2020年中国科学院大学硕士研究生考试真题之数学专业综合.pdf
I 2020ca)K 8;n )L: 1. 150moO180 2. kY7L3K3Kv Ko540E!A!V KO90 )I3K?L150K(I XJKLUKKSg K8LK8cKKSN E 1. (10) N/N 2. (10) z4 5z + 1 = 0 3S9fz2C : 1 jzj 0). 4. (20) f(z) limr!+1M(r)rn .M = S1 D.D = S1 S1 No iii) mM=M u=m 4. (15) xT = S1 S1, T+I) 5. (15) mdydmkN+ 6. (15) y Brouwer :nu n 4N Dn YgN f : Dn ! Dn; f(x) = x; x2Dn;3) 1. (20) b X , f : X!0;1 , RXfd = c, 0 c : 1 e 0 1; c e = 1; 0 e 1 0, ,r 1, kfkr 1 . y: p!1, kfkp!kfk1: 3. (15) y: 3FAmH , z48Ek. 4. (15) bf : I!R1 AC , I = a;b, F(x) = sup NX i=1 jf(ti) f(ti 1)j (a x b); 8;n 12 5 , (.HkN 9kv a = t0 t1 0, y: Z +1 0 sinax Z +1 0 f(y)e xydydx = a Z +1 0 f(y) a2 +y2 dy: 1. (15) k = Fq qkV 4 k-5mKV k-fmk 2. (15) GNfNC+ (=kNmfN$|+)yG = S4o+T+3N 3 LAI 3. (20 ) (1) o (splitting eld) (2) pqpk = FpKLkXq X=Lqk AO/kqk 4. (20)yZXn (1) (0) (2) (p) (p) (3) (f(X) (f(X)ZX) (4) (p;f(X) ( pf(X)pFpX ) 5. (20 )pGk+P G Sylow p-f+N = NG(P)p NG(H) =fg2GjgHg 1 = Hg KN = NG(N) A 1. (15) (1)r = r(u;v) = (acosucosv; acosusinv; asinu); =2 u =2; 0 v 0;b6= 0 r = r(u;v) = (acosu; asinu; bv); 0 u 2 ; 1v +1; (t) = (acost; asint; bt); 0 t 2 : (1) L (2) t02(0;2 ). 3 (t0) ? 0(t0) (3) /nd 8;n 13 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 2 (0; =2). O C / kg HCkgds . 4. (30) M E3 Kzr = r(u;v). Kfru;rv;ngg,Ie| pn|?:p2MP W : TpM!TpM Weingarten C (1) CWv W ru rv ! = A ru rv ! : A.1!1/XL (2) ye D dv D duru D du D dvru = K jrurvj R 2 (ru); p Ddu; Ddv CKpdR 2 (ru)ru _= 2 = R 2 (ru) =jruj (rv) ? j(rv)?j; (rv)? := rv rv; rujr uj rujr uj : V 1. (10) ,kN Wzk|l l/3gm;:8ez r|lkgClV. 2. (10) =fr2Q : r20;1g0;1 kn8, A, z8kp Xe/8A:fr : a r bg, fr : a r bg, fr : a r bg, fr : a r bg;0 a b 1, XJB = nP i=1 Ai X/8Ai kKP(B) = nP i=1 (bi ai), ai;bi O8Ai m:, y8P(A);A2AAk8. 3. (15) - (x) = 1p2 e x22 ; 1x1; g(x) = 8 : cos(x); jxj 0; jxj ; p(x;y) = (x) (y) + 12 e 2g(x)g(y); 1x;yS, (3)X0, (4) S, (5). 8;n 14 5 4. (15) e 1; 2; ; n C, l, p(x), Ok n: E 1 + 2 + + k 1 + 2 + + n : 5. (15) kk , zg81i Vpi, kP i=1 pi = 1, zg8m p. e8 n , o a(=8n , 1 a) 6. (15) LfNt;t 2 R+gr PoissonL(;mY), eP(N0 = 0) = 1, fNt;t 2 R+gOL, 8st, Nt Nsl (t s)Poisson. WrP1r au) , =fWr tgu)L1r ayuut, l fWr tg=fNt rg. CWr F(t) = P(Wr t) 9Vp(t). 7. (10) ,(:U)C,: f(x) = 2x; 0 x 1: , =#O.Xi L1i, Sn = nP i=1 XiL1n , rXe: r = limn!1 nS n : bCfXi;i 1gp, r. 8;n 15 5