Question 8.2.3: Find the Taylor series expansion and radius of convergence o...

If we were to attempt to find the series via the definition by taking derivatives, we would find that the process becomes laborious after computing f (t )=1/(1+t^{2}), since differentiating will involve both the chain and quotient rules. Instead, we observe that

f^{\prime}(t ) = \frac{1}{1+t^{2}}

itself has a series expansion that is not difficult to find. Similar to our work in example 8.2.2, we use the final result in (8.2.4) and substitute −t 2 for t to write

e^{t} = 1+t + \frac {t^{2}}{2!} + \frac {t^{3}}{3!}+· · ·+ \frac {t^{n}}{n!}+· · ·                       R=∞
sin t = t − \frac {t^{3}}{3!}+ \frac {t^{5}}{5!}−· · ·+(−1)^{n+1} \frac {t^{2n+1}}{(2n+1)!}+· · ·                   R=∞
cos t = 1− \frac {t^{2}}{2!}+ \frac {t^{4}}{4!}−· · ·+(−1)^{n} \frac {t^{2n}}{(2n)!}+· · ·                     R=∞
\frac {1}{1−t}= 1+t +t^{2} +t^{3}+· · ·+t^{n} +· · ·                R = 1                          (8.2.4)

f^{\prime}(t ) = \frac {1}{1−(−t^{2})}= 1+(−t^{2})+(−t^{2})^{2} +(−t^{2})^{3}+· · ·+(−t^{2})^{n} +· · ·
= 1−t^{2} +t^{4} −t^{6}+· · ·+(−1)^{n}t^{2n} +· · ·           (8.2.7)

Because we now have a series expansion for f^{\prime}(t ), it is natural to integrate both sides of (8.2.7) to find the series for f (t ). Doing so, we see that

f (t ) = arctan t = C +t − \frac{1}{3}t^{3} + \frac{1}{5}t^{5} − \frac {1}{7}t^{7}+· · ·+ \frac{(−1)^{n}}{2n+1}t^{2n+1} · · ·        (8.2.8)

It is a straightforward exercise to use the Ratio Test to show that (8.2.8) converges for all t such that |t | < 1. Moreover, since arctan(0) = 0, it follows that C = 0.