Ｔa ｙl ｏr ＇s Ｆormu ｌa an ｄ the St ｕdy ｏf Ext ｒema1. Ｔay ｌｏr's Ｆorm ｕｌa for M ａpping ｓTh ｅｏre ｍ 1. If ａ mapping Y U f →: from a ｎeigh ｂo ｒho ｏｄ ()x U U = of a point x in a no ｒm ｅd ｓｐａce X into a no ｒmed spac ｅ Y ha ｓ d ｅｒiva ｔiv ｅs up ｔo o ｒde ｒ n -１ in ｃlus ｉve ｉｎ U an ｄ h ａｓ ａn n-th ｏrde ｒ deriv ａti ｖe ()()x f n at t ｈe p ｏint x, ｔhe ｎ()()()()()⎪⎭⎫ ⎝⎛++++=+n n n h o h x f n h x f x f h x f !1,(１）as 0→h ．Equ ａli ｔy （１) i ｓ one of the ｖariet ｉe ｓ ｏf Ｔa ｙlor's fo ｒmu ｌａ, writte ｎ here for ra ｔｈer ge ｎｅral c ｌas ｓｅs of ma ｐpings.Proof ． We p ｒｏv ｅ T ａy ｌｏr's ｆo ｒmu ｌa ｂｙ ｉnd ｕｃｔi ｏn. For 1=n it is tru ｅ ｂｙ d ｅｆinition ｏf ()x f ,.Ａs ｓume form ｕl ａ (1) is ｔrue for ｓo ｍe N n ∈-1.Ｔｈen b ｙ the mean －val ｕｅ th ｅo ｒｅm, fo ｒm ｕla （12) o ｆ Sec ｔ. 10.5, ａnd the ｉn ｄuc ｔion h ｙｐot ｈesis ， we obtain.()()()()()()()()()()()()()⎪⎭⎫ ⎝⎛=⎪⎭⎫ ⎝⎛=⎪⎪⎭⎫ ⎝⎛-+++-+≤⎪⎭⎫ ⎝⎛+++-+--<<n n n n n n h o h h o hh x f n h x f x f h x f h x f n h f x f h x f 11,,,,10!11sup !1x θθθθθ ，ａs 0→h .W ｅ s ｈall ｎot take t ｈｅ t ｉｍe h ｅre t ｏ di ｓcuss o ｔh ｅr v ｅr ｓi ｏns of Ｔayl ｏr ＇ｓ f ｏｒmu ｌａ, ｗhic ｈ are so ｍeti ｍes quite useful ． Ｔhey ｗer ｅ discu ｓse ｄ earlier in ｄｅｔail ｆｏr ｎume ｒica ｌ fun ｃti ｏｎs ． At thi ｓ point w ｅ le ａv ｅ it ｔｏ th ｅ reader to derive ｔhem (s ｅe, for e ｘample ， Pro ｂlem 1 below)．2. Ｍeth ｏd ｓ of St ｕdying Ｉｎte ｒｉor Ｅxt ｒemaUsin ｇ Ta ｙlor'ｓ formul ａ, we sha ｌｌ ｅｘhib ｉt nece ｓsar ｙ condi ｔio ｎs a ｎｄ a ｌso suffic ｉe ｎt c ｏndi ｔions for a ｎ ｉｎｔe ｒior l ｏcal ex ｔre ｍum ｏf r ｅａl －ｖal ｕe ｄ fun ｃtions def ｉned on ａn ｏp ｅn su ｂse ｔ ｏf ａ nor ｍed sp ａce. Ａs we sha ｌl ｓee ， ｔｈｅse cond ｉt ｉo ｎs are an ａｌogous to t ｈe diffe ｒｅnti ａl conditi ｏｎs alr ｅady known t ｏ ｕs fo ｒ ａｎ ex ｔr ｅmum o ｆ a ｒeal －ｖal ｕe ｄ ｆunc ｔion of a ｒe ａｌ ｖariabl ｅ.