数学系2002级三、四班数学分析补充材料(四) (导数、多元函数等杂题,2002年12月) 设0<b≤a.则 (a-2≤2+b-、a≤如-62 2、证明:函数f(x)=ln(1+e-x),g(x)=√a+x2(c>0)为严凸函数 3、证明不等式 ≥-(0<x≤ > V/cos x(0<x≤a); (0 (x>0); (0≤x<1) 5、证明 (1+x)°≥1+ar(x≥-1,0<a<1):(1+x)°≤1+ar(x≥-1,a>1,ora<0) 6、证明:求最小的和最大的a使得Ⅶ∈N+有 (1+)+a≤e≤(1+2) 7、证明:m∑(√k+2-2Vk+1+k)存在 8、证明斯托兹( Stolz)定理(离散型的洛必塔求极限法则) (1)(型)设数列{an)是超于零的,数列{n}是严格进减超于零的,如果=an-b1 存在或为+x,则m动也存在或为+∞且m=m如二物杜 (2)(受型)设数列{hn}严格单调增加,且Imhn=+∞.如果mb-bn+1存在或为 im也存在或为+∝ →。 bn -bn Hint of Proof Only prove that conclusions(1)and(2)hold for the cases with the finite limit, the proof of the other cases are similar. The proof of (1). Suppose th nbn-b.=l∈R.Then,v>0, there exists N1∈N+such hold for all n> N1. Further, the fact that sequence 6n is strictly decreasing implies that
1 2002 2002 12 1 0 < b ≤ a. (a − b)2 8a ≤ a + b 2 − √ ab ≤ (a − b)2 8b . 2 f(x) = ln(1 + e−x), g(x) = √c2 + x2 (c > 0) 3 (1) 1 > sin x x ≥ 2 π (0 < x ≤ π 2 ); (2) sin x x > √3 cos x (0 < x ≤ π 2 ); (3) (sin x)−2 ≤ x−2 + 1 − 4 π2 (0 < x ≤ π 2 ); (4) 2 2x + 1 < ln(1 + 1 x) < 1 √x2 + x (x > 0); (5) ln x 1 − x ≤ 1 √x (x > 0, x = 1); (6) 4x π(1 − x2) ≤ tan(πx 2 ) ≤ πx 2(1 − x2) (0 ≤ x < 1). 5 (1 + x) α ≥ 1 + αx (x ≥ −1, 0 <α< 1); (1 + x) α ≤ 1 + αx (x ≥ −1,α> 1, or α < 0). 6 β α ∀n ∈ N + (1 + 1 n) n+α ≤ e ≤ (1 + 1 n) n+β. 7 limn→∞ n k=1 ( √ k + 2 − 2 √ k +1+ √ k) 8 (Stolz) ( ! " #). 1 ( 0 0 ) $ {an} !"#%$ {bn} !&'("#% $) limn→∞ an − an+1 bn − bn+1 * +∞, limn→∞ an bn %* +∞ & limn→∞ an bn = limn→∞ an − an+1 bn − bn+1 . 2 (∞ ∞ ) $ {bn} &+,'-& limn→∞ bn = +∞. $) limn→∞ an − an+1 bn − bn+1 * +∞, limn→∞ an bn %* +∞ & limn→∞ an bn = limn→∞ an − an+1 bn − bn+1 . Hint of Proof Only prove that conclusions (1) and (2) hold for the cases with the finite limit, the proof of the other cases are similar. The proof of (1). Suppose that limn→∞ an − an+1 bn − bn+1 = l ∈ R. Then, ∀ε > 0, there exists N1 ∈ N + such that | an − an+1 bn − bn+1 − l| < ε hold for all n ≥ N1. Further, the fact that sequence {bn} is strictly decreasing implies that −ε(bm − bm+1) < am − am+1 − l(bm − bm+1) < ε(bm − bm+1),
