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1/((1-x)^2)展开为x的幂级数,
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1/((1-x)^2)展开为x的幂级数,
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答案和解析
f(x) = 1/((1-x)^2),在x=0进行泰勒展开?
f(0) = 1,
f'(x) = 2/((1-x)^3),so f'(0) = 2,
f''(x) = 2*3/((1-x)^4),so f''(0) = 6,
f(x)的n阶导数= (n+1)!/((1-x)^(n+2)),so f(x)的n阶导数在0点取值 = (n+1)!,
f(x) = Sigma[(n+1)!*(x^n)],
f(0) = 1,
f'(x) = 2/((1-x)^3),so f'(0) = 2,
f''(x) = 2*3/((1-x)^4),so f''(0) = 6,
f(x)的n阶导数= (n+1)!/((1-x)^(n+2)),so f(x)的n阶导数在0点取值 = (n+1)!,
f(x) = Sigma[(n+1)!*(x^n)],
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