Expand 1/√((1+6 x)) in ascending powers of x, up to and including the term in x^3, simplifying the coefficients.

Expand $\frac{1}{\sqrt[3]{(1+6 x)}}$ in ascending powers of $x$, up to and including the term in $x^3$, simplifying the coefficients. [4]

高等数学 · 已解决 · 大学数学级数 · 级数展开
提问于2023年05月04日 · 阅读 52

解答

$$ \begin{aligned} & 设 f(x)=\frac{1}{\sqrt[3]{1+6 x}}, 则 f^{\prime}(x)=-\frac{2}{(1+6 x)^{\frac{8}{3}}} \\ & f^{\prime \prime}(x)=\frac{16}{(1+6 x)^{7 / 3}}, f^{\prime \prime \prime}(x)=-\frac{224}{(1+6 x)^{10 / 3}} \\ & \text { 故 } f(0)=1, \quad f^{\prime}(0)=-2 \\ & f^{\prime \prime}(0)=16, \quad f^{\prime \prime \prime}(0)=-224 \text {. } \\ & \text { 从而 } f(x)=1-\frac{2}{1 !} x+\frac{16}{2 !} x^2-\frac{224}{3 !} x^3+o\left(x^3\right) \\ & =1-2 x+8 x^2-\frac{112}{3} x^3+o\left(x^3\right) \\ & \end{aligned} $$

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