2012中国科学院大学考研真题之数学分析.pdf

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AXAAARAUASABAIAU2012AFAVALA9A8AMAJARAPASABAIAGARAOATADAKAKANBIBYBWAJA6CJD6AZCWACAHAQAWA41. AACFBFBPAXCS150AXA1C6AFBGCFCBBCA6BA180AXA5A02. CLDBAOA8ACD3D2DFAOCNA2C8A1D2DFCFCNA2C8B5AHB0A2C8D8BOCUD1A01 (ABCOBQAY 30AYA2BRD0CO 15AY) BBCKB7CYA5(1) limnn3parenleftbigg2sin 1n sin 2nparenrightbigg. (2) limnparenleftBiggradicalbiggcos 1x2parenrightBiggx4.2 (ABCOBQAY 30AYA2BRD0CO 15AY) BBCKB6AYA5(1) I =integraldisplay pi/20dx1 + tg3x. (2) J =integraldisplayintegraldisplaySxparenleftbig1 + yf(x2 + y2)parenrightbigdxdy,BZA4 S CTDAC5CZ y = x3,y = 1,x = 1CMCQAKASC4DDA2 f(x)CTCDA1BLD4B4CIA33 (ABCOBQAY 15AY) C3CXBNBSB9CIASCGBMDDA5summationdisplayn=1xn1 + 12 + + 1n .4 (ABCOBQAY 15AY) A0BUA5B4CIBN sn(x) = x1 + n2x2(n 1)DGC4BD(,+)C9D9A3CGBMA7B4CIBN tn(x) =nx1 + n2x2(n 1)DGC4BD (0,1) C9D9A3CGBMA35 (ABCOBQAY15AY) CADGC4BDa,bC9A2f(x)BLD4A2g(x)BJB6A2ADC1f(x) 0,g(x) 0.A0BUlimnparenleftbiggintegraldisplay bafn(x)g(x)dxparenrightbigg1/n= maxaxbf(x).6 (ABCOBQAY15AY) CADGC4BD0,aC9A2f(x)AVANBJAQA2ADC1 |f(x)| 1,|f(x)| 1,DHAPx 0,aCCA2|f(x)| 2a + a2.7 (ABCOBQAY 15AY) CA nCED9B1DJDICIA3A0BUA5AWAL xn + nx 1 = 0DCCRD9ASDJCDB2xn,ADC1AP 1CCA2B9CIsummationtextn=1xn CGBMA38 (ABCOBQAY 15 AY) CA (x,y,z) CEDEAU O ARCPC2BTx22 +y22 + z2 = 1ASC9A9AGAY(B8BQA7 z 0ASAGAY) ASC7D9AU (x,y,z)AMASC0BTASBEBKA2C3B6AYintegraldisplayintegraldisplayz(x,y,z)dS.BHBXBVAIA4CHD5AXCV AT1D7A1B31D7
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