Faa de Bruno's Theorem

Faa de Bruno's Theorem
作者：E. B. Elliott
1924年4月
译者：曾炯

刊于《国立武昌师范大学数理化杂志》第十二期（十周年纪念号），民国十三年四月，武昌师范大学数理化学会出版。原文取自E. B. Elliott著《An introduction to the algebra of quantics》（1895年初版，1913年二版）。

姊妹计划: 数据项

曾　炯

lm'-l'm

乘幂之行列式

　因欲直接证明多种函数之为行列式形者，为二元形Binary quantic之不变式invariants及共变式covariants，有初简单定理对于某种行列式极为有助，即Faa de Bruno's Theorem是也．此定理之最初三类为：

　　 ${\begin{vmatrix}l,&m\\l',&m'\end{vmatrix}}=lm'-l'm$

　　 ${\begin{vmatrix}l^{2}&lm&m^{2}\\2ll'&lm'+l'm,&2mm'\\l'^{2}&l'm'&m'^{2}\end{vmatrix}}=(lm'-l'm)^{3}$

　　 ${\begin{vmatrix}l^{3}&l^{2}m&lm^{2}&m^{3}\\3l^{2}l',&2ll'm+l^{2}m',&2lmm'+l'm^{2}&3m^{2}m'\\3ll'^{2},&l'^{2}m+2ll'm',&lm^{2}+2l'mm',&3mm'^{2}\\l'^{3}&l'^{2}m'&l'm'^{2}&m'^{2}\end{vmatrix}}=(lm'-l'm)^{6}$

而此一般定理General Theorem,乃一行列式,其第一列由下列各元素而成

　　 $l^{r},l^{r-1}m,l^{r-2}m^{2},\cdots \cdots \cdots \cdots lm^{r-1},m^{r}$

而其各列可由第一列继续的in succession与

$l'{\frac {\partial }{\partial l}}+m'{\frac {\partial }{\partial m}},{\frac {1}{1\cdot 2}}\left(l'{\frac {\partial }{\partial l}}+m'{\frac {\partial }{\partial m}}\right)^{2},\cdots \cdots \cdots \cdots {\frac {1}{r!}}\left(l'{\frac {\partial }{\partial l}}+m'{\frac {\partial }{\partial m}}\right)^{r},$

演算Operating而得之,此行列式为 $lm'-l'm$ 之乘宽即 ${\frac {1}{2}}r(r+1)$ th幂也.

　此又有显然易见者,即可同样的先书下其最末一列:

　　 $l'^{r},l'^{r-1}m',l'^{r-2}m'^{2},\cdots \cdots \cdots \cdots l'm'^{r-1},m'^{r}$

而继续的向上upwards 由此列与

　　 $l{\frac {\partial }{\partial l'}}+m{\frac {\partial }{\partial m'}},{\frac {1}{1\cdot 2}}\left(l{\frac {\partial }{\partial l'}}+m{\frac {\partial }{\partial m'}}\right)^{2},\cdots \cdots \cdots \cdots {\frac {1}{r!}}\left(l{\frac {\partial }{\partial l'}}+m{\frac {\partial }{\partial m'}}\right)^{r}$

演算而得其馀各列也.因由Taylor's Theorem,在第 $(s+1)$ 行之元素向下读之乃

　　 $(l+tl')^{r-s}(m+tm')^{s}$

之展开 $t$ 之各幂Various powers之系数;而同样向上读之,乃

　　 $(\tau l+l')^{r-s}(\tau m+m')^{s}$

之展开 $\tau$ 之各幂之系数也.

　此两种形成行列式之方法;谓之第一种写法及第二种写法.

　此定理之第一类即可证明,其第二类之证明:以 $-m/m'$ 乘第二列, $m^{2}/m'^{2}$ 乘第三列,加之于第一列,即可得证明之,第三类之证明亦可由同样之方法易于求之.此一般定理,乃线偏微分方程式Lagrange解法之定理The theory of Lagrange's solution of linear partial differential equations中,一简易习题,兹将进而讨论之.

　由乘积Products微分之普通之规则,可知微分 $r$ 次行列式之结果,可书为 $r$ 个行列式之和,其每行列式由微分原行列式中一列之元素,而遗留其馀各列元素不动而得之.试思此原行列式为第一种写法,以 $l'{\frac {\partial }{\partial l}}+m'{\frac {\partial }{\partial m}}$ 演算之.此结果乃 $r$ 个行列式之和,而此等列行式皆消灭为零Vanish因演算任何列之结果,除最末一列外,皆发生为其下一列following row之数值之倍数a numerical multiple而最末一列演算之结果为一列零.于是若 $D$ 表示此原行列式,则得

　　 $l'{\frac {\partial D}{\partial l}}+m'{\frac {\partial D}{\partial m}}=0,$

故由Lagrange定理仅含有 $l$ 及 $m$ 于 $lm'-l'm$ 关系之中.

