Find the Taylor series of 1/x about x = 0, or explain why it does not exist.

Step-by-step solution

1. 1

   Recall what a Taylor series requires. The expansion $\sum_{n\ge0} \frac{f^{(n)}(a)}{n!}(x-a)^n$ needs $f$ to be defined at the centre $a$ and to have derivatives of every order there. Both conditions are about the single point $a$ — a function can behave perfectly well nearby and still fail.

2. 2

   Check the very first coefficient. The constant term is $f(0) = \tfrac{1}{0}$, which is undefined. The series cannot even be started, so there is no Taylor (Maclaurin) series of $1/x$ about $x = 0$.

3. 3

   Confirm the failure is essential, not removable. $\lim_{x\to 0^{+}} \tfrac1x = +\infty$ while $\lim_{x\to 0^{-}} \tfrac1x = -\infty$, so no value can be assigned at $0$ to patch the function. The singularity is a genuine pole of order $1$, not a removable hole like the one in $\tfrac{\sin x}{x}$.

4. 4

   Give the expansion that does exist. Allowing negative powers gives a Laurent series, and for this function it is already in final form:

   $\frac{1}{x} = x^{-1}$

   valid on the punctured disc $0 < |x| < \infty$. A single $x^{-1}$ term with no others is what "simple pole at the origin" means.

5. 5

   Contrast with a legitimate centre. About $a = 1$ the function is perfectly analytic:

   $\frac{1}{x} = \frac{1}{1 + (x-1)} = \sum_{n=0}^{\infty} (-1)^n (x-1)^n, \qquad |x-1| < 1$

   The radius $1$ is exactly the distance from the centre $1$ to the singularity at $0$ — the general rule for how far a Taylor series of a rational function can reach.

6. 6

   Sanity-check that expansion. At $x = 1.2$ the first six terms give $1 - 0.2 + 0.04 - 0.008 + 0.0016 - 0.00032 = 0.83328$, converging toward $\tfrac{1}{1.2} = 0.8\overline{3}$.