2019最新高等数学期末考试试题（含答案）

一、解答题

1．设f(a)?f(c)?f(b)，且a?c?b,f??(x)在[a，b]内存在，证明：在（a，b）内至

少有一点?，使f??(?)?0.

证明：f??(x)在[a，b]内存在，故f(x)在[a，b]上连续，在（a，b）内可导，且

f(a)?f(c)?f(b)，故由罗尔定理知，??1?(a,c)，使得f?(?1)?0，??2?(c,b)，

使得f?(?2)?0，又f?(x)在[?1,?2]上连续，在(?1,?2)内可导，由罗尔定理知，

???(?1,?2)，使f??(?)?0，即在（a，b）内至少有一点?，使f??(?)?0.

2．设f(x)是周期为2的周期函数，它在[-1,1]上的表达式为f(x)=e，试将f(x)展成傅里叶级数的复数形式.

解：函数f(x)在x≠2k+1，k=0,±1,±2处连续.

nπ?ix1l11cn??f?x?eldx??e?xe?inπxdx2l?l2?11???e??1?nπi?x?1?12?1?nπi?-xe?e1???1?n?21?nπi1?nπi?sinh1???1?n?1??nπ?2?故f(x)的傅里叶级数的复数形式为

??1

??1?n??1?inπ?inπxf?x??sinh1?e (x≠2k+1，k=0,±1,±2,…) 21??nπ?n???

3．将函数f (x) = x-1(0≤x≤2)展开成周期为4的余弦级数.

解：将f(x)作偶延拓，作周期延拓后函数在(-∞,+∞)上连续,则有bn=0 (n=1,2,3,…)

212fxdx????0?x?1?dx?0 2??2a0?212nπxnπxfxcosdx?x?1cosdx?????02??2224n?22[??1??1] nπn?2,4,6,?0,???8?22,n?1,3,5,??nπan?8故f?x???2π?n?1?1?2n?1?2?cos?2n?1?πx2 (0≤x≤2)

4．设f (x) = x+1(0≤x≤π)，试分别将f (x)展开为正弦级数和余弦级数. 解：将f (x)作奇延拓，则有an=0 (n=0,1,2,…)

2π2πbn??f?x?sinnxdx???x?1?sinnxdxπ0π0?21???1??1?π??πnnn

2?1???1??1?π?从而f?x???sinnx (0

πn?1n若将f(x)作偶延拓，则有bn=0 (n=1,2,…)

an?2π2πfxcosnxdx????x?1?cosnxdx ??00ππn?2,4,6?0,????4,n?1,3,5,??n2π1π2πa0??f?x?dx???x?1?dx?π?2

π?ππ0π?24?cos?2n?1?x??从而f?x?? (0≤x≤π) 22πn?1?2n?1?

5．将下列函数f(x)展开为傅里叶级数： (1)f?x??πx?42??π?x?π?

(2)f?x??sinx解：(1) a0??0?x?2π?

1π1π?πx?π fxcosnxdx??dx???????-π?πππ?42?21π?πx?1π1πan?????cosnxdx??cosnxdx?xcosnxdxπ?π?42?4-π2π?-π

1π??sinnx??π?0?0?n?1,2,?4nbn?1π?πx?1π1π?sinnxdx?sinnxdx?xsinnxdx??????π-π-ππ?42?42π

n1???1??nπ?nsinnx故f?x??????1? (-π

4n?1n(2)所给函数拓广为周期函数时处处连续， 因此其傅里叶级数在[0,2π]上收敛于f(x)，注意到f(x)为偶函数，有bn=0,a0?1π1πfxcos0xdx?sinxdx ?????π?πππ2π4??sinxdx?π0π2π2πan??f?x?cosnxdx??sinxcosnxdxπ0π?π1π???sin?n?1?x?sin?n?1?x??dxπ0??2?n??1??1??2??π?n?1?n?1,3,5,?0,????4,n?2,4,6,?π?n2?1??所以

2??4cos2nx (0≤x≤2π) f?x????2πn?1π?4n?1?

6．将函数F?x??x?0arctantdt展开成x的幂级数． tnt2n?1解：由于arctant????1?

2n?1n?0?所以F?x???x02nx?arctantntdt?????1?dt0t2n?1n?0xn（|x|≤1）

?t2nx2n?1n?????1?dt????1?02n?1?2n?1?2n?0n?0?

7．证明，若

?Un收敛，则?2n?1?Un绝对收敛． n?1n?证：∵

?Un1??Un?nn2?Un2?1n2?1U2?1?1

n222n2而由

?Un收敛，?n?11收敛，知 2nn?1