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[家里蹲大学数学杂志]第034期中山大学2008年数学分析考研试题参考解答

时间:2014-05-22 06:10:57      阅读:261      评论:0      收藏:0      [点我收藏+]

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1  (每小题6分,共48分)

 (1) 求limbubuko.com,布布扣x0+bubuko.com,布布扣xbubuko.com,布布扣xbubuko.com,布布扣;bubuko.com,布布扣

解答: 

 原式bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣bubuko.com,布布扣limbubuko.com,布布扣x0+bubuko.com,布布扣ebubuko.com,布布扣xlnxbubuko.com,布布扣=limbubuko.com,布布扣x0+bubuko.com,布布扣ebubuko.com,布布扣bubuko.com,布布扣lnxbubuko.com,布布扣bubuko.com,布布扣1/xbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=ebubuko.com,布布扣limbubuko.com,布布扣x0+bubuko.com,布布扣bubuko.com,布布扣lnxbubuko.com,布布扣bubuko.com,布布扣1/xbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣Lbubuko.com,布布扣bubuko.com,布布扣Hospitalbubuko.com,布布扣ebubuko.com,布布扣limbubuko.com,布布扣x0+bubuko.com,布布扣bubuko.com,布布扣1/xbubuko.com,布布扣bubuko.com,布布扣?1/xbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣ebubuko.com,布布扣?limbubuko.com,布布扣x0+bubuko.com,布布扣xbubuko.com,布布扣=ebubuko.com,布布扣0bubuko.com,布布扣=1.bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
 

(2) 求xbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣sinxbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣dx;bubuko.com,布布扣

解答:  

 原式bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣t=xbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣bubuko.com,布布扣2tbubuko.com,布布扣2bubuko.com,布布扣sintdt=?2tbubuko.com,布布扣2bubuko.com,布布扣d(cost)=?2tbubuko.com,布布扣2bubuko.com,布布扣cost+4tcostdtbubuko.com,布布扣?2tbubuko.com,布布扣2bubuko.com,布布扣cost+4td(sint)=?2tbubuko.com,布布扣2bubuko.com,布布扣cost+4tsint?4sintdtbubuko.com,布布扣?2tbubuko.com,布布扣2bubuko.com,布布扣cost+4tsint+4cost+Cbubuko.com,布布扣?2xcosxbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣+4xbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣sinxbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣+4cosxbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣+C(其中C是任意常数).bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣

 (3) 求bubuko.com,布布扣ebubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣dxbubuko.com,布布扣bubuko.com,布布扣x(2+lnbubuko.com,布布扣2bubuko.com,布布扣x)bubuko.com,布布扣bubuko.com,布布扣;bubuko.com,布布扣

解答: 

 原式bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣t=lnxbubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣dtbubuko.com,布布扣bubuko.com,布布扣2+tbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣dtbubuko.com,布布扣bubuko.com,布布扣1+bubuko.com,布布扣tbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣d(bubuko.com,布布扣tbubuko.com,布布扣bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣)bubuko.com,布布扣bubuko.com,布布扣1+(bubuko.com,布布扣tbubuko.com,布布扣bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣)bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣arctan(bubuko.com,布布扣tbubuko.com,布布扣bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣)|bubuko.com,布布扣1bubuko.com,布布扣0bubuko.com,布布扣=bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣arctanbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣.bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
 

(4) 求bubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣xebubuko.com,布布扣?xbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣(1+ebubuko.com,布布扣?xbubuko.com,布布扣)bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣dx;bubuko.com,布布扣

解答:  

 原式bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣Lbubuko.com,布布扣bubuko.com,布布扣Hospitalbubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣ebubuko.com,布布扣xbubuko.com,布布扣=tbubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣xebubuko.com,布布扣xbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣(ebubuko.com,布布扣xbubuko.com,布布扣+1)bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣dx=?bubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣xd(bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣ebubuko.com,布布扣xbubuko.com,布布扣+1bubuko.com,布布扣bubuko.com,布布扣)bubuko.com,布布扣?xbubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣ebubuko.com,布布扣xbubuko.com,布布扣+1bubuko.com,布布扣bubuko.com,布布扣|bubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣+bubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣ebubuko.com,布布扣xbubuko.com,布布扣+1bubuko.com,布布扣bubuko.com,布布扣dxbubuko.com,布布扣?limbubuko.com,布布扣x+bubuko.com,布布扣bubuko.com,布布扣xbubuko.com,布布扣bubuko.com,布布扣ebubuko.com,布布扣xbubuko.com,布布扣+1bubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣ebubuko.com,布布扣xbubuko.com,布布扣+1bubuko.com,布布扣bubuko.com,布布扣dxbubuko.com,布布扣?limbubuko.com,布布扣x+bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣ebubuko.com,布布扣xbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣ebubuko.com,布布扣xbubuko.com,布布扣+1bubuko.com,布布扣bubuko.com,布布扣dx=bubuko.com,布布扣+bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣ebubuko.com,布布扣xbubuko.com,布布扣+1bubuko.com,布布扣bubuko.com,布布扣dxbubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣t(t+1)bubuko.com,布布扣bubuko.com,布布扣dt=bubuko.com,布布扣+bubuko.com,布布扣1bubuko.com,布布扣(bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣tbubuko.com,布布扣bubuko.com,布布扣?bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣t+1bubuko.com,布布扣bubuko.com,布布扣)dtbubuko.com,布布扣[lnt?ln(t+1)]|bubuko.com,布布扣+bubuko.com,布布扣1bubuko.com,布布扣=ln(bubuko.com,布布扣tbubuko.com,布布扣bubuko.com,布布扣t+1bubuko.com,布布扣bubuko.com,布布扣)|bubuko.com,布布扣+bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣limbubuko.com,布布扣t+bubuko.com,布布扣ln(bubuko.com,布布扣tbubuko.com,布布扣bubuko.com,布布扣t+1bubuko.com,布布扣bubuko.com,布布扣)?lnbubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣=ln2.bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
 