　再试思 $D$ 为第二种写法,以 $l{\frac {\partial }{\partial l'}}+m{\frac {\partial }{\partial m'}}$ 演算之,同样得

　　 $l{\frac {\partial D}{\partial l'}}+m{\frac {\partial D}{\partial m'}}=0.$

故 $D$ 仅含有 $l'$ 及 $m'$ 于 $lm'-l'm$ 关系之中.

　是故此行列式 $D$ ,仅为 $lm'-l'm$ 之函数,且为同次式Homogeneous而必为 $lm'-l'm$ 之一单幂single power带有一可能的数值的因数而成者也.但此数值的因数为 $1$ ,例如取 $l=m'=1$ , $l'=m=0$ ,则 $lm'-l'm$ 为 $1$ ,而 $D$ 为一主对角线A principal diagonal之 $1$ ,及其他各元素零而成.

　由是可知 $D$ 为 $l,m,l',m'$ 之 $r(r+1)$ 次元Dimension,更由此事实而知 $D$ 为 $lm'-l'm$ 之 ${\frac {1}{2}}r(t+1)$ th幂矣.

　试证明

　　 ${\begin{vmatrix}{\frac {\partial ^{4}u}{\partial x^{4}}},&{\frac {\partial ^{4}u}{\partial x^{3}\partial y}},&{\frac {\partial ^{4}u}{\partial x^{2}\partial y^{2}}}\\{\frac {\partial ^{4}u}{\partial x^{3}\partial y}},&{\frac {\partial ^{4}u}{\partial x^{2}\partial y^{2}}},&{\frac {\partial ^{4}u}{\partial x\partial y^{3}}}\\{\frac {\partial ^{4}u}{\partial x^{2}\partial y^{2}}},&{\frac {\partial ^{4}u}{\partial x\partial y^{3}}},&{\frac {\partial ^{4}u}{\partial y^{4}}}\end{vmatrix}}$

为一个二元形 $u$ 之一共变式，于特别情形,若 $u$ 为四次式quartic,则为一不变式.以为本定理之标准应用。

　由上述 $(lm'-l'm)^{3}$ 即 $M^{3}$ 之行列式，行与行乘此式二次。

　第一次乘法之结果，因

　　 $\left(l{\frac {\partial }{\partial x}}+l'{\frac {\partial }{\partial y}}\right)^{2}={\frac {\partial ^{2}}{\partial X^{2}}},\left(l{\frac {\partial }{\partial x}}+l'{\frac {\partial }{\partial y}}\right)\left(m{\frac {\partial }{\partial x}}+m'{\frac {\partial }{\partial y}}\right)={\frac {\partial ^{2}}{\partial X\partial Y}}$

　　 $\left(m{\frac {\partial }{\partial x}}+m'{\frac {\partial }{\partial y}}\right)\equiv {\frac {\partial ^{2}}{\partial Y^{2}}}$

　　 ${\begin{vmatrix}{\frac {\partial ^{2}}{\partial X^{2}}}\cdot {\frac {\partial ^{2}u}{\partial x^{2}}},&{\frac {\partial ^{2}}{\partial X\partial Y}}\cdot {\frac {\partial ^{2}u}{\partial x^{2}}},&{\frac {\partial ^{2}}{\partial Y^{2}}}\cdot {\frac {\partial ^{2}u}{\partial x^{2}}}\\{\frac {\partial ^{2}}{\partial X^{2}}}\cdot {\frac {\partial ^{2}u}{\partial x\partial y}},&{\frac {\partial ^{2}}{\partial X\partial Y}}\cdot {\frac {\partial ^{2}u}{\partial x\partial y}},&{\frac {\partial ^{2}}{\partial Y^{2}}}\cdot {\frac {\partial ^{2}u}{\partial x\partial y}}\\{\frac {\partial ^{2}}{\partial X^{2}}}\cdot {\frac {\partial ^{2}u}{\partial y^{2}}},&{\frac {\partial ^{2}}{\partial X\partial Y}}\cdot {\frac {\partial ^{2}u}{\partial y^{2}}},&{\frac {\partial ^{2}}{\partial Y^{2}}}\cdot {\frac {\partial ^{2}u}{\partial y^{2}}}\end{vmatrix}}$

　第二次乘法变易其每元素之微分次数,则

　　 ${\begin{vmatrix}{\frac {\partial ^{4}u}{\partial X^{4}}},&{\frac {\partial ^{4}u}{\partial X^{3}\partial Y}},&{\frac {\partial ^{4}u}{\partial X^{2}\partial Y^{2}}}\\{\frac {\partial ^{4}u}{\partial X^{3}\partial Y}},&{\frac {\partial ^{4}u}{\partial X^{2}\partial Y^{2}}},&{\frac {\partial ^{4}u}{\partial X\partial Y^{3}}}\\{\frac {\partial ^{4}u}{\partial X^{2}\partial Y^{2}}},&{\frac {\partial ^{4}u}{\partial X\partial Y^{3}}}&{\frac {\partial ^{4}u}{\partial Y^{4}}}\end{vmatrix}}$