历年北大直博试题

2011!!!2012cccŒŒŒêêꆆ†ÆÆÆÁÁÁKKKPenny1.f(x)´½Â3[0;1]þ]¼ê§²L:(0;0);(1;0);13;2§¿…f(x)÷vZ10f(x)dx=1§ÁŠÑy=f(x)ã”2.(1)f(x)2C[0;1]§f(0)=f(1)§y²µé801;12N;92[0;1]s:t:f(+)=f()(2)y²µé801;162N;9g(x)2C[0;1];g(0)=g(1)s:t:8x2[0;1]Ñkg(x+)g(x)6=03.f(x)3x0,‡+SëY§±e`{´Ä¤áº(1)D+f(x0)3§KD0f(x0)3(2)D0f(x0)3§KDf(x0)3Ù¥D+f(x0)=limx1;x2!x0(x1x0)(x2x0)0f(x1)f(x2)x1x2D0f(x0)=limx!x0f(x)f(x0)xx0Df(x0)=limx1;x2!x0(x1x0)(x2x0)0f(x1)f(x2)x1x24.an0;+1Xn=1a2nuѧy²µ3fbng;fcng÷vbn;cn0;+1Xn=1b2n+1;+1Xn=1c2n+1§¦+1Xn=1anbn+1;+1Xn=1ancnuÑ5.f(x)=xn+a1xn1+


+an1x+an;g(x)=xm+b1xm1+


+bm1x+bm;Am+nm+n§cm1´f(x)Xê§n1´g(x)Xê§=µ0BBBBBBBBBBBBBB@1a1a2


an1a1a2


an............1a1a2


an1b1


bm1b1


bm1b1


bm.........1b1


bm1CCCCCCCCCCCCCCAy²µf(x)Úg(x)ŒúϪgêum+nrank(Am+n)6.U´ABX=0Ä:)X§Ù¥A´mnݧB´npݧX´p1ݧW´Rnf˜m§W=fY=BXjX2Ug§y²µW‘ê´rank(B)rank(AB)§é?¿ÝA;B;C§krank(AB)+rank(BC)rank(B)+rank(ABC)7.A;B´n§A;BŒ†§B˜Ý§y²µA+BÚAkƒÓAõ‘ª8.:x2+y2z2=1§²¡()L†‚x=1z=0§†²¡z=0Y(1)l0=2ëYCzž§()\L«­‚´=«a.º(2)y²µe²¡Ax+By+Cz+D=0†‚ü^†‚§KA2+B2=C2+D29.f(x)3[a;+1)þk½Â§Z+1af(x)dxÂñ§gye^‡´ÄUíÑlimx!+1f(x)(1)f(x)2C1(2)Z+10jf(x)j3dxÂñ(3)Z+10jf0(x)j2dxÂñ(4)Z+10jf00(x)jdxÂñ10.¦e¡AÛNNȵ1x1;x2;


;xn1;1x1+x2+


+xn111.é?¿n‘•þ;;0A=0…=0A=0§y²µA顽‡é¡12.²¡þ†‚AA0;BB0;CC0²1§ABÚA0B0u:D§BCÚB0C0u:E§CAÚC0A0u:F§y²µD;E;F
‚1.Rn¥8ÜS´;8duS´k.482.LãÚy²õ¼ê‘Lagrange{‘TaylorÐmª3.Z10dxZpxxsinyydy4.b½nm1§¯´Ä3ëYŒNF:Rn!Rn=f0g;p!F(p)§¦FJacobiݝ??m§limp!1F(p)=15.fn(x)´¢¶Rþ˜˜—k.ëY¼ê§÷vé?¿4«m[a;b]R;limn!1Zbafn(x)dx=0§y²µéu¢¶¥?¿4«m[c;d]R±9[c;d]þýéŒÈ¼êg(x)ðklimn!1Zdcg(x)
fn(x)dx=06.F(x;y;z)ÚG(x;y;z)Ñ´«D2R3þëYŒ¼ê§÷vN(x;y;z)!(F(x;y;z);G(x;y;z))Jacobiݝ3Dþ??1§¼êF(x;y;z)Fݕþgrad(F)3Dþ??؏0.b½p0=(x0;y0;z0)2D;F(x0;y0;z0)=G(x0;y0;z0)=0.y²µ3p0+U2D§¦3Uþ¼ê§|F(x;y;z)=G(x;y;z)=0†¼ê§F(x;y;z)=0Ó)§§‚3Dþ´ÄÓ)Qº7.f(x);g(x)Ñ´«m(a;b)þnŒ¼ê§e3:x02(a;b)þf(x);g(x)0–kêу§K¡f(x);g(x)3x0?kƒƒ.f(x);g(x)3(a;b)¥p؃Ó:x1;


