Question

第四题！求答案！

Answer 1

(uv)'=u'v+uv'
(uv)''=((uv)')'=(u'v+uv')'= u''v+u'v'+u'v'+uv''=u''v+2u'v'+uv''
(uv)'''=(u''v+2u'v'+uv'')'=(u'''v+u''v'')+(2u''v'+2u'v'')+(uv'''+u'v'')=u'''v+3u''v'+3u'v''+uv'''
(uv)(n) = C(0,n)u(0)v(n)+C(1,n)u(1)v(n-1)+C(2,n)u(2)v(n-2)+.+C(n,n)u(n)v(0)
因为电脑不太好打n阶导的符号，和C排列的符号，所以用C(0,n),C(1,n),C(2,n)表示排列组合,u(n),v(n)表示n阶导数
因此你的题目中
u=x^2 ,v=sin2x
u'=2x,u''=2,u'''=0,因此,u的三阶导数以上都是零了,上面的展开式只需要求前面含有的u的零阶、一阶和二阶导数的项C(0,100)u(0)v(100)、C(1,100)u(1)v(99)、C(2,100)u(2)v(98)就可以了
u(0)=x^2 u(1)=2x u(2)=2
sin(kx)(n)=k^nsin(kx+0.5nπ)
v(100)=2^100sin(2x+50π)=2^100sin2x
v(99)=2^99sin(2x+49.5π)=-2^99sin(2x+0.5π)= 2^99cos2x
v(48)=2^98sin(2x+49π)=-2^98sin2x
C(0,100)=1,C(1,100)=100,C(2,100)=100*99/2=4950
y(100)=C(0,100)u(0)v(100) + C(1,100)u(1)v(99) + C(2,100)u(2)v(98)
=x^2·(2^100sin2x) + 200x·(2^99cos2x) - 4950×2^99sin2x
把x=0带进去，y(100)(0)=0