Faa de Bruno's Theorem

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

刊於《國立武昌師範大學數理化雜誌》第十二期（十週年紀念號），民國十三年四月，武昌師範大學數理化學會出版。原文取自E. B. Elliott著《An introduction to the algebra of quantics》（1895年初版，1913年二版）。

姊妹計劃: 數據項

曾　炯

lm'-l'm

乘冪之行列式

　因欲直接證明多種函數之爲行列式形者，爲二元形Binary quantic之不變式invariants及共變式covariants，有初簡單定理對於某種行列式極爲有助，即Faa de Bruno's Theorem是也．此定理之最初三類爲：

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

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

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

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繼續的in succession與

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

演算Operating而得之,此行列式爲 $lm'-l'm$ 之乘寬即 ${\frac {1}{2}}r(r+1)$ th冪也.

　此又有顯然易見者,卽可同樣的先書下其最末一列:

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

而繼續的向上upwards 由此列與

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

演算而得其餘各列也.因由Taylor's Theorem,在第 $(s+1)$ 行之元素向下讀之乃

　　 $(l+tl')^{r-s}(m+tm')^{s}$

之展開 $t$ 之各冪Various powers之係數;而同樣向上讀之,乃

　　 $(\tau l+l')^{r-s}(\tau m+m')^{s}$

之展開 $\tau$ 之各冪之係數也.

　此兩種形成行列式之方法;謂之第一種寫法及第二種寫法.

　此定理之第一類卽可證明,其第二類之證明:以 $-m/m'$ 乘第二列, $m^{2}/m'^{2}$ 乘第三列,加之於第一列,卽可得證明之,第三類之證明亦可由同樣之方法易於求之.此一般定理,乃線偏微分方程式Lagrange解法之定理The theory of Lagrange's solution of linear partial differential equations中,一簡易習題,茲將進而討論之.

　由乘積Products微分之普通之規則,可知微分 $r$ 次行列式之結果,可書爲 $r$ 個行列式之和,其每行列式由微分原行列式中一列之元素,而遺留其餘各列元素不動而得之.試思此原行列式爲第一種寫法,以 $l'{\frac {\partial }{\partial l}}+m'{\frac {\partial }{\partial m}}$ 演算之.此結果乃 $r$ 個行列式之和,而此等列行式皆消滅爲零Vanish因演算任何列之結果,除最末一列外,皆發生爲其下一列following row之數值之倍數a numerical multiple而最末一列演算之結果爲一列零.於是若 $D$ 表示此原行列式,則得

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

故由Lagrange定理僅含有 $l$ 及 $m$ 於 $lm'-l'm$ 關係之中.

　再試思 $D$ 爲第二種寫法,以 $l{\frac {\partial }{\partial l'}}+m{\frac {\partial }{\partial m'}}$ 演算之,同樣得

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

故 $D$ 僅含有 $l'$ 及 $m'$ 於 $lm'-l'm$ 關係之中.

　是故此行列式 $D$ ,僅爲 $lm'-l'm$ 之函數,且爲同次式Homogeneous而必爲 $lm'-l'm$ 之一單冪single power帶有一可能的數值的因數而成者也.但此數值的因數爲 $1$ ,例如取 $l=m'=1$ , $l'=m=0$ ,則 $lm'-l'm$ 爲 $1$ ,而 $D$ 爲一主對角線A principal diagonal之 $1$ ,及其他各元素零而成.

　由是可知 $D$ 爲 $l,m,l',m'$ 之 $r(r+1)$ 次元Dimension,更由此事實而知 $D$ 爲 $lm'-l'm$ 之 ${\frac {1}{2}}r(t+1)$ th冪矣.

　試證明

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

爲一個二元形 $u$ 之一共變式，於特別情形,若 $u$ 爲四次式quartic,則爲一不變式.以爲本定理之標準應用。

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

　第一次乘法之結果，因

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

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

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

　第二次乘法變易其每元素之微分次數,則

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