关于实函数泰勒展开式的一个注记
A note on Taylor expansion of real function
摘要：
设$f(x)$是在区间$[a, b]$上具有$(n+1)$阶导数的实函数.对任意点$x_0\in (a, b)$和任意整数$0\leq k\leq n$，记$S_{k,x_0}(x)$为$f(x)$在$x_0$上的泰勒展开式的$k$阶截断，即$$S_{k,x_0}(x)=\sum_{i=0}^k\frac{f^{(i)}(x_0)}{i!}(x-x_0)^i.$$本文考虑$f(x)$的$L_2$阶多项式$\leq k$逼近，证明了$S_{k,x_0}(x)$是$f(x)$在$[x_0-\varepsilon, x_0+\varepsilon]$上的最佳逼近的极限为$\varepsilon\to 0$ .
英文摘要：
Let $f(x)$ be a real function which has $(n+1)$-th derivative on an interval $[a, b]$. For any point $x_0\in (a, b)$ and any integer $0\leq k\leq n$, denote by $S_{k,x_0}(x)$ the $k$-th truncation of the Taylor expansion of $f(x)$ at $x_0$, i.e. $$S_{k,x_0}(x)=\sum_{i=0}^k\frac{f^{(i)}(x_0)}{i!}(x-x_0)^i.$$ In this note, we consider the $L_2$-approximation of $f(x)$ by polynomials of degree $\leq k$, we show that $S_{k,x_0}(x)$ is the limit of the best approximations of $f(x)$ on $[x_0-\varepsilon, x_0+\varepsilon]$ as $\varepsilon\to 0$.
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