(5) 方程z=f(x,xy)+φ(y+z)bubuko.com,布布扣 确定函数z=z(x,y)bubuko.com,布布扣 , 求全微分dzbubuko.com,布布扣 ;

解答: 在方程z=f(x,xy)+φ(y+z)bubuko.com,布布扣 左右两边分别关于x,ybubuko.com,布布扣 求偏导,可得

zbubuko.com,布布扣xbubuko.com,布布扣=fbubuko.com,布布扣1bubuko.com,布布扣(x,xy)+yfbubuko.com,布布扣2bubuko.com,布布扣(x,xy)+φbubuko.com,布布扣bubuko.com,布布扣(y+z)zbubuko.com,布布扣xbubuko.com,布布扣?zbubuko.com,布布扣xbubuko.com,布布扣=bubuko.com,布布扣fbubuko.com,布布扣1bubuko.com,布布扣(x,xy)+yfbubuko.com,布布扣2bubuko.com,布布扣(x,xy)bubuko.com,布布扣bubuko.com,布布扣1?φbubuko.com,布布扣bubuko.com,布布扣(y+z)bubuko.com,布布扣bubuko.com,布布扣,bubuko.com,布布扣
zbubuko.com,布布扣ybubuko.com,布布扣=xfbubuko.com,布布扣2bubuko.com,布布扣(x,xy)+φbubuko.com,布布扣bubuko.com,布布扣(y+z)(1+zbubuko.com,布布扣ybubuko.com,布布扣)?zbubuko.com,布布扣ybubuko.com,布布扣=bubuko.com,布布扣xfbubuko.com,布布扣2bubuko.com,布布扣(x,xy)+φbubuko.com,布布扣bubuko.com,布布扣(y+z)bubuko.com,布布扣bubuko.com,布布扣1?φbubuko.com,布布扣bubuko.com,布布扣(y+z)bubuko.com,布布扣bubuko.com,布布扣,bubuko.com,布布扣
从而全微分
dz=bubuko.com,布布扣fbubuko.com,布布扣1bubuko.com,布布扣(x,xy)+yfbubuko.com,布布扣2bubuko.com,布布扣(x,xy)bubuko.com,布布扣bubuko.com,布布扣1?φbubuko.com,布布扣bubuko.com,布布扣(y+z)bubuko.com,布布扣bubuko.com,布布扣dx+bubuko.com,布布扣xfbubuko.com,布布扣2bubuko.com,布布扣(x,xy)+φbubuko.com,布布扣bubuko.com,布布扣(y+z)bubuko.com,布布扣bubuko.com,布布扣1?φbubuko.com,布布扣bubuko.com,布布扣(y+z)bubuko.com,布布扣bubuko.com,布布扣dy.bubuko.com,布布扣
 

(6) 求曲线ybubuko.com,布布扣2bubuko.com,布布扣=xbubuko.com,布布扣2bubuko.com,布布扣(4?x)bubuko.com,布布扣 所围图形的面积;

解答: 由分析可知,曲线ybubuko.com,布布扣2bubuko.com,布布扣=xbubuko.com,布布扣2bubuko.com,布布扣(4?x)bubuko.com,布布扣 关于xbubuko.com,布布扣 轴对称;又由于x4,y(0)=y(4)=0,bubuko.com,布布扣 x0bubuko.com,布布扣 时,y(x)bubuko.com,布布扣 单调递减且limbubuko.com,布布扣x?bubuko.com,布布扣y(x)=+bubuko.com,布布扣 ,故所求面积 

Sbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣bubuko.com,布布扣2bubuko.com,布布扣4bubuko.com,布布扣0bubuko.com,布布扣xbubuko.com,布布扣2bubuko.com,布布扣(4?x)bubuko.com,布布扣?bubuko.com,布布扣?bubuko.com,布布扣?bubuko.com,布布扣?bubuko.com,布布扣?bubuko.com,布布扣?bubuko.com,布布扣?bubuko.com,布布扣?bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣dx=2bubuko.com,布布扣4bubuko.com,布布扣0bubuko.com,布布扣x4?xbubuko.com,布布扣?bubuko.com,布布扣?bubuko.com,布布扣?bubuko.com,布布扣?bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣dx=bubuko.com,布布扣4?xbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=tbubuko.com,布布扣4bubuko.com,布布扣2bubuko.com,布布扣0bubuko.com,布布扣(4?tbubuko.com,布布扣2bubuko.com,布布扣)tbubuko.com,布布扣2bubuko.com,布布扣dtbubuko.com,布布扣4(bubuko.com,布布扣4bubuko.com,布布扣bubuko.com,布布扣3bubuko.com,布布扣bubuko.com,布布扣tbubuko.com,布布扣3bubuko.com,布布扣?bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣5bubuko.com,布布扣bubuko.com,布布扣tbubuko.com,布布扣5bubuko.com,布布扣)|bubuko.com,布布扣2bubuko.com,布布扣0bubuko.com,布布扣=bubuko.com,布布扣256bubuko.com,布布扣bubuko.com,布布扣15bubuko.com,布布扣bubuko.com,布布扣.bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
 

(7) 计算二重积分bubuko.com,布布扣Dbubuko.com,布布扣(bubuko.com,布布扣xbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣abubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣ybubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣)dxdy,bubuko.com,布布扣 其中D={(x,y)|xbubuko.com,布布扣2bubuko.com,布布扣+ybubuko.com,布布扣2bubuko.com,布布扣1};bubuko.com,布布扣

解答:  

 原式bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣x=ρcosθ,y=ρsinθbubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣2πbubuko.com,布布扣0bubuko.com,布布扣(bubuko.com,布布扣ρbubuko.com,布布扣2bubuko.com,布布扣cosbubuko.com,布布扣2bubuko.com,布布扣θbubuko.com,布布扣bubuko.com,布布扣abubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣ρbubuko.com,布布扣2bubuko.com,布布扣sinbubuko.com,布布扣2bubuko.com,布布扣θbubuko.com,布布扣bubuko.com,布布扣bbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣)ρdρdθbubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣0bubuko.com,布布扣ρbubuko.com,布布扣3bubuko.com,布布扣dρ?bubuko.com,布布扣2πbubuko.com,布布扣0bubuko.com,布布扣(bubuko.com,布布扣ρbubuko.com,布布扣2bubuko.com,布布扣cosbubuko.com,布布扣2bubuko.com,布布扣θbubuko.com,布布扣bubuko.com,布布扣abubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣ρbubuko.com,布布扣2bubuko.com,布布扣sinbubuko.com,布布扣2bubuko.com,布布扣θbubuko.com,布布扣bubuko.com,布布扣bbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣)dθbubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣4bubuko.com,布布扣bubuko.com,布布扣ρbubuko.com,布布扣4bubuko.com,布布扣|bubuko.com,布布扣1bubuko.com,布布扣0bubuko.com,布布扣?bubuko.com,布布扣2πbubuko.com,布布扣0bubuko.com,布布扣(bubuko.com,布布扣1+cos2θbubuko.com,布布扣bubuko.com,布布扣2abubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣1?cos2θbubuko.com,布布扣bubuko.com,布布扣2bbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣)dθbubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣4bubuko.com,布布扣bubuko.com,布布扣[(bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣2abubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣2bbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣)θ+bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣(bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣2abubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣?bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣2bbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣)sin2θ]|bubuko.com,布布扣2πbubuko.com,布布扣0bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣πbubuko.com,布布扣bubuko.com,布布扣4bubuko.com,布布扣bubuko.com,布布扣(bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣abubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣bbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣).bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
 

(8) 判别级数bubuko.com,布布扣n=1bubuko.com,布布扣bubuko.com,布布扣ububuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣 的敛散性,其中ububuko.com,布布扣nbubuko.com,布布扣=bubuko.com,布布扣1!+2!+3!+?+n!bubuko.com,布布扣bubuko.com,布布扣(2n)!bubuko.com,布布扣bubuko.com,布布扣,n=1,2,?bubuko.com,布布扣 .

解答: 方法一:由于

ububuko.com,布布扣nbubuko.com,布布扣=bubuko.com,布布扣1!+2!+3!+?+n!bubuko.com,布布扣bubuko.com,布布扣(2n)!bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣n?n!bubuko.com,布布扣bubuko.com,布布扣(2n)!bubuko.com,布布扣bubuko.com,布布扣<bubuko.com,布布扣n?n!bubuko.com,布布扣bubuko.com,布布扣n!?n?n?nbubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣nbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣,bubuko.com,布布扣
而级数bubuko.com,布布扣n=1bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣nbubuko.com,布布扣2bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣 收敛,因而由正项级数的比较原则可知,级数bubuko.com,布布扣n=1bubuko.com,布布扣bubuko.com,布布扣ububuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣 收敛.