Ｔｈeorem 2. Let R U f →: be ａ ｒeal-v ａl ｕed functio ｎ d ｅｆｉned o ｎ ano ｐｅｎ set Ｕ in a ｎorm ｅd ｓpa ｃｅ X a ｎd ｈaving ｃontinuous derivatives up to ｏrder 11≥-k inclu ｓiv ｅ ｉn a neig ｈb ｏｒh ｏod of a point U x ∈ an ｄ a d ｅｒｉvative()()x f k of order ｋ at th ｅ p ｏi ｎt ｘ i ｔself. If ()()()0,,01,==-x f x f k an ｄ ()()0≠x f k , ｔhen for x ｔo be ａn e ｘt ｒｅm ｕm of ｔｈｅ fu ｎｃt ｉｏn ｆ ｉt ｉs:n ｅces ｓary ｔhat k be ev ｅn and th ａt the form()()k k h x f b ｅ s ｅmid ｅfini ｔｅ,a ｎd ｓｕfficient ｔh ａt the v ａlues o ｆ t ｈe ｆor ｍ()()k k h x f ｏn th ｅ ｕnit s ｐｈere 1=h be b ｏund ｅｄ away ｆr ｏm ze ｒo; ｍo ｒeov ｅｒ, x i ｓ a lo ｃal ｍinimum ｉｆ the i ｎｅquali ｔi ｅs()()0>≥δk k h x f ,h ｏｌd o ｎ that ｓｐｈere, and a ｌo ｃal ma ｘimum if()()0<≤δk k h x f ，Ｐroof. For the pr ｏof we con ｓi ｄe ｒ the Taylor ｅｘｐansio ｎ (1) of f i ｎ a ｎeigh ｂo ｒhood of x ． Ｔh ｅ ass ｕmptions e ｎable u ｓ ｔo w ｒite()()()()()k k k h h h x f k x f h x f α+=-+!1where ()h α is a re ａl-valued f ｕnc ｔio ｎ, and ()0→h α as 0→h ．W ｅ fi ｒs ｔ prove th ｅ neces ｓａｒｙ co ｎditions.Since ()()0≠x f k , th ｅr ｅ ｅxist ｓ a v ｅctor 00≠h on w ｈic ｈ()()00≠k k h x f . The ｎ for values of th ｅ real pa ｒam ｅｔer ｔ suffici ｅntly close to ｚero ，()()()()()()k k k th th th x f k x f th x f 0000!1α+=-+()()()kk k k t h th h x f k ⎪⎭⎫ ⎝⎛+=000!1αa ｎd the ex ｐr ｅｓsion in the outer pa ｒent ｈｅｓes h ａs the same sign a ｓ()()k k h x f 0.For ｘ ｔo be an e ｘtre ｍum i ｔ i ｓ neces ｓary for the left-hand ｓid ｅ (an ｄ h ｅnce ａlso ｔｈｅ ｒight-h ａｎd side) ｏf this last e ｑuality to be of co ｎｓｔan ｔ sign ｗｈｅｎ t c ｈan ｇes sig ｎ. B ｕt th ｉｓ is pos ｓible only ｉf k is ev ｅn ．Thi ｓ reaso ｎing sho ｗs that if ｘ ｉｓ an ex ｔre ｍum, ｔhen the sign o ｆ the diffe ｒen ｃe ()()x f th x f -+0 ｉs th ｅ ｓａme a ｓ th ａt of ()()k k h x f 0 fo ｒ ｓuffi ｃi ｅｎt ｌy sma ｌl t; hen ｃe in that ca ｓｅ th ｅｒｅ ｃａｎn ｏt be two vect ｏrs0h , 1hａt whi ｃｈ th ｅ for ｍ ()()x f k ａs ｓum ｅs val ｕes ｗi ｔh o ｐpo ｓi ｔｅ s ｉgns. We ｎow ｔurn to t ｈe ｐro ｏf ｏｆ t ｈｅ su ｆficie ｎcy ｃon ｄit ｉo ｎs. For ｄe ｆi ｎitene ｓs we consider the ｃa ｓｅ w ｈｅn()()0>≥δk k h x f for 1=h ． Then()()()()()k k k h h h x f k x f h x f α+=-+!1()()()k k k h h h h x f k ⎪⎪⎪⎭⎫ ⎝⎛+⎪⎪⎭⎫ ⎝⎛=α!1 ()k h h k ⎪⎭⎫ ⎝⎛+≥αδ!1ａnd ， sinc ｅ ()0→h α as 0→h , the l ａs ｔ ter ｍ i ｎ this ine ｑuality ｉs ｐos ｉtive for all vect ｏr ｓ 0≠h suff ｉciently clo ｓe to zero. Thu ｓ， for al ｌ su ｃh ｖe ｃtors h, ()()0>-+x f h x f ,t ｈat i ｓ, x is a ｓtri ｃt ｌoc ａｌ m ｉnimu ｍ.Th ｅ ｓuf ｆｉcie ｎｔ cond ｉｔio ｎ for a strict lo ｃal m ａx ｉmum i ｓ v ｅrified sim ｉl ｉarly ．Remark 1. If t ｈｅ space X is ｆi ｎite-dimensi ｏｎal, t ｈe u ｎit sp ｈere ()1;x S with ｃente ｒ at X x ∈， bein ｇ a ｃlo ｓed ｂo ｕnd ｅd ｓubset of X, is c ｏmpac ｔ. T ｈe ｎ the co ｎtinuous ｆun ｃtio ｎ ()()()()k k i i i i k k h h x f h x f ⋅⋅∂= 11 (ａ k-f ｏrm) h ａs bｏt ｈ ａ maxim ａｌ ａnd a minimal value ｏn ()1;x S . Ｉｆ thes ｅ ｖal ｕes are of opposite s ｉgn, t ｈｅn f ｄoes no ｔ ｈａve a ｎ ex ｔr ｅmu ｍ a ｔ x. If they are both of the ｓame sig ｎ， ｔhen, as wa ｓ ｓh ｏw ｎ in Theorem ２, ｔｈere is ａn extr ｅmum. Ｉn ｔhe ｌatter case ， a s ｕｆficient con ｄition ｆo ｒ an extr ｅm ｕm ca ｎ ob ｖiously be state ｄ as the equi ｖale ｎt req ｕirement that the for ｍ ()()k k h x f ｂe eithe ｒ p ｏsitiv ｅ－ o ｒ negativ ｅ-ｄefi ｎit ｅ．It was t ｈis form ｏf ｔhe ｃonditio ｎ tha ｔ we en ｃoun ｔe ｒe ｄ in ｓtudying realvalu ｅd functi ｏns o ｎn R .Ｒｅm ａrk ２． As w ｅ h ａｖe seen in the ｅxam ｐle of function ｓR R fn →:, t ｈｅ se ｍi-de ｆｉni ｔｅness of th ｅ form ()()k k h x f e ｘhibit ｅd i ｎ th ｅ nec ｅss ａｒy ｃondit ｉo ｎs ｆor ａn ｅx ｔr ｅm ｕｍ is not a suffi ｃie ｎt ｃrit ｅｒi ｏn for a ｎ extremum.R ｅｍａr ｋ 3. In practice, when ｓｔudying extrema ｏf dif ｆer ｅntiabl ｅ fu ｎct ｉon ｓ o ｎe no ｒmally u ｓes only the ｆirst ｏr se ｃｏn ｄ dif ｆeren ｔia ｌｓ. Ifｔｈe un ｉq ｕene ｓs ａnd type o ｆ extremum ａｒe ob ｖｉous from th ｅ mea ｎing o ｆ th ｅ ｐrobl ｅm being ｓｔud ｉed, one ｃan r ｅs ｔri ｃt atte ｎt ｉon ｔo the first di ｆf ｅrential ｗh ｅn s ｅeking a ｎ extr ｅmum, si ｍp ｌy fin ｄｉｎｇ the point ｘ w ｈｅr ｅ()0,=x f3. S ｏme Ｅxam ｐl ｅsE ｘample 1. Ｌe ｔ()()R R C L ;31∈ ａnd ()[]()R b a C f ;,1∈. Ｉn oth ｅr ｗords, ()()321321,,,,u u u L u u u i ｓ a ｃo ｎｔi ｎuousl ｙ di ｆｆe ｒenti ａble r ｅa ｌ-v ａlued ｆｕn ｃt ｉｏn d ｅf ｉned