Let $\mathbb{R}^+$ denote the set of positive real numbers. Find all functions $f : \mathbb{R}^+\to\mathbb{R}^+$ that satisfy \[ \Big(1+yf(x)\Big)\Big(1-yf(x+y)\Big)=1\] for all $x,y\in\mathbb{R}^+$.
Problem
Source: Czech-Polish-Slovak Match, 2009
Tags: function, symmetry, algebra unsolved, algebra
Dijkschneier
21.08.2011 23:51
Equivalent to $f(x)=f(x+y)+yf(x)f(x+y)$, and putting z=x+y it becomes equivalent to $f(x)=f(z)+(z-x)f(x)f(z)$ for z>x>0 or $f(x)(1+xf(z)) = f(z)(1+zf(x))$ for all x,z > 0 (due to the symmetry of the last expression). Letting z=1, we get : $f(x)=\frac{a}{b+ax}$ where a and b are two constants, which is conversely a solution.
bah_luckyboy
22.08.2011 02:41
Dijkschneier wrote: Equivalent to $f(x)=f(x+y)+yf(x)f(x+y)$, and putting z=x+y it becomes equivalent to $f(x)=f(z)+(z-x)f(x)f(z)$ for z>x>0 or $f(x)(1+xf(z)) = f(z)(1+zf(x))$ for all x,z > 0 (due to the symmetry of the last expression). Letting z=1, we get : $f(x)=\frac{a}{b+ax}$ where a and b are two constants, which is conversely a solution. But you've said that $z=x+y$. How if we set a value $x+y$ such that $x+y\neq1$?
bah_luckyboy
22.08.2011 03:05
Then I'll post my solution here
Expanding the LHS, we'll get \[1+yf(x)-yf(x+y)-y^2f(x)f(x+y)=1\], and hence $yf(x)-yf(x+y)-y^2f(x)f(x+y)=0$. Dividing the last equation by $y$, we get $f(x)-f(x+y)-yf(x)f(x+y)=0$. Then $f(x+y)(1+yf(x))=f(x)$, and \[f(x+y)=\frac{f(x)}{1+yf(x)}\]By each side dividing 1, the equation becomes \[\frac{1}{f(x+y)}=\frac{1+yf(x)}{f(x)}\rightarrow \frac{1}{f(x+y)}=\frac{1}{f(x)}+y\]But note also that \[\frac{1}{f(y)}+x=\frac{1}{f(y+x)}=\frac{1}{f(x+y)}=\frac{1}{f(x)}+y\]hence $\frac{1}{f(x)}-x=\frac{1}{f(y)}-y$. Hence the value of $\frac{1}{f(x)}-x$ is equal to a constant $C$. Then $\frac{1}{f(x)}=x+C$, and therefore \[f(x)=\frac{1}{x+C}\]for all $x\in\mathbb{R}^+$ and a constant $C\in\mathbb{R}^+_0$
leosen
26.08.2011 00:02
$(1+yf(x))(1-yf(x+y))=1$, $x=0$ and $y$ is any real number different from $0$. \[(1+yf(0))(1-yf(0+y))=1\] \[1-yf(y) = \frac{1}{1+yf(0)}\] \[1-\frac{1}{1+yf(0)} = yf(y)\] \[\frac{1+ yf(0) - 1}{1+yf(0)} = yf(y)\] \[ \frac{yf(0)}{1+yf(0)} = y f(y)\] \[\frac{f(0)}{1+yf(0)}= f(y)\] If $f(0)=K$, then \[f(x) = \frac{k}{1+kx}\]
prasanna1712
26.08.2011 00:39
@leosen: it's on $ \mathbb{R}^{+} $ so you can't use $x=0$.
abhinavzandubalm
26.08.2011 10:51
A very simple solution ( based on Dijkschneier's idea ). Assume that $f$ is non constant. We get $1+yf(x)-yf(x+y)-y^2f(x)f(x+y)=1$. And $y\ne0$. $\implies f(x)=f(x+y)+yf(x)f(x+y)$. Put $x+y=z$. $\implies f(x)-f(z)=(z-x)f(x)f(z)$. Put $g(x)=\frac{1}{f(x)}$.
Suppose that $f(a)=0$ for any one $a$ ,
then putting $x=a$ in the original equation we get $yf(a+y)=0$.
Hence for all numbers $f(x)=0$.
Hence $f$ is constant, but our assumption states otherwise.
We get $\implies g(z)-g(x)=z-x$. $\implies g(x)-x=a$. $\implies f(x)=\frac{1}{x+a}$. And $f(x)=0$ is also another solution Which are the only solutions and are easily verified.
Dijkschneier
27.08.2011 00:16
bah_luckyboy wrote: Dijkschneier wrote: Equivalent to $f(x)=f(x+y)+yf(x)f(x+y)$, and putting z=x+y it becomes equivalent to $f(x)=f(z)+(z-x)f(x)f(z)$ for z>x>0 or $f(x)(1+xf(z)) = f(z)(1+zf(x))$ for all x,z > 0 (due to the symmetry of the last expression). Letting z=1, we get : $f(x)=\frac{a}{b+ax}$ where a and b are two constants, which is conversely a solution. But you've said that $z=x+y$. How if we set a value $x+y$ such that $x+y\neq1$? ??
abhinavzandubalm
28.08.2011 13:49
I made some slight errors which I have just edited. So my solution should be fine now.