Power Series Expansion for Reciprocal of Square Root of 1 - x/Proof 1

Theorem

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

Then:

\(\ds \dfrac 1 {\sqrt {1 - x} }\)

\(=\)

\(\ds \sum_{k \mathop = 0}^\infty \frac {\paren {2 k}!} {\paren {2^k k!}^2} x^k\)

\(\ds \)

\(=\)

\(\ds 1 + \frac 1 2 x + \frac {1 \times 3} {2 \times 4} x^2 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} x^3 + \cdots\)

Proof

\(\ds \frac 1 {\sqrt {1 + x} }\)

\(=\)

\(\ds 1 - \frac 1 2 x + \frac {1 \times 3} {2 \times 4} x^2 - \frac {1 \times 3 \times 5} {2 \times 4 \times 6} x^3 + \cdots\)

Power Series Expansion for $\dfrac 1 {\sqrt {1 + x} }$

\(\ds \leadsto \ \ \)

\(\ds \frac 1 {\sqrt {1 - x} }\)

\(=\)

\(\ds 1 - \frac 1 2 \paren {-x} + \frac {1 \times 3} {2 \times 4} \paren {-x}^2 - \frac {1 \times 3 \times 5} {2 \times 4 \times 6} \paren {-x}^3 + \cdots\)

substituting $x \gets -x$

\(\ds \)

\(=\)

\(\ds 1 + \frac 1 2 x + \frac {1 \times 3} {2 \times 4} x^2 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} x^3 + \cdots\)

simplifying

$\blacksquare$