2 for all m> NI. When n>N1, by adding the above inequalities in order from m= n to n+p, it is easy to see that l(6n -bm+p)< E(bn -bn+p), =(1、bn+p、an_an+2 l(1 Since lim an=0, and lim bn=0, then, by letting p- oo in(i), we get (i) for any n>Ni. The conclusion(1)follows(ii) The proof of (2). Suppose that lim -n+isleR. Then, Ve >0, there exists N,E N+ such l hold for all n 2 N1. Further, the fact that sequence (bn) is strictly increasing implies that E(bm+1-bm)<am+1-am-I(bm1-bm)<E(bm+1-bm), for all m 2 N1. When n>N1, by adding the above inequalities in order from m= Ni to n, it is easy to see ∈(bn-bN1)<an-aN1-l(bn-bN1)<ε(bn-b and hence a(-)<--1 )<ε(1 (iii) Since lim bn =+oo, then, there exists N2 E N+ such that 17-1< E, and |-< minE, 1 hold for any n 2 N2. Further, from(iii), we get 2(1+|)≤--1≤2(1+, for any n>mar(N1, N2). The conclusion(2)follows(iv) (3)令x1∈(0,1)及xn+1=xn(1-xn),n=1,2,3,…证明: lim nIn=1 (4)设函数列f1(x)=sinx,fn+1(x)= sin fn(x),n=1,2,…若sinx>0 证明:lim oo van()=1 g”、(1)(、型斯托兹( Stolz)定理的推广)设T为正常数.若函数g(x),f(x),x∈l,+∞)满足 (i)g(a+r)>g(z),I (i)limg(x)=+∞,且函数f(x),g(x)在[a,+∞)的任意子区间上有界; f(r+r)-f(a) nx+)-= ∫(x) (2)(型斯托兹( Stolz)定理的推广)设T为正常数.若函数g(x),f(x),x∈l,+∞)满足
2 for all m ≥ N1. When n>N1, by adding the above inequalities in order from m = n to n + p, it is easy to see that −ε(bn − bn+p) < an − an+p − l(bn − bn+p) < ε(bn − bn+p), and hence −ε(1 − bn+p bn ) < an bn − an+p bn − l(1 − bn+p bn ) < ε(1 − bn+p bn ). (i) Since limn→∞ an = 0, and limn→∞ bn = 0, then, by letting p → ∞ in (i), we get −ε ≤ an bn − l ≤ ε, (ii) for any n>N1. The conclusion (1) follows (ii). The proof of (2). Suppose that limn→∞ an − an+1 bn − bn+1 = l ∈ R. Then, ∀ε > 0, there exists N1 ∈ N + such that | an − an+1 bn − bn+1 − l| < ε hold for all n ≥ N1. Further, the fact that sequence {bn} is strictly increasing implies that −ε(bm+1 − bm) < am+1 − am − l(bm+1 − bm) < ε(bm+1 − bm), for all m ≥ N1. When n>N1, by adding the above inequalities in order from m = N1 to n, it