"What function does ∑n=0∞(−1)n(2n)!(1−2n)n!24nxny2n correspond to?" was answered by students like you

ymhonnolfq

Answered question

2022-02-26

What function does

\sum_{n = 0}^{\infty} \frac{{(- 1)}^{n} (2 n)!}{(1 - 2 n) n!^{2} 4^{n}} \frac{x^{n}}{y^{2 n}}

correspond to?

asserena3wx

Beginner2022-02-27Added 7 answers

A good series to know offhand is

\frac{1}{\sqrt{1 - 4 t}} = \sum_{n = 0}^{\infty} (\begin{matrix} 2 n \\ n \end{matrix}) t^{n}

which almost gets us the series we desire with

t = - \frac{x}{4 y^{2}}

\frac{1}{\sqrt{1 + \frac{x}{4 y^{2}}}} = \sum_{n = 0}^{\infty} (\begin{matrix} 2 n \\ n \end{matrix}) {(- \frac{x}{4 y^{2}})}^{n}

except for the offending term

1 - 2 n

in the denominator. However a small ajustment to the original series with

t = z^{- 2}

\frac{1}{\sqrt{1 + \frac{4}{z^{2}}}} = \frac{| z |}{\sqrt{z^{2} + 4}} = \sum_{n = 0}^{\infty} (\begin{matrix} 2 n \\ n \end{matrix}) {(- 1)}^{n} z^{- 2 n}

gives us the series we want via integration

\sqrt{z^{2} + 4} sgn (z) = \sum_{n = 0}^{\infty} (\begin{matrix} 2 n \\ n \end{matrix}) \frac{z^{1 - 2 n}}{1 - 2 n} \frac{\sqrt{z^{2} + 4}}{| z |}

= \sum_{n = 0}^{\infty} (\begin{matrix} 2 n \\ n \end{matrix}) \frac{z^{- 2 n}}{1 - 2 n}

by substituting

z = \frac{2 y}{\sqrt{x}}

\sum_{n = 0}^{\infty} (\begin{matrix} 2 n \\ n \end{matrix}) \frac{1}{1 - 2 n} {(- \frac{x}{4 y^{2}})}^{n} = \frac{\sqrt{y^{2} + x}}{| y |} = \sqrt{1 + \frac{x}{y^{2}}}

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