An introduction to the mathematical theory of attraction ...

Examples. 255
Substituting in (h) and reducing, we obtain
y = log2 - 2 {1 ( 3)3S3 + i (i)5S + &c.} = 0-57712.
By (27), Art. 124,
Lab 1 1
E~a b = nb + 1 n-\~] a a+ b + - n?-l
since - - b. Hence by (d) and (e) we have
Agai, =- - = - ia y - L log r (1 - t ) }
+b:dp b ni a+ b du )
ab 7rb
= 7rL L c Cot
a+ b a + b\
5. If two spheres (A) and (B) intersect, and JT be the point which is the
image in (B) of A the centre
of (A), If the image of Ji
in (-A), J2 the image of ri
^-^./ \in B, and so on, and if a
C \ charge e be placed at A,
and charges, which are the
successive images of e in
L 4 J B a(B) and (A), at Ji, Il, J2,
A J Ji I2B / fI2,&c., find expressions for
\/1 /y- Ethe sum /a, of the charges
\, < /r at A, I1, I2, &c., and the
sumn Eb of the charges at
J1, J2, &c.
If we denote the charges
at I1, Ji, &c. by il, j, &c.,
e, -il il -j2
we have = - = CI = _ -&.,
C-A CJî C.i CTi
where C is a point common to the two spheres.
Again, if Jn, 1,, 7n+,? and I,,4, be successive images, it is easy to see that
BJn+C = BOCI = BCA - ACI = B -A - AJCnG,
and that
AI.+lC = ACJ,+l = BGA - BCJn.+ = BUA - BInC,