# Geometric Series – Far Out

One of the great things – at least I like it – about Taylor series is that they are unique. There is only one Taylor series for any function centered at a given point, What that means is that any way you get it, it’s right.

Faced with writing the power series for, say, $\displaystyle f\left( x \right)=\frac{3x}{1-2x}$, instead of cranking out a bunch of derivatives, we can say this looks a lot like the formula for the sum of a geometric series, $\displaystyle \sum\limits_{k=1}^{\infty }{a{{r}^{k-1}}}=\frac{a}{1-r}$. Taking  a = 3x and r = 2x, the series is $\displaystyle \frac{3x}{1-2x}=3x+6{{x}^{2}}+12{{x}^{3}}+24{{x}^{4}}+\cdots =3\cdot \sum\limits_{k=1}^{\infty }{{{2}^{k-1}}{{x}^{k}}}$.

Furthermore, since a geometric series converges only when $\left| r \right|<1$, the interval of convergence for this series is $\left| 2x \right|<1$ or $-\tfrac{1}{2} and we don’t even have to check the endpoints.

There are other choices as well.  We could write $\displaystyle f\left( x \right)=3x{{\left( 1-2x \right)}^{-1}}$ and then expand the binomial using the binomial theorem. Or we could use the technique of long division of polynomials to divide 3x by (1 – 2x) – leaving the divisor as written here.

This works even in more complicated situations. Let $\displaystyle g\left( x \right)=\frac{3x}{{{x}^{2}}-4}$. Begin by dividing each term by –4. This gives $\displaystyle g\left( x \right)=\frac{-\tfrac{3}{4}x}{1-\tfrac{1}{4}{{x}^{2}}}$. Then treating this as a geometric series $\displaystyle g\left( x \right)=\sum\limits_{k=1}^{\infty }{-\tfrac{3}{4}x{{\left( \tfrac{1}{4}{{x}^{2}} \right)}^{k-1}}=-\frac{3}{4}x-\frac{3}{16}{{x}^{3}}-\frac{3}{64}{{x}^{5}}-\frac{3}{256}{{x}^{7}}-\cdots }$

The interval of convergence is $\displaystyle \left| \tfrac{1}{4}{{x}^{2}} \right|<1$, or –2 < x < 2

Now the fun part

I once heard of a student making one of those great “mistakes.” For the series above, she divided by (–x2) and found that $\displaystyle g\left( x \right)=\frac{-\frac{3}{x}}{1-\frac{4}{{{x}^{2}}}}$ and then wrote: $\displaystyle g\left( x \right)=\sum\limits_{k=1}^{\infty }{\left( -\frac{3}{x} \right){{\left( \frac{4}{{{x}^{2}}} \right)}^{k-1}}=-\frac{3}{x}-\frac{12}{{{x}^{3}}}-\frac{48}{{{x}^{5}}}-\frac{192}{{{x}^{7}}}-\cdots }$

So, what’s wrong with that?

Nothing actually.

Okay, it’s not a Taylor Series since a Taylor series is allowed only non-negative exponents, but it’s still a geometric series. Let’s take a look at its interval of convergence: $\displaystyle \left| \frac{4}{{{x}^{2}}} \right|<1$, or $\displaystyle \left| \frac{{{x}^{2}}}{4} \right|>1$,  or the union of $x>2$ and $x<-2$, Whoa, that’s different and not even an interval.

The graph will make things clear (as usual): The original function graphed as a rational expression is shown in black. The Taylor polynomial (4 terms) is shown in blue; it approximates the function well between –2 and 2 as we should expect. The red graph is the student’s series (4 terms) and it is a good approximation of the series outside of the interval (–2, 2), far outside! Way Cool!

Of course, this kind of series is not studied in beginning calculus. It may make a good topic for a report or project for someone in your class.

## One thought on “Geometric Series – Far Out”

1. Jim says:

It’s called a Laurent series. It is a power series centered at oo.

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