# 微分與泰勒級數

『如果我想用一個多項式來逼近函數，應該如何做呢？』

(1)
\begin{align} f(x) = c_0 + c_1 x + c_2 x^2 + ...+ c_k x^k+...=\sum_{k=0}^\infty c_k x^k \end{align}

(2)
\begin{eqnarray} f'(x) & = & \frac{d f(x)}{dx} &=& c_1+c_2*2*x+c_3*3*x^2+c_4*4*x^3+... \\ f''(x) & = & \frac{d f'(x)}{dx} &=& c_2*2*1+c_3*3*2*x+c_4*4*3*x^2+... \\ f'''(x) & = & \frac{d f''(x)}{dx} &=& c_3*3*2*1+c_4*4*3*2*x+... \\ ... \\ f^k(x) & = & \frac{d f^{k-1}(x)}{dx} \;\;\; &=& c_k k!+c_{k+1} (k+1)! x+... \end{eqnarray}

(3)
\begin{align} c_k = \frac{f^k(0)}{k!} \end{align}

(4)
\begin{align} f(x) = f(0) + \frac{f'(0)}{1!} x +...+ \frac{f^k (0)}{k!} x^k+...=\sum^{\infty}_{k=0} \frac{f^k(0)}{k!} x^k \end{align}

(5)
\begin{align} f(x) = f(a) + \frac{f'(a)}{1!} (x-a) +...+ \frac{f^{k(a)}}{k!} (x-a)^k+...= \sum^\infty_{k=0} \frac{f^k(a)}{k!} (x-a)^k \end{align}

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page revision: 17, last edited: 12 Sep 2012 23:53
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