac{f(x)(x+1)}{x(x-1)}$. Thus if $f(x) \neq 0$, then we have $(x+1)(x-2) = x(x-1)$, a contradiction. Thus $f(x) = 0$ for all $x \neq 0, 1, 2$ and $f(0) = 0$, therefore $f(x) = 0$ for all $x$. Hence $f(x) = ax^2 + bx$ for some $a, b$. $\blacksquare$
20.01.2025 02:49
Easily $f(0)=0$. Take $P$ a polynomial such as $P(x)=(\frac{f(2)}{2}-f(1))x^2+(2f(1)-\frac{f(2)}{2})x$, we have $P=f$ over $\{0,1,2\}$. One can easily verify that $P$ is a solution to our fe. We claim that $P=f$ over $\mathbb{R}$, We use induction to prove that $P=f$ over $\mathbb{N}_0$ ($x=1, y=n$). Take $g=f-P$, we get that $g$ is a solution to our fe such $g(\mathbb{N}_0)=\{0\}$ Take $y=n-x$, thus $g(x)=g(n-x)=g(m-(n-x))=g((m-n)+x)=g(p+x)$ with $p\in\mathbb{Z}$ Take $y=p+x$ we get $g(p+2x)=0 \Rightarrow g(x)=0 \Rightarrow g=0 \Rightarrow f=P$ as desired. Solutions are all such $P$ with $f(1),f(2)$ arbitrary real constants.
22.01.2025 07:21
oops obfuscated solution because I forgot about vector spaces Setting $x=-y$ yields $f(0) = 0$. Note that applying the assertion to $(x, c-x)$, we have \[\frac{f(x) - f(c-x)}{2x-c} = \frac{f(c)}c = g(c)\]for some $r_c \in \mathbb R$ defined for all $c \neq 0$. Claim: $r$ is Jensen. Proof: Notice that \begin{align*} f(x) - f(2a-b-x) &= 2x r_{2a-b} - (2a-b)r_{2a-b} \\ f(x) - f(a-x) &= 2x r_a - ar_a \\ f(x) - f(b-x) &= 2x r_b - br_b. \end{align*}Subtracting the first two equations yields \[f(a-x) - f(2a-b-x) = 2(r_{2a-b} - r_a) x + ar_a - (2a-b) r_{2a-b}.\]Subtracting the second two equations yields \[f(b-x) - f(a-x) = 2(r_a-r_b) x + br_b - ar_a.\]On the other hand, the substitution $x \to x + (a-b)$ in the first equation transforms it into \[f(b-x) - f(a-x) = 2(r_{2a-b}-r_a)(x+a-b) + ar_a - (2a-b) r_{2a-b}.\]Since both equations hold for all $x$, we have $r_b + r_{2a-b} = 2r_a$, as needed. $\blacksquare$ Now we define a function $g(x)$ such that $g(x) = r_x$ on all $x \neq 0$. So $g(x)-g(0)$ is Cauchy, and it follows that \[\frac{f(x)}x + \frac{f(y)}y =\frac{f(x+y)}{x+y} + g(0) = \frac{f(x)-f(y)}{x-y} + g(0).\]This expands to \[xy(x-y) g(0) = x^2 f(y) - y^2 f(x) \iff \frac{f(y)}{y^2} - \frac{f(x)}{x^2} = \frac{g(0)}x - \frac{g(0)}y.\]So the function $\frac{f(x)}{x^2} + \frac{g(0)}x = c$ for some constant $c$, i.e. $f(x) = -g(0) x + cx^2$. Indeed, all functions of the form $f(x) = ax^2+bx$ work, so we are done.