is easy to see that −ε(bn − bN1 ) < an − aN1 − l(bn − bN1 ) < ε(bn − bN1 ), and hence −ε(1 − bN1 bn ) < an bn − aN1 bn − l(1 − bN1 bn ) < ε(1 − bN1 bn ). (iii) Since limn→∞ bn = +∞, then, there exists N2 ∈ N + such that | aN1 bn | < ε, and | bN1 bn | < min{ε, 1} hold for any n ≥ N2. Further, from (iii), we get −2(1 + |l|)ε ≤ an bn − l ≤ 2(1 + |l|)ε, (iv) for any n > max(N1, N2). The conclusion (2) follows (iv). (3) . x1 ∈ (0, 1) / xn+1 = xn(1 − xn), n = 1, 2, 3, ··· . limn→∞ nxn = 1 (4) $ f1(x) = sin x, fn+1(x) = sin fn(x), n = 1, 2, ··· . ( sin x > 0, limn→∞ n 3 fn(x)=1. 9∗ (1) (∞ ∞ (Stolz) )0) T *1( g(x), f(x), x ∈ [a, +∞) 2+ (i) g(x + T ) > g(x), x ∈ [a, +∞); (ii) lim x→+∞ g(x)=+∞, & f(x), g(x) [a, +∞) ,-./3045 (iii) lim x→+∞ f(x + T ) − f(x) g(x + T ) − g(x) = l. lim x→+∞ f(x) g(x) = l. (2) ( 0 0 (Stolz) )0) T *1( g(x), f(x), x ∈ [a, +∞) 2+
3 (i)g(x+T)>g(x)>0,x∈[a,+∞) (ii) lim g(a=0,E lim f(a)=0 (ii) lim f(a+T)-f(r)=l x-+∞9(x+T)-9(x) f(r) x→+∞g(x) 10、设函数∫满足:(i)-∞<a≤f(x)≤b<+∞,(a≤x≤b);(i)|f(x)-f(y)≤kx-y,(0< k<1,x,y∈la,b)设r1∈la,b,并定义序列{xn}:rn+1=f(xn),n=1,2, 试证明: lim n=x存在,且x=f(x) 1、设函数∫满足:(i)-∞<a≤f(x)≤b<+∞,(a≤x≤b);(i)f(x)-f(y)≤|x-y,(x,y∈ ab)设x1∈la,b,并定义序列{xn}:xn+1=7(xn+f(xn),n=1,2 试证明: lim n=x存在,且x=f(x) 12.设a=0,定义an+1=1+si(n-1),n=0,1,2,…试求:lim∑ (Hint: Let bn an -1. Then, it is easy to prove that bn <0 and bn bn+1 for any n E N+) 13*.设函数∫连续不减,取定x0,由此定义rn=f(xn-1),n=1,2,…。若ro=a时,{xn} 收敛,试证明当min(a,f(a)≤xo≤mar(a,f(a)时,{xn}也收敛 14*、若ε0,ε1,ε2,…中的每一个均取自-1,0,1三数之一,证明 an =:EoV2+E1V2+.+Env2=2sin/o 24 对每个n成立 15*、定义数列xn=yne12n,n=n!n2en,n=1,2,…试证区间(xk,)包含了区间 (xk+1,3k+1) 1(1)数列{n=(+)a+单调减少的充分必要条件是2 (2)数列{n=(1+ x)1+n 单调减少的充分必要条件是0<x≤2 17、设∫(x)在[a上二阶可导,f(a)<0,f(b)>0,对一切x∈[ab],f'(x)≥0>0.,f"(x)≥0 (1)今1=f(n+1=xm-(x,n=1,2,3…证明{x}收敛于f在a,列上的零点 (2)令y (b-a)f(b) (b-yn)f(yn) f(b)-f)3n+1=--(m)=1,2,3…证明{n}也收敛于f在 a,b上的零点 (3)证明由(1)、(2)求∫的零点等价于求()=x-()且(x)=x-了()-/的不动点 (b-a)f(r f'(a) (Hint of(2): Let y=f(6)J(6)-f(a (a-b). Find out the point of intersection I1 of the straight line with T-axes. After that, find out the point of intersection r2 of the new straight line with z-axes 18、设∫是定义在[a,b上的实值函数,又设∫在o处可微分,其中a<xo<b,并设数列{an}
