泰勒级数测验（含答案）：多项式、收敛半径与误差

Question 1

求 $\ln(1+x)$ 的麦克劳林级数（写成幂级数形式）。

Accepted Answer

$\displaystyle \sum_{n=0}^{\infty} (-1)^n \frac{x^{n+1}}{n+1}$

Answer

$\displaystyle \sum_{n=0}^{\infty} (-1)^n \frac{x^n}{n+1}$

Answer

$\displaystyle \sum_{n=1}^{\infty} \frac{x^n}{n}$

Answer

$\displaystyle \sum_{n=1}^{\infty} (-1)^{n+1} \frac{x^{n-1}}{n}$

Question 2

求 $\arctan x$ 的麦克劳林级数。

Accepted Answer

$\displaystyle \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}$

Answer

$\displaystyle \sum_{n=1}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1}$

Answer

$\displaystyle \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{2n+1}$

Answer

$\displaystyle \sum_{n=0}^{\infty} \frac{x^{2n+1}}{2n+1}$

Question 3

用泰勒定理估计误差：用 $\ln(1+x)$ 的三次麦克劳林多项式近似 $\ln(1.2)$（即取 $x=0.2$）时，给出误差上界。

Accepted Answer

$\displaystyle |R_3(0.2)| \le \frac{6(0.2)^4}{4!}$

Answer

$\displaystyle |R_3(0.2)| \le \frac{(0.2)^4}{4}$

Answer

$\displaystyle |R_3(0.2)| \le \frac{(0.2)^3}{3!}$

Answer

$\displaystyle |R_3(0.2)| \le \frac{6(0.2)^4}{3!}$

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