Page:EB1911 - Volume 01.djvu/666

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

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

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

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

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

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

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

For, writing as before,

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

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

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

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

The law of reciprocity shows that

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

viz.: a linear function of symmetric functions symbolized by the ${\displaystyle k}$ specifications; and that ${\displaystyle \theta _{st}=\theta _{ts}.}$ A table may be formed expressing the ${\displaystyle k}$ expressions ${\displaystyle {\text{P}}^{(1)},{\text{P}}^{(2)},\ldots {\text{P}}^{(k)}}$ as linear functions of the ${\displaystyle k}$ expressions ${\displaystyle (m_{1s}^{\mu _{1s}}m_{2s}^{\mu _{2s}}m_{3s}^{\mu _{3s}}\ldots )}$, ${\displaystyle s=1,2,\ldots k}$, and the numbers ${\displaystyle \theta _{st}}$ occurring therein possess row and column symmetry. By solving ${\displaystyle k}$ 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 ${\displaystyle (m_{1s}^{\mu _{1s}}m_{2s}^{\mu _{2s}}m_{3s}^{\mu _{3s}}\ldots )}$ whose partition is a specification of a separation of the function symbolized by ${\displaystyle (l_{1}^{\lambda _{1}}l_{2}^{\lambda _{2}}l_{3}^{\lambda _{3}}\ldots )}$ is expressible as a linear function of symmetric functions symbolized by separations of ${\displaystyle (l_{1}^{\lambda _{1}}l_{2}^{\lambda _{2}}l_{3}^{\lambda _{3}}\ldots )}$ and a symmetrical table may be thus formed.” It is now to be remarked that the partition ${\displaystyle (l_{1}^{\lambda _{1}}l_{2}^{\lambda _{2}}l_{3}^{\lambda _{3}}\ldots )}$ can be derived from ${\displaystyle (m_{1s}^{\mu _{1s}}m_{2s}^{\mu _{2s}}m_{3s}^{\mu _{3s}}\ldots )}$ by substituting for the numbers ${\displaystyle m_{1s},m_{2s},m_{3s},\ldots }$ 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.

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

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

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

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

Theorem of Symmetry.—If we form the separation function

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

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

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

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

The Differential Operators.—Starting with the relation

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

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

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

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

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

where

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

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

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

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

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

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

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

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

Comparing coefficients of like powers of ${\displaystyle \mu }$ we obtain

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

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

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

and the operations are evidently commutative.

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

We have obtained the equivalent operations

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

where ${\displaystyle {\overline {exp}}}$ denotes (by the rule over ${\displaystyle exp}$) that the multiplication of operators is symbolic as in Taylor’s theorem. ${\displaystyle d_{1}^{s}}$ denotes, in fact, an operator of order ${\displaystyle s,}$ 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

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

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

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

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

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

${\displaystyle \xi }$ being an undetermined algebraic quantity.

Hence we derive the particular cases

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

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

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

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

and now expanding and equating coefficients of like powers of ${\displaystyle \mu }$

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

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

Writing these results

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