Power Series Expansion for Logarithm of x/Formulation 1

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Theorem

\(\ds \ln x\) \(=\) \(\ds 2 \paren {\sum_{n \mathop = 0}^\infty \frac 1 {2 n + 1} \paren {\frac {x - 1} {x + 1} }^{2 n + 1} }\)
\(\ds \) \(=\) \(\ds 2 \paren {\frac {x - 1} {x + 1} + \frac 1 3 \paren {\frac {x - 1} {x + 1} }^3 + \frac 1 5 \paren {\frac {x - 1} {x + 1} }^5 + \cdots}\)

valid for all $x \in \R$ such that $-1 < x < 1$.


Proof

From Power Series Expansion for $\dfrac 1 2 \map \ln {\dfrac {1 + x} {1 - x} }$:

$(1): \quad \ds \frac 1 2 \map \ln {\frac {1 + x} {1 - x} } = \sum_{n \mathop = 0}^\infty \frac {x^{2 n + 1} } {2 n + 1}$

for $-1 < x < 1$.


Let $z = \dfrac {1 + x} {1 - x}$.

Then:

\(\ds z\) \(=\) \(\ds \dfrac {1 + x} {1 - x}\)
\(\ds \leadsto \ \ \) \(\ds z \paren {1 - x}\) \(=\) \(\ds 1 + x\)
\(\ds \leadsto \ \ \) \(\ds z - z x\) \(=\) \(\ds 1 + x\)
\(\ds \leadsto \ \ \) \(\ds z - 1\) \(=\) \(\ds z x + x\)
\(\ds \leadsto \ \ \) \(\ds x \paren {z + 1}\) \(=\) \(\ds z - 1\)
\(\ds \leadsto \ \ \) \(\ds x\) \(=\) \(\ds \frac {z - 1} {z + 1}\)


Then we have:

$\ds \lim_{x \mathop \to 1^-} \dfrac {1 + x} {1 - x} \to +\infty$

and:

$\ds \lim_{x \mathop \to -1^+} \dfrac {1 + x} {1 - x} \to 0$

Thus when $x \in \openint {-1} 1$ we have that $z \in \openint 0 \to$.


Thus, substituting $z$ for $\dfrac {1 + x} {1 - x}$ in $(1)$ gives the result.

$\blacksquare$


Sources

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