PAL 
n he a oe a 
es Je (Ifa ee 
0 
Comparing this result, with the former, we obtain the interesting 
formula . 
= tn Set glt f 
EERE MA LEL ee (14) 
(Ltr! EE 
0 0) 
which is evident if we put n= 0. 
From (13) we see also that 
ma go 1 4 1 ss @ 
2S iy, be ae 2 de fer dit 
0 1+¢ 9 \l+t 
0 
0 
which shows, that the expansion 
il “ 
—_ = = n Pn AC 
pps Dore) 
holds for # = 0. 
5. As a second example we will expand a discontinuous function. 
Supposing f(z) —1. from:2=0 to e= and /@)=0iorz a 
we have 
J (c) =a, +4, p, (2) + a, p, (©) + … 
i 
Ay = las Pr (a) da. 
0 
This coefficient may be determined in the following way. From 
the differential equation 
where 
d 
ma epa EN Ane Ep, (ay ne 
a“ 
it appears that 
T 
zet pn (x) + „fe pr (x) de = 0 
J 
therefore, putting w=1, we have 
1 
an) «> 
. ne . 
or, according to (5) 
1 
US Ee [Pu—1 (1) — gn (1)] (n > 0) 
