04-05-3高等数学电类期末试卷参考答案 05.6.23
一. 填空题(本题共5小题，每小题4分，满分2 0分) 1.
x?14?y?22?z3;2.??2,2?;3.?dx?012?xxf?x,y?dy;4.2?;5.??1,??3.
二. 单项选择题(本题共4小题，每小题4分，满分1 6分) 1.C;2.B;3.D;4.B. 三. (本题共5小题，每小题7分，满分3 5分)
1. 令F?x,y,z??x?2z?f?z?xFxFz?2x3f??22?y2?3z?,则Fx?2x,Fy??2yf?,Fz??2?3f?,
???,
?z?y??FyFz?2yf?3f??2??1,2y?z?x?3x?z?y??2?2xy
2. 由条件得
?P?y??Q?x , 即 4?xy?6???1?x,??3,
????x0,0?3,1?2?4xy3?dx??6x2?132223?y?2y?dy??x?y?2xy??3??3,1??0,0??26
3.f?ln?x?2??x?1??ln?x?2??ln?x?1?
x?2??而 ln?x?2??ln4?ln?1???ln4?4?????n?1??1?nn?1?x?2???,?2?x?6 4??nln?x?1?????1?n?1?n?1?x?2?nn?1n,1?x?3
f(x)?ln4??n?1??1?n1?n?1?x?2 1?x?3 ???n?4??4.(1) 将f(x)作奇延拓,再作周期延拓. an?0(n?0,1,2,?),
bn??20f(x)sin?n?x2dx??10xsinn?x2dx??2n?cosn?2?4n?22sinn?2,
f(x)?2n?4??cos???n?22n?n?1?2sinn??n?xsin,x??0,1???1,2?, ?2?2(2)S(1)?12, (1分) S?11?7??1??S??? ???222????5. (1)f(z)?(z?1)(z?1)?1z?1?1z?1??2?1??2??? ??(?1)n?0n(z?1)2n?1n?1
f(z)?(2)
?1?11?1?11???????2?z?1z?1?2?(z?2)?3(z?2)?1??n??13????n?2(z?2)n?0?z?2???n?01?z?2?zn?1???2?
1(z?2)n?1??n?0?3?1?n四．(本题满分7分)f(z)?(z?1)(z?1)22 在z?2内有奇点:z?1(二级极点),
??1z??i(一级极点) (1分)原式??2?iRes?f(z),???2?iRes?f???z??1?,0?2? ?z?z?1?1f???2?,z?0是可去奇点,留数为0,故原积分=0. 22zz(1?z)(1?z)???五．(本题满分8分) 收敛域为??1,1?,设S(x)??n?1(?1)n?11n(2n?1)x2n,??1,1?
则
12?S(x)??n?1(?1)n?112n(2n?1)x2n,12?S??(x)??n?1(?1)n?1x2n?2?11?x2
S?(x)?2arctanx S(x)?2xarctanx?ln(1?x),x???1,1?
21六．(本题满分8分) en?1?1n?1?1???????n??? 2n?n?2n1n??e??1n1?因此绝对值级数发散. , 又?(?1)n?1n为Leibniz型级数,故收敛.
1而en??1单调减少,且lim?en????1n??n?1??0, 所以?(?1)?n?1???1?收敛. ??由收敛级数性质知原级数收敛. 故原级数条件收敛.
??k七．(本题满分6分)f(x)??ak?1xk?1,x??0,1?,由于级数?ak收敛,所以limak?0,因而
k?1k???ak?有界.设 ak?1??M(k?1,2,?), 则 f????n????k?1ak1nk?1??M?k?11nk?1.
?1?当n?2时,f???M?n??k?11nk?1?Mn(n?1)?, 又级数?n?21n(n?1)收敛,由比较判别法知原
级数绝对收敛,故原级数收敛.