;xt©Ok11;


;kt1ƒƒ§n=k1+


+kt§y²µé8x2(a;b);9y2(a;b)s:t:f(x)=g(x)+f(n)(y)g(n)(y)n!(xx1)k1(xx2)k2


(xxt)kt8.V´êPþk‘‚5˜m§…Vþ‚5C†A,‡zõ‘ªg(x)Œ©)˜gϪ˜¦È/ªµg(x)=mYi=1(xi)ti(i6=j)i6=j)y²µe9h(x)2P(x)s:t:(xi)tij(h(x)i);8i=1;2;


;m§K(1)h(A)Œéz(2)h(A)A´˜C†9.A;Bn§D=ABBA;DA=AD§y²µDn=010.VêPþn‘‚5˜m§W1;


;WkVýf˜m§y²µ(1)V¥3ƒØáu?¿˜‡Wj;j=1;2;


;k(2)3V˜|Ä1;2;


;n÷vi62[kj=1Wj;8i=1;


;n(3)UVýf˜m§-CU,fWjWVf˜m…WU=Vg§KCU¥¹káõ‡ƒ11.Vn‘m§A2L(v)´C†(1)y²µVŒ©)˜˜‘½‘üüAf˜m†Ú(2)1;


;n†1;


;nVü|IOħ…(1;


;n)=(1;


;n)A§y²µA=0BBB@cos(1;1)cos(2;1)


cos(n;1)cos(1;2)cos(2;2)


cos(n;2)............cos(1;n)cos(2;n)


cos(n;n)1CCCA(3)(½‘mþ1˜aÝ91aݘ„/ª§¿‰Ñn‘mþC†ƒqIO.£=,«{zL«/ª¤12.¦†n†‚l1:yx=0z1=0l2:y+x=0z+1=0l3:y=0z=0󆂤)­¡§13.‰Ñ•C†8:x0=x+y+3zy0=x+5y+zz0=3x+y+z¦3‡pƒ•þ§¦3dC†eEC3‡pƒ•þ.¿ò¤‰•C†L«˜‡C†Ú©Oé3‡pƒ•–1 C†¦È14.y²µ©OáuV­Ô¡x2a2y2b2=2zþüxpƒR††1‚:;,´V­Ô¡†²¡2z=b2a2‚§¿…´˜^V­‚2012Œê†ÆVANILLA1©Û1.Rn¥8ÜS´;8duS´k.48.2.LãÚy²õ¼ê‘Lagrange{‘TaylorЪ.3.R10dxRpxxsinyydy:4.b½nm1,¯´Ä3ëYŒNF:Rn!Rnf0g;p!F(p),¦FJacobiݝ??m,limp!1F(p)=1,why?5.fn(x)´¢¶Rþ˜˜—k.ëY¼ê,÷v8[a;b]2R,klimn!1Rbafn(x)dx=0.y²:é8[c;d]2R±9[c;d]þýéŒÈ¼êg(x),ðklimn!1Rbag(x)fn(x)dx=0.6.F(x;y;z)ÚG(x;y;z)Ñ´«D2R3þëYŒ¼ê,÷vN(x;y;z)!(F(x;y;z);G(x;y;z))Jacobiݝ3Dþ??1,¼êF(x;y;z)Fݕþgrad(F)3Dþ??؏0.b½3:p0=(x0;y0;z0)2D,F(x0;y0;z0)=G(x0;y0;z0)=0.y²:3:p0U2D,¦3Uþ¼ê§|F(x;y;z)=G(x;y;z)=0†¼ê§F(x;y;z)=0Ó).¯3Dþùü|§´ÄÓ),why?7.f(x);g(x)Ñ´«m(a;b)þnŒ¼ê,e3:x02(a;b)þf(x);g(x)0–kêу,K¡f(x);g(x)3x0?kƒƒ.yf(x);g(x)3(a;b)¥p؃Ó:x1;:::;xt©Ok11;:::;kt1ƒƒ,n=k1+


