Find all functions $f: (1,\infty)\text{to R}$ satisfying $f(x)-f(y)=(y-x)f(xy)$ for all $x,y>1$.
HIDE: hint you may try to find $f(x^5)$ by two ways and then continue the solution. I have also solved by using this method.By finding $f(x^5)$ in two ways I found that $f(x)=xf(x^2)$ for all $x>1$.Problem
Source: Czech and Slovak match
Tags: function, limit, algebra proposed, algebra
09.02.2006 22:24
Let $c=f(t_{t\to 1})$; $x\to 1$, $c=yf(y)$, so the graph is a hyperbolas. Can we do that?
09.02.2006 22:44
Unfortunately, we don't necessarily know that $c=\lim_{x\to 1}f(x)$ exists.
22.08.2022 15:42
Let $P(x,y)$ denote the given assertion. $P(x,2)\implies f(x)-f(2)=(2-x)f(2x)$ $P(x,4)\implies f(x)-f(4)=(4-x)f(4x)$ $P(2x,2)\implies f(2x)-f(2)=(2-2x)f(4x)$ We can do some algebra and derive $f(x)\equiv k/x,$ which indeed fits.
22.08.2022 16:12
ZETA_in_olympiad wrote: Let $P(x,y)$ denote the given assertion. $P(x,2)\implies f(x)-f(2)=(2-x)f(2x)$ $P(x,4)\implies f(x)-f(4)=(4-x)f(4x)$ $P(2x,2)\implies f(2x)-f(2)=(2-2x)f(4x)$ We can do some algebra and derive $f(x)\equiv k/x,$ which indeed fits. uh how Calculation gives us \[ f(x) = \frac{2f(4)(x^2 - 3x + 2) - f(2)(x^2 - 7x + 12)}{x(2x - 5)} \]
22.08.2022 17:55
@above it is provable that $2f(4)=f(2)$ and so $k=2f(2).$
22.08.2022 21:23
ZETA_in_olympiad wrote: @above it is provable that $2f(4)=f(2)$ and so $k=2f(2).$ How?
22.08.2022 21:38
RevolveWithMe101 wrote: ZETA_in_olympiad wrote: @above it is provable that $2f(4)=f(2)$ and so $k=2f(2).$ How? Easy. Use your formula to calculate $f(8)$ and then $P(2,4).$