方法二:

limbubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣ububuko.com,布布扣n+1bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣ububuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣=bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣1!+2!+3!+?+n!+(n+1)!bubuko.com,布布扣bubuko.com,布布扣(2(n+1))!bubuko.com,布布扣bubuko.com,布布扣?bubuko.com,布布扣(2n)!bubuko.com,布布扣bubuko.com,布布扣1!+2!+3!+?+n!bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣limbubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣1!+2!+3!+?+n!+(n+1)!bubuko.com,布布扣bubuko.com,布布扣(2n+1)(2n+2)(1!+2!+3!+?+n!)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣limbubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣1!+2!+3!+?+n!bubuko.com,布布扣bubuko.com,布布扣(2n+1)(2n+2)(1!+2!+3!+?+n!)bubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣limbubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣(n+1)!bubuko.com,布布扣bubuko.com,布布扣(2n+1)(2n+2)(1!+2!+3!+?+n!)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣limbubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣(2n+1)(2n+2)bubuko.com,布布扣bubuko.com,布布扣+limbubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣n!bubuko.com,布布扣bubuko.com,布布扣2(2n+1)(1!+2!+3!+?+n!)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣0+0=0<1,bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
因而由正项级数的d‘Alembert判别法或比式判别法可知,级数bubuko.com,布布扣n=1bubuko.com,布布扣bubuko.com,布布扣ububuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣 收敛.

注记: 由于

0bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣n!bubuko.com,布布扣bubuko.com,布布扣2(2n+1)(1!+2!+3!+?+n!)bubuko.com,布布扣bubuko.com,布布扣<bubuko.com,布布扣n!bubuko.com,布布扣bubuko.com,布布扣2(2n+1)(n!)bubuko.com,布布扣bubuko.com,布布扣=bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣2(2n+1)bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣0,(n),bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣
因此,由夹逼准则可知
limbubuko.com,布布扣nbubuko.com,布布扣bubuko.com,布布扣n!bubuko.com,布布扣bubuko.com,布布扣2(2n+1)(1!+2!+3!+?+n!)bubuko.com,布布扣bubuko.com,布布扣=0.bubuko.com,布布扣

 

2 (16分) 求函数f(x)=|x|ebubuko.com,布布扣?|x?1|bubuko.com,布布扣bubuko.com,布布扣 的导函数,以及函数f(x)bubuko.com,布布扣 的极值.

解答: 由题意可知

1) 当x<0bubuko.com,布布扣 时,此时f(x)=?xebubuko.com,布布扣x?1bubuko.com,布布扣bubuko.com,布布扣 , 从而fbubuko.com,布布扣bubuko.com,布布扣(x)=?ebubuko.com,布布扣x?1bubuko.com,布布扣?xebubuko.com,布布扣x?1bubuko.com,布布扣=?(x+1)ebubuko.com,布布扣x?1bubuko.com,布布扣;bubuko.com,布布扣

2) 当0<x<1bubuko.com,布布扣 时,此时f(x)=xebubuko.com,布布扣x?1bubuko.com,布布扣bubuko.com,布布扣 , 从而fbubuko.com,布布扣bubuko.com,布布扣(x)=ebubuko.com,布布扣x?1bubuko.com,布布扣+xebubuko.com,布布扣x?1bubuko.com,布布扣=(x+1)ebubuko.com,布布扣x?1bubuko.com,布布扣;bubuko.com,布布扣

3) 当x>1bubuko.com,布布扣 时,此时f(x)=xebubuko.com,布布扣?(x?1)bubuko.com,布布扣bubuko.com,布布扣 , 从而fbubuko.com,布布扣bubuko.com,布布扣(x)=ebubuko.com,布布扣?(x?1)bubuko.com,布布扣?xebubuko.com,布布扣?(x?1)bubuko.com,布布扣=(1?x)ebubuko.com,布布扣1?xbubuko.com,布布扣;bubuko.com,布布扣

4) 当x=1bubuko.com,布布扣 时,此时

fbubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣(1)=limbubuko.com,布布扣x1bubuko.com,布布扣+bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣f(x)?f(1)bubuko.com,布布扣bubuko.com,布布扣x?1bubuko.com,布布扣bubuko.com,布布扣=limbubuko.com,布布扣x1bubuko.com,布布扣+bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣xebubuko.com,布布扣?(x?1)bubuko.com,布布扣?1bubuko.com,布布扣bubuko.com,布布扣x?1bubuko.com,布布扣bubuko.com,布布扣=limbubuko.com,布布扣x1bubuko.com,布布扣+bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣(1?x)ebubuko.com,布布扣1?xbubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣=0,bubuko.com,布布扣
fbubuko.com,布布扣bubuko.com,布布扣?bubuko.com,布布扣(1)=limbubuko.com,布布扣x1bubuko.com,布布扣?bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣f(x)?f(1)bubuko.com,布布扣bubuko.com,布布扣x?1bubuko.com,布布扣bubuko.com,布布扣=limbubuko.com,布布扣x1bubuko.com,布布扣?bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣xebubuko.com,布布扣(x?1)bubuko.com,布布扣?1bubuko.com,布布扣bubuko.com,布布扣x?1bubuko.com,布布扣bubuko.com,布布扣=limbubuko.com,布布扣x1bubuko.com,布布扣?bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣(x+1)ebubuko.com,布布扣x?1bubuko.com,布布扣bubuko.com,布布扣bubuko.com,布布扣1bubuko.com,布布扣bubuko.com,布布扣=2,bubuko.com,布布扣
从而f‘_{-}(1) \ne f‘_{+}(1),fbubuko.com,布布扣bubuko.com,布布扣?bubuko.com,布布扣(1)fbubuko.com,布布扣bubuko.com,布布扣+bubuko.com,布布扣(1),bubuko.com,布布扣 f(x) x = 1 时导数不存在;

