Power Series Expansion for Cosecant Function

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Theorem

The cosecant function has a Laurent series expansion:

\(\ds \csc x\) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \dfrac {\paren {-1}^{n - 1} 2 \paren {2^{2 n - 1} - 1} B_{2 n} \, x^{2 n - 1} } {\paren {2 n}!}\)
\(\ds \) \(=\) \(\ds \frac 1 x + \frac x 6 + \frac {7 x^3} {360} + \frac {31 x^5} {15 \, 120} + \cdots\)


where $B_n$ denotes the Bernoulli numbers.


This converges for $0 < \size x < \pi$.


Proof

\(\ds \sin x\) \(=\) \(\ds 2 \sin \dfrac x 2 \cos \dfrac x 2\) Double Angle Formula for Sine
\(\ds \leadstoandfrom \ \ \) \(\ds \dfrac 1 {\sin x}\) \(=\) \(\ds \dfrac 1 {2 \sin \dfrac x 2 \cos \dfrac x 2}\) taking the reciprocal of both sides
\(\ds \leadstoandfrom \ \ \) \(\ds \csc x\) \(=\) \(\ds \dfrac 1 2 \csc \dfrac x 2 \sec \dfrac x 2\) Cosecant is Reciprocal of Sine, Secant is Reciprocal of Cosine
\(\ds \) \(=\) \(\ds \dfrac 1 2 \csc \dfrac x 2 \dfrac 2 {e^{i \frac x 2} + e^{-i \frac x 2} }\) Euler's Secant Identity
\(\ds \) \(=\) \(\ds \csc \dfrac x 2 \dfrac {e^{-i \frac x 2} } {1 + e^{-i x} }\)
\(\ds \leadstoandfrom \ \ \) \(\ds \csc x \paren {1 + e^{-i x} }\) \(=\) \(\ds \csc \dfrac x 2 e^{-i \frac x 2}\) multiplying both sides by $1 + e^{-i x}$
\(\ds \leadstoandfrom \ \ \) \(\ds \csc x\) \(=\) \(\ds \csc \dfrac x 2 e^{-i \frac x 2} - \csc x e^{-i x}\) subtracting $\csc x e^{-i x}$ from both sides
\(\ds \) \(=\) \(\ds \dfrac {2 i} {e^{i \frac x 2} - e^{-i \frac x 2} } e^{-i \frac x 2} - \dfrac {2 i} {e^{i x} - e^{-i x} } e^{-i x}\) Euler's Cosecant Identity
\(\ds \) \(=\) \(\ds 2 \dfrac i {e^{i x} - 1} - \dfrac {2 i} {e^{2 i x} - 1}\)
\(\ds \) \(=\) \(\ds \dfrac 1 x \paren {2 \dfrac {i x} {e^{i x} - 1} - \dfrac {2 i x} {e^{2 i x} - 1} }\)
\(\ds \) \(=\) \(\ds \dfrac 1 x \sum_{n \mathop = 0}^\infty \paren {2 \dfrac {\paren {i x}^n B_n} {n!} - \dfrac {\paren {2 i x}^n B_n} {n!} }\) Definition of Bernoulli Numbers
\(\ds \) \(=\) \(\ds \dfrac 1 x \sum_{n \mathop = 0}^\infty \dfrac {B_n \paren {2 \paren {i x}^n - \paren {2 i x}^n} } {n!}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \dfrac {B_n i^n x^{n - 1} 2 \paren {1 - 2^{n - 1} } } {n!}\)
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \dfrac {B_{2 n} i^{2 n} x^{2 n - 1} 2 \paren {1 - 2^{2 n - 1} } } {\paren {2 n}!}\) Odd Bernoulli Numbers Vanish and the term $n = 1$ vanishes
\(\ds \) \(=\) \(\ds \sum_{n \mathop = 0}^\infty \dfrac {B_{2 n} \paren {-1}^{n - 1} x^{2 n - 1} 2 \paren {2^{2 n - 1} - 1} } {\paren {2 n}!}\)

$\Box$


Convergence

By Combination Theorem for Limits of Real Functions we can deduce the following.

\(\ds \lim_{n \mathop \to \infty} \size {\frac {\frac {B_{2 n + 2} \paren {-1}^n x^{2 n + 1} 2 \paren {2^{2 n + 1} - 1} } {\paren {2 n + 2}!} } {\frac {B_{2 n} \paren {-1}^{n - 1} x^{2 n - 1} 2 \paren {2^{2 n - 1} - 1} } {\paren {2 n}!} } }\) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {\frac {2^{2 n + 1} - 1} {2^{2 n - 1} - 1} \frac 1 {\paren {2 n + 1} \paren {2 n + 2} } \frac {B_{2 n + 2} } {B_{2 n} } } x^2\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {\frac {2^{2 n + 1} - 1} {2^{2 n - 1} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } \frac 1 2 x^2\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {\frac {2^{2 n + 1} } {2^{2 n - 1} - 1} - \frac 1 {2^{2 n - 1} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } \frac 1 2 x^2\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {4 \frac {2^{2 n - 1} - 1 + 1} {2^{2 n - 1} - 1} - \frac 1 {2^{2 n - 1} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } \frac 1 2 x^2\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {4 + \frac 4 {2^{2 n - 1} - 1} - \frac 1 {2^{2 n - 1} - 1} } \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } \frac 1 2 x^2\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {B_{2 n + 2} } {B_{2 n} } } 2 x^2\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {\frac 1 {\paren {2 n + 1} \paren {n + 1} } \frac {\paren {-1}^{n + 2} 4 \sqrt {\pi \paren {n + 1} } \paren {\frac {n + 1} {\pi e} }^{2 n + 2} } {\paren {-1}^{n + 1} 4 \sqrt {\pi n} \paren {\frac n {\pi e} }^{2 n} } } 2 x^2\) Asymptotic Formula for Bernoulli Numbers
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {\frac {\paren {n + 1}^2} {\paren {2 n + 1} \paren {n + 1} } \sqrt {\frac {n + 1} n} \paren {\frac {n + 1} n}^{2 n} } \frac 2 {\pi^2 e^2} x^2\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {\paren {\frac {n + 1} n }^{2 n} } \frac 1 {\pi^2 e^2} x^2\)
\(\ds \) \(=\) \(\ds \lim_{n \mathop \to \infty} \size {\paren {\paren {1 + \frac 1 n}^n}^2} \frac 1 {\pi^2 e^2} x^2\)
\(\ds \) \(=\) \(\ds \frac {e^2} {\pi^2 e^2} x^2\) Definition of Euler's Number as Limit of Sequence
\(\ds \) \(=\) \(\ds \frac 1 {\pi^2} x^2\)

This is less than $1$ if $\size x < \pi$.

Hence by the Ratio Test, the outer radius of convergence is $\pi$

The principal part of the Laurent series is finite so converges for $x \ne 0$.

Thus we have the annulus of convergence to be $0 < \size x < \pi$.

$\blacksquare$


Sources

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