同济高数 第9章 第9-9-3题

[AI解答]

[AI解答]

我们要求函数 $ f(x, y) = \sin x \sin y $ 在点 $\left( \frac{\pi}{4}, \frac{\pi}{4} \right)$ 的二阶泰勒展开。

首先，记展开点为 $$ (x_0, y_0) = \left( \frac{\pi}{4}, \frac{\pi}{4} \right) $$

**第一步：计算函数值及一阶、二阶偏导数**

函数： $$ f(x, y) = \sin x \sin y $$

一阶偏导： $$ f_x = \cos x \sin y, \quad f_y = \sin x \cos y $$

二阶偏导： $$ f_{xx} = -\sin x \sin y, \quad f_{yy} = -\sin x \sin y, \quad f_{xy} = \cos x \cos y $$

**第二步：在点 $\left( \frac{\pi}{4}, \frac{\pi}{4} \right)$ 处取值**

已知： $$ \sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}, \quad \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2} $$

所以： $$ f\left( \frac{\pi}{4}, \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{2} $$

一阶偏导值： $$ f_x\left( \frac{\pi}{4}, \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{2} $$ $$ f_y\left( \frac{\pi}{4}, \frac{\pi}{4} \right) = \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2} = \frac{1}{2} $$

二阶偏导值： $$ f_{xx}\left( \frac{\pi}{4}, \frac{\pi}{4} \right) = -\frac{1}{2}, \quad f_{yy}\left( \frac{\pi}{4}, \frac{\pi}{4} \right) = -\frac{1}{2}, \quad f_{xy}\left( \frac{\pi}{4}, \frac{\pi}{4} \right) = \frac{1}{2} $$

**第三步：写出二阶泰勒公式**

二元函数在 $(x_0, y_0)$ 处的二阶泰勒展开为： $$ f(x, y) = f(x_0, y_0) + f_x (x-x_0) + f_y (y-y_0) + \frac{1}{2!} \left[ f_{xx} (x-x_0)^2 + 2 f_{xy} (x-x_0)(y-y_0) + f_{yy} (y-y_0)^2 \right] + R_2 $$

代入数值： $$ f(x, y) = \frac{1}{2} + \frac{1}{2}\left(x - \frac{\pi}{4}\right) + \frac{1}{2}\left(y - \frac{\pi}{4}\right) + \frac{1}{2} \left[ -\frac{1}{2} \left(x - \frac{\pi}{4}\right)^2 + 2\cdot \frac{1}{2} \left(x - \frac{\pi}{4}\right)\left(y - \frac{\pi}{4}\right) - \frac{1}{2} \left(y - \frac{\pi}{4}\right)^2 \right] + R_2 $$

化简二次项部分： $$ \frac{1}{2} \left[ -\frac{1}{2} (x - \frac{\pi}{4})^2 + (x - \frac{\pi}{4})(y - \frac{\pi}{4}) - \frac{1}{2} (y - \frac{\pi}{4})^2 \right] $$ $$ = -\frac{1}{4} (x - \frac{\pi}{4})^2 + \frac{1}{2} (x - \frac{\pi}{4})(y - \frac{\pi}{4}) - \frac{1}{4} (y - \frac{\pi}{4})^2 $$

因此二阶泰勒公式为： $$ \boxed{f(x, y) = \frac{1}{2} + \frac{1}{2}\left(x - \frac{\pi}{4}\right) + \frac{1}{2}\left(y - \frac{\pi}{4}\right) - \frac{1}{4}\left(x - \frac{\pi}{4}\right)^2 + \frac{1}{2}\left(x - \frac{\pi}{4}\right)\left(y - \frac{\pi}{4}\right) - \frac{1}{4}\left(y - \frac{\pi}{4}\right)^2 + R_2} $$

其中 $R_2$ 为拉格朗日余项或佩亚诺余项。

难度：★★☆☆☆