中国科学院2010年数学分析参考解答

èÆk) · · · · · · uia.china@gmail.com· · · · · ·›Š 1 ¥‰�2010ê© To my parents 1 OŽ: (1) limx→0 ∫ sin2 x 0 ln(1+t)dt√ 1+x4−1 ; (2) ∫∫ |x|+|y|≤1 |xy|dxdy. )‰. (1) lim x→0 ∫ sin2 x 0 ln(1 + t)dt√ 1 + x4 − 1 = limx→0 ∫ sin2 x 0 ln(1 + t)dt x4 (√ 1 + x4 + 1 ) = 2 lim x→0 ln(1 + sin2 x) · 2 sinx cosx 4x3 = lim x→0 ln ( 1 + sin2 x ) sin2 x · ( sinx x )3 cosx = 1; (2) ∫∫ |x|+|y|≤1 |xy|dxdy = 4 ∫∫ x+y≤1; x,y≥0 xydxdy = 4 ∫ 1 0 xdx ∫ 1−x 0 ydy = 4 ∫ 1 0 x (1− x)2 2 dx = 2 ∫ 1 0 x(1− x)2dx = 1 6 . 2 (1) - f(x) = { x2 sin 1 x , x 6= 0; 0, x = 0. ¦f ′(0),¿y²f ′(x)3x = 0?ØëY. èÆk) · · · · · · uia.china@gmail.com· · · · · ·›Š 2 (2) eλ = n∑ k=1 1 k ,y² eλ > n+ 1. y². (1) f ′(0) = lim x→0 f(x)− f(0) x = lim x→0 x sin 1 x = 0. ��x 6= 0ž, f ′(x) = 2x sin 1 x − cos 1 x , xn = 1 2npi → 0, f ′(xn) = −19 0, � n→∞. dd,f ′(x)3x = 0?ØëY. (2) ^êÆ8B{.w,�,e1 = 2.7 · · · > 1 + 1. yb�i = nžØ�ª ¤á,K�i = n+ 1ž, e ∑n+1 k=1 1 k = e ∑n k=1 1 k e 1 n+1 = (n+ 1)e 1 n+1 > (n+ 1) [( 1 + 1 n+ 1 )n+1] 1n+1 = n+ 2. 5P. • �˜Ú´Ï( 1 + 1 n )n = 1 · ( 1 + 1 n )n < ( 1 + n+1 n · n n+ 1 )n+1 = ( 1 + 1 n+ 1 )n+1 ↗ e. èÆk) · · · · · · uia.china@gmail.com· · · · · ·›Š 3 • 1(2)¯,yXe: ∀ k, ∃ ξk ∈ (k, k + 1), s.t. ln(k + 1)− ln k = 1 ξk < 1 k , ¦Úk ln(n+ 1) < n∑ k=1 1 k , =eλ > n+ 1. 3 ef(x)3[0, 1]þëY,3(0, 1)þ�gŒ‡,¿… f(0) = f ( 1 4 ) = 0, ±9 ∫ 1 1 4 f(y)dy = 3 4 f(1). y²,3ξ ∈ (0, 1),¦�f ′′(ξ) = 0. y². P F (x) = ∫ x 1 4 [f(y)− f(1)] dy, x ∈ [ 1 4 , 1 ] . KdK¿, F ( 1 4 ) = 0 = F (1). dLagrange¥Š½n, ∃ η ∈ ( 1 4 , 1 ) , s.t. f(η) = 0. fkn‡ØÓ":.|^ügRolle½n,k ∃ ξ ∈ (0, 1), s.t. f ′′(ξ) = 0. 4 ¦?ê ∞∑ n=1 n (n+ 1)! �Ú. èÆk) · · · · · · uia.china@gmail.com· · · · · ·›Š 4 )‰. � f(x) = ∞∑ n=1 nxn−1 (n+ 1)! , ÙýéÂñu[−R,R],é∀ R ∈ (0,∞). ¤¦f(1). � f(x) = ∞∑ n=1 nxn−1 (n+ 1)! = [ ∞∑ n=1 xn (n+ 1)! ]′ = [ 1 x ∞∑ n=1 xn+1 (n+ 1)! ]′ = [ ex − 1− x x ]′ = (x− 1)ex + 1 x2 , � ¤¦ = f(1) = 1. 5P. ,yXe: ∞∑ n=1 n (n+ 1)! = ∞∑ n=1 (n+ 1)− 1 (n+ 1)! = ∞∑ n=1 1 n! − ∞∑ n=1 1 (n+ 1)! = 1. 