5) 当x = 0 时,此时f‘_{+}(0) = \lim\limits_{x \to 0^{+}}\cfrac{f(x) - f(0)}{x - 0} = \lim\limits_{x \to 0^{+}}\cfrac{xe^{(x - 1)} - 0}{x - 0} = \lim\limits_{x \to 0^{+}}e^{x - 1} = e^{-1},

f‘_{-}(0) = \lim\limits_{x \to 0^{-}}\cfrac{f(x) - f(0)}{x - 0} = \lim\limits_{x \to 0^{-}}\cfrac{-xe^{(x - 1)} - 0}{x - 0} = -\lim\limits_{x \to 0^{-}}e^{x - 1} = -e^{-1},
从而f‘_{-}(0) \ne f‘_{+}(0), f(x) x = 0 时导数不存在;

综上可知,所求f(x) 的导函数为f‘(x) = \left\{ \begin{array}{ll} (1 - x)e^{1 - x}, & x > 1 \\ \textrm{不存在}, & x = 1 \\ (x + 1)e^{x - 1}, & 0 < x < 1 \\ \textrm{不存在}, & x = 0 \\ -(x + 1)e^{x - 1}, & x < 0 \end{array} \right. ; 1\,^{\circ} , x > 1, f‘(x) < 0, 0 < x < 1, f‘(x) > 0 \Longrightarrow f(x) x = 1 处取得极大值f(1) = 1; 2\,^{\circ} , 0 < x < 1, f‘(x) > 0, -1 < x < 0, f‘(x) < 0 \Longrightarrow f(x) x = 0 处取得极小值f(0) = 0;  3\,^{\circ} , -1 < x < 0, f‘(x) < 0, x < -1, f‘(x) > 0 \Longrightarrow f(x) x = -1 处取得极大值f(-1) = e^{-2};

 综上可知,f(x) 的极大值为1和e^{-2} , 极小值为0.

 

3 (10分) 设f(x) [0,1] 上有一阶连续导数,且f(0) = f(1) = 0, M = \max\limits_{0 \le x \le 1}|f‘(x)|, 求证:|\int_{0}^{1}f(x)\rd x|\le \cfrac{1}{4}M.

证明:

方法一: \begin{eqnarray*}|\int_{0}^{1}f(x)\rd x| & \le & |\int_{0}^{\cfrac{1}{2}}f(x)\rd x| + |\int_{\cfrac{1}{2}}^{1}f(x)\rd x|\\ & = & |\int_{0}^{\cfrac{1}{2}}[f(x) - f(0)]\rd x| + |\int_{\cfrac{1}{2}}^{1}[f(x) -f(1)]\rd x| \\ & = & |\int_{0}^{\cfrac{1}{2}}f‘(\xi)(x-0)\rd x| + |\int_{\cfrac{1}{2}}^{1}f‘(\eta)(x-1)\rd x| \\ & \le & \int_{0}^{\cfrac{1}{2}}|f‘(\xi)(x-0)|\rd x + \int_{\cfrac{1}{2}}^{1}|f‘(\eta)(x-1)|\rd x \\ & \le & M\int_{0}^{\cfrac{1}{2}}x \rd x + M\int_{\cfrac{1}{2}}^{1}(1 - x)\rd x \\ & = & M\cfrac{x^2}{2}|_{0}^{\cfrac{1}{2}} + M(x - \cfrac{x^2}{2})|_{\cfrac{1}{2}}^{1} = \cfrac{1}{4}M, (\textrm{其中}\xi \in (0,\cfrac{1}{2}), \eta \in (\cfrac{1}{2},1)). \end{eqnarray*}

方法二: \begin{eqnarray*}|\int_{0}^{1}f(x)\rd x| & \stackrel{t = x - \cfrac{1}{2}}{=} & |\int_{-\cfrac{1}{2}}^{\cfrac{1}{2}}f(t + \cfrac{1}{2})\rd t| \stackrel{\textrm{ 分步积分}}{=} |t f(t+\cfrac{1}{2})|_{\cfrac{1}{2}}^{\cfrac{1}{2}} - \int_{-\cfrac{1}{2}}^{\cfrac{1}{2}}t f‘(t + \cfrac{1}{2})\rd t|\\ & = & |\int_{-\cfrac{1}{2}}^{\cfrac{1}{2}}t f‘(t + \cfrac{1}{2})\rd t| \le \int_{-\cfrac{1}{2}}^{\cfrac{1}{2}}|t f‘(t + \cfrac{1}{2})|\rd t\\ & \le & M\int_{-\cfrac{1}{2}}^{\cfrac{1}{2}}|t|\rd t = 2M\int_{0}^{\cfrac{1}{2}}|t|\rd t = 2M\int_{0}^{\cfrac{1}{2}}t \rd t \\ & = & 2M\cdot \cfrac{t^2}{2}|_{0}^{\cfrac{1}{2}} = \cfrac{1}{4}M.\end{eqnarray*}

