Thence we obtain
$\{1-\!a_{1}y+\!a_{2}y^{2}\!-\!\ldots \!+\!(-)^{n}\!a_{n}y^{n}\!\}(1+\!h_{1}y+\!h_{2}y^{2}\!+\!\ldots \!+\!h_{w}y^{w}\!\!+\!\ldots )\!=\!1$.

Since this is an identity we may multiply out the left-hand side and
equate the coefficients of the successive powers of $y$ to zero; obtaining
${\begin{array}{l}h_{1}-a_{1}=0{\text{,}}\\h_{2}-a_{1}h_{1}+a_{2}=0{\text{,}}\\h_{3}-a_{1}h_{2}+a_{2}h_{1}-a_{3}=0{\text{,}}\\\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \\h_{n}-a_{1}h_{n-1}+a_{2}h_{n-2}-\ldots +(-)^{n}a_{n}=0{\text{,}}\\h_{n+1}-a_{1}h_{n}+a_{2}h_{n-1}-\ldots +(-)^{n}a_{n}h_{1}=0{\text{,}}\\h_{n+2}-a_{1}h_{n+1}+a_{2}h_{n}-\ldots +(-)^{n}a_{n}h_{2}=0{\text{,}}\\\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \end{array}}$
relations which enable us to express any function $h_{w}$ in terms of
members of the series $a_{1},a_{2},a_{3},\ldots a_{n}$.

7. In the applications to combinatory analysis it usually happens that
we may regard $n$ as being indefinitely great and then the relations are
simply ${\begin{array}{l}h_{1}-a_{1}=0{\text{,}}\\h_{2}-a_{1}h_{1}+a_{2}=0{\text{,}}\\h_{3}-a_{1}h_{2}+a_{2}h_{1}-a_{3}=0{\text{,}}\\\ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \end{array}}$ continued indefinitely.

The before-written identity now becomes
$(1\!-\!a_{1}y\!+\!a_{2}y^{2}\!\!-\!a_{3}y^{3}\!\!+\!\ldots \!{\mathit {ad\ inf.}})$$(1\!+\!h_{1}y\!+\!h_{2}y^{2}\!\!+\!h_{3}y^{3}\!\!+\!\ldots \!{\mathit {ad\ inf.}})\!\equiv \!1$,
and herein writing $-y$ for $y$ and transposing the factors we find
$(1\!-\!h_{1}y\!+\!h_{2}y^{2}\!\!-\!h_{3}y^{3}\!\!+\!\ldots \!{\mathit {ad\ inf.}})$$(1\!+\!a_{1}y\!+\!a_{2}y^{2}\!\!+\!a_{3}y^{3}\!\!+\!\ldots \!{\mathit {ad\ inf.}})\!\equiv \!1$,
an identity which is derivable from the former by interchange of the
symbols $a$ and $h$.

There is thus perfect symmetry between the symbols and it follows
as a matter of course that in any relation connecting the quantities
$a_{1},a_{2},a_{3},\ldots$ with the quantities $h_{1},h_{2},h_{3},\ldots$ we are at liberty to
interchange the symbols $a$, $h$. This interesting fact can be at once verified in
the case of the relations $h_{1}-a_{1}=0$, etc.

Solving these equations we find
${\begin{array}{l}h_{1}=a_{1}{\text{,}}\\h_{2}={a_{1}}^{2}-a_{2}{\text{,}}\\h_{3}={a_{1}}^{3}-2a_{1}a_{2}+a_{3}{\text{,}}\end{array}}$