The power series expansion of the function f(x)= an(x- c)n is given as,
∑0∞an(x - c)n=a0+ a1(x- c)+ a2(x- c)2+.....
Here, c is the constant and an the coefficient of the nth term.
From the equation 7.6.7
x(t,ε) = (1- ε2)-1/2e-εtsin[(1-ε2)1/2t]
By differentiating the above equation with respect to ε,
dx(t,ε)dε = d((1- ε2)-1/2e-εtsin[(1-ε2)1/2t])dε
dx(t,ε)dε = e-εtsin[(1-ε2)1/2t]d((1- ε2)-1/2)dε+(1- ε2)-1/2d(e-εtsin[(1-ε2)1/2t])dε
dx(t,ε)dε = e-εtsin[(1-ε2)1/2t](−12(1- ε2)-3/2(−2ε))+(1- ε2)-1/2d(e-εtsin[(1-ε2)1/2t])dε
dx(t,ε)dε = e-εtsin[(1-ε2)1/2t](ε)(1- ε2)-3/2+(1- ε2)-1/2d(e-εtsin[(1-ε2)1/2t])dε
dx(t,ε)dε = e-εtsin[(1-ε2)1/2t](ε)(1- ε2)-3/2+(1- ε2)-1/2e-εtd(sin[(1-ε2)1/2t])dε+(1- ε2)-1/2sin[(1-ε2)1/2t]d(e-εt)dε
dx(t,ε)dε = e-εtsin[(1-ε2)1/2t](ε)(1- ε2)-3/2+(1- ε2)-1/2e-εtcos[(1- ε2)1/2t]t12(1-ε2)-1/2(−2ε) + (1- ε2)-1/2sin[(1-ε2)1/2t]e-εt(-t)
dx(t,ε)dε = e-εtsin[(1-ε2)1/2t](ε)(1- ε2)-3/2+(1- ε2)-1/2e-εtcos[(1- ε2)1/2t]t(1-ε2)-1/2(−ε) + (1- ε2)-1/2sin[(1-ε2)1/2t]e-εt(-t)
dx(t,ε)dε = e-εtsin[(1-ε2)1/2t](ε)(1- ε2)-3/2+(1- ε2)-1e-εtcos[(1- ε2)1/2t]t(1-ε2)-1/2(−ε) + (1- ε2)-1/2sin[(1-ε2)1/2t]e-εt(-t)
dx(t,ε)dε = ε(1- ε2)-3/2e-εtsin[(1-ε2)1/2t]−εt(1- ε2)-1e-εtcos[(1- ε2)1/2t] - t(1- ε2)-1/2e-εtsin[(1-ε2)1/2t]
The value of x(t,ε) at ε→0 is calculated as,
Substitute 0 for ε in the above expression x(t,ε)
x(t,0) = sin t
Value of dx(t,ε)dε, as ε→0
d(x(t, 0))dε=-t sin t
The power series expansion of x(t,ε) is given as:
x(t,τ)= x(t,ε)|ε=0+εdx(t,ε)dε|ε=0+Ο(ε2)
x(t,τ) = x(t,0)+εdx(t, 0)dε+Ο(ε2)
x(t,τ) = sin t+ε(-t sin t)+Ο(ε2)
x(t,τ) = sin t - εt sin t+Ο(ε2)
Hence, the power series expansion is x(t,τ) = sin t - εt sin t+Ο(ε2) for the given equation x(t,ε) = (1- ε2)-1/2e-εtsin[(1-ε2)1/2t].