中国科学院大学硕士研究生入学考试统一命题科目试题数学专业综合.pdf

《中国科学院大学硕士研究生入学考试统一命题科目试题数学专业综合.pdf》由会员分享，可在线阅读，更多相关《中国科学院大学硕士研究生入学考试统一命题科目试题数学专业综合.pdf（5页珍藏版）》请在得力文库 - 分享文档赚钱的网站上搜索。

1、III? 2020cccaaa )KKK 888;nnn )LLL: 1. ?150?moO180 2. kY7L?3K?3Kv? Ko540E!“!A!V K?O90 )I3KKK?LLL150?K(I XJ?KL?UK?KSg? K8L?K8?cK?K?SN EEE 1. (10) ?N?/N? 2. (10) z4 5z + 1 = 0 3 ?S9? z C : 1 |z| 0). 4. (20) ? f(z) ? lim r+ M(r) rn . M = S1 D ?. D = S1? S1? N?o- iii) m M/M ?u=m 4. (15) x T = S1 S1, ?T-?

2、+I)? 5. (15) ?m?d?y?d?mk?N+ 6. (15) y Brouwer :nu n 4N Dn?YgN? f : Dn Dn, f(x) = x, x Dn, 3) 1. (20) b? X ?, f : X 0, ?, R X f d = c, 0 c , . y: lim n Z X nlog1 + (f/n)d = e 0 1, ce = 1, 0e 1 0, , r , kfkr . y: ? p , kfkp kfk. 3. (15) y: 3F?Am H , z48 E ?k?. 4.(15) b? f : I R1 AC ?, I = a,b, F(x) =

3、sup N X i=1 |f(ti) f(ti1)|(a x b), 8;n12?5 , (.?Hk? N 9kv a = t0 t1 0, y: Z + 0 sinax Z + 0 f(y)exydydx = a Z + 0 f(y) a2+ y2 dy. “ 1. (15) ? k = Fq q ?k V 4 k-5mK V ? k-fmk? 2. (15) ? G ?N?fNC+ (=k?N?mfN$|?+)y G= S4o?+?T+3?N? 3 L?A?I 3. (20 ) (1) o? (splitting fi eld) (2) ? p q p ? k = FpK L k Xq X

4、 ?=? L q ?k AO/k q ?k? 4. (20 ) y?“ ZX ?n (1) (0) (2) (p) (p ) (3) (f(X) (f(X) ZX ?) (4) (p,f(X) ( p f(X) ? p ? FpX ?) 5. (20 )? p G k+ P G ? Sylow p-f+ N = NG(P)p NG(H) = g G | gHg1= H K N = NG(N) AAA 1. (15) (1) ? r = r(u,v) = (acosucosv, acosusinv, asinu), /2 u /2, 0 v 0,b 6= 0 ?- r = r(u,v) = (a

5、cosu, asinu, bv), 0 u 2, v +, ?- (t) = (acost, asint, bt), 0 t 2. (1) - ?-L (2) ?t0 (0,2). -3 (t0) ? 0(t0) ?- (3) ?- -?/nd 8;n13?5 3. (20)3 r = r(u,v) = (acosucosv, acosusinv, asinu), /2 u /2, 0 v 2 ? C(v) := r(u0,v), 0 v 2 , p u0 (0,/2). O- C /- kg? H C kgds . 4. (30) ? M E3?K-z r = r(u,v). K ru,rv

6、,n g,Ie| p n |?: p MP W : TpM TpM Weingarten C (1) ?C W v W ru rv ! = A ru rv ! . ?A.1!1?/?XL (2) ye D dv D duru D du D dv ru= K |ru rv| R 2(ru), p D du, D dv ?C? K pd-R 2(ru) ru _?= 2 ?= R 2(ru) = |ru| (rv) |(rv)|, (rv):= rv ? rv, ru |ru| ? ru |ru|. VVV 1. (10) ,kN Wzk|l?l?/?3gm;:8ez ?r|l?k?gC?l?V.

7、 2. (10) ? = r Q : r 0,1 0,1 ?kn8, A “, z8kp ?Xe/?8A ?:r : a r b, r : a r b, r : a r b, r : a r b,0 a b 1, XJB = n P i=1 AiX/8Ai?k?KP(B) = n P i=1(b i ai), ai,bi O8Ai?m:, y8P(A),A A A ?k8?. 3. (15) - (x) = 1 2e x2 2, x , g(x) = cos(x),|x| 0,|x| , p(x,y) = (x)(y) + 1 2 e 2g(x)g(y), x,y S?, (3)X0, (4)

8、 S, (5)?. 8;n14?5 4. (15) e1,2, ,n?C, l, p(x), Ok n : E ? 1+ 2+ + k 1+ 2+ + n ? . 5. (15) ?kk ? , zg8?1i ?Vpi, k P i=1 pi= 1, zg8m p?. e?8?n , o ?a?(=8?n ?, ? 1? a?) 6. (15) LNt,t R+r?PoissonL(;?mY), eP(N0= 0) = 1, Nt,t R+OL, s t, Nt Nsl(t s)?Poisson. WrP1r a?u) ?, =Wr t u)L1r a?y?u?ut, l?Wr t = Nt r. CWr?F(t) = P(Wr t) 9Vp(t). 7. (10) ?,?( :U)C,: f(x) = 2x,0 x 1. ?”?, =#?O.XiL1i ?, Sn= n P i=1 XiL1n ? ”?, ”?r Xe: r = lim n n Sn . b?CXi,i 1 p, ”?r. 8;n15?5