Parte fractionara

[Marius Mainea]

Pentru \( n\in\mathbb{N^{\ast}} \), definim functia \( f_n:[0,1]\rightarrow\mathbb{R} \),

\( f_n(x)=\{nx\}(1-\{nx\}) \).

Sa se demonstreze ca:
a) Valoarea minima ,m si valoarea maxima ,M ale lui \( f_n \) nu depind de n;
b) Ecuatia \( f_n(x)=M \) are n solutii distincte \( x_{n,1},x_{n,2},...,x_{n,n} \);
c) Pentru \( m,n\in \mathbb{N^{\ast}} \), are loc egalitatea

\( \frac{x_{n,1}+x_{n,2}+...+x_{n,n}}{x_{m,1}+x_{m,2}+...+x_{m,m}}=\frac{n}{m} \)

D. Piciu, ,,Gh.Titeica'' 2005

[Mateescu Constantin]

a) Cum \( \{nx\}\in[0,\ 1),\ (\forall)x\in[0,\ 1] \) si \( (\forall)n\in\mathbb{N}^{\ast}, \) obtinem ca \( f_n(x)\geq0=f_n(0)=f_n(1),\ (\forall)x\in[0,\ 1]\ \c{s}i \ (\forall)n\in\mathbb{N}^{\ast}, \) deci \( m=0 \), care nu depinde de \( n \).

Din \( x\in[0,\ 1] \) si \( n\in\mathbb{N}^{\ast} \) avem ca \( nx\in[0,\ n]=\bigcup_{k=0}^{n-1}\left\[k,k+1\right\)\ \cup\ \left\{ n \right\}\ \Longrightarrow\ [0\ ,\ 1]=\bigcup_{k=0}^{n-1}\left\[\frac{k}{n},\ \frac{k+1}{n}\right\)\ \bigcup\ \left\{1\right\}. \)

Folosind \( \{x\}=x-[x], \) \( (\forall)x\in\mathbb{R}, \) obtinem usor ca daca \( x=1 \) atunci \( f_n(1)=0\ si\ f_n(x)=\frac{1}{4}-\left\(nx-\frac{2k+1}{2}\right\)^{2}, \) daca \( x\in\left\[\frac{k}{n},\ \frac{k+1}{n}\right\)\ ,\ k=\overline{0,\ n-1}. \)

Prin urmare, \( f_n(x)\leq\frac{1}{4},\ (\forall)x\in[0,\ 1],\ (\forall)n\in\mathbb{N}^{\ast}, \) deci \( M=\frac{1}{4}. \)

b) Folosind a), obtinem ca \( f_n(x)=\frac{1}{4}\ \Longleftrightarrow x\in\{x_{n,1},\ x_{n,2},\ \dots,\ x_{n,n}|x_{n,k+1}=\frac{2k+1}{2n},\ k={\overline {0,\ n-1}\}. \)

c) Calcul direct, folosind b).