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2011!!!2012cccêêêÆÆÆÁÁÁKKKPenny1.f(x)´½Â3[0;1]þ]¼ê§²L:(0;0);(1;0);13;2§¿ f(x)÷vZ10f(x)dx=1§ÁÑy=f(x)ã2.(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§KD f(x0)3Ù¥D+f(x0)=limx1;x2!x0(x1 x0)(x2 x0)0f(x1) f(x2)x1 x2D0f(x0)=limx!x0f(x) f(x0)x x0D f(x0)=limx1;x2!x0(x1 x0)(x2 x0)0f(x1) f(x2)x1 x24.an0;+1Xn=1a2nuѧy²µ3fbng;fcng÷vbn;cn0;+1Xn=1b2n+1;+1Xn=1c2n+1§¦+1Xn=1anbn+1;+1Xn=1ancnuÑ5.f(x)=xn+a1xn 1++an 1x+an;g(x)=xm+b1xm 1++bm 1x+bm;Am+nm+n§cm1´f(x)Xê§ n1´g(x)Xê§=µ0BBBBBBBBBBBBBB@1a1a2an1a1a2an............1a1a2an1b1bm1b1bm1b1bm.........1b1bm1CCCCCCCCCCCCCCAy²µf(x)Úg(x)úϪgêum+n rank(Am+n)6.U´ABX=0Ä:)X§Ù¥A´mnݧB´npݧX´p1ݧW´Rnfm§W=fY=BXjX2Ug§y²µWê´rank(B) rank(AB)§é?¿ÝA;B;C§krank(AB)+rank(BC)rank(B)+rank(ABC)7.A;B´n§A;B§Bݧy²µA+BÚAkÓAõª8. :x2+y2 z2=1§²¡()Lx=1z=0§²¡z=0Y(1)l0=2ëYCz§()\ 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¡AÛNNȵ 1x1;x2;;xn1; 1x1+x2++xn111.é?¿nþ;;0A=0 =0A=0§y²µA顽é¡12.²¡þAA0;BB0;CC0²1§ABÚA0B0u:D§BCÚB0C0u:E§CAÚC0A0u:F§y²µD;E;F1.Rn¥8ÜS´;8duS´k.482.LãÚy²õ¼êLagrange{TaylorÐmª3.Z10dxZpxxsinyydy4.b½nm1§¯´Ä3ëYNF:Rn!Rn=f0g;p!F(p)§¦FJacobiÝ??m§limp!1F(p)=15.fn(x)´¢¶Rþk.ëY¼ê§÷vé?¿4«m[a;b]R;limn!1Zbafn(x)dx=0§y²µéu¢¶¥?¿4«m[c;d]R±9[c;d]þýéȼêg(x)ðklimn!1Zdcg(x)fn(x)dx=06.F(x;y;z)ÚG(x;y;z)Ñ´«D2R3þëY¼ê§÷vN(x;y;z)!(F(x;y;z);G(x;y;z))JacobiÝ3Dþ??1§¼êF(x;y;z)FÝþgrad(F)3Dþ??Ø0.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)=0¼ê§F(x;y;z)=0Ó)§§3Dþ´ÄÓ)Qº7.f(x);g(x)Ñ´«m(a;b)þn¼ê§e3:x02(a;b)þf(x);g(x)0kêѧK¡f(x);g(x)3x0?k.f(x);g(x)3(a;b)¥pØÓ:x1;;xt©Ok1 1;;kt 1§n=k1++kt§y²µé8x2(a;b);9y2(a;b)s:t:f(x)=g(x)+f(n)(y) g(n)(y)n!(x x1)k1(x x2)k2(x xt)kt8.V´êPþk5m§ Vþ5CA,zõªg(x)©)gϪ¦È/ªµg(x)=mYi=1(x i)ti(i6=j)i6=j)y²µe9h(x)2P(x)s:t:(x i)tij(h(x) i);8i=1;2;;m§K(1)h(A)éz(2)h(A) A´C9.A;Bn§D=AB BA;DA=AD§y²µDn=010.VêPþn5m§W1;;WkVýfm§y²µ(1)V¥3Øáu?