8 
PROFESSOR K. PEARSON ON MATHEMATICAL CONTRIBUTIONS 
_ gP 2 f g-jptanfl - i-sec 6? C JQ ~| 
Jo ■ 
re ! 
_ Q W j g-I(i- tan e - 7i see S) 2 
(xxii.), 
if r = sin 6 . 
Now either of the quantities under the sign of integration in (xxii.) can be expanded 
in powers of 6 by Maclaurin’s theorem. Thus let 
X 
e 
—l(k tan 6 — h sec 0) 2 
— Xo + 
tl x 
cie 
/d?x\ ej 
0 + wm +■ ■ • + 
d n x\ 0 n 
d6 7o 
+ • 
Then 
d x\ & , f d2 X\ V s 
£ - el '1 * 0 ® + ( f (,) 0 \2 + [ w ) 0 J 3 + • ' ' + Oft 
d» x \ 6 n+[ 
dd n l Q |^ + 1 
+ 
f cl n y 
and it remains to find ( ~~' a ) . 
Now 
Hence 
log y = — (k tan 6 — h sec 6) 2 . 
cos 3 0 Cl f e = - x [(A 3 + A 2 ) sin 6 - hk (|-l cos 20)}. 
Differentiating n — 1 times by Leibnitz’s theorem, and putting 6 = 0, 
' d n y\ , 
41 m - +■■■ 
+ 
(n — 1) . . . (n — r + 1) 
I r — 1 
21 »• 
+ sinTz-U-lf^l + . . . = 0 
‘ 2 ! \ dd n ~ r n 
cos r -Z\- --(3 + 3') - 4(/i 2 -hF) 
(xxiii.). 
Clearly y 0 = e - h \ then we rapidly find 
d A = hk ^ 
(%) = - (/r+F-7« 2 )e-!'- 
V de 2 / 
(*x\ 
\ dey 
; = hk [hW - 3(A 2 + ¥■) + 5}. 
Or, finally 
c = 0 + \hW - (A 2 + A 2 - A 2 A 2 ) ^ + hk{h z k 2 - 3 (A 2 + A 2 ) + 5} £ + . . . (xxiv.), 
where more terms if required can be found by (xxiii.). If 6 be fairly small, 6 ° will be 
negligible. Or if h and k be small, the lowest term in the next factor wall be A 2 -f A 2 , 
