Page:Scientific Memoirs, Vol. 1 (1837).djvu/470

$Q_{i}={\frac {1}{r}}Q_{i}^{(1)}-{\frac {1}{i}}{\frac {dQ_{i}^{(1)}}{dr}}$

the last of these quantities will satisfy the equation (4), and will be its complete integral.

If the successive substitutions are performed, and, for brevity's sake, we make

$a_{i}^{(i)}={\frac {(-1)^{i}}{1.2.3..i}};\qquad a_{i}^{(i-\prime )}=\left[-{\frac {1}{i}}\right]^{i}\Sigma {\frac {\left(1+{\frac {\prime -1}{i+1}}\right)a_{(i-1)}^{(i-\prime )}}{\left[-{\frac {1}{i+1}}\right]^{i+1}}}\colon$[1]

which gives, in the particular case of $\prime =i$, $a_{i}^{(0)}=\left[1+{\frac {i-1}{i}}\right]^{i}$, we shall have

$Q_{i}={\frac {a_{i}^{(0)}}{r^{i}}}Q_{i}^{(i)}+{\frac {a_{i}^{(1)}}{r^{i-1}}}{\frac {dQ_{i}^{(i)}}{dr}}+{\frac {a_{i}^{(2)}}{r^{i-2}}}{\frac {d^{2}Q_{i}^{(1)}}{dr^{2}}}\cdots \cdots \cdots +a_{i}^{(i)}{\frac {d^{i}Q_{i}^{(i)}}{dr^{i}}}.$

Now if we make

${\begin{aligned}\Omega _{i}(r')&=\left\{{\frac {a_{i}^{(0)}}{r^{\prime i}}}+{\frac {a_{i}^{(i)}}{r^{\prime i-1}}}\alpha +{\frac {a^{(2)}}{r^{\prime i-1}}}\alpha ^{2}\cdots \cdots +{\frac {a_{i}^{(i)}}{r^{\prime 0}}}\alpha ^{i}\right\}e^{\alpha r'},\\\Omega '_{i}(r')&=\left\{{\frac {a_{i}^{(0)}}{r^{\prime i}}}-{\frac {a_{i}^{(1)}}{r^{\prime i-1}}}\alpha +{\frac {a_{i}^{(2)}}{r^{\prime i-2}}}\alpha ^{2}\cdots \cdots +{\frac {a_{i}^{(i)}}{r^{\prime 0}}}\alpha ^{i}\right\}e^{-\alpha r'},\end{aligned}}$

where $\alpha$ is put instead of ${\sqrt {\frac {4\pi f}{k}}}$ we shall have

$Q_{i}^{'}=T_{i}^{'}\Omega _{i}(r')+V_{i}^{'}\Omega '(r'),$

and the expression for $F$ may take the form

(5)

${\begin{aligned}F&=\Sigma _{0}^{\infty }\iint f\left\{{\frac {1}{r^{\prime n+1}}}\int _{0}^{r}\Sigma _{0}^{\infty }\Omega _{i}(r')T'_{i}r^{\prime n+1}\,dr'\right.\\&+\left.r^{n}\int _{0}^{\infty }\Sigma _{0}^{\infty }{\frac {\Omega _{i}(r')T'_{i}\,dr'}{r^{\prime n}}}\right\}P_{n}\sin \theta '\,d\theta '\,d\psi '\end{aligned}}$

1. The brackets are here employed in the same way as in Vandermonde's notation.