Page:EB1911 - Volume 01.djvu/666

$(l_{1}^{\lambda _{1}}l_{2}^{\lambda _{2}}\ldots )$ and it is seen intuitively that the number $\theta$ remains unaltered when the first two of these partitions are interchanged (see Combinatorial Analysis). Hence the theorem is established.

Putting $x_{1}=1$ and $x_{2}=x_{3}=x_{4}=\ldots =0,$ we find a particular law of reciprocity given by Cayley and Betti,

$${\begin{aligned}(1^{m_{1}})^{\mu _{1}}(1^{m_{2}})^{\mu _{2}}(1^{m_{3}})^{\mu _{3}}\ldots &=\ldots +\theta (s_{1}^{\sigma _{1}}s_{2}^{\sigma _{2}}s_{3}^{\sigma _{3}}\ldots )+\ldots ,\\(1^{s_{1}})^{\sigma _{1}}(1^{s_{2}})^{\sigma _{2}}(1^{s_{3}})^{\sigma _{3}}\ldots &=\ldots +\theta (m_{1}^{\mu _{1}}m_{2}^{\mu _{2}}m_{3}^{\mu _{3}}\ldots )+\ldots ;\end{aligned}}$$

and another by putting $x_{1}=x_{2}=x_{3}=\ldots =1$, for then $\mathrm {X} _{m}$ becomes $h_{m}$, and we have

$$h_{m_{1}}^{\mu _{1}}h_{m_{2}}^{\mu _{2}}h_{m_{3}}^{\mu _{3}}\ldots =\ldots +\theta ^{\prime }(s_{1}^{\sigma _{1}}s_{2}^{\sigma _{2}}s_{3}^{\sigma _{3}}\ldots )+\ldots ,$$ $$h_{s_{1}}^{\sigma _{1}}h_{s_{2}}^{\sigma _{2}}h_{s_{3}}^{\sigma _{3}}\ldots =\ldots +\theta ^{\prime }(m_{1}^{\mu _{1}}m_{2}^{\mu _{2}}m_{3}^{\mu _{3}}\ldots )+\ldots ,$$

Theorem of Expressibility.—“If a symmetric function be symboilized by $(\lambda \mu \nu \ldots )$ and $(\lambda _{1}\lambda _{2}\lambda _{3}\ldots ),$ $(\mu _{1}\mu _{2}\mu _{3}\ldots ),$ $(\nu _{1}\nu _{2}\nu _{3}\ldots )\ldots$ be any partitions of $\lambda ,\mu ,\nu \ldots$ respectively, the function $(\lambda \mu \nu \ldots )$ is expressible by means of functions symbolized by separation of

$$(\lambda _{1}\lambda _{2}\lambda _{3}\ldots \mu _{1}\mu _{2}\mu _{3}\ldots \nu _{1}\nu _{2}\nu _{3}\ldots ).{\text{”}}$$

For, writing as before,

$$\mathrm {X} _{m_{1}}^{\mu _{1}}\mathrm {X} _{m_{2}}^{\mu _{2}}\mathrm {X} _{m_{3}}^{\mu _{3}}\ldots =\Sigma \Sigma \theta (s_{1}^{\sigma _{1}}s_{2}^{\sigma _{2}}s_{3}^{\sigma _{3}}\ldots )x_{l_{1}}^{\lambda _{1}}x_{l_{2}}^{\lambda _{2}}x_{l_{3}}^{\lambda _{3}}\ldots ,$$ $$=\Sigma {\text{P}}x_{l_{1}}^{\lambda _{1}}x_{l_{2}}^{\lambda _{2}}x_{l_{3}}^{\lambda _{3}}\ldots ,$$

${\text{P}}$ is a linear function of separations of $(l_{1}^{\lambda _{1}}l_{2}^{\lambda _{2}}l_{3}^{\lambda _{3}}\ldots )$ of specification $(m_{1}^{\mu _{1}}m_{2}^{\mu _{2}}m_{3}^{\mu _{3}}\ldots ),$ and if $\mathrm {X} _{s_{1}}^{\sigma _{1}}\mathrm {X} _{s_{2}}^{\sigma _{2}}\mathrm {X} _{s_{3}}^{\sigma _{3}}\ldots =\Sigma {\text{P}}^{\prime }x_{l_{1}}^{\lambda _{1}}x_{l_{2}}^{\lambda _{2}}x_{l_{3}}^{\lambda _{3}}\ldots ,{\text{P}}^{\prime }$ is a linear function of separations of $(l_{1}^{\lambda _{1}}l_{2}^{\lambda _{2}}l_{3}^{\lambda _{3}}\ldots )$ of specification $(s_{1}^{\sigma _{1}}s_{2}^{\sigma _{2}}s_{3}^{\sigma _{3}}\ldots ).$ Suppose the separations of $(l_{1}^{\lambda _{1}}l_{2}^{\lambda _{2}}l_{3}^{\lambda _{3}}\ldots )$ to involve $k$ different specifications and form the $k$ identities

