en>Mathematician757: /* Factorization */

2013-07-13T14:19:27Z

Factorization

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	~~Friends call him Royal~~. ~~Delaware is our beginning location~~. ~~The preferred hobby~~ for ~~him~~ and ~~his children is to generate and now he is trying to earn cash with it~~. ~~Interviewing is what she does in her working day job but quickly her husband and her will begin their own business~~.<br><br>~~My weblog~~ - ~~extended car warranty (~~[~~http:~~//~~Www~~.~~Thenewhireguide.com~~/~~ActivityFeed~~/~~MyProfile~~/~~tabid~~/61/~~UserId~~/~~2133~~/~~Default~~.~~aspx address here~~])		{{dablink\|This article is not about [[Hermite's cotangent identity]].}}
			In [[mathematics]], '''Hermite's identity''', named after [[Charles Hermite]], gives the value of a [[summation]] involving the [[floor function]]. It states that for every [[real number]] ''x'' and for every positive [[integer]] ''n'' the following [[Identity (mathematics)\|identity]] holds:<ref>{{citation\|title=Mathematical Miniatures\|volume=43\|series=New Mathematical Library\|first1=Svetoslav\|last1=Savchev\|first2=Titu\|last2=Andreescu\|publisher=[[Mathematical Association of America]]\|year=2003\|isbn=9780883856451\|chapter=12 Hermite's Identity\|pages=41–44}}.</ref><ref>{{citation
			\| last = Matsuoka \| first = Yoshio
			\| doi = 10.2307/2311413
			\| issue = 10
			\| journal = [[The American Mathematical Monthly]]
			\| mr = 1533020
			\| page = 1115
			\| title = Classroom Notes: On a Proof of Hermite's Identity
			\| volume = 71
			\| year = 1964}}.</ref>

			: <math>\sum_{k=0}^{n-1}\left\lfloor x+\frac{k}{n}\right\rfloor=\lfloor nx\rfloor .</math>

			==Proof==
			Split <math>x</math> into its [[integer part]] and [[fractional part]], <math>x=\lfloor x\rfloor+\{x\}</math>. There is exactly one <math>k'\in\{1,\ldots,n\}</math> with
			:<math>\lfloor x\rfloor=\left\lfloor x+\frac{k'-1}{n}\right\rfloor\le x<\left\lfloor x+\frac{k'}{n}\right\rfloor=\lfloor x\rfloor+1.</math>
			By subtracting the same integer <math>\lfloor x\rfloor</math> from inside the floor operations on the left and right sides of this inequality, it may be rewritten as
			:<math>0=\left\lfloor \{x\}+\frac{k'-1}{n}\right\rfloor\le \{x\}<\left\lfloor \{x\}+\frac{k'}{n}\right\rfloor=1.</math>
			Therefore,
			:<math>1-\frac{k'}{n}\le \{x\}<1-\frac{k'-1}{n} ,</math>
			and multiplying both sides by <math>n</math> gives
			:<math>n-k'\le n\, \{x\}<n-k'+1.</math>

			Now if the summation from Hermite's identity is split into two parts at index <math>k'</math>, it becomes
			: <math>\sum_{k=0}^{n-1}\left\lfloor x+\frac{k}{n}\right\rfloor
			=\sum_{k=0}^{k'-1} \lfloor x\rfloor+\sum_{k=k'}^{n-1} (\lfloor x\rfloor+1)=n\, \lfloor x\rfloor+n-k'
			=n\, \lfloor x\rfloor+\lfloor n\,\{x\}\rfloor=\left\lfloor n\, \lfloor x\rfloor+n\, \{x\} \right\rfloor=\lfloor nx\rfloor.
			</math>

			==References==
			{{reflist}}

			{{DEFAULTSORT:Hermite's Identity}}
			[[Category:Mathematical identities]]
			[[Category:Articles containing proofs]]

en>BudgieJane: Expanded the factorization section

2012-03-11T13:10:23Z

Expanded the factorization section

New page

Friends call him Royal. Delaware is our beginning location. The preferred hobby for him and his children is to generate and now he is trying to earn cash with it. Interviewing is what she does in her working day job but quickly her husband and her will begin their own business.<br><br>My weblog - extended car warranty ([http://Www.Thenewhireguide.com/ActivityFeed/MyProfile/tabid/61/UserId/2133/Default.aspx address here])

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en>Mathematician757: /* Factorization */

en>BudgieJane: Expanded the factorization section