- #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)?
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.
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).
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.
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.
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.