$$\mathrm {X} _{m_{1s}}^{\mu _{1s}}\mathrm {X} _{m_{2s}}^{\mu _{2s}}\mathrm {X} _{m_{3s}}^{\mu _{3s}}\ldots =\Sigma {\text{P}}^{(s)}x_{l_{1}}^{\lambda _{1}}x_{l_{2}}^{\lambda _{2}}x_{l_{3}}^{\lambda _{3}}\ldots (s=1,2,\ldots k),$$

where $m_{1s}^{\mu _{1s}}m_{2s}^{\mu _{2s}}m_{3s}^{\mu _{3s}}\ldots )$ is one of the $k$ specifications.

The law of reciprocity shows that

$${\text{P}}^{(s)}={\overset {t=k}{\underset {t=1}{\Sigma \theta _{st}}}}(m_{1t}^{\mu _{1t}}m_{2t}^{\mu _{2t}}m_{3t}^{\mu _{3t}}\ldots ),$$

viz.: a linear function of symmetric functions symbolized by the $k$ specifications; and that $\theta _{st}=\theta _{ts}.$ A table may be formed expressing the $k$ expressions ${\text{P}}^{(1)},{\text{P}}^{(2)},\ldots {\text{P}}^{(k)}$ as linear functions of the $k$ expressions $(m_{1s}^{\mu _{1s}}m_{2s}^{\mu _{2s}}m_{3s}^{\mu _{3s}}\ldots )$, $s=1,2,\ldots k$, and the numbers $\theta _{st}$ occurring therein possess row and column symmetry. By solving $k$ linear equations we similarly express the latter functions as linear functions of the former, and this table will also be symmetrical.

Theorem.—“The symmetric function $(m_{1s}^{\mu _{1s}}m_{2s}^{\mu _{2s}}m_{3s}^{\mu _{3s}}\ldots )$ whose partition is a specification of a separation of the function symbolized by $(l_{1}^{\lambda _{1}}l_{2}^{\lambda _{2}}l_{3}^{\lambda _{3}}\ldots )$ is expressible as a linear function of symmetric functions symbolized by separations of $(l_{1}^{\lambda _{1}}l_{2}^{\lambda _{2}}l_{3}^{\lambda _{3}}\ldots )$ and a symmetrical table may be thus formed.” It is now to be remarked that the partition $(l_{1}^{\lambda _{1}}l_{2}^{\lambda _{2}}l_{3}^{\lambda _{3}}\ldots )$ can be derived from $(m_{1s}^{\mu _{1s}}m_{2s}^{\mu _{2s}}m_{3s}^{\mu _{3s}}\ldots )$ by substituting for the numbers $m_{1s},m_{2s},m_{3s},\ldots$ certain partitions of those numbers (vide the definition of the specification of a separation).

Hence the theorem of expressibility enunciated above. A new statement of the law of reciprocity can be arrived at as follows:—Since.

$${\text{P}}^{(s)}=\mu _{1s}!\mu _{2s}!\mu _{3s}!\ldots \sum {\frac {({\text{J}}_{1})^{j_{1}}({\text{J}}_{2})^{j_{2}}({\text{J}}_{3})^{j_{3}}\ldots }{j_{1}!j_{2}!j_{3}!\ldots }},$$

where $({\text{J}}_{1})^{j_{1}}({\text{J}}_{2})^{j_{2}}({\text{J}}_{3})^{j_{3}}\ldots$ is a separation of $(l_{1}^{\lambda _{1}}l_{2}^{\lambda _{2}}l_{3}^{\lambda _{3}}\ldots )$ of specification $(m_{1s}^{\mu _{1s}}m_{2s}^{\mu _{2s}}m_{3s}^{\mu _{3s}}\ldots ),$ placing $s$ under the summation sign to denote the specification involved,

