題目

Problem

2. (a) Let $f(x, y) = (x + 2)^{y+2}$ . Then $\left. \frac{\mathrm{d}^2}{\mathrm{d}t^2} f(t, t^2) \right|_{t=0} = \underline{\quad (4) \quad} \,.$

(b) Let $g(x) = \frac{1}{\sqrt{x^2+2x+5}}$ . Then $g^{(6)}(-1) = \underline{\quad (5) \quad} \,.$

解答

展開

展開

設 $F(t) = f(t, t^2) = (t+2)^{t^2+2}$ 。我們寫成指數形式：

F(t) = e^{(t^2+2)\ln(t+2)}

令 $h(t) = (t^2+2)\ln(t+2)$ ，則 $F(t) = e^{h(t)}$ 。對其求一階與二階導數：

h'(t) = 2t\ln(t+2) + \frac{t^2+2}{t+2}

h''(t) = 2\ln(t+2) + \frac{2t}{t+2} + \frac{2t(t+2) - (t^2+2)}{(t+2)^2} = 2\ln(t+2) + \frac{2t}{t+2} + \frac{t^2+4t-2}{(t+2)^2}

在 $t=0$ 處代入：

h(0) = 2\ln 2, \qquad h'(0) = 0 + \frac{2}{2} = 1, \qquad h''(0) = 2\ln 2 + 0 + \frac{-2}{4} = 2\ln 2 - \frac{1}{2}

由複合函數求導法則：

F'(t) = e^{h(t)}h'(t) \implies F''(t) = e^{h(t)}(h'(t))^2 + e^{h(t)}h''(t) = F(t) \left[ (h'(t))^2 + h''(t) \right]

將 $t=0$ 代入，注意 $F(0) = e^{2\ln 2} = 4$ ：

F''(0) = F(0) \left[ (h'(0))^2 + h''(0) \right] = 4 \left[ 1^2 + 2\ln 2 - \frac{1}{2} \right] = 4 \left[ \frac{1}{2} + 2\ln 2 \right] = \boxed{2 + 8\ln 2}

展開

展開

將 $g(x)$ 改寫以 $x+1$ 為基底的形式：

g(x) = \frac{1}{\sqrt{(x+1)^2+4}} = \frac{1}{2\sqrt{1 + \frac{(x+1)^2}{4}}} = \frac{1}{2} \left( 1 + \frac{(x+1)^2}{4} \right)^{-1/2}

利用廣義二項式展開式 $(1+u)^{-1/2}$ ：

(1+u)^{-1/2} = 1 - \frac{1}{2}u + \frac{3}{8}u^2 - \frac{5}{16}u^3 + O(u^4)

令 $u = \frac{(x+1)^2}{4}$ ，我們只關注含有 $(x+1)^6$ 的項（即 $u^3$ 項）：

\begin{align*} g(x) =&\, \frac{1}{2} \left[ 1 - \frac{1}{2}\left(\frac{(x+1)^2}{4}\right) + \frac{3}{8}\left(\frac{(x+1)^2}{4}\right)^2 - \frac{5}{16}\left(\frac{(x+1)^2}{4}\right)^3 + \dots \right] \\[4mm] =&\, \dots + \frac{1}{2} \left( -\frac{5}{16} \cdot \frac{(x+1)^6}{64} \right) + \dots = \dots - \frac{5}{2048}(x+1)^6 + \dots \end{align*}

由於泰勒展開式之定義， $(x+1)^6$ 的係數為 $\frac{g^{(6)}(-1)}{6!}$ ，故：

\frac{g^{(6)}(-1)}{6!} = -\frac{5}{2048} \implies g^{(6)}(-1) = -\frac{5 \cdot 6!}{2048} = -\frac{5 \cdot 720}{2048} = \boxed{-\frac{225}{128}}