Power Series Expansion for Real Area Hyperbolic Cosine

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Theorem

The (real) area hyperbolic cosine function has a Taylor series expansion:

\(\ds \arcosh x\) \(=\) \(\ds \map \ln {2 x} - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} x^{2 n} } }\)
\(\ds \) \(=\) \(\ds \map \ln {2 x} - \paren {\dfrac 1 {2 \times 2 x^2} + \dfrac {1 \times 3} {2 \times 4 \times 4 x^4} + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6 \times 6 x^6} + \cdots}\)

for $x \ge 1$.


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

$\Box$


We have that the (real) area hyperbolic cosine is defined for $x \ge 1$.

Let $z = \dfrac 1 x$.

Then we have:

$0 < \dfrac 1 z \le 1$

Now we consider:

\(\ds \map \arcosh {\dfrac 1 z} + \map \ln {2 z}\) \(=\) \(\ds \map \ln {2 z} + \map \ln {\dfrac 1 z + \sqrt {\dfrac 1 {z^2} - 1} }\) Definition of Real Area Hyperbolic Cosine
\(\ds \) \(=\) \(\ds \map \ln {2 z \paren {\dfrac 1 z + \dfrac {\sqrt {1 - z^2} } z} }\) Sum of Logarithms and simplification
\(\ds \) \(=\) \(\ds \map \ln {2 \paren {1 + \sqrt {1 - z^2} } }\) simplifying
\(\ds \) \(=\) \(\ds \ln 2 + \map \ln {1 + \sqrt {1 - z^2} }\) Sum of Logarithms
\(\ds \leadsto \ \ \) \(\ds \map \arcosh {\dfrac 1 z}\) \(=\) \(\ds -\map \ln {2 z} + \ln 2 + \map \ln {1 + \sqrt {1 - z^2} }\) rearranging
\(\ds \) \(=\) \(\ds \map \ln {\dfrac 1 {2 z} } + \ln 2 + \map \ln {1 + \sqrt {1 - z^2} }\) Logarithm of Reciprocal
\(\ds \) \(=\) \(\ds \map \ln {\dfrac 1 z} + \map \ln {1 + \sqrt {1 - z^2} }\) Sum of Logarithms
\(\ds \) \(=\) \(\ds \map \ln {\dfrac 1 z} + \ln 2 - \paren {\dfrac 1 2 \cdot \dfrac {z^2} 2 + \dfrac {1 \times 3} {2 \times 4} \cdot \dfrac {z^4} 4 + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} \cdot \dfrac {z^6} 6 + \cdots}\) Lemma $2$
\(\ds \) \(=\) \(\ds \map \ln {\dfrac 2 z} - \paren {\dfrac 1 2 \cdot \dfrac {z^2} 2 + \dfrac {1 \times 3} {2 \times 4} \cdot \dfrac {z^4} 4 + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} \cdot \dfrac {z^6} 6 + \cdots}\) Sum of Logarithms
\(\ds \leadsto \ \ \) \(\ds \arcosh x\) \(=\) \(\ds \map \ln {2 x} - \paren {\dfrac 1 {2 \times 2 x^2} + \dfrac {1 \times 3} {2 \times 4 \times 4 x^4} + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6 \times 6 x^6} + \cdots}\) substituting $x \gets \dfrac 1 z$

$\blacksquare$


Also presented as

Some sources present this result in the form:

\(\ds \cosh^{-1} x\) \(=\) \(\ds \map \pm {\map \ln {2 x} - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} x^{2 n} } } }\) \(\ds \begin {cases} \text {$+$ if $x \ge 1$} \\ \text {$-$ if $x \le -1$} \end {cases}\)
\(\ds \) \(=\) \(\ds \map \pm {\map \ln {2 x} - \paren {\dfrac 1 {2 \times 2 x^2} + \dfrac {1 \times 3} {2 \times 4 \times 4 x^4} + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6 \times 6 x^6} + \cdots} }\) \(\ds \begin {cases} \text {$+$ if $x \ge 1$} \\ \text {$-$ if $x \le -1$} \end {cases}\)

This takes into account the interpretation that $\cosh^{-1} x$ is a multifunction arising from the fact that $\cosh x = \map \cosh {-1}$ for $\size x \ge 1$.


Also see


Sources

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