3 (i) g(x + T ) > g(x) > 0, x ∈ [a, +∞); (ii) lim x→+∞ g(x)=0, & lim x→+∞ f(x) = 0; (iii) lim x→+∞ f(x + T ) − f(x) g(x + T ) − g(x) = l. lim x→+∞ f(x) g(x) = l. 10 f 2+(i) −∞ < a ≤ f(x) ≤ b < +∞, (a ≤ x ≤ b); (ii) |f(x)−f(y)| ≤ k|x−y|, (0 < k < 1, x, y ∈ [a, b]). x1 ∈ [a, b] 6 12$ {xn}: xn+1 = f(xn), n = 1, 2, ··· . 3 lim n→+∞ xn = x & x = f(x). 11 f 2+(i) −∞ < a ≤ f(x) ≤ b < +∞, (a ≤ x ≤ b); (ii) |f(x)−f(y)|≤|x−y|, (x, y ∈ [a, b]). x1 ∈ [a, b] 6 12$ {xn}: xn+1 = 1 2 xn + f(xn) , n = 1, 2, ··· . 3 lim n→+∞ xn = x & x = f(x). 12. a0 = 0, 1 an+1 = 1 + sin(an − 1), n = 0, 1, 2, ··· . 3 lim n→+∞ n k=0 ak n . (Hint: Let bn = an − 1. Then, it is easy to prove that bn < 0 and bn < bn+1 for any n ∈ N +.) 13∗. f 74(Æ x0 58 1 xn = f(xn−1), n = 1, 2, ··· ( x0 = a 6 {xn} 793: min(a, f(a)) ≤ x0 ≤ max(a, f(a)) 6 {xn} %79 14∗ ( ε0, ε1, ε2, ··· 8;9<=Æ: −1, 0, 1 ;<9 an =: ε0 2 + ε1 2 + ··· + εn √ 2 = 2 sin π 4 n k=0 ε0ε1 ··· εk 2k =: bn >;< n ?@ 15∗ 1$ xn = yne − 1 12n , yn = n!n −n− 1 2 en, n = 1, 2, ··· . 3/3 (xk, yk) ABC/3 (xk+1, yk+1). 16∗ 1 $ an = 1 + 1 n n (1 + x n) +,(=DE!>?F! x ≥ 1 2 . 2 $ an = 1 + x n 1+n +,(=DE!>?F! 0 < x ≤ 2. 17 f(x) [a, b] 0GHI f(a) < 0, f(b) > 0, >9@ x ∈ [a, b], f (x) ≥ δ > 0, f(x) ≥ 0. (1) . x1 = b − f(b) f (b) , xn+1 = xn − f(xn) f (xn) , n = 1, 2, 3, ··· . {xn} 79# f [a, b] 0%J ξ. (2) . y1 = b − (b − a)f(b) f(b) − f(a) , yn+1 = yn − (b − yn)f(yn) f(b) − f(yn) , n = 1, 2, 3, ··· . {yn} %79# f [a, b] 0%J ξ. (3) 5 (1) (2) f %JK# g(x) = x− f(x) f (x) & h(x) = x− (b − x)f(x) f(b) − f(x) LJ (Hint of (2): Let y = f(b) + f(b) − f(a) b − a (x− b). Find out the point of intersection x1 of the straight line with x−axes. After that, find out the point of intersection x2 of the new straight line with x−axes by using x1 replecing a, and repeat continuously the program. What is the signification of geometry?) 18 f ! 1 [a, b] 0ABC f x0 MIDEE8 a<x0 < b 6 $ {an},
