# Different series than Taylor series for a function

## Main Question or Discussion Point

I am trying to solve an integral that has $\frac{1}{1+x^2}$ as a factor in the integrand. In my book it is claimed that if we use $\displaystyle \frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-1)^n \frac{1}{x^{2n+2}}$ the problem can be solved immediately. But, I am confused as to where this series representation comes from... It certainly has nothing to do with Taylor series it seems.

haushofer
That doesn't seem right to me.

jasonRF
Gold Member
The expression the OP gave is valid for $|x|>1$. It is the generalized binomial as mathman said, although I think the equation he posted isn't quite correct.

mathwonk
Homework Helper
have you ever done any complex analysis? in that setting we often find it convenient to complete the complex plane by adding in a point at infinity, and then one can make a taylor expansion centered at infinity.

in your case set w = 1/x and then 1/(1+x^2) becomes w^2/(1+w^2) which has a taylor expansion centered at w= 0, or x = infinity:

namely 1/(1+x^2) = w^2/(1+w^2) = (geometric series) w^2 (1 -w^2 + w^4 ±.....) = (1/x^2)(1 - 1/x^2 + 1/x^4 ± .....) = 1/x^2 -1/x^4 + 1/x^6 ±.......

(in particular mathman's equation seems off by a factor of 1/x^2.)

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haushofer
The expression the OP gave is valid for $|x|>1$. It is the generalized binomial as mathman said, although I think the equation he posted isn't quite correct.
Ah, yes, I implicitly assumed the integration interval is between 0 and 1, but the exercise is probably an integral which has an integration interval outside of this. My bad.

FactChecker
Gold Member
One can obtain this series from the Taylor series $\frac 1 {1-r} = 1+r+r^2+r^3+\cdots$, which is valid for $|r|<1$.
$$\frac 1 {1+x^2} = x^{-2} \frac 1 {x^{-2}+1} = x^{-2} \frac 1 {1 - (-x^{-2})}$$
$$= x^{-2} (1+(-x^{-2})+(-x^{-2})^2+(-x^{-2})^3+(-x^{-2})^4+\cdots)$$ which is valid for $|x^{-2}|<1$.

If you study power series and analytic functions, the series $\frac 1 {1-r} = 1+r+r^2+r^3+\cdots$, $|r|<1$ will be fundamental.

mathwonk
Homework Helper
that is of course exactly what i said above in #5, but maybe looks more elementary by not mentioning the point at infinity where x^(-2) = 0. note too that the series is also valid at x = infinity since both sides equal zero. indeed this is the reason the expansion works.

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have you ever done any complex analysis? in that setting we often find it convenient to complete the complex plane by adding in a point at infinity, and then one can make a taylor expansion centered at infinity.

in your case set w = 1/x and then 1/(1+x^2) becomes w^2/(1+w^2) which has a taylor expansion centered at w= 0, or x = infinity:

namely 1/(1+x^2) = w^2/(1+w^2) = (geometric series) w^2 (1 -w^2 + w^4 ±.....) = (1/x^2)(1 - 1/x^2 + 1/x^4 ± .....) = 1/x^2 -1/x^4 + 1/x^6 ±.......

(in particular mathman's equation seems off by a factor of 1/x^2.)
I have not studied complex analysis yet. But, does this have anything to do with what's called a Laurent series? I'm asking because on WolframAlpha, when I input $\frac{1}{1+x^2}$ it gives the Maclaurin series expansion at 0 and also gives $\displaystyle \sum_{n=0}^{\infty} (-1)^n \frac{1}{x^{2n+2}}$ and calls it the Laurent series expansion at $\infty$.

FactChecker
Yes. I hadn't noticed that you had already given the same answer till later. When I realized that, I decided not to delete my post because I think that the expansion of $\frac 1 {1-z}$ is so important and I wanted to state that.