Power Series Expansion for Real Area Hyperbolic Cosine/Lemma 2

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Lemma for Power Series Expansion for Real Area Hyperbolic Cosine

$\map \ln {1 + \sqrt {1 - x^2} } = \ln 2 - \paren {\dfrac 1 2 \cdot \dfrac {x^2} 2 + \dfrac {1 \times 3} {2 \times 4} \cdot \dfrac {x^4} 4 + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} \cdot \dfrac {x^6} 6 + \cdots}$

This holds for $x \in \R: \size x < 1$.


Proof

Lemma 1

\(\ds \dfrac 1 {\sqrt {1 - x^2} }\) \(=\) \(\ds 1 + \frac 1 2 x^2 + \frac {1 \times 3} {2 \times 4} x^4 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} x^6 + \cdots\)
\(\text {(1)}: \quad\) \(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2} x^{2 n}\)

for $x \in \R: -1 < x < 1$.

$\Box$


\(\ds \map {\dfrac \d {\d x} } {\map \ln {1 + \sqrt {1 - x^2} } }\) \(=\) \(\ds \dfrac 1 {1 + \sqrt {1 - x^2} } \cdot \dfrac 1 {2 \sqrt {1 - x^2} } \cdot \paren {-2 x}\)
\(\ds \) \(=\) \(\ds \dfrac {-x} {1 - x^2 + \sqrt {1 - x^2} }\) simplifying and multiplying out
\(\ds \) \(=\) \(\ds \dfrac {-x \paren {1 - x^2 - \sqrt {1 - x^2} } } {\paren {1 - x^2 + \sqrt {1 - x^2} } \paren {1 - x^2 - \sqrt {1 - x^2} } }\) multiplying top and bottom by $1 - x^2 - \sqrt {1 - x^2}$
\(\ds \) \(=\) \(\ds \dfrac {x \paren {1 - x^2 - \sqrt {1 - x^2} } } {\paren {1 - x^2}^2 - \paren {1 - x^2} }\) Difference of Two Squares
\(\ds \) \(=\) \(\ds \dfrac {x \paren {1 - x^2 - \sqrt {1 - x^2} } } {\paren {1 - x^2} \paren {1 - x^2 - 1} }\) factoring $1 - x^2$ in denominator
\(\ds \) \(=\) \(\ds \dfrac {x \paren {1 - x^2} - x \sqrt {1 - x^2} } {x^2 \paren {1 - x^2} }\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 x \paren {1 - \dfrac 1 {\sqrt {1 - x^2} } }\) simplifying
\(\ds \) \(=\) \(\ds \dfrac 1 x \paren {1 - \paren {1 + \frac 1 2 x^2 + \frac {1 \times 3} {2 \times 4} x^4 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} x^6 + \cdots} }\) Lemma $1$: for $0 < \size x < 1$
\(\ds \) \(=\) \(\ds -\paren {\frac 1 2 x + \frac {1 \times 3} {2 \times 4} x^3 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} x^5 + \cdots}\) simplifying
\(\ds \leadsto \ \ \) \(\ds \map \ln {1 + \sqrt {1 - x^2} }\) \(=\) \(\ds C - \paren {\frac 1 2 \cdot \dfrac {x^2} 2 + \frac {1 \times 3} {2 \times 4} \cdot \dfrac {x^4} 4 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} \cdot \dfrac {x^6} 6 + \cdots}\) Integrating with respect to $x$: Power Series is Termwise Integrable within Radius of Convergence

It remains to determine the value of the arbitrary constant $C$.

Let $x \to 0$.

Then we have:

\(\ds \lim_{x \mathop \to 0} \map \ln {1 + \sqrt {1 - x^2} }\) \(=\) \(\ds \map {\lim_{x \mathop \to 0} } {C - \paren {\frac 1 2 \cdot \dfrac {x^2} 2 + \frac {1 \times 3} {2 \times 4} \cdot \dfrac {x^4} 4 + \frac {1 \times 3 \times 5} {2 \times 4 \times 6} \cdot \dfrac {x^6} 6 + \cdots} }\)
\(\ds \leadsto \ \ \) \(\ds \ln 2\) \(=\) \(\ds C\)

$\blacksquare$