5 y²: 2n 3 √ n < n∑ k=1 √ k < ( 2n 3 + 1 2 )√ n. èÆk) · · · · · · uia.china@gmail.com· · · · · ·›Š 5 y². • 2n 3 √ n < n∑ k=1 √ k. dLagrange¥Š½n,∀ k ∈ N, ∃ ξk ∈ (k − 1, k), s.t. k 3 2 − (k − 1) 32 = 3 2 ξ 1 2 k < 3 2 k 1 2 , ¦Ú k n 3 2 < 3 2 n∑ k=1 k 1 2 . • n∑ k=1 √ k < ( 2n 3 + 1 2 )√ n. ·‚^êÆ8B{y². F d1 < 2 3 + 1 2 l = 1ž¤á; F b��i = nž¤á,wi = n+ 1ž�œ/,d n+1∑ k=1 k 1 2 = n∑ k=1 k 1 2 + (n+ 1) 1 2 < ( 2n 3 + 1 2 ) n 1 2 + (n+ 1) 1 2 , L( 2n 3 + 1 2 ) n 1 2 + (n+ 1) 1 2 < ( 2(n+ 1) 3 + 1 2 ) (n+ 1) 1 2 ⇐ ( 2n 3 + 1 2 ) n 1 2 < ( 2n 3 + 1 6 ) (n+ 1) 1 2 ⇐ (4n+ 3)2n < (4n+ 1)2(n+ 1) ⇐ 16n3 + 24n2 + 9n < 16n3 + 24n2 + 9n+ 1 ⇐ 0 < 1, ù´é�. 6 OŽ ∫∫∫ V (x3 + y3 + z3)dxdydz, Ù¥VL«­¡x2 + y2 + z2 − 2a(x + y + z) + 2a2 = 0(a > 0) ¤Œ¤ �«. èÆk) · · · · · · uia.china@gmail.com· · · · · ·›Š 6 )‰. • d ∂V = { (x, y, z) ∈ R3; x2 + y2 + z2 − 2a(x+ y + z) + 2a2 = 0} = { (x, y, z) ∈ R3; (x− a)2 + (y − a)2 + (z − a)2 = a2} , V´±(a, a, a)%,aŒ»�¥. • �ª = 3 ∫∫∫ V x3dxdydz = 3 ∫ 2a 0 x3dx ∫∫ (y−a)2+(z−a)2≤a2−(x−a)2 dydz = 3 ∫ 2a 0 x3pi [ a2 − (x− a)2] dx = 3pi ∫ 2a 0 x3(2ax− x2)dx = 3pi ( 2a 5 x5 − 1 6 x6 )∣∣∣∣2a 0 = 3pi ( 2a 5 · 32a5 − 64 6 a6 ) = 32 5 pia6. 7 A^GreenúªOŽÈ© I = ∮ L ex(x sin y − y cos y)dx+ ex(x cos y + y sin y)dy x2 + y2 , Ù¥L´Œ�:�{ü1w4­‚,_ž�•. )‰. �ε¿©�,¦�Bε(0)¹uLS§ dGreenúª9 ∂x [ ex(x cos y + y sin y) x2 + y2 ] − ∂y [ ex(x sin y + y sin y) x2 + y2 ] = 0,∀ x, y 6= 0, k �ª = ∮ Bε(0) ex(x sin y − y cos y)dx+ ex(x cos y + y sin y)dy x2 + y2 èÆk) · · · · · · uia.china@gmail.com· · · · · ·›Š 7 = 1 ε2 ∫ 2pi 0 { ex [ (x sin y − y cos y)(−y) +(x cos y + y sin y)x ]}∣∣∣∣∣ x=ε cos θ y=ε sin θ dθ = 1 ε2 ∫ 2pi 0 eε cos θε2 cos(ε sin θ)dθ = ∫ 2pi 0 eε cos θ cos(ε sin θ)dθ, �ª = lim ε→0+ ∫ 2pi 0 eε cos θ cos(ε sin θ)dθ = 2pi. 8 �f(x)½Â3(−∞,∞)þ,…3x = 0ëY,¿…é¤kx, y ∈ (−∞,∞),k f(x+ y) = f(x) + f(y). y²,f(x)3(−∞,∞)þëY,…f(x) = f(1)x. y². • f(x)3(−∞,∞)þëY. lim y→x f(y) = lim y→x [f(x) + f(y − x)] = f(x)+ lim |y−x|→0 