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- Thread starter Mr Davis 97
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- #2

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http://mathworld.wolfram.com/NegativeBinomialSeries.html

- #3

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That doesn't seem right to me.

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- #5

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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.)

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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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.

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$$\frac 1 {1+x^2} = x^{-2} \frac 1 {x^{-2}+1} = x^{-2} \frac 1 {1 - (-x^{-2})}$$

$$= x^{-2} (1+(-x^{-2})+(-x^{-2})^2+(-x^{-2})^3+(-x^{-2})^4+\cdots)$$ 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.

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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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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##.

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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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.

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