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。