Example 12-10. Determination of series
Find a power series expansion to represent the function y=y(x) = (h+x)n, where
his a constant.
Solution Differentiate the function y=y(x) = (h+x)nand show
dy
dx =y′(x) = n(h+x)n− (^1) (12.164)
Multiply equation (12.164) by (h+x) and show y is a solution of the differential
equation
(h+x)dy
dx =n y (12.165)with initial condition at x= 0 given by y(0) = hn.Assume y= (h+x)n has the power
series representation
y= (h+x)n=c 0 +c 1 x+c 2 x^2 +c 3 x^3 +···+cmxm+··· (12.166)with derivative
dy
dx =c^1 + 2 c^2 x+ 3c^3 x(^2) +··· +mcmxm+··· (12.167)
Note that the index mhas been selected for the general term of the series as the
value noccurs in the differential equations and we don’t want these values to become
confused with one another. Substitute the power series (12.166) and (12.167) into
the differential equation (12.165) to obtain
(h+x)[
c 1 + 2c 2 x+ 3c 3 x^2 +···+mcmxm−^1 +···] = n[c 0 +c 1 x+c 2 x^2 +··· +cmxm+···](12.168)Expand the lefthand side of equation (12.168) and show
hc 1 +2hc 2 x+ 3 hc 3 x^2 +···+hmcmxm−^1 +···
+c 1 x+ 2c 2 x^3 + 3c 3 x^3 +···+mcmxm+···
= n c 0 +n c 1 x+n c 2 x^2 +n c 3 x^3 +···+n cmxm+···(12.169)In equation (12.169) equate the coefficients of like powers of xto obtain a recurrence
relation or recurrence formula. One finds
hc 1 =n c 0
(2 hc 2 +c 1 ) = n c 1
(3 hc 2 + 2 c 2 ) = n c 2..
.
..
.
[(m+ 1)hcm+1 +ncm] = n cm(12.170)