Power Series Expansion for Reciprocal of 1 + x
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Theorem
Let $x \in \R$ such that $-1 < x < 1$.
Then:
| \(\ds \dfrac 1 {1 + x}\) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k x^k\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1 - x + x^2 - x^3 + x^4 - \cdots\) |
Proof 1
| \(\ds \frac 1 {1 + x}\) | \(=\) | \(\ds \frac 1 {1 - \paren {-x} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \paren {-x}^k\) | Sum of Infinite Geometric Sequence | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k x^k\) |
$\blacksquare$
Proof 2
| \(\ds \frac 1 {1 + x}\) | \(=\) | \(\ds \paren {1 + x}^{-1}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^{\underline k} } {k!} x^k\) | General Binomial Theorem | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {\paren {-1} - j} } {k!} x^k\) | Definition of Falling Factorial | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 0}^{k - 1} \paren {-\paren {j + 1} } } {\ds \prod_{j \mathop = 1}^k j} x^k\) | Definition of Factorial and simplifying | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\ds \prod_{j \mathop = 1}^k \paren {-j} } {\ds \prod_{j \mathop = 1}^k j} x^k\) | Translation of Index Variable of Product | |||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \frac {\paren {-1}^k \ds \prod_{j \mathop = 1}^k j} {\ds \prod_{j \mathop = 1}^k j} x^k\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \sum_{k \mathop = 0}^\infty \paren {-1}^k x^k\) | simplification |
$\blacksquare$