Different series than Taylor series for a function

  • #1
1,456
44

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.
 

Answers and Replies

  • #3
haushofer
Science Advisor
Insights Author
2,278
634
That doesn't seem right to me.
 
  • #4
jasonRF
Science Advisor
Gold Member
1,258
303
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.
 
  • #5
mathwonk
Science Advisor
Homework Helper
10,820
983
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.)
 
Last edited:
  • #6
haushofer
Science Advisor
Insights Author
2,278
634
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.
 
  • #7
FactChecker
Science Advisor
Gold Member
5,382
1,951
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.
 
  • #8
mathwonk
Science Advisor
Homework Helper
10,820
983
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.
 
Last edited:
  • #9
1,456
44
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##.
 
  • #10
FactChecker
Science Advisor
Gold Member
5,382
1,951
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.
 

Related Threads on Different series than Taylor series for a function

Replies
2
Views
1K
  • Last Post
Replies
12
Views
9K
Replies
1
Views
1K
Replies
4
Views
10K
Replies
14
Views
2K
Replies
5
Views
12K
Replies
7
Views
2K
Replies
7
Views
1K
Replies
2
Views
2K
Replies
6
Views
2K
Top