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Find the radius of convergence of the Maclaurin series for the function below f(x)=1/(1-x²)¹/².

asked
User Keelin
by
9.1k points

1 Answer

4 votes
to find the radius of convergence for the Maclaurin series of \(f(x) = \frac{1}{\sqrt{1 - x^2}}\), you can use the following simplified approach:

1. Start with the Maclaurin series for \(\frac{1}{\sqrt{1 - x^2}}\), which is known to be:

\[f(x) = \sum_{n=0}^{\infty} \frac{(2n)!}{2^{2n}(n!)^2}x^{2n}\]

2. To find the radius of convergence (\(R\)), use the ratio test:

\[
R = \lim_{n \to \infty} \left| \frac{a_{n}}{a_{n+1}} \right|
\]

where \(a_n\) is the coefficient of \(x^{2n}\) in the series.

3. Calculate \(a_n\) and \(a_{n+1}\) using the formula for \(f(x)\) above.

4. Take the limit as \(n\) approaches infinity to find \(R\).

This approach will give you the radius of convergence for the Maclaurin series.
answered
User Samuelabate
by
7.6k points
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