Power Series Expansion for Reciprocal of Square Root of 1 + x

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Theorem

Let $x \in \R$ such that $-1 < x \le 1$.

Then:

\(\ds \dfrac 1 {\sqrt {1 + x} }\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\paren {2 k}!} {\paren {2^k k!}^2} x^k\)
\(\ds \) \(=\) \(\ds 1 - \frac 1 2 x + \frac {1 \times 3} {2 \times 4} x^2 - \frac {1 \times 3 \times 5} {2 \times 4 \times 6} x^3 + \cdots\)


Proof

\(\ds \frac 1 {\sqrt {1 + x} }\) \(=\) \(\ds \paren {1 + x}^{-\frac 1 2}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \frac {\paren {-\frac 1 2}^{\underline k} } {k!} x^k\) General Binomial Theorem
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\paren {-\frac 1 2} - j} } {k!} x^k\) Definition of Falling Factorial
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\ds \prod_{j \mathop = 1}^k \paren {2 j - 1} } {2^k k!} x^k\) Translation of Index Variable of Product and simplifying
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\ds \prod_{j \mathop = 1}^k \paren {2 j - 1} \prod_{j \mathop = 1}^k \paren {2 j} } {2^k k! \ds \prod_{j \mathop = 1}^k \paren {2 j} } x^k\) multiplying top and bottom by $\ds \prod_{j \mathop = 1}^k \paren {2 j}$
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\ds \prod_{j \mathop = 1}^{2 k} j} {2^k k! \paren {2^k \ds \prod_{j \mathop = 1}^k j} } x^k\) simplifying
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\paren {2 k}!} {2^k k! \paren {2^k k!} } x^k\) Definition of Factorial
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k \frac {\paren {2 k}!} {\paren {2^k k!}^2} x^k\) simplifying

$\blacksquare$


Sources

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