高等數(shù)學(xué)習(xí)題全解第11章無(wú)窮級(jí)數(shù)

第五 函數(shù)展開(kāi)成冪級(jí)習(xí) 11-

ax(a0) (a0)asin(xπ); exe 4 shx 42 x23x

sin2x xsinxdx (xln

lnna解 a

n!x

xn1 xa a1 n0aπ

22 22sin(x )4

2

(1) 2 (2n

x n11ln(ax)ln[a(1)]lna

x(a,a]n nexe 1 n n shx 2[n! n!]2[1(1)]n!(2n1)! 1

4 x24 1{1(1)(1x2)1(1)(11)(1x2 ·1(1)(11)·(1n1)(1x2)n1

(2n1)!! n1x n1

11 x23x x x 1 212

1

xn(1

1

(

n n122n1sinx2(1cos2x)[1 (2n)!] 2

nn

x(,)

(1)(n1)x(1

xsin x1 x xdx0x(1)(2n dx0 (2n0 x 0

=

xdx

x(,)dex

x

dex

(n ) (n0 ) ( ) x0. dxn1n! n1dex (n

將下列函數(shù)在指定點(diǎn)處展開(kāi)成(xx0的冪級(jí)數(shù)lnx,x 1,x3 cosxxπ ,x cosx

x24x 201,x0 ln(x1x2),20

0x n1 n1(x解 令tx1,則lnx

nn

n 1

n(xn 令tx3,則xt t 令txπ,3

31 3

cosxcos(tπ)1cost3sin (xπ

π1(1)n[

(x

] 22令tx1,

(2n 1[11x24x (x3)(x (t4)(t 2t t 1 1 1 ]4

t

t]4n22

221 1

n0 8n0

)tn

1)(x

令tx1,1 (

)'((1)ntn)'

(t t

( (1)n1n(x1)n1(1)n(n1)(x1)n x(0,2) 1注意求函數(shù)的級(jí)數(shù)時(shí),往往通過(guò)變量代換轉(zhuǎn)為求函數(shù)的林級(jí)數(shù)1f(xln(x

),f'(x)(1x2)1(1)x21(1)(11)(x2

·1(1)(11)·(1n1)(x2)n 1(1)n(2n1)!!2n,

xf(x) f'(x)dxx(1)n(2n x2n1 xx

(2n)!!2nf(xanxn(RxR),試證f(x為奇函數(shù)時(shí),必有a2k0(k0,12,·f(x為偶函數(shù)時(shí),必有a2k10(k0,1,2,· 證 由題得RxR時(shí),f(x)an(1)nxn,故f(x)an(1)n1xn 所以由函數(shù)冪級(jí)數(shù)展式的唯一性知an(1)n1an(n0,1,2,·),因此當(dāng)n2k(k0,1,2,·時(shí),a2k0(k0,12,· 由題知當(dāng)RxR時(shí) f(x)an(1)nxn,從而f(x)an(1)nxn 故由函數(shù)的冪級(jí)數(shù)展式的唯一性知an(1)nan(n0,1, ,因此n2k1(k0,1,2,·時(shí),a2k10(k0,1,2,·利用冪級(jí)數(shù)展開(kāi)式的唯一性,f(xex2x0n階導(dǎo)數(shù) 2 解x(時(shí),f(xex(x)