新讲 第12章 多元微分学 第5题
📝 题目
例 5 设 $f\left( {x,y}\right)$ 是 $n$ 阶连续可微函数,并设
$$
\varphi \left( t\right) = f\left( {x + {th},y + {tk}}\right) .
$$
试计算函数 $\varphi \left( t\right)$ 的 $n$ 阶导数 ${\varphi }^{\left( n\right) }\left( t\right)$ .
💡 答案解析
解 设 $g\left( {x,y}\right)$ 是任意连续可微函数. 我们先对形状如
$$
\psi \left( t\right) = g\left( {x + {th},y + {tk}}\right)
$$
的函数, 证明一个求导公式. 运用复合函数求导的链式法则可得
$$
{\psi }^{\prime }\left( t\right) = h\frac{\partial }{\partial x}g\left( {x + {th},y + {tk}}\right)
$$
$$
+ k\frac{\partial }{\partial y}g\left( {x + {th},y + {tk}}\right)
$$
$$
= \left( {h\frac{\partial }{\partial x} + k\frac{\partial }{\partial y}}\right) g\left( {x + {th},y + {tk}}\right) .
$$
我们看到: 以 $\frac{\mathrm{d}}{\mathrm{d}t}$ 作用于 $\psi \left( t\right)$ ,相当于以微分算子
$$
\left( {h\frac{\partial }{\partial x} + k\frac{\partial }{\partial y}}\right)
$$
作用于
$$
g\left( {x + {th},y + {tk}}\right) \text{ . }
$$
对于 $n$ 阶连续可微函数 $f\left( {x,y}\right)$ ,我们来计算复合函数
$$
\varphi \left( t\right) = f\left( {x + {th},y + {tk}}\right)
$$
的各阶导数. 利用上面讨论的结果, 容易得到
$$
{\varphi }^{\prime }\left( t\right) = \left( {h\frac{\partial }{\partial x} + k\frac{\partial }{\partial y}}\right) f\left( {x + {th},y + {tk}}\right) ,
$$
$$
{\varphi }^{\prime \prime }\left( t\right) = {\left( h\frac{\partial }{\partial x} + k\frac{\partial }{\partial y}\right) }^{2}f\left( {x + {th},y + {tk}}\right) ,
$$
\_\_\_\_\_
$$
{\varphi }^{\left( n\right) }\left( t\right) = {\left( h\frac{\partial }{\partial x} + k\frac{\partial }{\partial y}\right) }^{n}f\left( {x + {th},y + {tk}}\right) .
$$
对于连续可微足够多次的函数,求偏导数的运算 $\frac{\partial }{\partial x}$ 与 $\frac{\partial }{\partial y}$ 可以交换次序. 涉及 $\frac{\partial }{\partial x}$ 与 $\frac{\partial }{\partial y}$ 这些算子的相加、相乘以及乘以实数的运算,遵循多项式代数中关于文字符号的运算法则. 因此, 算子二项式
$$
{\left( h\frac{\partial }{\partial x} + k\frac{\partial }{\partial y}\right) }^{n}
$$
可以按照代数中的二项式定理展开:
$$
{\left( h\frac{\partial }{\partial x} + k\frac{\partial }{\partial y}\right) }^{n} = \mathop{\sum }\limits_{{p = 0}}^{n}\left( \begin{array}{l} n \\ p \end{array}\right) {h}^{p}{k}^{n - p}\frac{{\partial }^{n}}{\partial {x}^{p}\partial {y}^{n - p}},
$$
这里
$$
\left( \begin{array}{l} n \\ p \end{array}\right) = \frac{n!}{p!\left( {n - p}\right) !}
$$
是二项式系数. 我们所得的结果可以写成
$$
{\varphi }^{\left( n\right) }\left( t\right) = {\left( h\frac{\partial }{\partial x} + k\frac{\partial }{\partial y}\right) }^{n}f\left( {x + {th},y + {tk}}\right)
$$
$$
= \mathop{\sum }\limits_{{p = 0}}^{n}\left( \begin{array}{l} n \\ p \end{array}\right) {h}^{p}{k}^{n - p}\frac{{\partial }^{n}}{\partial {x}^{p}\partial {y}^{n - p}}f\left( {x + {th},y + {tk}}\right) .
$$