将 sin x 展成 x - (pi)/(4) 的幂级数.A. sum_(n=1)^infty ((-1)^n)/((2n+1)!) ( x - (pi)/(4) )^2n+1 - sum_(n=1)^infty ((-1)^n)/((2n)!) ( x - (pi)/(4) )^2n, x in (-infty, +infty)B. (1)/(sqrt(2)) [ 1 - (1)/(2!) ( x - (pi)/(4) )^2 + (1)/(4!) ( x - (pi)/(4) )^4 - ... ]C. (1)/(sqrt(2)) [ ( x - (pi)/(4) ) - (1)/(3!) ( x - (pi)/(4) )^3 + (1)/(5!) ( x - (pi)/(4) )^5 + ... ]D. (1)/(sqrt(2)) ( 1 + ( x - (pi)/(4) ) - (1)/(2!) ( x - (pi)/(4) )^2 - (1)/(3!) ( x - (pi)/(4) )^3 + ... )
将 $\sin x$ 展成 $x - \frac{\pi}{4}$ 的幂级数.
A. $\sum_{n=1}^{\infty} \frac{(-1)^n}{(2n+1)!} \left( x - \frac{\pi}{4} \right)^{2n+1} - \sum_{n=1}^{\infty} \frac{(-1)^n}{(2n)!} \left( x - \frac{\pi}{4} \right)^{2n}$, $x \in (-\infty, +\infty)$
B. $\frac{1}{\sqrt{2}} \left[ 1 - \frac{1}{2!} \left( x - \frac{\pi}{4} \right)^2 + \frac{1}{4!} \left( x - \frac{\pi}{4} \right)^4 - \cdots \right]$
C. $\frac{1}{\sqrt{2}} \left[ \left( x - \frac{\pi}{4} \right) - \frac{1}{3!} \left( x - \frac{\pi}{4} \right)^3 + \frac{1}{5!} \left( x - \frac{\pi}{4} \right)^5 + \cdots \right]$
D. $\frac{1}{\sqrt{2}} \left( 1 + \left( x - \frac{\pi}{4} \right) - \frac{1}{2!} \left( x - \frac{\pi}{4} \right)^2 - \frac{1}{3!} \left( x - \frac{\pi}{4} \right)^3 + \cdots \right)$
题目解答
答案
将 $\sin x$ 展开为 $x - \frac{\pi}{4}$ 的幂级数，可使用泰勒级数公式。设 $y = x - \frac{\pi}{4}$，则 $\sin x = \sin\left(y + \frac{\pi}{4}\right)$。利用和角公式得：
$\sin\left(y + \frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}(\sin y + \cos y).$
将 $\sin y$ 和 $\cos y$ 的泰勒展开式相加：
$\sin y = y - \frac{y^3}{3!} + \frac{y^5}{5!} - \cdots,$
$\cos y = 1 - \frac{y^2}{2!} + \frac{y^4}{4!} - \cdots.$
相加得：
$\sin y + \cos y = 1 + y - \frac{y^2}{2!} - \frac{y^3}{3!} + \cdots.$
代入 $y = x - \frac{\pi}{4}$，得：
$\sin x = \frac{1}{\sqrt{2}}\left[1 + \left(x - \frac{\pi}{4}\right) - \frac{1}{2!}\left(x - \frac{\pi}{4}\right)^2 - \frac{1}{3!}\left(x - \frac{\pi}{4}\right)^3 + \cdots\right].$
该级数在 $(-\infty, +\infty)$ 内收敛，与选项 D 相符。
答案： $\boxed{D}$