1908. No. 1. ON THE GRAPHIC SOLUTION OF DYNAMICAL PROBLEMS. 7 
n 
we also obtain for tan $ the value å If we compare these two expres- 
sions for tan 3, we obtain 
op =pD Q.E.D. 
It is easy to find an expression for the accuracy attained, for, accord- 
ing to Taylor's formula, we have 
r —f( raved 2 BEE, +0.40) : 
mee 4U)—7(U,)+-f' (U) A0 + SS 0) (AUS SE u (AU)3 
and 
rv 0) 40420 cup — TUI A) go, 
X 2 1X2X3 
where 4 and # are numbers between o and 1. From this we get 
LAU) f(U,—4U)=2f’(U,) AU +4 f(y +0.4U) +f" (U,—0'.AU)](A0)%. 
If we now choose U, =? and JU = = we obtain 
BO I m re wer p—0 
et Bie [ | n )+/ | n ilk 
e=pD+e, 
where we have put 
‘111 + 6 wer =. 0 
ae PE 
p 
4 PF. r == ===" I . 
Now if f’’(U) in the interval eat == , lies between — m 
n 
and —+ m, then 
and thus 
NS As 
NG 378 m. 
3 
In order to find m, f’” (U) may be expressed by ae va and Ger - 
As E = f(U), we have 
fe a dU d8U 
3 d= des 
