高数 49题求教
2个回答
展开全部
∫xsinx dx
=-∫xdcosx
=-xcosx +∫cosx dx
=-xcosx +sinx + C
an
= ∫(0->nπ) x|sinx| dx
= ∫(0->π) xsinx dx -∫(π->2π) xsinx dx+...+ (-1)^(n-1). ∫((n-1)π->nπ) xsinx dx
=[-xcosx +sinx]||(0->π)- [-xcosx +sinx]||(π->2π)
+... +(-1)^(n-1).[-xcosx +sinx]||((n-1)π->nπ)
= π -(-2π-π) +...+(-1)^(n-1).[ (-1)^(n+1). nπ - (-1)^(n+1) . (n-1)π ]
=(0+1)π + (2+1)π +...+ (n+(n-1) )π
= [n(n+1)/2 + (n-1)n/2]π
=n^2. π
consider
1/(1-x)=∑(n: 0->∞) x^n
1/(1-x)^2 = ∑(n: 1->∞) nx^(n-1) ( 两边取导数)
x/(1-x)^2 = ∑(n: 1->∞) nx^n (两边乘以x )
(1+x)/(1-x)^3 = ∑(n: 1->∞) n^2.x^(n-1) ( 两边取导数)
x(1+x)/(1-x)^3 = ∑(n: 1->∞) n^2.x^n (两边乘以x )
x =1/2
∑(n: 1->∞) n^2./2^n
=(1/2)(3/2)/(1/2)^3
=24
an = ∑(n: 1->∞) an/2^n
= π∑(n: 1->∞) n^2./2^n
= 24π
=-∫xdcosx
=-xcosx +∫cosx dx
=-xcosx +sinx + C
an
= ∫(0->nπ) x|sinx| dx
= ∫(0->π) xsinx dx -∫(π->2π) xsinx dx+...+ (-1)^(n-1). ∫((n-1)π->nπ) xsinx dx
=[-xcosx +sinx]||(0->π)- [-xcosx +sinx]||(π->2π)
+... +(-1)^(n-1).[-xcosx +sinx]||((n-1)π->nπ)
= π -(-2π-π) +...+(-1)^(n-1).[ (-1)^(n+1). nπ - (-1)^(n+1) . (n-1)π ]
=(0+1)π + (2+1)π +...+ (n+(n-1) )π
= [n(n+1)/2 + (n-1)n/2]π
=n^2. π
consider
1/(1-x)=∑(n: 0->∞) x^n
1/(1-x)^2 = ∑(n: 1->∞) nx^(n-1) ( 两边取导数)
x/(1-x)^2 = ∑(n: 1->∞) nx^n (两边乘以x )
(1+x)/(1-x)^3 = ∑(n: 1->∞) n^2.x^(n-1) ( 两边取导数)
x(1+x)/(1-x)^3 = ∑(n: 1->∞) n^2.x^n (两边乘以x )
x =1/2
∑(n: 1->∞) n^2./2^n
=(1/2)(3/2)/(1/2)^3
=24
an = ∑(n: 1->∞) an/2^n
= π∑(n: 1->∞) n^2./2^n
= 24π
2016-05-14
展开全部
这么快啊,1000题都做到第6章了
已赞过
已踩过<
评论
收起
你对这个回答的评价是?
推荐律师服务:
若未解决您的问题,请您详细描述您的问题,通过百度律临进行免费专业咨询