Vreau si eu o demonstratie la identitatea lui Hermite :
\( [x]+[x+\frac{1}{n}]+[x+\frac{2}{n}]+...+[x+\frac{n-1}{n}]=[nx] \)
Identitatea lui Hermite
Moderator: Marius Dragoi
Cea mai simpla pe care o stiu :
Notam \( [x]=k\in \mathbb{Z} \Rightarrow \)\( x\in [k;k+1)=[k;k+\frac{1}{n})\cup [k+\frac{1}{n};k+\frac{2}{n})\cup ... \)\( \cup [k+\frac{n-1}{n};k+\frac{n}{n}) \) , deci \( (\exists)i \in \{0;1;...;n-1\} \) cu \( x\in [k+\frac{i}{n};k+\frac{i+1}{n}) \) .
Deci \( S=\sum_{p=0}^{n-1}[x+\frac{p}{n}]=\sum_{p=0}^{n-i-1}[x+\frac{p}{n}]+\sum_{p=n-1}^{n}[x+\frac{p}{n}] \) , unde \( [x+\frac{p}{n}]=k \) , daca \( p=\overline{0;n-i-1} \) si \( [x+\frac{p}{n}]=k+1 \) , daca \( p= \overline{n-1;n} \)
Deci \( S=(n-1)k+i(k+1)=nk+i \) , iar \( [nx]=nk+i \) .
Notam \( [x]=k\in \mathbb{Z} \Rightarrow \)\( x\in [k;k+1)=[k;k+\frac{1}{n})\cup [k+\frac{1}{n};k+\frac{2}{n})\cup ... \)\( \cup [k+\frac{n-1}{n};k+\frac{n}{n}) \) , deci \( (\exists)i \in \{0;1;...;n-1\} \) cu \( x\in [k+\frac{i}{n};k+\frac{i+1}{n}) \) .
Deci \( S=\sum_{p=0}^{n-1}[x+\frac{p}{n}]=\sum_{p=0}^{n-i-1}[x+\frac{p}{n}]+\sum_{p=n-1}^{n}[x+\frac{p}{n}] \) , unde \( [x+\frac{p}{n}]=k \) , daca \( p=\overline{0;n-i-1} \) si \( [x+\frac{p}{n}]=k+1 \) , daca \( p= \overline{n-1;n} \)
Deci \( S=(n-1)k+i(k+1)=nk+i \) , iar \( [nx]=nk+i \) .
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Laurian Filip
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Fie \( m=[x] \) => \( x=m+\frac{y}{n} \), unde \( y\in \mathbb{R}, 0 \leq y < n \).
Inlocuim in relatie, scoatem intregii si identitatea devine
\( nx+ [\frac{y}{n}]+[\frac{y}{n}+\frac{1}{n}]+...+[\frac{y}{n}+\frac{n-1}{n}]=nx+[y] \)
\( [\frac{y}{n}]+[\frac{y}{n}+\frac{1}{n}]+...+[\frac{y}{n}+\frac{n-1}{n}]=[y] \)
Fie k natural a.i. \( k\leq y < k+1 \).
Primii n-k termenii ai membrului stang vor fi nuli, iar ultimii k vor fi egali cu 1 (y<n), de unde \( LHS=k \) iar \( RHS=[y]=k \).
Deci am ajuns la concluzia ca identitatea de la inceput este echivalenta cu \( k=k \), deci este adevarata.
Inlocuim in relatie, scoatem intregii si identitatea devine
\( nx+ [\frac{y}{n}]+[\frac{y}{n}+\frac{1}{n}]+...+[\frac{y}{n}+\frac{n-1}{n}]=nx+[y] \)
\( [\frac{y}{n}]+[\frac{y}{n}+\frac{1}{n}]+...+[\frac{y}{n}+\frac{n-1}{n}]=[y] \)
Fie k natural a.i. \( k\leq y < k+1 \).
Primii n-k termenii ai membrului stang vor fi nuli, iar ultimii k vor fi egali cu 1 (y<n), de unde \( LHS=k \) iar \( RHS=[y]=k \).
Deci am ajuns la concluzia ca identitatea de la inceput este echivalenta cu \( k=k \), deci este adevarata.
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Virgil Nicula
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Re: Identitatea lui Hermite
Clasic. Se observa ca functia \( f(x)=\sum_{k=0}^{n-1}\left[x+\frac{k}{n}\right]-[nx]\ ,\ x\in\mathbb R \) este periodica ( \( T=1 \) )
si se arata ca \( f(x)=0 \) pentru orice \( x \) care apartine lui \( [0,1)=\bigcup_{s=0}^{n-1}\left[\frac sn,\frac {s+1}{n}\right) \) (partitie !). Intr-adevar,
\( (\forall )\ n\in\mathbb N \) , \( n\ge 2 \) si \( (\forall )\ x\in[0,1) \) exista si este unic \( s=[nx]\in\overline {0,n-1} \) astfel incat \( x\in\left[\frac sn,\frac {s+1}{n}\right) \) etc.
si se arata ca \( f(x)=0 \) pentru orice \( x \) care apartine lui \( [0,1)=\bigcup_{s=0}^{n-1}\left[\frac sn,\frac {s+1}{n}\right) \) (partitie !). Intr-adevar,
\( (\forall )\ n\in\mathbb N \) , \( n\ge 2 \) si \( (\forall )\ x\in[0,1) \) exista si este unic \( s=[nx]\in\overline {0,n-1} \) astfel incat \( x\in\left[\frac sn,\frac {s+1}{n}\right) \) etc.
Last edited by Virgil Nicula on Thu Feb 12, 2009 10:01 pm, edited 1 time in total.
Re: Identitatea lui Hermite
De fapt, perioada principala este \( T=\frac1n \) si \( f(x)=0 \) pentru orice \( x\in [0,\frac1n) \).Virgil Nicula wrote:Clasic. Se observa ca functia \( f(x)=\sum_{k=0}^{n-1}\left[x+\frac{k}{n}\right]-[nx]\ ,\ x\in\mathbb R \) este periodica ( \( T=1 \) )
si se arata ca \( f(x)=0 \) pentru orice \( x \) care apartine lui \( [0,1)=\bigcup_{s=0}^{n-1}\left[\frac sn,\frac {s+1}{n}\right) \).
Bogdan Enescu