A . ‘ F Ay d , 
vields that value of , for which -— assumes the given value 2. So 
IX 
we find: 
ee ae is er (a AD) (4,4 b) 
eee a, —Ab, (a,—A b,)? | 
If: 
(a,,—4 b,,)? — (a,—4 b,) (a,—2 b,) i 0, 
v . . “ye 
the quantity is complex. This cannot be the case if the value 
th 
nae 3 ° a, a, . fe ° . 
of 2 lies between — and —. It can only occur if 2 is chosen either 
), ) 
a, a, 
smaller than — and — or greater than both these values. 
) ) 
1 2 
If 2 is chosen such that 
Kie) (Hg SADE erheen OE A er ze AS 
then this equation will yield the minimum value of 2. If we put in 
this equation either: 
qe 
N a, 
eS ont = 
b, b, 
then the first member will be negative. If we put: 
then the sien of the first member will be the same as that of 
Cc 
. . . . “ye a . 
(a, 26,)(a,—ab,). The second member is positive, if a is smaller 
ia 
a a, Hi 
than — and also smaller than i Consequently a value of 4 must 
a My 
; : 3 In : ay Skee 
exist for which (1) =O and for which therefore 5, assumes a mini- 
) 
& 
: 4 a a vds AGE 
mum value. This value lies between — and —; or if “<= 
bs b, b, b, 
Ge. a, 
between — and —. 
bi, b, 
a a a 
15 . 2 . 
In the case that — is both greater than — and than —, the first 
ia b, b, 
2 
member of equation (1) changes its sign in the same way, and the 
value of 2 for which the first member vanishes, lies also between 
O19 eine So ot yy ‘ : at 
and — or if bn, between ; and . So for a minimum 
) D b, dy 2 ia 