$${\begin{aligned}\mu _{1s}!\mu _{2s}!\mu _{3s}\ldots &\sum _{s}{\frac {({\text{J}}_{1})^{j_{1}}({\text{J}}_{2})^{j_{2}}({\text{J}}_{3})^{j_{3}}\ldots }{j_{1}!j_{2}!j_{3}!\ldots }}=\sum _{t=1}^{t=k}\theta _{st}(m_{1s}^{\mu _{1s}}m_{2s}^{\mu _{2s}}m_{3s}^{\mu _{3s}}\ldots ),\\\mu _{15}!\mu _{2t}!\mu _{3t}!&\ldots \sum _{t}{\frac {({\text{J}}_{1})^{j_{1}}({\text{J}}_{2})^{j_{2}}({\text{J}}_{3})^{j_{3}}\ldots }{j_{1}!j_{2}!j_{3}\ldots }}=\sum _{s=1}^{s=k}\theta _{ts}(m_{1s}^{\mu _{1s}}m_{2s}^{\mu _{2s}}m_{3s}^{\mu _{3s}}\ldots ),\end{aligned}}$$

where $\theta _{st}=\theta _{ts}$.

Theorem of Symmetry.—If we form the separation function

$$\sum _{s}{\frac {({\text{J}}_{1})^{j_{1}}({\text{J}}_{2})^{j_{2}}({\text{J}}_{3})^{j_{3}}\ldots }{j_{1}!j_{2}!j_{3}!\ldots }}$$

appertaining to the function $(l_{1}^{\lambda _{1}}l_{2}^{\lambda _{2}}l_{3}^{\lambda _{3}}\ldots ),$ each separation having a specification $(m_{1s}^{\mu _{1s}}m_{2s}^{\mu _{2s}}m_{3s}^{\mu _{3s}}\ldots )$, multiply by $\mu _{1s}!\mu _{2s}!\mu _{3s}!\ldots ,$ and take therein the coefficient of the function $(m_{1t}^{\mu _{1t}}m_{2t}^{\mu _{2t}}m_{3t}^{\mu _{3t}}\ldots ),$ we obtain the same result as if we formed the separation function in regard to the specification $(m_{1t}^{\mu _{1t}}m_{2t}^{\mu _{2t}}m_{3t}^{\mu _{3t}}\ldots ),$ multiplied by $\mu _{1t}!\mu _{2t}!\mu _{3t}!\ldots$ and took therein the coefficient of the function $(m_{1s}^{\mu _{1s}}m_{2s}^{\mu _{2s}}m_{3s}^{\mu _{3s}}\ldots ).$

Ex. gr., take $(l_{1}^{\lambda _{1}}l_{2}^{\lambda _{2}}\ldots )=(21^{4});(m_{1s}^{\mu _{1s}}m_{2s}^{\mu _{2s}}\ldots )=(321);(m_{1t}^{\mu _{1t}}m_{2t}^{\mu _{2t}}\ldots )=(31_{3});$ we find

$${\begin{aligned}(21)(1^{2})(1)+(1^{3})(2)(1)&=\ldots +13(31^{3})+\ldots ,\\(21)(1)^{3}&=\ldots +13(321)+\ldots \end{aligned}}$$

The Differential Operators.—Starting with the relation

$$(1+\alpha _{1}x)(1+\alpha _{2}x)\ldots (1+a_{n}x)=1+a_{1}x+a_{2}x^{2}+\ldots +a_{n}x^{n}$$

multiply each side by $1+\mu x,$ thus introducing a new quantity $\mu ;$ we obtain

$$(1+a_{1}x)(1+a_{2}x)\ldots (1+a_{n}x)(1+\mu x)=1+(a_{1}+\mu )x+(a_{2}+\mu a_{1})x^{2}+\ldots$$

so that $f(a_{1},a_{2},a_{3},\ldots a_{n})=f,$ a rational integral function of the elementary functions, is converted into

$$f(a_{1}+\mu ,a_{2}+\mu a_{1},\ldots a_{n}+\mu a_{n-1})=f+\mu d_{1}f+{\frac {\mu ^{2}}{2!}}{\overline {d_{1}^{2}}}f+{\frac {\mu ^{3}}{3!}}{\overline {d_{1}^{3}}}f+\ldots$$

where

$$d_{1}={\frac {\delta }{\delta a_{1}}}+a_{1}{\frac {\delta }{\delta a_{2}}}+a_{2}{\frac {\delta }{\delta a_{3}}}+\ldots +a_{n-1}{\frac {\delta }{\delta a_{n}}}$$

and ${\overline {d_{1}^{s}}}$ denotes, not $s$ successive operations of $d_{1},$ but the operator of order $s$ obtained by raising $d_{1}$ to the $s^{th}$ power symbolically as in Taylor’s theorem in the Differential Calculus.

