Different series than Taylor series for a function

[Mr Davis 97]

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.

 

[mathman]

It is an expression using the generalized binomial expansion. (1+x^2)^{-1}=1-x^{-2}+x^{-4}...
http://mathworld.wolfram.com/NegativeBinomialSeries.html

 

[haushofer]

That doesn't seem right to me.

 

[jasonRF]

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]

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.)

 

[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]

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]

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.

 

[Mr Davis 97]

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]

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.

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.