Let $f:\mathbb{R}^+\to\mathbb{R}^+$ be a function that satisfies for all $x>y>0$ \[f(x+y)-f(x-y)=4\sqrt{f(x)f(y)}\] a) Prove that $f(2x)=4f(x)$ for all $x>0$; b) Find all such functions. Nikolai Nikolov, Oleg Mushkarov
Problem
Source: Bulgarian National Olympiad 2006 First day problem 2
Tags: function, limit, induction, inequalities, algebra proposed, algebra
21.05.2006 22:45
I denote : $f(u-) = \lim_{(t\rightarrow u)\& (t<u)}f(t) ,\ f(u+)= \lim_{(t\rightarrow u)\& (t>u)}f(t)$ $f$ is increasing , so $f(0+),f(x-),f(x+)$ exist for all $x\in ]0,+\infty[$ You make $x\rightarrow 0 , y\rightarrow 0, x > y$ in the definition relation ship, and you obtain : $f(0+)-f(0+) = 4\sqrt{f^2(0+)}$, so $f(0+)=0$ . You make $y\rightarrow 0$ in the definition relation ship, and you obtain : $f(x+)-f(x-) = 4\sqrt{f(x+)f(0+)} =0$, so $f(x+)=f(x-)$, so $f$ is continous because $f$ is increasing. By induction you prove $f(nx)=n^2f(x)$ where $n$ is natural. From $m^2f(1)=f(m)=f(n\frac mn)=n^2f(\frac mn)$ you obtain $f(\frac mn)=\frac{m^2}{n^2}f(1)$ for all $\frac mn \in Q^+$(positive rational numbers) Finally $f(x)=x^2f(1)$, because of $f$ is continous . Do you follow me ?
22.05.2006 18:54
Diogene wrote: I denote : $f(u-) = \lim_{(t\rightarrow u)\& (t<u)}f(t) ,\ f(u+)= \lim_{(t\rightarrow u)\& (t>u)}f(t)$ $f$ is increasing , so $f(0+),f(x-),f(x+)$ exist for all $x\in ]0,+\infty[$ You make $x\rightarrow 0 , y\rightarrow 0, x > y$ in the definition relation ship, and you obtain : $f(0+)-f(0+) = 4\sqrt{f^2(0+)}$, so $f(0+)=0$ . You make $y\rightarrow 0$ in the definition relation ship, and you obtain : $f(x+)-f(x-) = 4\sqrt{f(x+)f(0+)} =0$, so $f(x+)=f(x-)$, so $f$ is continous because $f$ is increasing. By induction you prove $f(nx)=n^2f(x)$ where $n$ is natural. From $m^2f(1)=f(m)=f(n\frac mn)=n^2f(\frac mn)$ you obtain $f(\frac mn)=\frac{m^2}{n^2}f(1)$ for all $\frac mn \in Q^+$(positive rational numbers) Finally $f(x)=x^2f(1)$, because of $f$ is continous . Do you follow me ? How do you guess to get it in $lim$ or you do it by setting something inseat of $x$ there is some problems that are solved by $log$ or$ln$ or $lg$so i am stopped there I understood but i think i cannot do in new problems , can you give some advices. Abdurashid
26.12.2014 11:43
bilarev wrote: Let $f:\mathbb{R}^+\to\mathbb{R}^+$ be a function that satisfies for all $x>y>0$ \[f(x+y)-f(x-y)=4\sqrt{f(x)f(y)}\] a) Prove that $f(2x)=4f(x)$ for all $x>0$; b) Find all such functions. Nikolai Nikolov, Oleg Mushkarov a) Let $f\left( x \right) = a,f\left( {2x} \right) = b$. We have $f\left( {3x} \right) - f\left( x \right) = 4\sqrt {f\left( {2x} \right)f\left( x \right)} \Rightarrow f\left( {3x} \right) = a + 4\sqrt {ab} $ $f\left( {4x} \right) - f\left( {2x} \right) = 4\sqrt {f\left( {3x} \right)f\left( x \right)} \Rightarrow f\left( {4x} \right) = b + 4\sqrt {a\left( {a + 4\sqrt {ab} } \right)} $ $f\left( {5x} \right) - f\left( x \right) = 4\sqrt {f\left( {3x} \right)f\left( {2x} \right)} \Rightarrow f\left( {5x} \right) = a + 4\sqrt {b\left( {a + 4\sqrt {ab} } \right)} $ $f\left( {5x} \right) - f\left( {3x} \right) = 4\sqrt {f\left( {4x} \right)f\left( x \right)} \Rightarrow f\left( {5x} \right) = a + 4\sqrt {ab} + 4\sqrt {a\left( {b + 4\sqrt {a\left( {a + 4\sqrt {ab} } \right)} } \right)} $ So we have $\begin{array}{l} a + 4\sqrt {b\left( {a + 4\sqrt {ab} } \right)} = a + 4\sqrt {ab} + 4\sqrt {a\left( {b + 4\sqrt {a\left( {a + 4\sqrt {ab} } \right)} } \right)} \\ \Leftrightarrow \sqrt {b\left( {a + 4\sqrt {ab} } \right)} = \sqrt {ab} + \sqrt {a\left( {b + 4\sqrt {a\left( {a + 4\sqrt {ab} } \right)} } \right)} \end{array}$ $\begin{array}{l} \Leftrightarrow {\left( {\sqrt {b\left( {a + 4\sqrt {ab} } \right)} } \right)^2} = {\left( {\sqrt {ab} + \sqrt {a\left( {b + 4\sqrt {a\left( {a + 4\sqrt {ab} } \right)} } \right)} } \right)^2}\\ \Leftrightarrow b\left( {a + 4\sqrt {ab} } \right) = ab + a\left( {b + 4\sqrt {a\left( {a + 4\sqrt {ab} } \right)} } \right) + 2\sqrt {ab} .\sqrt {a\left( {b + 4\sqrt {a\left( {a + 4\sqrt {ab} } \right)} } \right)} \\ \Leftrightarrow 4b\sqrt b = b\sqrt a + 4a\sqrt {a + 4\sqrt {ab} } + 2\sqrt b .\sqrt {a\left( {b + 4\sqrt {a\left( {a + 4\sqrt {ab} } \right)} } \right)} \end{array}$ Use assess inequality If $a > \frac{1}{4}b$ right-hand side > left - hand side, and If $a < \frac{1}{4}b$ right-hand side > left - hand side. So we have $a = \frac{1}{4}b \Rightarrow f\left( {2x} \right) = 4f\left( x \right)$