Question 8.2.4: Find the first four terms of the Taylor series expansion abo...

Because f is the quotient of two functions that are analytic everywhere and the denominator is never zero, it follows that f is analytic everywhere. In particular, f is analytic at a = 0 and, therefore, has a Taylor series expansion there of the form

\frac {t}{e^{t} +1}= b_{0} +b_{1}t +b_{2}t^{2} +b_{3}t^{3}+· · ·              (8.2.12)

We know from the standard expansion of e^{t} that

e^{t} +1 = 2+t + \frac {t^{2}}{2!}+ \frac {t^{3}}{3!}+· · ·

Multiplying both sides of (8.2.12) by this expression for e^{t}+1, we obtain the identity

t=(2+t + \frac {t^{2}}{2!}+ \frac {t^{3}}{3!}+· · ·)( b_{0} +b_{1}t +b_{2}t^{2} +b_{3}t^{3}+· · ·)

Distributing to multiply these two series, we find that

t = 2b_{0} +(2b_{1} +b_{0})t +(2b_{2} +b_{1} + \frac {b_{0}}{2})t^{2} +(2b_{3} +b_{2} + \frac {b_{1}}{2}+ \frac {b_{0}}{6})t^{3}+· · ·

In order for this identity to hold, the uniqueness of Taylor series expansions established in theorem 8.2.2 implies that all of the coefficients of powers of t on the left must equal the corresponding coefficients of powers of t on the right. In particular, it must be the case that

0 = 2b_{0}
1 = 2b_{1} +b_{0}
0 = 2b_{2} +b_{1} + \frac {1}{2}b_{0}
0 = 2b_{3} +b_{2} + \frac {1}{2}b_{1} + \frac {1}{6}b_{0}

From this sequence of equalities, it follows that b_{0} = 0, b_{1} = 1/2, b_{2} =−1/4, and  b_{3} = 0, so that

f(t)=\frac {t}{e^{t} +1}=\frac {1}{2}t − \frac {1}{4}t^{2} + 0t^{3}+· · ·