 

4 (18分) 设函数f(x,y) = \left\{ \begin{array}{ll} x - y + \cfrac{(xy)^2}{(x^2 + y^2)^{3/2}}, & (x,y) \ne (0,0) \\ 0, & (x,y) = (0,0)\end{array} \right. , 证明:

 (1) f(x,y) 在原点处连续;

 (2) f(x,y) 在原点的偏导数f_x(0,0) f_y(0,0) 存在;

 (3) f(x,y) 在原点不可微.

解答:  (1)   \begin{eqnarray*}\mbox{原极限} & \stackrel{x=\rho\cos\theta,y=\rho\sin\theta}{=} & \lim\limits_{\rho \to 0}\left[\rho\cos\theta - \rho\sin\theta + \cfrac{(\rho\cos\theta\rho\sin\theta)^2}{\rho^3}\right] \\ & = & \lim\limits_{\rho \to 0}\rho\left[\cos\theta - \sin\theta + (\cos\theta\sin\theta)^2\right] \\ & = & 0 = f(0,0),\end{eqnarray*}

从而f(x,y) 在原点处连续;

 (2) f_x(0,0) = \lim\limits_{x \to 0}\cfrac{f(x,0) - f(0,0)}{x - 0} = \lim\limits_{x \to 0}\cfrac{x}{x} = 1,

f_y(0,0) = \lim\limits_{y \to 0}\cfrac{f(0,y) - f(0,0)}{y - 0} = \lim\limits_{x \to 0}\cfrac{-y}{y} = -1,
从而f(x,y) 在原点的偏导数f_x(0,0) f_y(0,0) 存在;

 (3)   \begin{eqnarray*}& & \lim\limits_{(\Delta x,\Delta y) \to (0,0)}\cfrac{f(\Delta x,\Delta y) - f(0,0) - f_x(0,0)\Delta x - f_y(0,0)\Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}}\\ & = & \lim\limits_{(\Delta x,\Delta y) \to (0,0)}\cfrac{\Delta x - \Delta y + \cfrac{(\Delta x \Delta y)^2}{((\Delta x)^2 + (\Delta y)^2)^{3/2}} - \Delta x + \Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} \\ & = & \lim\limits_{(\Delta x,\Delta y) \to (0,0)}\cfrac{(\Delta x\Delta y)^2}{((\Delta x)^2 + (\Delta y)^2)^2}\\ & \stackrel{\Delta y=k \Delta x}{=} & \lim\limits_{\Delta x \to 0}\cfrac{(k(\Delta x)^2)^2}{((\Delta x)^2 + k^2(\Delta x)^2)^2} \\ & = & \cfrac{k^2}{(1 + k^2)^2}(\textrm{随着}k\textrm{的值的变化而变化}),\end{eqnarray*}

从而极限\lim\limits_{(\Delta x,\Delta y) \to (0,0)}\cfrac{f(\Delta x,\Delta y) - f(0,0) - f_x(0,0)\Delta x - f_y(0,0)\Delta y}{\sqrt{(\Delta x)^2 + (\Delta y)^2}} 不存在,故f(x,y) 在原点不可微.

 

5 (16分) 求曲面z = xy -1 上与原点最近的点的坐标.

解答: 首先构造拉格朗日函数F(x,y,z,\lambda) = x^2 + y^2 + z^2 + \lambda(xy - z -1) , 于是有\left\{ \begin{array}{l} F_x = 2x + \lambda y = 0\\ F_y = 2y + \lambda x = 0\\ F_z = 2z - \lambda = 0\\ F_{\lambda} = xy - z - 1 =0\end{array} \right. \Longrightarrow \left\{ \begin{array}{l} x = 0\\ y = 0\\ z = -1\\ \lambda = -2,\end{array} \right. 由于(0,0,-1) 是此问题的唯一驻点 (稳定点) ,而此问题一定有最小值,故(0,0,-1) 为所求点.

 

6 (16分)  设\vec{F} = \cfrac{y\vec{i} - x\vec{j}}{x^2 + y^2}, 曲线L 由圆x^2 + y^2 = 1 和椭圆\cfrac{x^2}{4} + y^2 = 1 组成,方向均为逆时针方向,求\int_{L}\vec{F}\rd \vec{s}.

