Power Series Expansion for Cube Root of 1 + x
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Theorem
Let $x \in \R$ such that $-1 < x \le 1$.
Then:
- $\sqrt [3] {1 + x} = 1 + \dfrac 1 3 x - \dfrac 2 {3 \times 6} x^2 + \dfrac {2 \times 5} {3 \times 6 \times 9} x^3 - \cdots$
Proof
\(\ds \sqrt [3] {1 + x}\) | \(=\) | \(\ds \paren {1 + x}^{\frac 1 3}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\paren {\frac 1 3}^{\underline k} } {k!} x^k\) | General Binomial Theorem | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \sum_{k \mathop = 1}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\frac 1 3 - j} } {k!} x^k\) | Definition of Falling Factorial and extracting $k = 0$ | |||||||||||
\(\ds \) | \(=\) | \(\ds 1 + \sum_{k \mathop = 1}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {1 - 3 j} } {3^k k!} x^k\) | simplifying | |||||||||||
\(\ds \) | \(=\) | \(\ds \sqrt [3] {1 + x} = 1 + \dfrac 1 3 x - \dfrac 2 {3 \times 6} x^2 + \dfrac {2 \times 5} {3 \times 6 \times 9} x^3 - \cdots\) |
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 20$: Binomial Series: $20.14$