Taylor Series for 1/(1+x^2) without Substitution

In summary, the Taylor series of 1/(1+x^2) is an infinite series representation of the function 1/(1+x^2) around a given point, typically x = 0. The first few terms of the series are 1 - x^2 + x^4 - x^6 + x^8 - ... and the general term can be written as (-1)^n * x^(2n). The interval of convergence for the series is -1 < x < 1, meaning it is valid for all values of x between -1 and 1 and can be used to approximate the function within this range. The Taylor series is related to the Maclaurin series, which is a special case
  • #1
sparkle123
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Homework Statement



How do we get that the Taylor Series of 1/(1+x^2) around x= 0
is 1 - x^2 + x^4 + ... + (-1)^n x^{2n} + ... for |x|<1, without using a substitution of x=-x^2 into the Taylor series for 1/(1-x)?

Homework Equations


The Attempt at a Solution

 
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  • #2
You could go back to the definition of the Taylor series. Calculate the derivatives, etc.
 

What is the Taylor series of 1/(1+x^2)?

The Taylor series of 1/(1+x^2) is an infinite series representation of the function 1/(1+x^2) around a given point, typically x = 0. It is a way to express a complicated function as a sum of simpler terms.

What are the first few terms of the Taylor series of 1/(1+x^2)?

The first few terms of the Taylor series of 1/(1+x^2) are 1 - x^2 + x^4 - x^6 + x^8 - ... The general term in the series can be written as (-1)^n * x^(2n).

What is the interval of convergence for the Taylor series of 1/(1+x^2)?

The interval of convergence for the Taylor series of 1/(1+x^2) is -1 < x < 1. This means that the series is valid for all values of x between -1 and 1, and may or may not converge at the endpoints.

How can the Taylor series of 1/(1+x^2) be used to approximate the function?

The Taylor series of 1/(1+x^2) can be used to approximate the function for values of x within its interval of convergence. The more terms that are included in the series, the more accurate the approximation will be. This can be useful for evaluating the function at values that may be difficult to calculate directly.

How does the Taylor series of 1/(1+x^2) relate to the Maclaurin series?

The Maclaurin series is a special case of the Taylor series, where the given point is x = 0. Therefore, the Maclaurin series of 1/(1+x^2) is the same as its Taylor series. However, in general, the Taylor series can be centered at any point, while the Maclaurin series is always centered at x = 0.

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