1
$\begingroup$

It is well known that $\displaystyle\dfrac {1}{1-x} = \sum_{n=0}^{\infty}(x-0)^n$ when $x \in (-1, 1)$. If I want a power series representation on the interval $(2, 4)$, can I simply find a Taylor series for $f(x) = \dfrac {1}{1-x}$ centered at $3$? Can I continue moving the center around until I cover $\mathbb{R}$ \ $\{-1, 1\}$? If not, why would we ever use a center that is not $0$?

$\endgroup$
1
$\begingroup$

Yes. Observe that: $\frac{1}{1-x}=\frac{-1}{2}\frac{1}{1+(x-3)/2}=\frac{-1}{2}\sum_{n=0}(-1)^{n}\frac{1}{2^n}(x-3)^n$, as long as $x\in(1,5)$ since the radius of convergence of the power series is 2. In fact, for any $a\in\mathbb{R}-\{1\}$, You can write $1/(1-x)$ as a power series with center at $x=a$, and the power series converges to $1/(1-x)$ as long as $x\in(a-|a-1|,a+|a-1|)$.

$\endgroup$
0
$\begingroup$

The sum $$ \sum_{n=1}^{M} -x^{-n} = \frac{1/x-(1/x)^{M+1}}{1/x-1} = \frac{1}{1-x} - \frac{x^{-M}}{1-x} $$ converges to $\frac{1}{1-x}$ for $\lvert x \rvert > 1$. This is akin to a Taylor series at $\infty$, but is called a Laurent series because it contains powers of $1/x$.

One can find the Taylor series of $\frac{1}{1-x}$ about the point $x=a$ by writing it as $$ \frac{1}{1-a-(x-a)} = \frac{1}{1-a} \frac{1}{1-(x-a)/(1-a)} = \sum_{n=0}^{\infty} \frac{(x-a)^n}{(1-a)^{n+1}} $$ by again using the geometric series formula (Clearly $a=0$ recovers the original series about $x=0$). This series converges if $$ \left| \frac{x-a}{1-a} \right| < 1. $$ You therefore find, for example, if $a=3$, the series converges if $-2<x-3<2$, i.e. $1<x<5$. This will work for any $a \neq 1$. The general principle is that the series cannot converge on an interval that includes the singularity at $x=1$. In this case, this is the only obstacle to convergence.

The usefulness of the series expansion generally lies in the ability to approximate the function as closely as we like over a finite interval by only taking a finite number of terms (if you happen to know that $12^2=144$, you can get a pretty good estimate for $\sqrt{145}$, to take a very simple example). Hence expanding about a point closer to points you are interested in provides a better estimate for less cost than a further away point (such as trying to estimate $\sqrt{145}$ based on the series expansion of $\sqrt{x}$ at $100$ instead of $144$: it's pretty obvious you won't get the same level of accuracy without taking more terms).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.