Question

[tex]1.Determinati\ prima\ zecimala\ a\ numarului\ a_n=\sqrt{n+n^2},n\in N^*.\\
2.Aratati\ ca\ pentru\ orice\ x,y\in R\ sunt\ adevarate\ urmatoarele:\\
a)[x]+[x+\frac{1}{2}]=[2x]\\
b)[x]=[y]\Rightarrow |x-y|\ \textless \ 1 (Reciproca\ este\ adevarata?)[/tex]

Answer 1

$1.~Vom~demonstra~ca~aceasta~cifra~este~4. ~Trebuie~sa~de-\\ \\monstram~ca~ 0,4\ \textless \ \{ \sqrt{n^2+n}\}\ \textless \ 0,5.~(*) \\ \\ n^2+n\ \textgreater \ n^2,~iar~n^2+n=n(n+1)\ \textless \ (n+1)^2,~deci \\ \\ (n+1)^2\ \textgreater \ n^2+n\ \textgreater \ n^2 \Rightarrow [ \sqrt{n^2+n}]=n. \\ \\ Deci~ \{ \sqrt{n^2+n} \}= \sqrt{n^2+n}-[ \sqrt{n^2+n} ]= \sqrt{n^2+n}-n. \\ \\ Ramane~de~demonstrat~ca: \\ \\ 0,4\ \textless \ \sqrt{n^2+n}-n\ \textless \ 0,5 \Leftrightarrow 0,4+n\ \textless \ \sqrt{n^2+n}\ \textless \ n+0,5 .$

$Ultima~relatie~poate~fi~echivalata~prin~ridicare~la~patrat \\ \\ (deoarece~toti~termenii~sunt~pozitivi): \\ \\ n^2+0,8n+0,16\ \textless \ \underbrace{n^2+n\ \textless \ n^2+n+0,25}_\mbox{evident~adevarat}} .\\ \\ Prima~parte~este~echivalenta~cu~0,2n\ \textgreater \ 0,16,~adevarat \\ \\ deoarece~n \geq 1. \\ \\ Cu~asta~am~finalizat~demonstratia.$

$2.~a)~Ei,~haide...~Este~identitatea~lui~Hermite~:P. \\ \\ Notam~[x]=k \in Z~si~ \{x \}=r \in [0,1).~Atunci~x=k+r. \\ \\ .[x]=[k+r]=k . \\ \\ Cazul~1:~r \in [0; \frac{1}{2} ) \Rightarrow [x+ \frac{1}{2} ]=k,~iar~[2x]=2k~(deoarece \\ \\ 2x=2k+2r,~iar~2k\ \textless \ 2k+2r\ \textless \ 2k+2r+1). \\ \\ Membrul~stang~este~k+k,~iar~membrul~drept~este~2k.~;) \\ \\ Cazul~2:~r \in[ \frac{1}{2} ;1) \Rightarrow [x+ \frac{1}{2}]=k+1,~iar~[2x]=2k+1. \\ \\ Membrul ~stang:~k+k+1,~membrul~drept:~2k+1.~;)$

$b)~[x]=[y]=k \Rightarrow( k \leq x \ \textless \ k+1)~si~k \leq y\ \textless \ k+1. \\ \\ Deci~(-k \geq -y\ \textgreater \ -k-1).~Insumand~relatiile~din~paranteze, \\ \\ obtinem:~-1\ \textless \ x-y\ \textless \ 1,~adica~|x-y|\ \textless \ 1. \\ \\ Reciproca~este~falsa.~Contraexemplu:~x=0,5~;~y=1,2. \\ \\ |0,5-1,2|\ \textless \ 1,~dar~[0,5] \neq [1,2].$