in 3R and ()x f x a smoot ｈ r ｅal-va ｌｕｅd f ｕnction ｄｅfined o ｎ ｔhe closed interv ａl []R b a ⊂,.C ｏnsid ｅr the fu ｎｃtion()[]()R R b a C F →;,:1(2)d ｅf ｉne ｄ by t ｈe re ｌat ｉon()[]()()f F R b a C f ;,1∈()()()R dx x f x f x L ba ∈=⎰,,,(3）Thus ， (2） i ｓ a r ｅa ｌ-ｖａｌｕe ｄ fun ｃti ｏn ａl ｄｅｆin ｅd on the set of functi ｏns ()[]()R b a C ;,1.Ｔhe b ａsi ｃ v ａr ｉati ｏn ａｌ pri ｎｃipl ｅs ｃonnect ｅd ｗith motio ｎ are ｋno ｗn ｉn physics and ｍｅｃhan ｉc ｓ. Accord ｉng t ｏ thes ｅ pri ｎci ｐle ｓ， th ｅ a ｃt ｕal motion ｓ ａre ｄistinguishe ｄ ａｍong al ｌ the c ｏnceiva ｂle ｍｏtions in that they proc ｅed alo ｎｇ ｔｒa ｊe ｃtor ｉes alon ｇ whic ｈ cert ａi ｎ ｆｕnctio ｎals have a ｎ e ｘtremu ｍ. Que ｓt ｉons ｃｏnne ｃte ｄ wi ｔh ｔhe extrem ａ of f ｕn ｃt ｉｏnals are ce ｎtral i ｎ opt ｉｍal co ｎtrol theory ． Ｔhu ｓ, finding a ｎd st ｕdying t ｈe e ｘｔrem ａ ｏf functio ｎals is a p ｒob ｌｅmof int ｒｉｎsic impo ｒta ｎce, an ｄ th ｅ t ｈeo ｒｙ a ｓsoc ｉate ｄ with it is ｔh ｅ subjec ｔ of a larg ｅ are ａ of anal ｙｓis - ｔhe calculu ｓ ｏｆ va ｒiatio ｎs. We ha ｖe alr ｅａdy done a fe ｗ ｔh ｉn ｇs to ｍake the tr ａn ｓi ｔion ｆrom ｔｈe ａnalysis o ｆ t ｈｅ e ｘtre ｍａ ｏｆ n ｕmer ｉcal f ｕnctions t ｏ t ｈe problem of finding an ｄ ｓｔud ｙi ｎg ｅｘtr ｅma of ｆｕnctiona ｌs s ｅem ｎatura ｌ to t ｈe r ｅａｄer. Ho ｗｅｖe ｒ, w ｅ ｓｈall no ｔ ｇo ｄe ｅply into the sp ｅcial probl ｅms o ｆ variation ａl ｃa ｌｃul ｕs, but ｒa ｔｈｅr use ｔhe e ｘa ｍｐle of ｔhe f ｕnctional (３) t ｏ ｉｌｌu ｓｔra ｔe on ｌy the ｇｅne ｒal ide ａs o ｆ ｄi ｆferen ｔiat ｉon a ｎd study ｏf ｌｏcal ｅxtrema considered ａbove ．We sh ａll s ｈow tha ｔ th ｅ func ｔｉon ａl (３) is ａ ｄi ｆfe ｒe ｎt ｉate ma ｐpi ｎg an ｄ f ｉn ｄ i ｔs diffe ｒｅntial.W ｅ remark t ｈat th ｅ ｆunc ｔi ｏn (3） can b ｅ ｒegarde ｄ as the c ｏmp ｏs ｉtion o ｆ ｔhe m ａｐping ｓ()()()()()x f x f x L x f F ,1,,= (4)de ｆｉne ｄ ｂy the formul ａ()[]()[]()R b a C R b a C F ;,;,:11→(5）f ｏｌlo ｗed by ｔhe mapping[]()()()R dx x g g F R b a C g ba ∈=∈⎰2;, (６)B ｙ ｐrop ｅr ｔｉｅs of ｔhe integ ｒａl, the map ｐi ｎg 2F ｉｓ ob ｖｉously linear ａn ｄ c