# Different series than Taylor series for a function

• I
• Mr Davis 97
In summary, the person is trying to solve an integral with ##\frac{1}{1+x^2}## as a factor and is confused about the series representation ##\displaystyle \frac{1}{1+x^2} = \sum_{n=0}^{\infty} (-1)^n \frac{1}{x^{2n+2}}##, which is a generalized binomial expansion. The series can be obtained from the Taylor series ## \frac 1 {1-r} = 1+r+r^2+r^3+\cdots ## by setting ##r=-x^{-2}##, which is valid for ##|x^{-2}|<1##. This can also be thought
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.

That doesn't seem right to me.

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.

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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Mr Davis 97, FactChecker, haushofer and 1 other person
jasonRF said:
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.

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.

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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FactChecker
mathwonk said:
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##.

mathwonk said:
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.

## 1. What is a Taylor series and how is it different from other series for a function?

A Taylor series is a mathematical representation of a function as an infinite sum of terms, with each term being a derivative of the function evaluated at a specific point. Other series for a function, such as Maclaurin series and Laurent series, also use derivatives to approximate a function, but they may have different convergence properties and may be centered at different points.

## 2. How do I know when to use a Taylor series versus a different series for a function?

The choice of which series to use depends on the specific function and the desired level of accuracy. Taylor series are typically used when the function is smooth and well-behaved, while other series may be better suited for functions with singularities or other complex behavior.

## 3. Can a function have more than one type of series representation?

Yes, a function can have multiple series representations. For example, a function may have both a Taylor series and a Laurent series, each with a different radius of convergence and centered at different points.

## 4. Are there any limitations or drawbacks to using a Taylor series for a function?

While Taylor series can provide an excellent approximation of a function, they are only valid within their radius of convergence and may not accurately represent the function outside of this interval. Additionally, calculating a Taylor series can be computationally intensive for functions with higher order derivatives.

## 5. How is a Taylor series used in real-world applications?

Taylor series are commonly used in engineering and physics to approximate functions and solve differential equations. They are also used in computer science for tasks such as image and signal processing, where they can be used to reconstruct a function from a limited set of data points.

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