+kt.y²:é8x2(a;b);9y2(a;b)¦f(x)=g(x)+f(n)g(n)n!(xx1)k1:::(xxt)kt:2“ê1.V´êPþk‘‚5˜m,…Vþ‚5C†A,‡zõ‘ªg(x)Œ©)˜gϪ˜¦È/ª:g(x)=mYi=1(xi)ti(i6=jKi6=j)y²:e3h(x)2P(x)¦(xi)tij(h(x)i);8i=1;:::;m,K(i)h(A)Œéz;(ii)h(A)A´˜C†.12.A;Bn,D=ABBA…DA=AD,y²Dn=0.3.VêPþn‘‚5˜m,W1;:::;WkVýf˜m,y²:(a)3V¥ƒØáu?ۘ‡Wj;j=1;:::;k.(b)3V˜|Ä1;:::;n÷vi=2Skj=1Wj;8i=1;:::;n.(c)UVýf˜m,-CU,fWjWVf˜m…WU=Vg,KCU¥¹káõ‡ƒ.4.Vn‘m,A2L(V)´C†.(a)y²:VŒ©)˜˜‘½‘üüAf˜m†Ú.(b)1;:::;n;1;:::;nVü|IOÄ,…(1;:::;n)=(1;:::;n)A.y²:A=0BBBBB@cosh1;1icosh2;1i:::coshn;1icosh1;2icosh2;2i:::coshn;2i............cosh1;nicosh2;ni:::coshn;ni1CCCCCA(c)(½‘mþ1˜aÝ91aݘ„/ª,¿‰Ñn‘mþC†ƒqIO.(=,«{zL«/ª).3AÛ1.¦†n†‚l1:8:yx=0z1=0;l2:8:y+x=0z+1=0;l3:8:y=0z=0󆂤)­¡§.2.‰Ñ•C†8:x0=x+y+3zy0=x+5y+zz0=3x+y+z¦3‡pƒ•þ,¦3dC†eEC3‡pƒ•þ;¿ò¤‰•C†L«˜‡C†Ú©Oé3‡pƒ•–1 C†¦È.3.y²:©OáuV­Ô¡x2a2y2b2=2z(1)þüxpƒR††1‚:;,´­¡(1)†²¡2z=b2a2‚˜^V­‚.2北京大学2012年直博生入学考试试题考试科目：数学分析（满分150分）1.（25分）nR中的集合S称为紧集，如果对于S的开覆盖U，都存在U中的有限个元素也覆盖S.证明：nRS为紧集的充分必要条件是S是nR中的有界闭集.2.（25分）表述和证明多元函数带Lagrange（拉格朗日）余项的二阶Taylor（泰勒）展开公式（余项为二阶）.3.（20分）计算积分10sinxxdyyydx.4.（20分）假定1mn，问是否存在连续可导的映射0-:mnRRF，(p)Fp，使得F的Jacobi（雅可比）矩阵的秩处处为m，而(p)limFp，为什么？这里0表示mR的原点.5.（20分）设xfn是实轴R上一列一致有界的连续函数，满足对于任意闭区间Rba,，成立0(x)dxli