n}满足条件。a<an<x0<bn<b且ma=1mbn2=x0,试证明:im)-a2 f(ao) 19设证数∫在有穷正无穷的区间(a,b)中的任意一点x处有有限的导数f'(x),且lim,f(x)= f(x).试证明在(a,b)中托在点c,使斯∫(c)=0 20设设f(x)=a1sinx+a2sin2x+…+ an sin nt,并且|f(x)≤ Isin z,vr∈R,a1,a2,…,an为 实常数兹求证:|a1+2a2+…+nan|≤1 21设求数列{n(e-(1+)}的极限兹 22设证明数列{n2(2n-5)}(n≥3)是递增数列兹 23设求数列{n2 } 中取最在值的项兹 设在半径为a(a>0)的球的内接圆柱中,求体积最在的圆柱兹 25设证明:若∫在整个定义域上是严凸(正严凹)的,则f在其定义域内至则有一个极值兹 26设求f(x)=x2ln(ax)(a>0)的拐点,当a变动时拐点的轨迹是什么? 27设证明恒为式时m=ac+一<x+ 28设设∫在+∞)内可微,且, f(r) 0,证明必有点列{n},5n→+∞(m→∞),使斯 lim f'(En)=0 29设求证数y 的n阶导数兹 注:在P中开区域有闭区域的定义,如推一个集合是道路连通(即,其中任何两点,都有完全 落在该集合的连续曲线将这两点连接起来)的开集叫开区域,开区域的闭包叫闭区域兹在R中凸集 合的定义,如推一个集合ECPn中的任意两点x1,x2,都有x=tx1+(1-t)x2∈E,t∈[0.,1,就 称集合E是凸集兹 30设设G1,G2是P任意开集,且G1∩G2=,试证明:G1∩G2=0 31设设ECF为任意集合,E表示E的全体聚点组成的集合,称为E的导集,试证E为闭 集兹 32设设A,BCRn为开集,A∩B=0.试证:O(AUB)= dAub. 33设设A,BCn为有界闭集,A∩B=0兹试证:彐开集W,V使斯AcW,BCV,且 ∩v= 34设确定下列证数的定义域兹 (1)u=√①-x2+√1-y2;
4 {bn} 2+?F a<an < x0 < bn < b & lim n→+∞ an = lim n→+∞ bn = x0. 3 lim n→+∞ f(bn) − f(an) bn − an = f (x0). 19 f F*GF/3 (a, b) 8,-9J x M f (x) & lim x→a+ f(x) = lim x→b− f(x). 3 (a, b) 8J c f (c)=0. 20 f(x) = a1 sin x + a2 sin 2x + ··· + an sin nx, 6& |f(x)|≤|sin x|, ∀x ∈ R, a1, a2, ··· , an A1 |a1 + 2a2 + ··· + nan| ≤ 1. 21 $ n e − 1 + 1 n n " 22 $ { √3 n2(2n − 5)}(n ≥ 3) !''$ 23 $ n2 1 2n+1 8ÆBH 24 ÆN a(a > 0) IJOKL8MPKL 25 ( f N< 1O0! *Q f E 1OJP9<"B 26 f(x) = x2 ln(ax) (a > 0) RJ: a SL6RJTU!QVW 27 X arctan x = arcsin x √1 + x2 , −∞ < x + ∞. 28 f [a, +∞) JID& lim x→+∞ f(x) x = 0 !J$ {ξn}, ξn → +∞ (n → ∞), limn→∞ f (ξn)=0. 29 y = x √3 1 + x n H R Rn 8Y/OZ/O 1$)9<[\!]^7S _E8,`aJbTU cd[\74VWeXaJ7OYfY[gY/OY/OZAgZ/O Rn 8[ \ 1$)9<[\ E ⊂ Rn 8,-aJ x1, x2 b x = tx1 + (1 − t)x2 ∈ E, ∀t ∈ [0, 1] h i[\ E ![ 30 G1, G2 ! Rn ,-Y[& G1 ∩ G2 = ∅, 3 G1 ∩ G2 = ∅. 31 E ⊂ Rn ,-[\ E jZ E UMkJ[?[\i E [3 E Z [ 32 A, B ⊂ Rn Y[ A ∩ B = ∅. 3 ∂(A ∪ B) = ∂A ∪ ∂B. 33 A, B ⊂ Rn 4Z[ A ∩ B = ∅ 3 ∃ Y[ W, V A ⊂ W, B ⊂ V , & W ∩ V = ∅ 34 \ ]$ 1O (1) u = √1 − x2 + 1 − y2; (2) u = 2x − x2 − y2 x2 + y2 − x ; (3) u = arcsin y x ; (4) u = ln(−1 − x2 − y2 + z2).
