Power Series Expansion for Real Area Hyperbolic Sine

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Theorem

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

\(\ds \arsinh x\) \(=\) \(\ds \begin {cases}

\ds \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n \paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1} & : \size x < 1 \\ \ds \ln 2 x + \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n \paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} x^{2 n} } } & : x \ge 1 \\ \ds -\ln \paren {-2 x} - \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n \paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} x^{2 n} } } & : x \le -1 \\ \end {cases}\)

\(\ds \) \(\) \(\ds \)
\(\ds \) \(=\) \(\ds \begin {cases}

x - \dfrac {x^3} {2 \times 3} + \dfrac {1 \times 3 x^5} {2 \times 4 \times 5} - \dfrac {1 \times 3 \times 5 x^7} {2 \times 4 \times 6 \times 7} + \cdots & : \size x < 1 \\ \ln 2 x + \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 & : x \ge 1 \\ -\ln \paren {-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} & : x \le -1 \\ \end {cases}\)


Proof

Lemma 1

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

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

$\Box$


From Power Series is Termwise Integrable within Radius of Convergence, $(1)$ can be integrated term by term:

\(\ds \int_0^x \frac 1 {\sqrt {t^2 + 1} } \rd t\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \int_0^x \frac {\paren {-1}^n \paren {2 n}!} {2^{2 n} \paren {n!}^2} t^{2 n} \rd t\)
\(\ds \leadsto \ \ \) \(\ds \arsinh x\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n \paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1}\) Derivative of Inverse Hyperbolic Sine

$\Box$


We will now prove that the series converges for $-1 < x < 1$.

By Stirling's Formula:

\(\ds \frac {\paren {-1}^n \paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1}\) \(\sim\) \(\ds \frac {\paren {-1}^n \paren {2 n}^{2 n} e^{-2 n} \sqrt {4 \pi n} } {2^{2 n} n^{2 n} e^{-2 n} 2 \pi n} \frac {x^{2 n + 1} } {2 n + 1}\)
\(\ds \) \(=\) \(\ds \frac {\paren {-1}^n} {\sqrt {\pi n} } \frac {x^{2 n + 1} } {2 n + 1}\)


Then:

\(\ds \size {\frac 1 {\sqrt {\pi n} } \frac {x^{2 n + 1} } {2 n + 1} }\) \(<\) \(\ds \size {\frac {x^{2 n + 1} } {n^{3/2} } }\)
\(\ds \) \(\le\) \(\ds \frac 1 {n^{3/2} }\)


Hence by Convergence of P-Series:

$\ds \sum_{n \mathop = 1}^\infty \frac 1 {n^{3/2} }$

is convergent.


So by the Comparison Test, the Taylor series is convergent for $-1 \le x \le 1$.

$\Box$


Another lemma:

Lemma $2$

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

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

$\Box$


Let $x \ge 1$.

Let $z = \dfrac 1 x$.

Then we have:

$0 < \dfrac 1 z \le 1$

Now we consider:

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


Now let $x \le -1$.

We have that Inverse Hyperbolic Sine is Odd Function.

That is:

$\arsinh x = -\map \arsinh {-x}$

Thus for $x \le -1$:

\(\ds \arsinh x\) \(=\) \(\ds -\map \arsinh {-x}\)
\(\ds \) \(=\) \(\ds -\map \ln {-2 x} - \paren {\dfrac 1 {2 \times 2 \paren {-x}^2} - \dfrac {1 \times 3} {2 \times 4 \times 4 \paren {-x}^4} + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6 \times 6 \paren {-x}^6} - \cdots}\)
\(\ds \) \(=\) \(\ds -\map \ln {-2 x} - \paren {\dfrac 1 {2 \times 2 x^2} - \dfrac {1 \times 3} {2 \times 4 \times 4 x^4} + \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6 \times 6 x^6} - \cdots}\)

Hence the result.

$\blacksquare$


Also presented as

This can also be presented as:

\(\ds \arsinh x\) \(=\) \(\ds \begin {cases}

\ds \sum_{n \mathop = 0}^\infty \frac {\paren {-1}^n \paren {2 n}!} {2^{2 n} \paren {n!}^2} \frac {x^{2 n + 1} } {2 n + 1} & : \size x < 1 \\ \ds \pm \paren {\ln \size {2 x} + \paren {\sum_{n \mathop = 1}^\infty \frac {\paren {-1}^n \paren {2 n}!} {2^{2 n} \paren {n!}^2 \paren {2 n} x^{2 n} } } } & : \size x \ge 1 \end {cases}\)

\(\ds \) \(\) \(\ds \)
\(\ds \) \(=\) \(\ds \begin {cases}

x - \dfrac 1 2 \dfrac {x^3} 3 + \dfrac {1 \times 3} {2 \times 4} \dfrac {x^5} 5 - \dfrac {1 \times 3 \times 5} {2 \times 4 \times 6} \dfrac {x^7} 7 + \cdots & : \size x < 1 \\ \pm \paren {\ln 2 x + \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} & : \size x \ge 1 \end {cases}\)

where $\pm$ is $+$ for $x \ge 1$ and $-$ for $x \le -1$.


Also see


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