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ID：40060847
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页数：7页
时间：2019-07-18
《2015中科大数学分析试题与解答》由会员上传分享，免费在线阅读，更多相关内容在行业资料-天天文库。
1、2015¥‰ŒêÆ©ÛÁK†)‰morrismodelJanuary11,20151.¦Z1xlimsinjsintjdt:x!+1x0).ZZ1x1xlimsinjsintjdt=limjsintjdt:x!+1x0x!+1x0n2、2+y2)dxdy;Ù¥D=(x;y):+1:a2b2D).4‹I“†:x=arcos;y=brsin;0r1;02:KkZZZ1Z2(x2+y2)dxdy=abrdr(a2r2cos2+b2r2sin2)dD0022=ab(a+b):43.¦(xyx+y1f(x;y)=p+p3þ•ŒŠ:1+x21+y2x;y0).N´•xg(x)=p3[0;1]þ4O;1+x2x1 xf(x;y)p+p=:h(x):1+x21+(1 x)202 32 3h(x)=(1+x)2 (3、1+(1 x))2:04、ZZ=(y2+x2+z2)dxdydzfx2+y2+z2R2;z>0gZZZ1222=(x+y+z)dxdydz2fx2+y2+z2R2g25=R:5ZZ213235(xy+x)dydz+yzdzdx+Rdxdy= R:T3¤±ZZ2132375(xy+x)dydz+yzdzdx+Rdxdy=R:S355.¦Z11dx;n>1:1+xn0).·y=1,Kk1+xnZ1Z01111 11dx= y( 1)ndy01+xn1nyy2Z11 11 1=yn(1 y)ndyn0111=B(;1 )nnn5、=:nsinn36.y²:Z14n10:2n+202ny.·y=tanx,KZZ1nZ1n4nyy1tanxdx=dydy=:01+y2022n+27.?Ø?êX1pp(n+1 n)cosnn=1^‡ÂñÚýéÂñ5.).0ž,ϑتu0,lØÂñ; >2ž,ppjcosnj j(n+1 n)cosnj=ppn2:(n+1+n)l?êýéÂñ;0< 2ž,Xksin1 sin2k+1122cos6、n=;2sin1sin1n=122pp1(n+1 n)=püN4~ª•u0;p(n+1+n)4l?êÂñ.´XkppXkpp(n+1 n)jcosnj(n+1 n)cos2nn=1n=1Xkpp1=(n+1 n)2n=1Xkpp1 (n+1 n)cos2n:2n=1´•X1ppX1pp(n+1 n)cos2nÂñ;(n+1 n)uÑ;n=1n=1lX1pp(n+1 n)jcosnjuÑ:n=1¤±0< 2ž,?ê^‡Âñ.8.2(0;1),fang•î‚4O‘ê,…fan+1 angk.,7、¦ lima a:n+1nn!1).XJfangÃ.,jan+1 anj8、dt:00KZxg0(x)=x2f(x) 2f(x)f(t)dt0Zx=f(x)x2 2f(t)dt0Zxf(x)x2 2tdt0=0:lZ1Z12x2f(x)dx f(x)dx00Z1=g(1) g(0)=g0(x)dx0:0 Ò¤á…=é?Ûx2(0;1),Zxf(x)x2 2f(t)dt=0:0XJf0,Kþªw,¤á.ÄK,•3x02(0;1),f(x0)6=0.KZx0x2 2f(t)dt)
2、2+y2)dxdy;Ù¥D=(x;y):+1:a2b2D).4‹I“†:x=arcos;y=brsin;0r1;02:KkZZZ1Z2(x2+y2)dxdy=abrdr(a2r2cos2+b2r2sin2)dD0022=ab(a+b):43.¦(xyx+y1f(x;y)=p+p3þ•ŒŠ:1+x21+y2x;y0).N´•xg(x)=p3[0;1]þ4O;1+x2x1 xf(x;y)p+p=:h(x):1+x21+(1 x)202 32 3h(x)=(1+x)2 (
