Faa de Bruno's Theorem

　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：

　　$\left|\begin{array}{cc}l,& m\\ l^{\prime },& m^{\prime }\end{array}\right|=l{m}^{\prime }-l^{\prime }m$

　　$\left|\begin{array}{ccc}{l}^2& lm& m^2\\ 2l{l}^{\prime }& l{m}^{\prime }+l^{\prime }m,& 2m{m}^{\prime }\\ l{\text{'}}^2& l^{\prime }{m}^{\prime }& m{\text{'}}^2\end{array}\right|=(l{m}^{\prime }-l^{\prime }m{)}^3$

　　$\left|\begin{array}{cccc}{l}^3& l^{2}m& l{m}^2& m^3\\ 3l^2{l}^{\prime },& 2l{l}^{\prime }m+l^2{m}^{\prime },& 2lm{m}^{\prime }+l^{\prime }{m}^2& 3m^2{m}^{\prime }\\ 3ll{\text{'}}^2,& l{\text{'}}^{2}m+2l{l}^{\prime }{m}^{\prime },& l{m}^2+2l^{\prime }m{m}^{\prime },& 3mm{\text{'}}^2\\ l{\text{'}}^3& l{\text{'}}^2{m}^{\prime }& l^{\prime }m{\text{'}}^2& m{\text{'}}^2\end{array}\right|=(l{m}^{\prime }-l^{\prime }m{)}^6$

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

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

而其各列可由第一列繼續的in succession與

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

演算Operating而得之,此行列式爲$l{m}^{\prime }-l^{\prime }m$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:

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

而繼續的向上upwards 由此列與

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

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

　　$(l+t{l}^{\prime }{)}^{r-s}(m+t{m}^{\prime }{)}^s$

之展開$t$之各冪Various powers之係數;而同樣向上讀之,乃

　　$(\tau l+l^{\prime }{)}^{r-s}(\tau m+m^{\prime }{)}^s$

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$-m/m^{\prime }$乘第二列,$m^2/m{\text{'}}^2$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$l^{\prime }\frac{\partial }{\partial l}+m^{\prime }\frac{\partial }{\partial m}$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

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

故由Lagrange定理僅含有$l$及$m$於$l{m}^{\prime }-l^{\prime }m$關係之中.

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

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

故$D$僅含有$l^{\prime }$及$m^{\prime }$於$l{m}^{\prime }-l^{\prime }m$關係之中.

　是故此行列式$D$,僅爲$l{m}^{\prime }-l^{\prime }m$之函數,且爲同次式Homogeneous而必爲$l{m}^{\prime }-l^{\prime }m$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$l=m^{\prime }=1$, $l^{\prime }=m=0$,則$l{m}^{\prime }-l^{\prime }m$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$l,m,l^{\prime },m^{\prime }$之$r(r+1)$次元Dimension,更由此事實而知$D$爲$l{m}^{\prime }-l^{\prime }m$之$\frac{1}{2}r(t+1)$th冪矣.

　試證明

　　$\left|\begin{array}{ccc}\frac{{\partial }^{4}u}{\partial x^4},& \frac{{\partial }^{4}u}{\partial x^3\partial y},& \frac{{\partial }^{4}u}{\partial x^2\partial y^2}\\ \frac{{\partial }^{4}u}{\partial x^3\partial y},& \frac{{\partial }^{4}u}{\partial x^2\partial y^2},& \frac{{\partial }^{4}u}{\partial x\partial y^3}\\ \frac{{\partial }^{4}u}{\partial x^2\partial y^2},& \frac{{\partial }^{4}u}{\partial x\partial y^3},& \frac{{\partial }^{4}u}{\partial y^4}\end{array}\right|$

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

　第一次乘法之結果，因

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

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

　　$\left|\begin{array}{ccc}\frac{{\partial }^2}{\partial X^2}\cdot \frac{{\partial }^{2}u}{\partial x^2},& \frac{{\partial }^2}{\partial X\partial Y}\cdot \frac{{\partial }^{2}u}{\partial x^2},& \frac{{\partial }^2}{\partial Y^2}\cdot \frac{{\partial }^{2}u}{\partial x^2}\\ \frac{{\partial }^2}{\partial X^2}\cdot \frac{{\partial }^{2}u}{\partial x\partial y},& \frac{{\partial }^2}{\partial X\partial Y}\cdot \frac{{\partial }^{2}u}{\partial x\partial y},& \frac{{\partial }^2}{\partial Y^2}\cdot \frac{{\partial }^{2}u}{\partial x\partial y}\\ \frac{{\partial }^2}{\partial X^2}\cdot \frac{{\partial }^{2}u}{\partial y^2},& \frac{{\partial }^2}{\partial X\partial Y}\cdot \frac{{\partial }^{2}u}{\partial y^2},& \frac{{\partial }^2}{\partial Y^2}\cdot \frac{{\partial }^{2}u}{\partial y^2}\end{array}\right|$

　第二次乘法變易其每元素之微分次數,則

　　$\left|\begin{array}{ccc}\frac{{\partial }^{4}u}{\partial X^4},& \frac{{\partial }^{4}u}{\partial X^3\partial Y},& \frac{{\partial }^{4}u}{\partial X^2\partial Y^2}\\ \frac{{\partial }^{4}u}{\partial X^3\partial Y},& \frac{{\partial }^{4}u}{\partial X^2\partial Y^2},& \frac{{\partial }^{4}u}{\partial X\partial Y^3}\\ \frac{{\partial }^{4}u}{\partial X^2\partial Y^2},& \frac{{\partial }^{4}u}{\partial X\partial Y^3}& \frac{{\partial }^{4}u}{\partial Y^4}\end{array}\right|$

　如是上述之事實證明矣

　注意I. 此篇譯自 E. B. Elliott's Algebras of Quantics 中16,17兩節

　注意II. The theory of Lagrange's solution of linear partial differential equations可參考A. R. Forsyth's Differential Euqations之187,189兩節PP.392-394.

　注意III. Faa de Bruno (1825-1888)