解答: 方法一:记圆x^2 + y^2 = 1 为曲线L_1 , 椭圆\cfrac{x^2}{4} + y^2 = 1 为曲线L_2 , 于是L = L_1 + L_2 ,又设圆x^2 + y^2 = a, (0 < a < 1, a \to 0) 为曲线L_3 , 方向为逆时针方向,于是L = (L_1 - L_3) + (L_2 - L_3) + 2 L_3 ,再记P(x,y) = \cfrac{-x}{x^2 + y^2}, Q(x,y) = \cfrac{y}{x^2 + y^2} ,于是在(L_1 - L_3) + (L_2 - L_3) 上, \cfrac{\partial P}{\partial x} = \cfrac{\partial Q}{\partial y} = \cfrac{x^2 - y^2}{(x^2 + y^2)^2}\textrm{且连续}, 由此可知 \begin{eqnarray*} \int_{L}\vec{F}\rd \vec{s} &=& \int_{(L_1 - L_3) + (L_2 - L_3) + 2 L_3}\vec{F}\rd \vec{s}\\ &=& \int_{L_1 - L_3}\vec{F}\rd \vec{s} + \int_{L_2 - L_3}\vec{F}\rd \vec{s} + \int_{2 L_3}\vec{F}\rd \vec{s}\\ &\equiv& I_1 + I_2 + I_3. \end{eqnarray*}

由格林公式立即可得I_1 = I_2 = \int\!\!\!\int\left[\cfrac{\partial P}{\partial x} - \cfrac{\partial Q}{\partial y}\right]\rd x\rd y = 0, I_3 = 2\int_{L_3}\vec{F}\rd \vec{s} \stackrel{x=a\cos\theta,y=a\sin\theta}{=} 2\int_{0}^{2\pi}\cfrac{a^2\cos^2\theta + a^2\sin^2\theta}{a^2}\rd \tt = 2\int_{0}^{2\pi}\rd \tt = 4\pi.
从而\int_{L}\vec{F}\rd \vec{s}= 4\pi.

方法二:记圆x^2 + y^2 = 1 为曲线L_1 , 椭圆\cfrac{x^2}{4} + y^2 = 1 为曲线L_2 , 于是L = L_1 + L_2 ,再记P(x,y) = \cfrac{-x}{x^2 + y^2}, Q(x,y) = \cfrac{y}{x^2 + y^2} ,于是在L_2 - L_1 上, \cfrac{\partial P}{\partial x} = \cfrac{\partial Q}{\partial y} = \cfrac{x^2 - y^2}{(x^2 + y^2)^2}\textrm{且连续}, 由此可知 \int_{L}\vec{F}\rd \vec{s} = \int_{(L_2 - L_1) + 2 L_1}\vec{F}\rd \vec{s} = \int_{L_2 - L_1}\vec{F}\rd \vec{s} + \int_{2 L_1}\vec{F}\rd \vec{s} = I_1 + I_2,

由格林公式立即可得I_1 = \int\!\!\!\int\left[\cfrac{\partial P}{\partial x} - \cfrac{\partial Q}{\partial y}\right]\rd x\rd y = 0, I_2 = 2\int_{L_1}\vec{F}\rd \vec{s} \stackrel{x=\cos\theta,y=\sin\theta}{=} 2\int_{0}^{2\pi}\cfrac{\cos^2\theta + \sin^2\theta}{1}\rd \tt = 2\int_{0}^{2\pi}\rd \tt = 4\pi.
从而\int_{L}\vec{F}\rd \vec{s}= 4\pi.

 

7 (16分) 求函数项级数\sum\limits_{n=1}^{\infty}\cfrac{x^2}{(1 + x^2)^n} 的和函数,并讨论在x \in (-\infty,+\infty) 上的一致收敛性.

解答: 记f_n(x) = \cfrac{x^2}{(1 + x^2)^n} , 函数项级数\sum\limits_{n=1}^{\infty}\cfrac{x^2}{(1 + x^2)^n} 的前n 项部分和函数为S_n(x) , 和函数为S(x) , 于是有

(1) 当x = 0 时,此时S_n(x) = 0 , 从而S(x) = \lim\limits_{n \to \infty}S_n(x) = \lim\limits_{n \to \infty}0 = 0;

 (2) 当x \ne 0 时,此时S_n(x) = \cfrac{\cfrac{x^2}{1 + x^2}\left[1 - (\cfrac{1}{1 + x^2})^n\right]}{1 - \cfrac{1}{1 + x^2}},

从而S(x) = \lim\limits_{n \to \infty}S_n(x) = \lim\limits_{n \to \infty}\left[1 - (\cfrac{1}{1 + x^2})^n\right] = 1;

综上可知, S(x) = \left\{ \begin{array}{ll} 1, & x \ne 0 \\ 0, & x = 0 \end{array} \right. ;

又由于S(x) 不连续,而f_n(x)(n = 1,2,\cdots) 每一项都连续,故\sum\limits_{n=1}^{\infty}\cfrac{x^2}{(1 + x^2)^n} 不一致连续.

 

8 (10分) 研究级数\sqrt{2} + \sqrt{2 - \sqrt{2}} + \sqrt{2 - \sqrt{2 + \sqrt{2}}} + \sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2}}}} + \cdots 的敛散性.