3、1+(1 x))2:04、ZZ=(y2+x2+z2)dxdydzfx2+y2+z2R2;z>0gZZZ1222=(x+y+z)dxdydz2fx2+y2+z2R2g25=R:5ZZ213235(xy+x)dydz+yzdzdx+Rdxdy= R:T3¤±ZZ2132375(xy+x)dydz+yzdzdx+Rdxdy=R:S355.¦Z11dx;n>1:1+xn0).·y=1,Kk1+xnZ1Z01111 11dx= y( 1)ndy01+xn1nyy2Z11 11 1=yn(1 y)ndyn0111=B(;1 )nnn5、=:nsinn36.y²:Z14n10:2n+202ny.·y=tanx,KZZ1nZ1n4nyy1tanxdx=dydy=:01+y2022n+27.?Ø?êX1pp(n+1 n)cosnn=1^‡ÂñÚýéÂñ5.).0ž,ϑتu0,lØÂñ; >2ž,ppjcosnj j(n+1 n)cosnj=ppn2:(n+1+n)l?êýéÂñ;0< 2ž,Xksin1 sin2k+1122cos6、n=;2sin1sin1n=122pp1(n+1 n)=püN4~ª•u0;p(n+1+n)4l?êÂñ.´XkppXkpp(n+1 n)jcosnj(n+1 n)cos2nn=1n=1Xkpp1=(n+1 n)2n=1Xkpp1 (n+1 n)cos2n:2n=1´•X1ppX1pp(n+1 n)cos2nÂñ;(n+1 n)uÑ;n=1n=1lX1pp(n+1 n)jcosnjuÑ:n=1¤±0< 2ž,?ê^‡Âñ.8.2(0;1),fang•î‚4O‘ê,…fan+1 angk.,7、¦ lima a:n+1nn!1).XJfangÃ.,jan+1 anj8、dt:00KZxg0(x)=x2f(x) 2f(x)f(t)dt0Zx=f(x)x2 2f(t)dt0Zxf(x)x2 2tdt0=0:lZ1Z12x2f(x)dx f(x)dx00Z1=g(1) g(0)=g0(x)dx0:0 Ò¤á…=é?Ûx2(0;1),Zxf(x)x2 2f(t)dt=0:0XJf0,Kþªw,¤á.ÄK,•3x02(0;1),f(x0)6=0.KZx0x2 2f(t)dt)
4、ZZ=(y2+x2+z2)dxdydzfx2+y2+z2R2;z>0gZZZ1222=(x+y+z)dxdydz2fx2+y2+z2R2g25=R:5ZZ213235(xy+x)dydz+yzdzdx+Rdxdy= R:T3¤±ZZ2132375(xy+x)dydz+yzdzdx+Rdxdy=R:S355.¦Z11dx;n>1:1+xn0).·y=1,Kk1+xnZ1Z01111 11dx= y( 1)ndy01+xn1nyy2Z11 11 1=yn(1 y)ndyn0111=B(;1 )nnn
5、=:nsinn36.y²:Z14n10:2n+202ny.·y=tanx,KZZ1nZ1n4nyy1tanxdx=dydy=:01+y2022n+27.?Ø?êX1pp(n+1 n)cosnn=1^‡ÂñÚýéÂñ5.).0ž,ϑتu0,lØÂñ; >2ž,ppjcosnj j(n+1 n)cosnj=ppn2:(n+1+n)l?êýéÂñ;0< 2ž,Xksin1 sin2k+1122cos
6、n=;2sin1sin1n=122pp1(n+1 n)=püN4~ª•u0;p(n+1+n)4l?êÂñ.´XkppXkpp(n+1 n)jcosnj(n+1 n)cos2nn=1n=1Xkpp1=(n+1 n)2n=1Xkpp1 (n+1 n)cos2n:2n=1´•X1ppX1pp(n+1 n)cos2nÂñ;(n+1 n)uÑ;n=1n=1lX1pp(n+1 n)jcosnjuÑ:n=1¤±0< 2ž,?ê^‡Âñ.8.2(0;1),fang•î‚4O‘ê,…fan+1 angk.,
7、¦ lima a:n+1nn!1).XJfangÃ.,jan+1 anj8、dt:00KZxg0(x)=x2f(x) 2f(x)f(t)dt0Zx=f(x)x2 2f(t)dt0Zxf(x)x2 2tdt0=0:lZ1Z12x2f(x)dx f(x)dx00Z1=g(1) g(0)=g0(x)dx0:0 Ò¤á…=é?Ûx2(0;1),Zxf(x)x2 2f(t)dt=0:0XJf0,Kþªw,¤á.ÄK,•3x02(0;1),f(x0)6=0.KZx0x2 2f(t)dt)
8、dt:00KZxg0(x)=x2f(x) 2f(x)f(t)dt0Zx=f(x)x2 2f(t)dt0Zxf(x)x2 2tdt0=0:lZ1Z12x2f(x)dx f(x)dx00Z1=g(1) g(0)=g0(x)dx0:0 Ò¤á…=é?Ûx2(0;1),Zxf(x)x2 2f(t)dt=0:0XJf0,Kþªw,¤á.ÄK,•3x02(0;1),f(x0)6=0.KZx0x2 2f(t)dt)
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