35设求下列证数的极限 (r, y)-(0.)cos a sin y n(x3+y3) 35设要下列证数f(y),证明2imo(y)不存在 1)f(x,y) (2)f(a, y) =2y2 36设间下列证数是否在全平面收续,为什么? (1)f(x,9)=)2+y2x2+y2≠0 ≠0 (2)∫(x,y) 0 +y2=0 r4 ≠0 (x2+y2),x2+y2≠0 (3)f(x,y) (4)f(x,y)= 0 0 x2+y2=0. 37设设证数∫(x,y)在开取平面x>0上收续,且要V,极限lim.f(x,y)=0(y0)存在。 自证数∫在y轴上补徽定义a(y)后,证明证数f在闭取平面x≥0上收续 38设证数f(xy)在开取平面x>0上一致收续证明求()限a)+mf(y)=m 存在由 (2)证数∫在y轴上补徵定义叭y)后所得证数在x≥0上一致收续 39设设u=f(x)在点ro∈Fn收续,且f(xo)>0,证明求存在xo的一之邻域U(xo,6)(6>0) 使得f在U(x0,6)上取正值 40设设E是Fn中任意点集。求证求d(x,E)(x到集合E的距离)在Pn上一致收续 41设设证数f在F上收续,要任意实数a,作集合G={叫|f(x)>a},F={x|f(x)≥a}求 证求G是R中的开集,F是R中的闭集 41设设x∈R",x=(x1,x2,…,xn)。求证求 (1)n>0.b>0,使得叫≤∑ll≤b (2)3>0.b>0.使得叫≤理阿≤ 42设设ΩCR为有续闭区域,∫是Ω到F内的足射且收续。求证求∫-1在f()上收续 43设设f(x,y)除直线x=a与y=b外有定义。序足 (1)limf(x,y)=(x)存在由 (2)limf(x,y)=v(y)一致存在(即,vE>0.,36(-)>0,自0<|-a<6时,vy≠b,有 f(x,y)-v(y)<);
5 35 ]$" (1) lim (x,y)→(0,0) ex + ey cos x + sin y ; (2) lim (x,y)→(0,0) x2y 3 2 x4 + y4 ; (3) lim (x,y)→(+∞,+∞) (x2 + y2)ex+y; (4) lim (x,y)→(0,0) sin(x3 + y3) x2 + y2 . 35 >]$ f(x, y), lim (x,y)→(0,0) f(x, y) (1) f(x, y) = x2 x2 + y2 ; (2) f(x, y) = x2y2 x3 + y3 . 36 ^]$!lU_m74QVW (1) f(x, y) = x2 − y2 x2 + y2 , x2 + y2 = 0, 0, x2 + y2 = 0; (2) f(x, y) = | sin (xy) x |, x = 0, y, x = 0; (3) f(x, y) = x2 y2 e − x4 y2 , y = 0, 0, y = 0; (4) f(x, y) = y2 ln(x2 + y2), x2 + y2 = 0, 0, x2 + y2 = 0. 37 f(x, y) YÆ_m x > 0 074&> ∀y0, " lim (x,y)→(0+,y0) f(x, y) = φ(y0) : f y `0nD 1 φ(y) o f ZÆ_m x ≥ 0 074 38 f(x, y) YÆ_m x > 0 09a74 (1) ∀y0, " lim (x,y)→(0+,y0) f(x, y) = φ(y0) 5 (2) f y `0nD 1 φ(y) ob x ≥ 0 09a74 39 u = f(x) J x0 ∈ Rn 74& f(x0) > 0, x0 9<pO U(x0, δ) (δ > 0) f U(x0, δ) 0Æ*B 40 E ! Rn 8,-J[ d(x, E) (x q[\ E r ) Rn 09a74 41 f Rn 074>,-A α c[\ G = {x f(x) > α}, F = {x f(x) ≥ α} G ! Rn 8Y[ F ! Rn 8Z[ 41 x ∈ Rn, x = (x1, x2, ··· , xn) (1) ∃a > 0,b> 0, a|x| ≤ n j=1 |xj | ≤ b|x|; (2) ∃a > 0,b> 0, a|x| ≤ max 1≤j≤n |xj | ≤ b|x|. 42 Ω ⊂ Rn 4Z/O f ! Ω q Rn J+d&74 f −1 f(Ω) 074 43 f(x, y) seW x = a f y = b g 1 2+ (1) limy→b f(x, y) = φ(x) 5 (2) limx→a f(x, y) = ψ(y) 9a (_ ∀ε > 0, ∃δ(ε) > 0, : 0 < |x − a| < δ 6 ∀y = b, |f(x, y) − ψ(y)| < ε);