# Different series representation for the same function

Ok, we know that $\frac{1}{1-x}=1+x+x^2+x^3+\cdots$. Now if we want to have a series representation of $y=\frac{1}{2-x}$ then there are two approaches: First $\frac{1}{2-x}=\frac{1}{1-(x-1)}=1+(x-1)+(x-1)^2+(x-2)^3+\cdots$ where $x-1$ takes the "role" of the $x$ in the original geometric series. The interval of convergence is thus "shifted" to (0,2). On the other hand $\frac{1}{2-x}=\frac{1}{2(1-\frac{x}{2})}=\frac{1}{2}(1+\frac{x}{2}+(\frac{x}{2})^2+(\frac{x}{2})^3+\cdots)$ But this series has an interval convergence of $(-2,2)$ which is different from the interval of convergence of the first series. I have two different series with two different interval convergences for the very same function. I can't rhyme this. Which one is correct? If both are right, can you come up with other series for the given function with again a different interval of convergence? I feel I am missing a fundamental concept here and it bugs me. Any input is welcome. Thanks.

• @TimRaczkowski He's not saying they're the same function ... – Zubin Mukerjee Nov 22 '15 at 21:59
• Note for radius of convergence !$$1+(x-1)+(x-2)^2+... \to |x-1|<1 \to 0<x<2$$but in $$1+\frac{x}{2}+\frac{x^2}{4}+... \to |\frac{x}{2}|<1 \to -2<x<2$$ – Khosrotash Nov 22 '15 at 22:00
• They are power-series around different points of expansion. I think your arguments are OK. Because they are expanded around different points, they have different formulas and different intervals of convergence. It's fine. – Leaning Nov 22 '15 at 22:00
• @TimRaczkowski No, I am not. I am introducing the geometric series to introduce my problem with $y=\frac{1}{2-x}$ – imranfat Nov 22 '15 at 22:02
• @ZubinMukerjee,@imranfat Sorry. I was me who misread the post. – Tim Raczkowski Nov 22 '15 at 22:06

## 1 Answer

In complex analysis, this is often called "analytic continuation". The function $\frac{1}{2-x}$ you are describing only has a single pole $x = 2$ on the complex plane, and otherwise well defined.

But to write an arbitrary function in terms of power series, then such formalism has limitations, namely one would in general decrease the actual radius of convergence.

For instance when we expand around $x=1$ then such power series is only defined for $|x-1|<1$. If we expand around $x=0$ then the radius of convergence is $|x|<2$.

In fact for $\frac{1}{1-x}$ you can also expand around $x=-1$ to get another representative. There are infinite ways to expand this function, actually.

In one word, power series expansion is only a "partial" representation of the original function.

• Ah, I am trying to understand what you are getting at. Are you saying that in the first case I derived a power series around $x=1$ and in the second case I set it up around $x=0$? – imranfat Nov 22 '15 at 22:09
• @imranfat Yes that's right. Something to keep in mind, when you find the Taylor expansion for a function around a certain point, the radius of convergence can be no bigger than the distance from the given point and the closest point at which the function is undefined. – Tim Raczkowski Nov 22 '15 at 22:13
• Allright. Thanks all, I just need to absorb this for while... – imranfat Nov 22 '15 at 22:14
• @RoryDaulton You are right. Thank you for pointing out. – Kevin Ye Nov 23 '15 at 3:31
• Also keep in mind that "the closest point at which the function is undefined" may be a complex number! If you expand $1/(1+x^2)$ around $x=0$, you will find the interval of convergence is $|x| < 1$ which might seem mysterious because the function is defined for all real numbers. But in the complex plane there are singularities at $\pm i$, which explains why there is no convergence around $x=0$ beyond radius 1. – Ted Nov 23 '15 at 4:16