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:
| \(\ds \sqrt {1 - x}\) | \(=\) | \(\ds 1 - \sum_{k \mathop = 1}^\infty \frac {\paren {2 \paren {k - 1} }!} {2^{2 k - 1} k! \paren {k - 1}!} x^k\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - \frac 1 2 x - \frac 1 {2 \times 4} x^2 - \frac {1 \times 3} {2 \times 4 \times 6} x^3 - \cdots\) |
Proof
| \(\ds \sqrt {1 + x}\) | \(=\) | \(\ds 1 + \frac 1 2 x - \frac 1 {2 \times 4} x^2 + \frac {1 \times 3} {2 \times 4 \times 6} x^3 - \cdots\) | Power Series Expansion for $\sqrt {1 + x}$ | |||||||||||
| \(\ds \leadsto \ \ \) | \(\ds \sqrt {1 - x}\) | \(=\) | \(\ds 1 + \frac 1 2 \paren {-x} - \frac 1 {2 \times 4} \paren {-x}^2 + \frac {1 \times 3} {2 \times 4 \times 6} \paren {-x}^3 - \cdots\) | substituting $x \gets -x$ | ||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - \frac 1 2 x - \frac 1 {2 \times 4} x^2 - \frac {1 \times 3} {2 \times 4 \times 6} x^3 - \cdots\) | simplifying |
$\blacksquare$
Sources
- 1976: K. Weltner and W.J. Weber: Mathematics for Engineers and Scientists ... (previous) ... (next): $8$. Taylor Series and Power Series: Exercises: $8.2$ Expansion of a Function in a Power Series: $1. \ \text {(a)}$