Power Series Expansion for Logarithm of x/Formulation 2
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Theorem
\(\ds \ln x\) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \dfrac 1 n \paren {\frac {x - 1} x}^n\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {x - 1} x + \frac 1 2 \paren {\frac {x - 1} x}^2 + \frac 1 3 \paren {\frac {x - 1} x}^3 + \cdots\) |
valid for all $x \in \R$ such that $x \ge \dfrac 1 2$.
Proof
From the corollary to Power Series Expansion for $\map \ln {1 + x}$:
\(\ds \map \ln {1 - x}\) | \(=\) | \(\ds -\sum_{n \mathop = 1}^\infty \frac {x^n} n\) | for $-1 \le x < 1$. | |||||||||||
\(\text {(1)}: \quad\) | \(\ds \leadsto \ \ \) | \(\ds \map \ln {\frac 1 {1 - x} }\) | \(=\) | \(\ds \sum_{n \mathop = 1}^\infty \frac {x^n} n\) | Logarithm of Reciprocal |
Let $z = \dfrac 1 {1 - x}$.
Then:
\(\ds z\) | \(=\) | \(\ds \dfrac 1 {1 - x}\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds z \paren {1 - x}\) | \(=\) | \(\ds 1\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds z - z x\) | \(=\) | \(\ds 1\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds z x\) | \(=\) | \(\ds z - 1\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x\) | \(=\) | \(\ds \frac {z - 1} z\) |
Then we have:
- $\ds \lim_{x \mathop \to 1^-} \dfrac 1 {1 - x} \to +\infty$
and:
- $\ds \lim_{x \mathop \to -1^+} \dfrac 1 {1 - x} \to \frac 1 2$
Thus when $x \in \hointr {-1} 1$ we have that $z \in \hointr {\dfrac 1 2} \to$.
Thus, substituting $z$ for $\dfrac 1 {1 - x}$ in $(1)$ gives the result.
$\blacksquare$
Sources
- 1968: Murray R. Spiegel: Mathematical Handbook of Formulas and Tables ... (previous) ... (next): $\S 20$: Series for Exponential and Logarithmic Functions: $20.20$