f(y−x) = f(x). • f(x) = f(1)x. F ∀ n ∈ N, f(n) = f(n− 1) + f(1) = · · · = nf(1); F ∀ n ∈ N, f(−n) = f(0)− f(n) = −n; F ∀ m,n ∈ N, f ( n m ) = 1 m f ( n m ·m ) = 1 m f(n) = n m f(1); F ∀ x ∈ R, f(x) = lim rn→x rn∈Q f(rn) = limrn→x rn∈Q [rnf(1)] = f(1)x. èÆk) · · · · · · uia.china@gmail.com· · · · · ·›Š 8 9 y² ∫ 1 0 dx xx = ∞∑ n=1 1 nn y². d x−x = e−x lnx = ∞∑ n=0 (−x lnx)n n! ,  ∫ 1 0 x−xdx = ∞∑ n=0 ∫ 1 0 (−x lnx)n n! dx = ∞∑ n=0 (−1)n ∫ 1 0 xn lnn xdx n! , Ù¥1˜‡�ª´Ï|x lnx| ≤ e−1, ∀ x ∈ [0, 1], ∞∑ n=0 (−x lnx)n n! ( ≤ ∞∑ n=0 e−n n! ) 3[0, 1]þ˜—Âñ. q∫ 1 0 xn lnn xdx = − n n+ 1 ∫ 1 0 xn lnn−1 xdx = · · · = (−1)n n! (n+ 1)n ∫ 1 0 xndx = (−1)n n! (n+ 1)n+1 , ∫ 1 0 dx xx = ∞∑ n=1 1 nn . 10 �¼êf(x)3[0, 1]þëY…f(x) > 0, ?ؼê g(y) = ∫ 1 0 yf(x) x2 + y2 dx 3(−∞,∞)þ�ëY5. èÆk) · · · · · · uia.china@gmail.com· · · · · ·›Š 9 )‰. P M = sup x∈R f(x) <∞, • �y 6= 0ž, |g(y)| ≤ |y| ∫ 1 0 f(x) x2 + y2 dx < |y| 1|y|2 ∫ 1 0 f(x)dx ≤ M|y| <∞, �g(y)3(−∞, 0) ∪ (0,∞) þëY(Ϗg(y)Œ�). • �y = 0ž,g(y) = 0.�g3y = 0?a�mä. ¯¢þ,é?¿�½ �ε > 0, F df3x = 0?ëY,i.e. ∃ 1 > η > 0, s.t. |x| < η ⇒ |f(x)− f(0)| < 2ε 3pi , ∣∣∣∣∫ η 0 yf(x) x2 + y2 − yf(0) x2 + y2 dx ∣∣∣∣ ≤ max0≤x≤η |f(x)− f(0)| · ∫ η 0 d x|y| 1 + ( x |y| )2 = max 0≤x≤η |f(x)− f(0)| · arctan η|y| ≤ ε 3 . F éþãη > 0,∣∣∣∣∫ 1 η yf(x) x2 + y2 dx ∣∣∣∣ ≤ |y| ·M · 1− η|η|2 ≤ ε 3 , � |y| < δ1 ≡ ε |η| 2 3M(1− η2) ž. F d ∫ η 0 yf(0) x2 + y2 dx = f(0) · sgn(y) · arctan η|y| , èÆk) · · · · · · uia.china@gmail.com· · · · · ·›Š 10 ∃ δ2 = δ2(η, ε) = δ2(ε), s.t. 0 > y > −δ2 ⇒ ∣∣∣∣∫ η 0 yf(0) x2 + y2 dx+ pi 2 f(0) ∣∣∣∣ < ε3 , 0 < y < δ2 ⇒ ∣∣∣∣∫ η 0 yf(0) x2 + y2 dx− pi 2 f(0) ∣∣∣∣ < ε3 . o( k� 0 > y > −min {δ1, δ2} ž, ∣∣∣∣∫ 1 0 yf(x) x2 + y2 dx+ pi 2 f(0) ∣∣∣∣ ≤ ∣∣∣∣∫ 1 η yf(x) x2 + y2 dx ∣∣∣∣+ ∣∣∣∣∫ η 0 yf(x) x2 + y2 − yf(0) x2 + y2 dx ∣∣∣∣ + ∣∣∣∣∫ η 0 yf(0) x2 + y2 dx+ pi 2 f(0) ∣∣∣∣ ≤ ε; � 0 < y < min {δ1, δ2} ž, ∣∣∣∣∫ 1 0 yf(x) x2 + y2 dx− pi 2 f(0) ∣∣∣∣ < ε. u´ g(0− 0) = −pi 2 f(0), g(0 + 0) = pi 2 f(0).