Question 8.2.1: Find the Taylor series of f (t ) = ln(1+t ) centered at t = ...

We begin by taking the first several derivatives of f and evaluating them at 0:

f (t ) = ln(1+t )            f (0) = ln(1) = 0
f\prime(t ) = (1+t )^{−1}                                f\prime (0) = 1
f\prime \prime (t ) = (−1)(1+t )^{−2}                     f \prime \prime(0) = −1
f\prime \prime \prime(t ) = (−2)(−1)(1+t )^{−3}           f \prime \prime \prime(0) = 2!
f^{(4)}(t ) = (−3)(−2)(−1)(1+t )^{−4}                 f^{(4)}(0) = −3!

From these calculations, we see that the fourth Taylor polynomial is

P(t ) = 0+1t − \frac {1}{2!} t^{2} + \frac {2!}{3!} t^{3} − \frac {3!}{4!} t^{4}
= t − \frac {1}{2}t^{2} + \frac {1}{3}t^{3} − \frac {1}{4}t^{4}

The established pattern implies that the Taylor series of f (t ) = ln(1+t) is

P(t)= t− \frac {1}{2}t^{2} + \frac {1}{3}t^{3} − \frac {1}{4}t^{4}+…=\sum\limits_{n=1}^{∞}{(−1)^{n+1} \frac {1}{n}t^{n}}

From calculus, the standard way to test a power series for convergence is to use the Ratio Test. Doing so here with a_{n} = (−1)^{n+1}(1/n)t^{n}, we observe that

\underset{n\longrightarrow \infty }{\lim }|\frac {a_{n+1}}{a_{n}}|=\underset{n\longrightarrow \infty }{\lim }|\frac {(−1)^{n}+2(1/n+1) t^{n+1}}{(−1)^{n+1}(1/n) t^{n}}|
=\underset{n\longrightarrow \infty }{\lim }|−1 · \frac {n}{n+1}· t|

= |t |

The Ratio Test states that a given series converges if l\underset{n\longrightarrow \infty }{\lim }|a_{n+1}/a_{n}| < 1.Thus, if |t | < 1, it follows that

ln(1+t ) =t− \frac {1}{2}t^{2} + \frac {1}{3}t^{3} − \frac {1}{4}t^{4}+…=\sum\limits_{n=1}^{∞}{(−1)^{n+1} \frac {1}{n}t^{n}}        (8.2.3)

converges.