Question

Salut! Am nevoie si eu de o demonstrare a identitatii lui Hermite , dar sa nu fie de pe net . Astept raspunsuri cat mai bune ! :)

Answer 1

$\displaystyle Identitatea~lui~Hermite:~Pentru~orice~numar~real~x~si~orice \\ \\ numar~natural~n \geq 2,~avem: \\ \\ \boxed{\left [ x \right]+ \left[ x + \frac{1}{n} \right ]+ \left[x+ \frac{2}{n} \right]+... + \left [ x+ \frac{n-1}{n} \right]=\left [nx \right ]}~.$

$\displaystyle Demonstratia~este~foarte~cunoscuta,~deci~ma~indoiesc~ca~vei \\ \\ gasi~o~solutie~diferita.~Se~demonstreaza~pentru~n=2,~si \\ \\ restul~rezulta~prin~analogie. \\ \\ \underline{Demonstratie~(n=2)}:~Notam~\left[ x \right]=k \in \mathbb{Z}~si~ \left \{ x\right \}=r \in \left[0;1 \right). \\ \\ Evident,~x=k+r. \\ \\ Distingem~doua~cazuri:$

$\displaystyle i)~r \in \left[0; \frac{1}{2} \right).~In~acest~caz~x+ \frac{1}{2}=k+r+ \frac{1}{2}\ \textless \ k+1.~Si~cum~ \\ \\ k+r+ \frac{1}{2}\ \textgreater \ k \Rightarrow x+ \frac{1}{2} \in \left[k;k+1 \right) \Rightarrow \left[x+ \frac{1}{2}\right]=k . \\ \\ Deci~ \left [ x \right]+ \left[ x+ \frac{1}{2} \right]=k+k=2k.~~~(*) \\ \\ 2x=2k+2r \geq 2k~si~2x=2k+2r\ \textless \ 2k+1 \Rightarrow 2x \in \left[2k;2k+1 \right) \Rightarrow \\ \\ \Rightarrow \left[ 2x \right]=2k.~~~(**) \\ \\ Din~(*)~si~(**)~rezulta~concluzia~in~acest~caz.$

$\displaystyle ii)~Daca~r \in \left[ \frac{1}{2} ;1 \right)~avem: \\ \\ x+ \frac{1}{2}=k+r + \frac{1}{2} \geq k+1~si~x+ \frac{1}{2}=k+r+ \frac{1}{2}\ \textless \ k+2. \\ \\ Deci~\left[ x\right]+ \left[ x+ \frac{1}{2}\right]=k+k+1=2k+1. \\ \\ 2x=2k+2r \geq 2k+1~si~2x=2k+2r\ \textless \ 2k+2 \Rightarrow [2x]=2k+1. \\ \\ Deci~identitatea~este~demonstrata~si~in~acest~caz.$

$\displaystyle Pentru~un~n~arbitrar~(n \geq 2)~analizam~n~cazuri: \\ \\ r \in \left[ 0; \frac{1}{n}\right)~;~ r \in \left[ \frac{1}{n}; \frac{2}{n} \right)~;~...~;~ r \in \left[ \frac{n-1}{n};1 \right).$