R eceived d ate :2006207217第24卷第4期大　学　数　学Vol.24,№.42008年8月COLL EGE MA T H EMA TICS Aug.2008So me New Ways to Prove Rolle ’s TheoremYA O J i n g 2s un(Dept.of Math.,Anhui Normal University ,Wuhu 241000,China ) Abstract :We give three new methods proving Rolle ’s Theorem.The second simple way is only dependent on the well 2known Heine 2Borel Covering Theorem.This implies that Rolle ’s Theorem is the direct consequence of completeness of real numbers.K ey w ords :Rolle ’s theorem ;completeness of real numbers ;f ull cover ;Heine Borel covering theorem ;δ2fine tagged partitionC LC Number :O171 Document Code :C Article ID :167221454(2008)0420131203The st udy on Rolle ’s Theorem as well as ot her mean value t heorems of differentials is a very att ractive issue and it was also involved in calculus reform in U SA.Many scholars have done a great deal of work during t he past decade [1-3].We know t hat if Rolle ’s Theorem is proved ,it can be used to p rove Lagrange Mean Value Theorem and Cauchy Mean Value Theorem so long as a corresponding auxiliary f unction is const ructed.Therefore ,it is better to say Rolle ’s Theorem is t he essence and basis of t he next two t heorems t han to say t he conclusions of t he next two t heorems seem to have wider applicability t han t hat of Rolle ’s Theorem.To make t hings simpler ,people lay emp hasis on discussing t he ways to p rove Rolle ’s Theorem.The articles of professor Xu Ji 2hong [4]and t he aut hor [5]respectively give a new way to p rove Rolle ’s Theorem.In t he paper ,we shall give some met hods p roving Rolle ’s Theorem by some forms of completeness of real numbers.Def inition 1　A collection C of clo sed subintervals of [a ,b]is a f ull cover of [a ,b]if to each x ∈[a ,b]t here corresponds a number δ(x )>0such t hat every closed subinterval of [a ,b ]t hat contains x and has lengt h less t hat δ(x )belongs to C [6].Lemm a 1　If C is a f ull cover of [a ,b],t hen C contains a partition of [a ,b],i.e.,t here exist a =x 0,x 1,…,x n =b such t hat x k -1<x k and I k =[x k -1,x k ]is in C for each k [6].Def inition 2　A partition of an interval [a ,b]is a finite collection of non 2overlapping closed intervals whose union is [a ,b].A tagged partition of [a ,b]is a partition of [a ,b]wit h one point ,referred to as t he tag ,cho sen f rom each interval comprising t he partition.A tagged partition of [a ,b]will be denoted by {(c i ,[x i -1,x i ]):1≤i ≤n},wherea =x 0<x 1<x 2<…<x n -1<x n =band c i ∈[x i -1,x i ]is t he tag of t he interval [x i -1,x i ]for each index i .Now let δbe a positive f unction defined on [a ,b].A δ2fine tagged partition of [a ,b]is a tagged partition {(c i ,[x i -1,x i ]):1≤i ≤n}of[a ,b]t hatsatisfies[x i-1,x i]<(c i-δ(c i),c i+δ(c i))for each index i.In words,t he positive f unctionδ(which is often referred to as a gauge)determines t he size of t he interval associated wit h a given tag[7].Lemm a2　Ifδis a positive f unction defined on t he interval[a,b],t he t here exist s aδ2fine tagged partition of[a,b][7].R olle’s Theorem　Let f(x)be a f unction continuous o n[a,b],differentiable on(a,b)and f(a) =f(b),t hen t here exist sξ∈(a,b)such t hat f′(ξ)=0.Proof　We can suppose wit hout loss generality t hat f(x)is not constant on[a,b].It implies t hat t here exist s y0∈(a,b)such t hat f(y0)≠f(a).Assuming t hat f(y0)>f(a)(A similar proof can be made for t he case f(y0)<f(a)),we shall show by using disproof t hat t here isξin[a,b]such t hat f(ξ)≥f(t)for any t∈[a,b].If not,for every x in[a,b]t here exist s t x∈[a,b]such t hat f(x)<f(t x).LetC={I|I is a clo sed subinterval of[a,b],t here is t I∈[a,b]wit h f(y)<f(t I)for all y in I}.We claim t hat C is a f ull cover of[a,b].(To see t his,let x∈[a,b].There is t x∈[a,b]wit h f(x)<f(t x).Since f is co ntinuous at x,t here exist sδ(x)>0such t hat f(y)<f(t x)for all y∈(x-δ(x),x+δ(x))∩[a,b],t hen for every closed subinterval I of[a,b]t hat contains x and has lengt h less t hanδ(x),we have I<(x-δ(x),x+δ(x))∩[a,b]so f(y)<f(t x)for every y∈I, t herefore,I is in C.)By t he Lemma1,C contains a partition of[a,b]:I1,I2,…,I n.For each I i,i=1,2,…,n,according to t he st ruct ure of C,we can find t Ii ∈[a,b]such t hat f(y)<f(t Ii)for any y∈I i.LetM=max{f(t Ii)|i=1,2,…,n},so f(y)<M for every y∈[a,b].It contradict s t he definitio n of M and t he fact t hat t Ii∈[a,b].This cont radiction shows t hat t here isξ∈[a,b]such t hat f(ξ)is t he maximum value of f on[a,b].Sincef(ξ)≥f(y0)>f(a)=f(b),t husξ∈(a,b).Combining t he conditio n t hat f′(ξ)exist s,we have t hat f′(ξ)=0.R emark　Modifing t he disproving part of t he above p roof as follows,we can get a new met hod to p rove Rolle’s Theorem using t he well2known Heine Borel Covering Theorem.If not,for every x in[a,b]t here exist s t x∈[a,b]such t hat f(x)<f(t x).Since f(x)is continuous on[a,b],t here isδ(x)>0,such t hatf(y)<f(t x)　for all　y∈(x-δ(x),x+δ(x))∩[a,b].