¿Wj;j=1;2;;k(2)3V|Ä1;2;;n÷vi62[kj=1Wj;8i=1;;n(3)UVýfm§-CU,fWjWVfm WU=Vg§KCU¥¹káõ11.Vnîªm§A2L(v)´C(1)y²µV©)½üüA fmÚ(2)1;;n1;;nVü|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)(½îªmþ1aÝ91aÝ/ª§¿ÑnîªmþCqIO.£=,«{zL«/ª¤12.¦nl1:y x=0z 1=0l2:y+x=0z+1=0l3:y=0z=0Ѥ)¡§13.ÑC8:x0=x+y+3zy0=x+5y+zz0=3x+y+z¦3pþ§¦3dCeEC3pþ.¿ò¤CL«CÚ©Oé3p1 C¦È14.y²µ©OáuVÔ¡x2a2 y2b2=2zþüxpR1:;,´VÔ¡²¡2z=b2 a2§¿ ´^V2012êÆVANILLA1©Û1.Rn¥8ÜS´;8duS´k.48.2.LãÚy²õ¼êLagrange{TaylorЪ.3.R10dxRpxxsinyydy:4.b½nm1,¯´Ä3ëYNF:Rn!Rn f0g;p!F(p),¦FJacobiÝ??m,limp!1F(p)=1,why?5.fn(x)´¢¶Rþk.ëY¼ê,÷v8[a;b]2R,klimn!1Rbafn(x)dx=0.y²:é8[c;d]2R±9[c;d]þýéȼêg(x),ðklimn!1Rbag(x)fn(x)dx=0.6.F(x;y;z)ÚG(x;y;z)Ñ´«D2R3þëY¼ê,÷vN(x;y;z)!(F(x;y;z);G(x;y;z))JacobiÝ3Dþ??1,¼êF(x;y;z)FÝþgrad(F)3Dþ??Ø0.b½3:p0=(x0;y0;z0)2D,F(x0;y0;z0)=G(x0;y0;z0)=0.y²:3:p0U2D,¦3Uþ¼ê§|F(x;y;z)=G(x;y;z)=0¼ê§F(x;y;z)=0Ó).¯3Dþùü|§´ÄÓ),why?7.f(x);g(x)Ñ´«m(a;b)þn¼ê,e3:x02(a;b)þf(x);g(x)0kêÑ,K¡f(x);g(x)3x0?k.yf(x);g(x)3(a;b)¥pØÓ:x1;:::;xt©Ok1 1;:::;kt 1,n=k1++kt.y²:é8x2(a;b);9y2(a;b)¦f(x)=g(x)+f(n) g(n)n!(x x1)k1:::(x xt)kt:2ê1.V´êPþk5m, Vþ5CA,zõªg(x)©)gϪ¦È/ª:g(x)=mYi=1(x i)ti(i6=jKi6=j)y²:e3h(x)2P(x)¦(x i)tij(h(x) i);8i=1;:::;m,K(i)h(A)éz;(ii)h(A) A´C.12.A;Bn,D=AB BA DA=AD,y²Dn=0.3.VêPþn5m,W1;:::;WkVýfm,y²:(a)3V¥Øáu?ÛWj;j=1;:::;k.(b)3V|Ä1;:::;n÷vi=2Skj=1Wj;8i=1;:::;n.(c)UVýfm,-CU,fWjWVfm WU=Vg,KCU¥¹káõ.4.Vnî¼m,A2L(V)´C.(a)y²:V©)½üüA fmÚ.(b)1;:::;n;1;:::;nVü|IOÄ, (1;:::;n)=(1;:::;n)A.y²:A=0BBBBB@cosh1;1icosh2;1i:::coshn;1icosh1;2icosh2;2i:::coshn;2i............cosh1;nicosh2;ni:::coshn;ni1CCCCCA(c)(½î¼mþ1aÝ91aÝ/ª,¿Ñnî¼mþCqIO.(=,«{zL«/ª).3AÛ1.¦nl1:8:y x=0z 1=0;l2:8:y+x=0z+1=0;l3:8:y=0z=0Ѥ)¡§.2.ÑC8:x0=x+y+3zy0=x+5y+zz0=3x+y+z¦3pþ,¦3dCeEC3pþ;¿ò¤CL«CÚ©Oé3p1 C¦È.3.y²:©OáuVÔ¡x2a2 y2b2=2z(1)þüxpR1:;,´¡(1)²¡2z=b2 a2^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1mn,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
本文标题:历年北大直博试题
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