Page:EB1911 - Volume 01.djvu/666

 $$(l^{\lambda_1}_1 l^{\lambda_2}_2 \ldots)$$ and it is seen intuitively that the number $$\theta$$ remains unaltered when the first two of these partitions are interchanged (see ). 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,

$$ \begin{align} (1^{m_1})^{\mu_1} (1^{m_2})^{\mu_2} (1^{m_3})^{\mu_3} \ldots &= \ldots + \theta (s^{\sigma_1}_1 s^{\sigma_2}_2 s^{\sigma_3}_3 \ldots) + \ldots, \\ (1^{s_1})^{\sigma_1} (1^{s_2})^{\sigma_2} (1^{s_3})^{\sigma_3} \ldots &= \ldots + \theta (m^{\mu_1}_1 m^{\mu_2}_2 m^{\mu_3}_3 \ldots) + \ldots; \end{align} $$

and another by putting $$x_1 = x_2 = x_3 = \ldots = 1$$, for then $$\Chi_m$$ becomes $$h_m$$, and we have

$$ h^{\mu_1}_{m_1} h^{\mu_2}_{m_2} h^{\mu_3}_{m_3} \ldots = \ldots + \theta^\prime (s^{\sigma_1}_1 s^{\sigma_2}_2 s^{\sigma_3}_3 \ldots) + \ldots,$$ $$ h^{\sigma_1}_{s_1} h^{\sigma_2}_{s_2} h^{\sigma_3}_{s_3} \ldots = \ldots + \theta^\prime (m^{\mu_1}_1 m^{\mu_2}_2 m^{\mu_3}_3 \ldots) + \ldots,$$

Theorem of Expressibility.—“If a symmetric function be symboilizedsymbolized [sic] 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,

$$ \Chi^{\mu_1}_{m_1} \Chi^{\mu_2}_{m_2} \Chi^{\mu_3}_{m_3} \ldots = \Sigma \Sigma \theta (s^{\sigma_1}_1 s^{\sigma_2}_2 s^{\sigma_3}_3 \ldots) x^{\lambda_1}_{l_1} x^{\lambda_2}_{l_2} x^{\lambda_3}_{l_3} \ldots,$$ $$ = \Sigma \text{P} x^{\lambda_1}_{l_1} x^{\lambda_2}_{l_2} x^{\lambda_3}_{l_3} \ldots,$$

$$\text{P}$$ is a linear function of separations of $$(l^{\lambda_1}_1 l^{\lambda_2}_2 l^{\lambda_3}_3 \ldots)$$ of specification $$(m^{\mu_1}_1 m^{\mu_2}_2 m^{\mu_3}_3 \ldots),$$ and if $$\Chi^{\sigma_1}_{s_1} \Chi^{\sigma_2}_{s_2} \Chi^{\sigma_3}_{s_3} \ldots = \Sigma \text{P}^\prime x^{\lambda_1}_{l_1} x^{\lambda_2}_{l_2} x^{\lambda_3}_{l_3} \ldots, \text{P}^\prime$$ is a linear function of separations of $$(l^{\lambda_1}_1 l^{\lambda_2}_2 l^{\lambda_3}_3 \ldots)$$ of specification $$( s^{\sigma_1}_1 s^{\sigma_2}_2 s^{\sigma_3}_3 \ldots).$$ Suppose the separations of $$(l^{\lambda_1}_1 l^{\lambda_2}_2 l^{\lambda_3}_3 \ldots)$$ to involve $$k$$ different specifications and form the $$k$$ identities

$$ \Chi^{\mu_{1s}}_{m_{1s}} \Chi^{\mu_{2s}}_{m_{2s}} \Chi^{\mu_{3s}}_{m_{3s}} \ldots = \Sigma \text{P}^{(s)} x^{\lambda_1}_{l_1} x^{\lambda_2}_{l_2} x^{\lambda_3}_{l_3} \ldots (s = 1, 2, \ldots k),$$

where $$m^{\mu_{1s}}_{1s} m^{\mu_{2s}}_{2s} m^{\mu_{3s}}_{3s} \ldots)$$ 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 $$(m^{\mu_{1s}}_{1s} m^{\mu_{2s}}_{2s} m^{\mu_{3s}}_{3s} \ldots)$$ whose partition is a specification of a separation of the function symbolized by $$(l^{\lambda_1}_1 l^{\lambda_2}_2 l^{\lambda_3}_3 \ldots)$$ is expressible as a linear function of symmetric functions symbolized by separations of $$(l^{\lambda_1}_1 l^{\lambda_2}_2 l^{\lambda_3}_3 \ldots)$$ and a symmetrical table may be thus formed.” It is now to be remarked that the partition $$(l^{\lambda_1}_1 l^{\lambda_2}_2 l^{\lambda_3}_3 \ldots)$$ can be derived from $$(m^{\mu_{1s}}_{1s} m^{\mu_{2s}}_{2s} m^{\mu_{3s}}_{3s} \ldots)$$ 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 $$(l^{\lambda_1}_1 l^{\lambda_2}_2 l^{\lambda_3}_3 \ldots)$$ of specification $$(m^{\mu_{1s}}_{1s} m^{\mu_{2s}}_{2s} m^{\mu_{3s}}_{3s} \ldots),$$ placing $$s$$ under the summation sign to denote the specification involved,

$$ \begin{align} \mu_{1s}! \mu_{2s}! \mu_{3s} \ldots &\sum_s \frac{(\text{J}_1)^{j_1} (\text{J}_2)^{j_2} (\text{J}_3)^{j_3} \ldots}{j_1! j_2! j_3! \ldots} = \sum^{t=k}_{t=1} \theta_{st} (m^{\mu_{1s}}_{1s} m^{\mu_{2s}}_{2s} m^{\mu_{3s}}_{3s} \ldots), \\ \mu_{15}! \mu_{2t}! \mu_{3t}! &\ldots \sum_t \frac{(\text{J}_1)^{j_1} (\text{J}_2)^{j_2} (\text{J}_3)^{j_3} \ldots}{j_1! j_2! j_3 \ldots} = \sum^{s=k}_{s=1} \theta_{ts} (m^{\mu_{1s}}_{1s} m^{\mu_{2s}}_{2s} m^{\mu_{3s}}_{3s} \ldots), \end{align} $$