ｏntinuou ｓ, so t ｈat its ｄifferent ｉab ｉｌity ｉs c ｌe ａr. We shall s ｈow that t ｈe ｍａppi ｎｇ1F is ａlso d ｉｆｆer ｅn ｔiabl ｅ， ａnd t ｈat()()()()()()()()()()x h x f x f x L x h x f x f x L x h f F ,,3,2,1.,,,∂+∂=(7）fo ｒ ()[]()R b a C h ;,1∈.Indeed, by ｔhe ｃｏr ｏｌl ａry to ｔh ｅ mea ｎ-value ｔheo ｒｅｍ, we c ａn wr ｉte in ｔhe pre ｓe ｎｔ ca ｓe()()()i i i u u u L u u u L u u u L ∆∂--∆+∆+∆+∑=32131321332211,,,,,,()()()()()()∆⋅∂-∆+∂∂-∆+∂∂-∆+∂≤<<u L u L u L u L u L u L 3312211110sup θθθθ ()()ii i i u L u u L i ∆⋅∂-+∂≤=≤≤=3,2,110max max 33,2,1θθ(8)whe ｒe ()321,,u u u u = a ｎd ()321,,∆∆∆=∆.Ｉf we no ｗ recall t ｈａt the norm()1c f of t ｈｅ ｆｕn ｃｔｉon f in ()[]()R b a C ;,1 is ⎭⎬⎫⎩⎨⎧c c f f ,,max (where c f i ｓ the max ｉｍum ab ｓｏlute value of ｔhe f ｕnct ｉo ｎ o ｎ the ｃlosed interval []b a ,）, ｔhe ｎ, ｓe ｔtin ｇx u =1,()x f u =2， ()x f u ,3=, 01=∆, ()x h =∆2, and ()x h ,3=∆, ｗe ob ｔain f ｒom ｉnequa ｌi ｔy （8)， taking a ｃｃount o ｆ th ｅ ｕni ｆorm ｃontin ｕity o ｆ ｔhe functi ｏns ()3,2,1,,,321=∂i u u u L i , on bo ｕｎｄｅｄ sub ｓｅｔs o ｆ3R , t ｈa ｔ ()()()()()()()()()()()()()()()()x h x f x f x L x h x f x f x L x f x f x L x h x f x h x f x L b x ,,3,2,,,0,,,,,,,,max ∂-∂--++≤≤ ()()1c h o = ａｓ ()01→c hBut thi ｓ ｍｅans ｔha ｔ Eq. (7) hold ｓ．B ｙ t ｈｅ c ｈain rule ｆor dif ｆerenti ａting a com ｐｏsite functi ｏn, ｗe ｎow c ｏn ｃlude that ｔｈe f ｕnctional (3) is in ｄeed di ｆｆer ｅntiab ｌe ， and()()()()()()()()()()⎰∂+∂=ba dx x h x f x f x L x h x f x f x L h f F ,,3,2,,,,,(9）We of ｔen cons ｉde ｒ the ｒe ｓtrict ｉon o ｆ ｔh ｅ fu ｎct ｉo ｎa ｌ (3) to ｔhe affine spac ｅ ｃons ｉsti ｎg ｏf the functio ｎs ()[]()R b a C f ;,1∈ that assum ｅ fixed v ａｌue ｓ ()A a f =, ()B b f = at the end ｐoi ｎｔs ｏf ｔhe closed inter ｖａl []b a ,. Ｉn thi ｓ case, the fu ｎction ｓ h in the ｔangent s ｐａc ｅ()1f TC ， ｍust ｈave ｔhe value ｚero at the endpoints o ｆ ｔhe ｃlo ｓed ｉｎｔerval []b a ,. Ｔａｋｉng t ｈi ｓ f ａct i ｎｔo account, ｗe ｍay ｉｎt ｅgrate ｂy parts in （９) ａnd bring ｉt ｉnt ｏ the for ｍ()()()()()()()()⎰⎪⎭⎫ ⎝⎛∂-∂=b a dx x h x f x f x L dx d x f x f x L h f F ,3,2,,,,,(1０)o ｆ course ｕn ｄer the assu ｍption ｔh ａt L and ｆ belo ｎｇ ｔo the co ｒrespon ｄing clas ｓ ()2C .In ｐarticula ｒ, if f is ａn e ｘtr ｅmum （ex ｔremal) of ｓuc ｈ a func ｔiona ｌ, ｔｈen b ｙ Theorem 2 ｗe ｈa ｖe ()0,=h f F f ｏr every ｆunction ()[]()R b a C h ;,1∈ s ｕch that ()()0==b h a h . Fro ｍ ｔｈｉs ａnd ｒe ｌation (1０) one c ａn easily co ｎclu ｄe （see P ｒobl ｅm 3 be ｌｏw) th ａt th ｅ ｆunction f mu ｓt sa ｔi ｓｆy ｔhe e ｑu ａt ｉo ｎ()()()()()()0,,,,,3,2=∂-∂x f x f x L dx d x f x f x L(11)This is ａ fr ｅq ｕen ｔl ｙ-enc ｏuntered form of the ｅｑu ａtion ｋn ｏw ｎ ｉｎ ｔhe calcul ｕｓ of var ｉat ｉｏns as ｔh ｅ Ｅuler －Ｌag ｒange e ｑuation. Le ｔ us now c ｏnsid ｅr ｓome sp ｅｃi ｆｉc ｅxa ｍｐles.Ｅxam ｐｌe 2. The ｓh ｏr ｔes ｔ－path pro ｂlemAmo ｎg all the c ｕr ｖｅs in ａ pl ａne joining tw ｏ f ｉxed p ｏi ｎts, find the cur ｖe ｔhat has ｍｉn ｉｍal le ｎｇth.The ａn ｓｗer ｉｎ this cas ｅ is o ｂvio ｕs ， a ｎd i ｔ r ａｔh ｅr serve ｓ as ａ ｃｈec ｋ o ｎ th ｅ forma ｌ ｃo ｍｐｕta ｔi ｏn ｓ ｗe wi ｌl b ｅ d ｏi ｎｇ later.Ｗe ｓｈal ｌ a ｓsume that a fixed Ｃart ｅs ｉan ｃｏord ｉnate sy ｓtem has ｂｅen ｃhosen i ｎ t ｈe pl ａne, in which th ｅ ｔw ｏ poin ｔs are, ｆor e ｘa ｍple, ()0,0 an ｄ ()0,1 . We confine our ｓelve ｓ ｔｏ jus ｔ the cu ｒves ｔｈa ｔ are the g ｒａp ｈs of funct ｉons ()[]()R C f ;1,01∈ assuming t ｈe va ｌue ｚero at both en ｄs of t ｈe c ｌo ｓe ｄ interval []1,0 ． T ｈｅ ｌｅｎgth of such a c ｕrve()()()⎰+=102,1dx x f f F(12)depe ｎｄｓ ｏn ｔｈｅ ｆun ｃｔion ｆ and is a func ｔiona ｌ of the t ｙｐｅ cons ｉdered in Ex ａｍple １. In thi ｓ c ａｓe the f ｕn ｃt ｉon L has the for ｍ ()()233211,,u u u u L +=ａｎｄ ｔｈer ｅｆo ｒe the n ｅcess ａry condition (11) for ａn e ｘtrem ａl ｈｅｒｅ re ｄｕces to th ｅ equation ()()()012,,=⎪⎪⎪⎭⎫ ⎝⎛+x f x f dx d ｆr ｏｍ whic ｈ ｉt fol ｌows that ()()()常数≡+x f x f 2,,1(13）on ｔhe clos ｅd i ｎt ｅrv ａl []1,0Since the ｆu ｎc ｔion21u u + i ｓ not ｃo ｎstant o ｎ any int ｅr ｖal, Eq. (13) is possi ｂｌe onl ｙ ｉf ()≡x f ,const on []b a ,． Thus a sm ｏot ｈ ex ｔr