解答: 方法一:设a_1 = \sqrt{2}, a_2 = \sqrt{2 - \sqrt{2}}, a_3 = \sqrt{2 - \sqrt{2 + \sqrt{2}}}, \cdots, 从而可得

a_1 = \sqrt{2} = 2\sin\cfrac{\pi}{4} = 2\cos\cfrac{\pi}{4}, a_2 = \sqrt{2 - \sqrt{2}} = \sqrt{2 - 2\cos\cfrac{\pi}{4}} = 2\sin\cfrac{\pi}{8},

a_3 = \sqrt{2 - \sqrt{2 + \sqrt{2}}} = \sqrt{2 - \sqrt{2 + 2\cos\cfrac{\pi}{4}}}= \sqrt{ 2 - 2\cos\cfrac{\pi}{8}} = 2\sin\cfrac{\pi}{16},
a_4= \sqrt{2 - \sqrt{2 + \sqrt{2 + \sqrt{2}}}} = \sqrt{2 - \sqrt{2 + \sqrt{2 + 2\cos\cfrac{\pi}{4}}}} = 2\sin\cfrac{\pi}{32},
\cdots, a_n = 2\sin\cfrac{\pi}{2^{n + 1}} ,于是猜想a_n = 2\sin\cfrac{\pi}{2^{n + 1}}(n=1,2,\cdots) , 下面用数学归纳法来证明

(1) 当n = 1 时,此时a_1=2\sin\cfrac{\pi}{4} 显然成立;

(2) 假设n = k 时,a_k 成立,即a_k = 2\sin\cfrac{\pi}{2^{k + 1}} , 下面证明当n = k + 1 时, a_{k+1} = \sqrt{2 - \sqrt{2 + 2 - a_k^2}} = \sqrt{2 - 2\cos\cfrac{\pi}{k+1}} = 2\sin\cfrac{\pi}{2^{k+2}}, 可知当n = k + 1 时也成立.

于是可得a_n = 2\sin\cfrac{\pi}{2^{n + 1}}(n=1,2,\cdots) , 而显然可得a_n \le 2\cfrac{\pi}{2^{n + 1}} = \cfrac{\pi}{2^{n}} , 而级数\sum\limits_{n=1}^{\infty}\cfrac{\pi}{2^{n}} 收敛,由正项级数的比较原则可知,所求原级数收敛.

 

方法二:设a_n = \sqrt{2 + \sqrt{2 + \cdots + \sqrt{2}}}, (n 个根号) ,满足a_{n + 1} = \sqrt{2 + a_n},

现在用数学归纳法来证明数列\{a_n\} 是有界的.

显然,a_1 = \sqrt{2} \in (0,2);

假设n = k 时,0 < a_k < 2,

则当n = k + 1 时,0 < a_{k + 1} = \sqrt{2 + a_k} < \sqrt{2 + 2} = 2, 所以0 < a_n < 2 (n = 1,2,\cdots), 数列\{a_n\} 有界的. 由于\cfrac{a_{n + 1}}{a_n} = \cfrac{\sqrt{2 + a_n}}{a_n} = \sqrt{\cfrac{2}{a_n^2} + \cfrac{1}{a_n}} >1,

因此数列\{a_n\} 单调递增.

由单调有界原理,数列\{a_n\} 有极限,记为a .由于a_{n + 1} = \sqrt{2 + a_n},

运用数列极限的四则运算法则,当n \to \infty 时有

a = \sqrt{2 + a} , \Longrightarrow a = 2 , 即\lim\limits_{n \to \infty}a_n = 2. 从而  \begin{eqnarray*}\lim\limits_{n \to \infty}\cfrac{\sqrt{2 - a_{n+1}}}{\sqrt{2 - a_n}} & = & \lim\limits_{n \to \infty}\sqrt{\cfrac{2 - a_{n+1}}{2 - a_n}} = \lim\limits_{n \to \infty}\sqrt{\cfrac{2 -\sqrt{2 + a_n}}{2 - a_n}} \\ & = & \lim\limits_{n \to \infty}\sqrt{\cfrac{(2 -\sqrt{2 + a_n})(2 +\sqrt{2 + a_n})}{(2 - a_n)(2 +\sqrt{2 + a_n})}} \\ & = & \lim\limits_{n \to \infty}\sqrt{\cfrac{2 - a_n}{(2 - a_n)(2 +\sqrt{2 + a_n})}}\\ & = & \lim\limits_{n \to \infty}\sqrt{\cfrac{1}{2 +\sqrt{2 + a_n}}} = \cfrac{1}{2} < 1.\end{eqnarray*}

因而由正项级数的d‘Alembert判别法或比式判别法可知,所求原级数收敛. 

[家里蹲大学数学杂志]第034期中山大学2008年数学分析考研试题参考解答,布布扣,bubuko.com

[家里蹲大学数学杂志]第034期中山大学2008年数学分析考研试题参考解答

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原文地址:http://www.cnblogs.com/zhangzujin/p/3738112.html

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