解答如下 问题:
(1)设{an },{bn}为两个数列,且Bn=b1+b2+⋯+bn,证明:
an bn=an Bn+
Bk (ak-ak-1)
(2)记1/n
kxk=A,证明:
1/n^2
k2 xk=A/2.
(1)由已知,b1=B1,bk=Bk-Bk-1
an bn =a1 b1+a2 b2+⋯+an bn
=a1 B1+a2 (B2-B1 )+⋯+an (Bn-Bn-1)
=a1 B1+a2 B2-a2 B1+⋯+an Bn-an Bn-1
=B1 (a1-a2 )+B2 (a2-a3 )+⋯+Bn-1 (an-1-an )+an Bn
=an Bn+Bk (ak-a_(k+1))
(2)使用Stolz公式:
(
kxk )/n=
(
kxk -
kxk )/(n-(n-1))=
nxn=A
(
k2 xk )/n2 =
k2 xk -
k2 xk )/(n2-(n-1)2 )=
(n2 xn)/(2n-1)=
(nxn)/(2-1/n)=A/2
已知函数f(x)为(A,B)上的连续函数,且有[a,b]⊂(a,b),证明:
1/h
[f(x+h)-f(x)]dx=f(b)-f(a)
1/h
[f(x+h)-f(x)]dx
=1/h [
f(x+h)dx-
f(x) dx]
=1/h [
f(x)dx-
f(x) dx]
=1/h [
f(x+h)dx-
f(x) dx]
=f(x+h)dx/h-lim┬(h→0)
f(x) dx/h
=f(b+h)-
f(a+h)
=f(b)-f(a)
已知an=a≠0,试用ε-N语言证明:
1/an =1/a.
由 an=a得
|an |=|a|.
根据极限的定义,对∀ε>0,∃N1>0,当n>N1时,有|an-a|<a2/2 ε.
由保号性知,对上述ε>0,∃N2>0,当n>N2时,有|an |>|a|/2.
令N=max{N1,N2},则当n>N时
|1/an -1/a|=|(a-an)/(an a)|<a2/2 ε∙2/a2 =ε
故,由ε-N定义知, 1/an =1/a.
解答如下问题:
(1)证明:(-1)n n(n+1)/(n(n+1) x2+2n)关于x∈(-∞,+∞)一致收敛.
(2)计算
(-1)n n(n+1)/(n(n+1) x2+2n ).
(1)∵ |(-1)n n(n+1)/(n(n+1) x2+2n )|≤n(n+1)/2n ,
而
=1/2<1
故n(n+1)/2n 收敛.
魏尔斯特拉斯判别法(M-判别法)知(-1)n n(n+1)/(n(n+1) x2+2n )关于x∈(-∞,+∞)一致收敛.
(2) (-1)n n(n+1)/(n(n+1) x2+2n )=
(-1/2)n n(n+1)
令S(x)=n(n+1) xn,x∈(-1,1)
S(x)=xn(n+1) xn-1=x
(xn+1)''=x(
xn+1 )''=x∙(x2/(1-x))''=2x/(1-x)3
故
S(-1/2)=2∙(-1/2)/(1+1/2)3 =-8/27
求(x2+y2+z2 )2=4(x2+y2-z2)所围立体的体积.
令x=rcosθsinφ,y=rsinθsinφ,z=rcos,则r=2,φ∈[π/4,π/2]
由于曲面关于三个坐标平面对称,所以
V=8dθ
dφ
r2 sinφ dr=32π/3
(-cos2φ)3/2 sinφ dφ
=32π/3(1-2 cos2φ )3/2 d(-cosφ)
令t=cosφ,上式化为:
V=32π/3 (1-2t2 )3/2 dt
令p2=1-2t2,上式化为:
V=(16√2 π)/3 p4/√(1-p2 ) dp
再令u=p2,上式化为:
V=(8√2 π)/3 u3/2 (1-u)-1/2 du=(8√2 π)/3 B(5/2,1/2)
=(8√2 π)/3∙Γ(5/2)Γ(1/2)/Γ(3) =(8√2 π)/3∙(3/4 Γ(1/2)2)/2!=√2 π2