# Power series expansion

1. Apr 17, 2017

### Mr Davis 97

1. The problem statement, all variables and given/known data
Show that $\displaystyle \frac{1}{1+x^2} = \frac{1}{x^2} - \frac{1}{x^4} + \frac{1}{x^6} - \frac{1}{x^8} + \cdots$

2. Relevant equations

3. The attempt at a solution
I know that the power series expansion of $\displaystyle \frac{1}{1+x^2}$ about $x=0$ is $1-x^2 + x^4 - x^6 + \cdots$, but I don't see how this can translate into the expressions as above.

2. Apr 17, 2017

### Staff: Mentor

Substitute $x=\frac{1}{y}$ and write everything in terms of $\frac{1}{y}$. At the end, re-substitute $\frac{1}{y}=x$.

3. Apr 17, 2017

### Mr Davis 97

I don't follow. If I do that then $\frac{1}{1+1/y^2} = 1-1/y^2 + 1/y^4 - 1/y^6 + \cdots$. I don't see where that gets me. Maybe I am missing what you mean by "write everything in terms of 1/y".

4. Apr 17, 2017

### Staff: Mentor

I hope, I didn't make a mistake this time:
$$\frac{1}{1+\frac{1}{y^2}}=\frac{y^2}{1+y^2}=y^2 \cdot (1-y^2+y^4-+ \ldots) = y^2 - y^4 +y^6-+\ldots = \frac{1}{\frac{1}{y^2}}-\frac{1}{\frac{1}{y^4}}+\frac{1}{\frac{1}{y^6}}-+\ldots$$

5. Apr 17, 2017

### Staff: Mentor

If you use long division to divide 1 by $x^2 + 1$ you'll get the expansion you show just above. Instead, if you divide 1 by $1 + x^2$ you'll get the expansion at the top of this post.

6. Apr 17, 2017

### vela

Staff Emeritus
That expansion only converges for $x^2<1$. You need a different expansion when $x^2>1$.

The way I'd do it is to factor $x^2$ out of the denominator:
$$\frac{1}{1+x^2} = \frac{1}{x^2[(1/x^2)+1]} = \frac{1}{x^2}\left[\frac{1}{1+(1/x^2)}\right]$$
Then expand the term in the square brackets in powers of $(1/x^2)$ the usual way. When $x^2>1$, you know that $(1/x^2)<1$ so the series will converge.

7. Apr 18, 2017

### pasmith

And the implication of a series in $x^{-2}$ is that you are not looking near $x = 0$.