ｅｍal of this pro ｂlem ｍｕst be ａ ｌinea ｒ f ｕn ｃｔion w ｈose ｇraph pass ｅs throu ｇｈ th ｅ p ｏints ()0,0 a ｎd ()0,1． Ｉt fo ｌlows that ()0≡x f , ａｎd w ｅ a ｒｒi ｖe at the ｃlo ｓed ｉnterva ｌ o ｆ th ｅ line ｊoining t ｈe two g ｉv ｅn ｐoi ｎt ｓ.Exa ｍpl ｅ 3. Ｔｈe ｂra ｃｈist ｏｃｈｒo ｎe problemTh ｅ class ｉc ａl ｂｒａch ｉsto ｃhrone prob ｌem, p ｏsed by J ｏha ｎｎ Bern ｏulli I in 1６96， was ｔｏ find ｔh ｅ sh ａpe of a track a ｌon ｇ ｗｈich a poi ｎｔ mass would pas ｓ from a prescr ｉbed po ｉnt 0P t ｏ another ｆi ｘed point 1P ａt ａｌowe ｒ ｌe ｖel under t ｈe ac ｔio ｎ of ｇra ｖity in the sh ｏrte ｓｔ t ｉｍe.Ｗe ｎｅglect ｆricti ｏn, of ｃo ｕrs ｅ． Ｉｎ addi ｔio ｎ， we sha ｌl a ｓs ｕme ｔhat ｔｈe tr ｉｖｉal case in ｗｈic ｈ bo ｔh po ｉｎt ｓ ｌi ｅ o ｎ the sam ｅ v ｅrt ｉｃa ｌ line is ex ｃｌuded.In ｔh ｅ v ｅr ｔｉcal p ｌane pa ｓsing throug ｈ the points 0P a ｎｄ1P we ｉｎtroduce a recta ｎg ｕl ａｒ c ｏo ｒd ｉn ａte system ｓｕch t ｈａt 0P ｉs ａt theori ｇi ｎ, the x －axis is d ｉrecte ｄ ve ｒt ｉｃall ｙ ｄo ｗnward, and th ｅ p ｏi ｎt1P has ｐｏｓitive ｃｏｏrdinate ｓ ()11,y x .We shal ｌ ｆind the ｓｈａpe ｏf t ｈe track amo ｎｇ ｔhe gr ａｐｈs ｏf s ｍooth ｆun ｃtio ｎｓ defined on th ｅ cl ｏｓed i ｎt ｅrval []1,0x an ｄ s ａtisfy ｉｎｇ the co ｎditi ｏｎ ()00=f ,()11y x f =. At the mo ｍent we shall n ｏt ｔa ｋｅ ｔime to d ｉsc ｕss this by no ｍean ｓ ｕｎcontrove ｒｓｉal ａｓsumption (see Pr ｏｂl ｅm 4 bel ｏw).I ｆ the particle be ｇａn i ｔs de ｓcen ｔ f ｒo ｍ t ｈe ｐoi ｎｔ 0P with ｚer ｏ vel ｏci ｔy, the la ｗ ｏf ｖar ｉａtion of its vel ｏcity in ｔh ｅse coor ｄin ａt ｅs c ａn be w ｒit ｔen asgx v 2=(１4)Rec ａl ｌｉn ｇ t ｈａt ｔhe di ｆferen ｔｉal of th ｅ ar ｃ le ｎgth ｉs ｃomp ｕted by the formula ()()()()dx x f dy dx ds 2,221+=+= (１5)we ｆi ｎｄ t ｈe tim ｅ ｏf de ｓcen ｔ()()()⎰+=102,121x dx x x f g f F(１6)ａlong the ｔr ａject ｏry defi ｎed by the g ｒaph of the fu ｎction()x f y = on ｔｈe clos ｅｄ in ｔerv ａl []1,0x .For the fu ｎc ｔional （1６) ()()1233211,,u u u u u L +=,and t ｈerefo ｒe the condition (１1) ｆo ｒ ａn ｅｘtre ｍu ｍ reduc ｅs i ｎ t ｈis case to the equ ａtion()()()012,,=⎪⎪⎪⎪⎪⎭⎫⎝⎛⎪⎭⎫ ⎝⎛+x f x x f dx d , fr ｏm ｗhic ｈ it fo ｌｌow ｓ that ()()()x c x f x f =+2,,1(17)wh ｅｒｅ ｃ i ｓ ａ nonzero cons ｔa ｎt, ｓince ｔhe po ｉnt ｓ a ｒe not both o ｎ t ｈe ｓａme ｖert ｉcal ｌi ｎe.Takin ｇ accoun ｔ ｏf (１5), we can rewr ｉｔe (17) i ｎ the ｆｏrmx c ds dy =(１8)Ｈowever, fr ｏm the g ｅom ｅｔric po ｉｎt ｏf vi ｅｗϕcos =ds dx ,ϕsin =ds dy(19)w ｈｅｒｅ ϕ is ｔhe angle be ｔｗeen t ｈｅ tan ｇent t ｏ the t ｒa ｊectory and ｔｈe po ｓｉtiv ｅ ｘ－axis.By ｃo ｍｐaring Eq ． (1８) wit ｈ ｔhe second e ｑｕation ｉn (19）, we ｆin ｄϕ22sin 1c x =（20)Ｂut it f ｏl ｌows f ｒｏm (1９） and (2０） th ａt dx dy d dy =ϕ,2222sin 2sin c c d d tg d dx tg d dx ϕϕϕϕϕϕϕ=⎪⎪⎭⎫ ⎝⎛==,ｆro ｍ whi ｃh ｗe ｆin ｄ()b c y +-=ϕϕ2sin 2212 (21)Ｓｅｔtinga c =221 and t =ϕ2， we wr ｉte r ｅｌa ｔions （20） a ｎd （2１)as ()()b t t a y t a x +-=-=sin cos 1(２２）Since 0≠a , it fo ｌｌow ｓ th ａｔ 0=x on ｌy f ｏr πk t 2=,Z k ∈. Ｉｔ f ｏl ｌow ｓ from th ｅ form ｏf t ｈe ｆunc ｔion (２2) ｔｈat ｗｅ m ａｙ a ｓsu ｍe w ｉth ｏｕｔ l ｏss of gener ａｌｉty tha ｔ the p ａrame ｔer value 0=t ｃorre ｓｐonds ｔo th ｅ po ｉｎｔ ()0,00=P . In t ｈis case Eq. (21) ｉｍｐlies 0=b , and we ar ｒive at the sｉmpler ｆorm()()t t a y t a x sin cos 1-=-= (23)fo ｒ ｔｈｅ parame ｔric ｄef ｉnition ｏｆ this cu ｒｖe.Thus th ｅ br ａch ｉs ｔo ｃhrone is a cyc ｌo ｉd havi ｎg ａ c ｕsp ａt the ini ｔｉa ｌ po ｉnt 0P where the tan ｇent is vertic ａl. The consta ｎt a, ｗhich is a sca ｌin ｇ co ｅf ｆic ｉｅnt, ｍus ｔ ｂｅ ch ｏsen ｓo that the ｃｕrve (23) also ｐa ｓse ｓ ｔｈrou ｇh ｔhe poi ｎt 1P ． Suc ｈ a c ｈｏic ｅ, ａｓ one c ａn ｓee ｂy ｓketc ｈｉｎg the c ｕrve （23), is by n ｏ ｍｅa ｎs a ｌw ａy ｓ un ｉq ｕe, and this s ｈows tha ｔ t ｈｅ ne ｃess ａｒy cond ｉｔion (１１) ｆo ｒ an extrem ｕm ｉs i ｎ general not su ｆｆicient. However, from physical cons ｉderati ｏｎs i ｔ is ｃle ａr w ｈic ｈ ｏf t ｈe poss ｉｂle v ａｌues of the para ｍeter ａ ｓh ｏuld ｂe pre ｆe ｒred (ａｎd thi ｓ， o ｆ cours ｅ, c ａｎ be confi ｒｍed by dir ｅct com ｐutati ｏn).泰勒公式和极值的研究1．映射的泰勒公式定理1 如果从赋范空间X 的点ｘ的邻域()x U U =到赋范空间Ｙ的映射Y U f →:在U 中有直到ｎ-1阶(包括n －1在内）的导数,而在点x 处有n 阶导数。