Write also ${\frac {1}{s!}}{\overline {d_{1}^{s}}}={\text{D}},$ so that

$$f(a_{1}+\mu ,a_{2}+\mu a_{1},\ldots a_{n}+\mu a_{n-1})=f+\mu {\text{D}}_{1}f+\mu ^{2}{\text{D}}_{2}f+\mu ^{3}{\text{D}}_{3}f+\ldots .$$

The introduction of the quantity $\mu$ converts the symmetric function $(\lambda _{1}\lambda _{2}\lambda _{3}\ldots )$ into

$$(\lambda _{1}\lambda _{2}\lambda _{3}+\ldots )+\mu ^{\lambda _{1}}(\lambda _{2}\lambda _{3}\ldots )+\mu ^{\lambda _{2}}(\lambda _{1}\lambda _{3}\ldots )+\mu ^{\lambda _{3}}(\lambda _{1}\lambda _{2}\ldots )+\ldots .$$

Hence, if $f(a_{1},a_{2},\ldots a_{n})=(\lambda _{1}\lambda _{2}\lambda _{3}\ldots ),$

$$(\lambda _{1}\lambda _{2}\lambda _{3}\ldots )+\mu ^{\lambda _{1}}(\lambda _{2}\lambda _{3}\ldots )+\mu ^{\lambda _{2}}(\lambda _{1}\lambda _{3}\ldots )+\mu ^{\lambda _{3}}(\lambda _{1}\lambda _{2}\ldots )+\ldots$$ $$=(1+\mu {\text{D}}_{1}+\mu ^{2}{\text{D}}_{2}+\mu ^{3}{\text{D}}_{3}+\ldots )(\lambda _{1}\lambda _{2}\lambda _{3}\ldots ).$$

Comparing coefficients of like powers of $\mu$ we obtain

${\text{D}}\lambda _{1}(\lambda _{1}\lambda _{2}\lambda _{3}\ldots )=(\lambda _{2}\lambda _{3}\ldots ),$

while ${\text{D}}_{s}(\lambda _{1}\lambda _{2}\lambda _{3}\ldots )=0$ unless the partition $(\lambda _{1}\lambda _{2}\lambda _{3}\ldots )$ contains a part $s.$ Further, if ${\text{D}}_{\lambda _{1}}{\text{D}}_{\lambda _{2}}$ denote successive operations of ${\text{D}}_{\lambda _{1}}$ and ${\text{D}}_{\lambda _{2}},$

$${\text{D}}\lambda _{1}{\text{D}}\lambda _{2}(\lambda _{1}\lambda _{2}\lambda _{2}\ldots )=(\lambda _{3}\ldots ),$$

and the operations are evidently commutative.

Also ${\text{D}}_{p_{1}}^{\pi _{1}}{\text{D}}_{p_{2}}^{\pi _{2}}{\text{D}}_{p_{3}}^{\pi _{3}}\ldots (p_{1}^{\pi _{1}}p_{2}^{\pi _{2}}p_{3}^{\pi _{3}}\ldots )=1,$ and the law of operation of the operators ${\text{D}}$ upon a monomial symmetric function is clear.

We have obtained the equivalent operations

$$1+\mu {\text{D}}_{1}+\mu ^{2}{\text{D}}_{2}+\mu ^{3}{\text{D}}_{3}+\ldots ={\overline {exp}}\mu d_{1}$$

where ${\overline {exp}}$ denotes (by the rule over $exp$) that the multiplication of operators is symbolic as in Taylor’s theorem. $d_{1}^{s}$ denotes, in fact, an operator of order $s,$ but we may transform the right-hand side so that we are only concerned with the successive performance of linear operations. For this purpose write

$$a_{s}=\partial _{a_{s}}+a_{1}\partial _{a_{s+1}}+a_{2}\partial _{a_{s+2}}+\ldots .$$

It has been shown (vide ”Memoir on Symmetric Functions of the Roots of Systems of Equations,” Phil. Trans. 1890, p. 490) that

$${\overline {exp}}(m_{1}d_{1}+m_{2}d_{2}+m_{3}d_{3}+\ldots )=exp(M_{1}d_{1}+M_{2}d_{2}+M_{3}d_{3}+\ldots ),$$

where now the multiplications on the dexter denote successive operations, provided that

$$exp({\text{M}}_{1}\xi +{\text{M}}_{2}\xi ^{2}+{\text{M}}_{3}\xi ^{3}+\ldots )=1+m_{1}\xi +m_{2}\xi ^{2}+m_{3}\xi ^{3}+\ldots ,$$

$\xi$ being an undetermined algebraic quantity.

