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00)为严凸函数 3、证明不等式 ≥-(0 V/cos x(00); (0≤x1,ora0, 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 0) 3 (1) 1 > sin x x ≥ 2 π (0 √3 cos x (0 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, or α 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)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)N1, by adding the above inequalities in order from m= Ni to n, it is easy to see ∈(bn-bN1)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) 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| N1, by adding the above inequalities in order from m = N1 to n, it is easy to see that −ε(bn − bN1 ) 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)-∞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) −∞ ;?F! x ≥ 1 2 . 2 $ an = 1 + x n 1+n +,(=DE!>?F! 0 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。a0)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 0) IJOKL8MPKL 25 ( f N 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 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);
6 (3)limv(y)=c存在 证明 (i)累次极限 lim lim f(x,y)=limo(x)=c存在; (i)累次极限 limlim f(x,y)=limv(y)=c存在; (i)二重极限limf(x,y)=c 44、设函数u=∫(x,y,2)在闭立方体a≤x≤b,a≤y≤b,a≤z≤b上连续,令叭x) mmf(x,y,2)试证明。φ在上连续 45、若函数f(x,y)在矩形D:-a≤x≤a,-b≤y≤b(a>0,b>0)上分别为x和y的连续函 数,而且f(0,0)=0。当x固定时,f(x,y)是y的严格递减函数,则有6>0,使对每个x∈(-6,6) 有y∈(-b,b)满足f(x,y)=0 Hint: Because the function f(0, y) is strictly decreasing on the interval -b, b, and f(0, 0)=0, then f(0, b)0. Further, consider to use the continuity of the function f 46、若f(x,y)函数在xy平面上处处连续,试证平面点集E={(x,y)f(x,y)=0}为闭集 47、设∫在P上连续,x≠O(中的原点)时,f(x)>0,且x∈Rn,及c>0有f(cx)=cf(x) 试证明:彐a,b>0,使得a≤f(x)≤b,(vx∈R) 48、设A是n×n矩阵,detA≠0.试证明:a>0,B>0使得∈R有x≤|Aarl≤a|r R,满足如下三条件: (1)vx∈Rn,f(x)≥0,且x=O←→f(x)=0: (2)VA∈R,有f(x)=|f(x); (3)vr,y∈Rn,f(x+y)≤f(x)+f(y) 试证明:()M>0,使得Vx∈R",有f(x)≤M|c (i)f满足 Lipschitz条件:即>0,使得|f(x)-f(列≤Lx-y(x,y∈R); (i)常数a>0使得∈R有ar≤f(x) 50、设f在Rn中点x0的邻域里有界,记Mf(xo,6)=sup{f()x∈Rn,|-xol<b} (x0,6)=inf{f(x)x∈P",|x-ol<}.则(1)极限r(xo)=:lim[M(xo,0)-mr(xo,6)存 在,并称之为∫在x处的振幅;(2)f(x)在xo连续的充分必要条件是ur(xo)=0 51、设集合ECRn是非闭的,函数f在E上一致连续。试证明:f只能惟一地连续延拓到E 上,即惟一地存在E上的函数F满足F(x)=f(x),x∈E,使得F在E上一致连续 52、设∫在R上连续,=m、f)存在。试证明f在P上一致连续
6 (3) limy→b ψ(y) = c (i) tu" limx→a lim y→b f(x, y) = limx→a φ(x) = c 5 (ii) tu" lim y→b limx→a f(x, y) = limy→b ψ(y) = c 5 (iii) Gh" lim (x,y)→(a,b) f(x, y) = c. 44 u = f(x, y, z) Z@vM a ≤ x ≤ b, a ≤ y ≤ b, a ≤ z ≤ b 074. φ(x) = max a≤y≤b { min a≤z≤b f(x, y, z)}. 3 φ [a, b] 074 45 ( f(x, y) wi D : −a ≤ x ≤ a, −b ≤ y ≤ b (a > 0,b > 0) 0Ex x y 74 y& f(0, 0) = 0 : x z 6 f(x, y) ! y &'( δ > 0 >; 0. Further, consider to use the continuity of the function f. 46 ( f(x, y) xy _m0MM743_mJ[ E = {(x, y)|f(x, y)=0} Z[ 47 f Rn 074x = O(Rn 8jJ) 6f(x) > 0, & ∀x ∈ Rn, / c > 0 f(cx) = cf(x). 3 ∃a, b > 0, a|x| ≤ f(x) ≤ b|x|,(∀x ∈ Rn). 48 A ! n × n wk detA = 0. 3 ∃α > 0,β > 0 ∀x ∈ Rn β|x|≤|Ax| ≤ α|x|. 49∗ f : Rn → R, 2+$];?F (1) ∀x ∈ Rn, f(x) ≥ 0, & x = O ⇐⇒ f(x) = 0; (2) ∀λ ∈ R, f(λx) = |λ|f(x); (3) ∀x, y ∈ Rn, f(x + y) ≤ f(x) + f(y). 3 (i) ∃M > 0, ∀x ∈ Rn, f(x) ≤ M|x|; (ii) f 2+ Lipschitz ?F_ ∃L > 0, |f(x) − f(y)| ≤ L|x − y|(∀x, y ∈ Rn); (iii) ∃ 1 α > 0 ∀x ∈ Rn α|x| ≤ f(x). 50 f Rn 8J x0 pO{4| Mf (x0, δ) = sup{f(x) x ∈ Rn, |x − x0| ?F! ωf (x0)=0. 51 [\ E ⊂ Rn !~Z f E 09a74 3 f mno974pqq E 0_o9 E 0 F 2+ F(x) = f(x), x ∈ E, F E 09a74 52 f Rn 074 lim |x|→+∞ f(x) 3 f Rn 09a74