📝 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 ${\varphi }^{\left( n\right) }\left( t\right)$ .
💡 答案与解析
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
$$ {\varphi }^{\left( n\right) }\left( t\right) = {\left( h\frac{\partial }{\partial x} + k\frac{\partial }{\partial y}\right) }^{n}f\left( {x + {th},y + {tk}}\right) $$
$$ = \mathop{\sum }\limits_{{p = 0}}^{n}\left( \begin{array}{l} n \\ p \end{array}\right) {h}^{p}{k}^{n - p}\frac{{\partial }^{n}}{\partial {x}^{p}\partial {y}^{n - p}}f\left( {x + {th},y + {tk}}\right) . $$