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 $$(l^{\lambda_1}_1 l^{\lambda_2}_2 l^{\lambda_3}_3 \ldots),$$ each separation having a specification $$(m^{\mu_{1s}}_{1s} m^{\mu_{2s}}_{2s} m^{\mu_{3s}}_{3s} \ldots)$$, multiply by $$\mu_{1s}! \mu_{2s}! \mu_{3s}! \ldots,$$ and take therein the coefficient of the function $$(m^{\mu_{1t}}_{1t} m^{\mu_{2t}}_{2t} m^{\mu_{3t}}_{3t} \ldots),$$ we obtain the same result as if we formed the separation function in regard to the specification $$(m^{\mu_{1t}}_{1t} m^{\mu_{2t}}_{2t} m^{\mu_{3t}}_{3t} \ldots),$$ multiplied by $$\mu_{1t}! \mu_{2t}! \mu_{3t}! \ldots$$ and took therein the coefficient of the function $$(m^{\mu_{1s}}_{1s} m^{\mu_{2s}}_{2s} m^{\mu_{3s}}_{3s} \ldots).$$

Ex. gr., take $$(l^{\lambda_1}_1 l^{\lambda_2}_2 \ldots) = (21^4); (m^{\mu_{1s}}_{1s} m^{\mu_{2s}}_{2s} \ldots) = (321); (m^{\mu_{1t}}_{1t} m^{\mu_{2t}}_{2t} \ldots) = (31_3);$$ we find

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

The Differential Operators.—Starting with the relation

$$ (1+\alpha_1x) (1+\alpha_2x) \ldots (1+a_nx) = 1 + a_1x + a_2x^2 + \ldots + a_nx^n $$

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

$$ (1+a_1x) (1+a_2x) \ldots (1+a_nx) (1+\mu x) = 1+(a_1+\mu)x + (a_2+\mu a_1)x^2 + \ldots $$

so that $$f(a_1, a_2, a_3, \ldots a_n) = f,$$ a rational integral function of the elementary functions, is converted into

$$ f(a_1 + \mu, a_2 + \mu a_1, \ldots a_n + \mu a_{n - 1}) = f + \mu d_1f + \frac{\mu^2}{2!} \overline{d^2_1}f + \frac{\mu^3}{3!} \overline{d^3_1}f + \ldots $$

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 $$\overline{d^s_1}$$ 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 $$\frac{1}{s!} \overline{d^s_1} = \text{D},$$ so that

$$ f(a_1+\mu, a_2+\mu a_1, \ldots a_n+\mu a_{n-1}) = f + \mu \text{D}_1f + \mu^2 \text{D}_2f + \mu^3 \text{D}_3f + \ldots. $$

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 $$\text{D}^{\pi_1}_{p_1} \text{D}^{\pi_2}_{p_2} \text{D}^{\pi_3}_{p_3} \ldots (p^{\pi_1}_1 p^{\pi_2}_2 p^{\pi_3}_3 \ldots) = 1,$$ 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. $$d^s_1$$ denotes, in fact, an operator of order $$s,$$ but we may transform the 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 ”“ [sic]Memoir on Symmetric Functions of the Roots of Systems of Equations,” Phil. Trans. 1890, p. 490) that

$$ \overline{exp} (m_1d_1 + m_2d_2 + m_3d_3 + \ldots) = exp (M_1d_1 + M_2d_2 + M_3d_3 + \ldots), $$

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

$$ \begin{align} (1 &+ \mu \text{D}_1 + \mu^2 \text{D}_2 + \ldots + \mu^s \text{D}_s + \ldots)f \\ = (1 &+ \mu \text{D}_1 + \mu^2 \text{D}_2 + \ldots + \mu^s \text{D}_s + \ldots)f_1 \\ \times \, (1 &+ \mu \text{D}_1 + \mu^2 \text{D}_2 + \ldots + \mu^s \text{D}_s + \ldots)f_2 \\ \times \, \quad &\cdot \qquad \quad \; \cdot \qquad \quad \; \cdot \qquad \quad \; \cdot \qquad \quad \; \cdot \\ \times \, (1 &+ \mu \text{D}_1 + \mu^2 \text{D}_2 + \ldots + \mu^s \text{D}_s + \ldots)f_m \end{align} $$

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

$$ \begin{align} &\text{D}_1f = \Sigma (\text{D}_1f_1) f_2 f_3 \ldots f_m, \\

&\text{D}_2f = \Sigma (\text{D}_2f_1) f_2 f_3 \ldots f_m + \Sigma (\text{D}_1f_1) (\text{D}_1 f_2) f_3 \ldots f_m, \\

&\text{D}_3f = \Sigma (\text{D}_3f_1) f_2 f_3 \ldots f_m + \Sigma (\text{D}_2f_1) (\text{D}_1 f_2) f_3 \ldots f_m + \Sigma (\text{D}_3f_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{align} $$

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

Writing these results

$$ \begin{align} &\text{D}_1 f = \text{D}_{(1)}f, \\ &\text{D}_2 f = \text{D}_{(2)}f + \text{D}_{(1^2)}f, \\ &\text{D}_3 f = \text{D}_{(3)}f + \text{D}_{(2^1)}f + \text{D}_{(1^2)}f, \end{align} $$