Hence we derive the particular cases

$${\overline {exp}}d_{1}=exp(d_{1}-{\frac {1}{2}}d_{2}+{\frac {1}{3}}d_{3}-\ldots );$$ $${\overline {exp}}\mu d_{1}=exp(\mu d_{1}-{\frac {1}{2}}\mu ^{2}d_{2}+{\frac {1}{3}}\mu ^{3}d_{3}-\ldots ),$$

and we can express ${\text{D}}_{s}$ in terms of $d_{1},d_{2},d_{3},\ldots ,$ products denoting successive operations, by the same law which expresses the elementary function $a_{s}$ in terms of the sums of powers $s_{1},s_{2},s_{3},\ldots$ Further, we can express $d_{s}$ in terms of ${\text{D}}_{1},{\text{D}}_{2},{\text{D}}_{3},\ldots$ by the same law which expresses the power function $s,$ in terms of the elementary functions $a_{1},a_{2},a_{3},\ldots$

Operation of ${\text{D}}_{s}$ upon a Product of Symmetric Functions.—Suppose $f$ to be a product of symmetric functions $f_{1}f_{2}\ldots f_{m}.$ If in the identity $f=f_{1}f_{2}\ldots f_{m}$ we introduce a new root $\mu$ we change $a_{s}$ into $a_{s}+\mu a_{s-1},$ and we obtain

$${\begin{aligned}(1&+\mu {\text{D}}_{1}+\mu ^{2}{\text{D}}_{2}+\ldots +\mu ^{s}{\text{D}}_{s}+\ldots )f\\=(1&+\mu {\text{D}}_{1}+\mu ^{2}{\text{D}}_{2}+\ldots +\mu ^{s}{\text{D}}_{s}+\ldots )f_{1}\\\times \,(1&+\mu {\text{D}}_{1}+\mu ^{2}{\text{D}}_{2}+\ldots +\mu ^{s}{\text{D}}_{s}+\ldots )f_{2}\\\times \,\quad &\cdot \qquad \quad \;\cdot \qquad \quad \;\cdot \qquad \quad \;\cdot \qquad \quad \;\cdot \\\times \,(1&+\mu {\text{D}}_{1}+\mu ^{2}{\text{D}}_{2}+\ldots +\mu ^{s}{\text{D}}_{s}+\ldots )f_{m}\end{aligned}}$$

and now expanding and equating coefficients of like powers of $\mu$

$${\begin{aligned}&{\text{D}}_{1}f=\Sigma ({\text{D}}_{1}f_{1})f_{2}f_{3}\ldots f_{m},\\&{\text{D}}_{2}f=\Sigma ({\text{D}}_{2}f_{1})f_{2}f_{3}\ldots f_{m}+\Sigma ({\text{D}}_{1}f_{1})({\text{D}}_{1}f_{2})f_{3}\ldots f_{m},\\&{\text{D}}_{3}f=\Sigma ({\text{D}}_{3}f_{1})f_{2}f_{3}\ldots f_{m}+\Sigma ({\text{D}}_{2}f_{1})({\text{D}}_{1}f_{2})f_{3}\ldots f_{m}+\Sigma ({\text{D}}_{3}f_{1})f_{2}f_{3}\ldots f_{m},\\&\cdot \qquad \quad \;\;\cdot \qquad \quad \;\cdot \qquad \quad \;\;\cdot \qquad \quad \;\;\cdot \qquad \quad \;\;\cdot \qquad \quad \;\cdot \qquad \quad \;\cdot \qquad \quad \;\;\cdot \end{aligned}}$$

the summation in a term covering every distribution of the operators of the type presenting itself in the term.

Writing these results

$${\begin{aligned}&{\text{D}}_{1}f={\text{D}}_{(1)}f,\\&{\text{D}}_{2}f={\text{D}}_{(2)}f+{\text{D}}_{(1^{2})}f,\\&{\text{D}}_{3}f={\text{D}}_{(3)}f+{\text{D}}_{(2^{1})}f+{\text{D}}_{(1^{2})}f,\end{aligned}}$$