Power Series Expansion for Square Root of 1 + x

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Theorem

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

Then:

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


Proof

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

$\blacksquare$


Sources

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