( 489 ) 
9 
: yw ae ; : . 
region of 2 positive or negative for the points of the tangent. For 
Vv 
dp 
the points above the tangent, however, ar 0 disappears where 
av 
dp. 
zond negative, and the reverse for the points below the tangent. 
ae 
2 
d 
But let us try to answer the question where en 
Ak 
= 0 disappears 
for arbitrary value of x. The reiation between n, 2, and y (Contri- 
bution III) is, indeed, a very intricate one — but to my astonishment 
it proved to be possible to find an answer by a comparatively simple 
reduction. If we start from equation (4) of Contribution III, we 
may write: 
1 a xv (le) 
n—l je agens 4H} 
and 
n x (le) 
Be ear el 
mn oe 
If we take the square of the first of these equations, and then 
divide by « — and the square of the second of these equations and 
then divide by 1 —, the sum of the two values obtained yields: 
ee t n° e(l—e) 
ra TE Sl maha 
wv (n—1) 1—z(n—1) (1— 22) 
yee CE js 
For the second member may also be written 1 sand the 
y 
2 
d EEP 
condition whether — is positive or negative for the point in which 
Vv 
d : 
ne =0 disappears, becomes then for the points below the tangent: 
av” 
1 PENS 
1 (1-4) SI 
eg 
derd 
In this equation we have a—=1 for the origin and a =0 for the 
tangent itself. With a—1 we find as condition: 
v +3) 2 —y)? 
1 
For OE which belongs to mo, the first member of the 
3 1 
inequality is 7 and the second member ris So, as we found above, 