(1) LetH={(x-δ(x),x+δ(x)),x∈[a,b]},then H covers[a,b]and Heine Borel Covering Theorem implies that there is H1={(x i-δ(x i),x i+δ(x i)) |i=1,2,…,n}<H such that H1is a finite covering of[a,b].LetM=max{f(t xi )|i=1,2,…,n}=f(t xi0),　1≤i0≤n,where t xi0in[a,b].It follows t hat t here exist s j:1≤j≤n such t hat t xi0∈(x j-δ(x j),x j+δ(x j)),according to(1)we havef(t xi0)<f(t xj)≤M=f(t xi0).It is cont radiction.To testify Rolle’s Theorem by t he Lemma2f rom anot her angle,we shall first show a p ropo sition as follows:Proposition　Let f(x)is a f unction differentiable on(a,b)and f′(x)≠0,t hen f(x)is st rictly monoto ne on(a,b).Proof　Let t1and t2be any two point s in t he open interval(a,b)wit h t1<t2,t hen for everyx∈[t1,t2],we have limy→x f(y)-f(x)y-x=f′(x)≠0,so t here exist sδ(x)>0such t hat y∈U0(x;δ(x))231大　学　数　学 第24卷∩[a ,b]implies f (y )-f (x )y -x>0,　if f ′(x )>0,(2)f (y )-f (x )y -x <0,　if f ′(x )<0.(3)It is clear that δ(x )is a positive function on [t 1,t 2].The Lemma 2implies that there is a δ2fine tagged partition {(c i ,[x i -1,x i ]),i =1,2,…,n}that is ,t 1=x 0<x 1<…<x n =t 2,c i ∈[x i -1,x i ]and [x i -1,x i ]<(c i -δ(c i ),c i +δ(c i ))for each i =1,2,…,n .We might as well suppose that c 1<c 2<…<c n .(4)(If not ,t here must be j :1≤j ≤n -1such t hat c j =c j +1.It follows t hat c j =x j =c j +1by c j ≤x j ≤c j +1,we obtain a new δ2fine tagged partition of [t 1,t 2]satisfying (4)so long as we merge [x j -1,x j ]wit h[x j ,x j +1]).Then we easily prove t hat for all i =1,2,…,n ,f ′(c i )have t he same sign (If not ,t here must be i satisfying 1≤i ≤n -1such t hat f ′(c i )・f ′(c i +1)<0.We might as well suppose that f ′(c i )>0and f ′(c i +1)<0.By (2)and (3),we have that f (y )>f (c i )and f (y )>f (c i +1)if y ∈(c i ,c i +δ(c i ))∩(c i +1-δ(c i +1),c i +1)∩[a ,b].For t his reason ,t here exist s y 0in (c i ,c i +1)such t hat f (y 0)is t hemaximum value of f on [c i ,c i +1],t herefore f ′(y 0)must be zero ,which cont radict s t he condition t hat f ′(x )≠0for all x in (a ,b )).Wit hout loss of generality we may assume t hat f ′(c i )>0for every i =1,2,…,n.Note t hat c i -δ(c i )<x i -1≤c i ≤x i <c i +δ(c i )and x i -1<x i by (2)we have eit her f (x i -1)<f (c i )≤f (x i )or f (x i -1)≤f (c i )<f (x i ),t hen f (x i -1)<f (x i )for each i =1,2,…,n t hus f (t 1)<f (t 2).Therefore f is st rictly monotone on (a ,b ).If Rolle ’s Theorem is faulty ,f (x )is st rictly monotone on (a ,b )by propo sition above.Thus f (x )is also st rictly monotone on [a ,b ]by t he condition t hat f (x )is continuous on [a ,b ].It cont radict s t he condition t hat f (a )=f (b ).This co nt radiction shows t hat Rolle ’s Theorem holds.[R eferences][1]　Tucker T W.Rethinking nigor in calculus :the Role of the mean value theorem[J ].Amer.Math.Monthly ,1997,104(3):231-240.[2]　Swann mentary on rethinking rigor in calculus :the Role of the mean value theorem [J ].Amer.Math.Monthly ,1997,104(3):241-245.[3]　Tong J.A converse of the mean value theorem[J ].Amer.Math.Monthly ,1997,104(12):939-942.[4]　Xu Ji 2hong.An alternative approach about several theorems in calculus[J ].Journal of Mathematical Research andExposition ,2006,26(1):63-66.[5]　YAO Jing 2sun.The application of f ull cover and tagged partition in analysis [J ].College Mathematics ,2006,22(4):109-112.[6]　Botsko M W.A unified treatment of various theorems in elementary analysis[J ].Amer.Math.Monthly ,1987,94(5):450-452.[7]　Russell a G ordon.The use of tagged partitions in elementary real analysis [J ].Amer.Math.Monthly ,1998,105(2):107-117.罗尔中值定理的一些新证法姚静荪(安徽师范大学数学计算机科学学院,芜湖241000)[摘　要]给出了罗尔微分中值定理的三种新的证明方法,其中第二种很简便的方法仅依赖于大家熟知的Heine 2Borel 有限覆盖定理.由此可见罗尔微分中值定理可以是实数的完备性的直接推论.[关键词]罗尔中值定理;实数的完备性;完全覆盖;有限覆盖定理;δ2精细加标分割331第4期 YAO Jing 2sun :Some New Ways to Prove Rolle ’sTheorem。