Hagen.) 100 [April 6, 
the equation (1) are to be considered, since our equation (18) of the mth 
degree admits of m solutions. And indeed, equation (13) being no more 
an identity, we cannot say, as we did in §8, that a and y will be zero at the 
same time. 
Consequently the condition (1) is be taken in its full extent and, since 
A, = 0, may be written this way : } 
r=m 
Di bey en wer (14) 
r= 1 
This equation is at once resolved into the following two : 
zr m. 
By == 0 amd 2 sae tel =e. 0, (14/) 
r=1 | 
thus showing that the solution of the equation of the mth degree is made de- 
pendent on the solution of an equation of the (m — 1)th degree, as has been 
remarked also by Prof. Schlémilch on page 26 of his article referred to. 
For each root By the formulas (7) and (8) will furnish a different set of co- 
éefticients B, By... . and consequently a different value of y, and the for- | 
mulas (9/) give at once the value of "4 for the root By) = 0: 
1 1 I 
=+-— a——; A, a (2 Aj? — A, A) a? — (5 A’— 
v A, AS Ag a} - ‘ 6 2 1 hg, Ay 2 
5 A, A, A; + Ay? Ay) aft... (15) 
As we have already noted in §2, this method is applicable especially 
to such series in which the first coefficient A, is large in comparison 
to the following ones, 
Hirst ecample.—Let the given equation be 
yo+ioy+1=0. 
Here we have A, = 1, A, = 10, a==-— 1, and from (14’) we get the 
two conditions By = 0 and B, =— 10. By the first we obtain from (15) 
1 i 2 5 
it gel an se 1010208 2... 
v1 i0 108 108 io? a 
By the second condition B, = — 10, we compute from (‘7) 
= — 10, 3, = 1, 3, = 0, ete.; 
and by means of these values from (8) or (9) 
uf 1 2 
a — i a i ow, 5, " \. 
= 07 Bs=-jor? Be iT ei jor? °° 
and finally we have from (13) 
1 1 2 5 
=—104+— — — + ,,.==—9,8980705... 
Ya ae ee he ge tee sdeeSeac as 
A proof of this calculation is found in that y, +- y, is equal to the nega- 
tive cofficient of y. 
Second ewample.—Let it be required to solve the equation of the fourth 
degree 
yt — Ay? —25 y? + 100 y + 1 = 0, 
Here we have A,=1, A, =—4, A, = — 25, A, = 100, a= —1, 
The equations (14) are now 
B, = Oand B,’ — 4 B,? — 26 B, + 100 = 0. 
