Cirrus83
58 lines
1.4 KiB
Markdown
58 lines
1.4 KiB
Markdown
+++
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date = '2025-12-17T17:29:22+08:00'
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draft = false
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title = '4 – 数院人的一天'
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tags = ['数学分析']
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categories = 'math'
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description = '我是数院的，数院的数学应该不差才对。'
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今天习题来自数学分析I 习题7.3 Taylor展开式。
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## 题目描述
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设函数$y=f(x)$在$x=0$处有五阶导数,其满足$x=y-3y^3+5y^5+o(y^5)(y\to0),f(0)=0$.求$f(x)$带Peano型余项的五阶Maclaurin展开式.
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## 解答（待定系数法）
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由 \( x = f^{-1}(y) = y - 3y^3 + 5y^5 + o(y^5) \)得：
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\[
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x = f(x) - 3f(x)^3 + 5f(x)^5 + o(f(x)^5)
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\]
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在 \( x \to 0 \) 时 \( f(x) \to 0 \)，则$o(f(x)^5)=o(x^5)$。
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待定系数法设：
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\[
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f(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5 + o(x^5), \\ a_0 = 0, a_1 = f'(0) = \frac{1}{f'^{-1}(0)}=1
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\]
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所以：
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$$
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\begin{align}
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f(x) &= x + a_2 x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5 + o(x^5)\\
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f^3(x)&=x^3+3a_2x^4+3(a_2+a_3)x^5+o(x^5)\\
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f^5(x)&=x^5+o(x^5)
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\end{align}
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$$
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代入方程$x = f(x) - 3f^3(x) + 5f^5(x) + o(x^5)$：
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$$
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\begin{align}
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x&=x + a_2 x^2 + a_3 x^3 + a_4 x^4 + a_5 x^5-3(x^3+3a_2x^4+3(a_2+a_3)x^5+)+5x^5+o(x^5)\\
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0&=a_2 x^2+(a_3-3)x^3+(a_4-9a_2)x^4+(a_5-9(a_2+a_3)+5)x^5+o(x^5)
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\end{align}
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$$
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也就是：
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$$
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\begin{cases}
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a_2=0\\
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a_3-3=0\\
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a_4-9a_2=0\\
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a_5-9(a_2+a_3)+5=0
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\end{cases}
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$$
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因此：
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\[
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a_2 = 0, \quad a_3 = 3, \quad a_4 = 0, \quad a_5 = 22
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\]
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所以：
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\[
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f(x) = x + 3x^3 + 22x^5 + o(x^5)
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\]
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## 注
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反函数求导只有一阶导有公式。 |