幂级数求和中,逐项求导或逐项积分到底该如何操作?
1个回答
①∑（n从1到正无穷）（-1)^(n-1) * (2x)^(2n-1) / (2n-1)
=∑（n从1到正无穷）（-1)^(n-1) ∫(2x)^(2n-2) dx（积分区间为0到x,下同）
=∑（n从1到正无穷）（-1)^(n-1)∫(4x²)^(n-1)dx
=∑（n从1到正无穷）∫（-4x²)^(n-1)dx
=∫[∑（n从1到正无穷）（-4x²)^(n-1)]dx
=∫[1/(1+4x²)]dx
=arctan2x
②∑（n从1到正无穷）n *（n＋2）X＾2n
=1/2∑（n从1到正无穷）2n(n+2)x^2n
=(1/2)x∑（n从1到正无穷）(n+2)2nx^(2n-1)
=(1/2)x∑（n从1到正无穷）(n+2)[x^(2n)]′
=(1/2)x[∑（n从1到正无穷）(n+2)x^(2n)]′
∑（n从1到正无穷）(n+2)x^(2n)
=1/(2x³)∑（n从1到正无穷）(2n+4)x^(2n+3)
=1/(2x³)∑（n从1到正无穷）[x^(2n+4)]′
=1/(2x³)[∑（n从1到正无穷）x^(2n+4)]′
=1/(2x³)[x^6/(1-x²)]′
=x²(3-2x²)/(1-x²)²
原式=(1/2)x[∑（n从1到正无穷）(n+2)x^(2n)]′
=(1/2)x[x²(3-2x²)/(1-x²)²]′
=x²(3-x²) /(1-x)³
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