Series Expansion for Pi Cotangent of Pi Lambda

From ProofWiki
Jump to navigation Jump to search





Theorem

Let $\lambda \in \R \setminus \Z$ be a real number which is not an integer.


Then:

$\ds \pi \cot \pi \lambda = \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \frac {2 \lambda} {\lambda^2 - n^2}$


Proof

Let $\map f x$ be the real function defined on $\openint 0 \pi$ as:

$\map f x = \cos \lambda x$


From Half-Range Fourier Cosine Series: $\cos \lambda x$ over $\openint 0 \pi$ its Fourier series can be expressed as:

$\ds \cos \lambda x \sim \frac {2 \lambda \sin \lambda \pi} \pi \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n x} {\lambda^2 - n^2} }$


Because of the nature of this expansion, we have that:

$\map f \pi = \map f {-\pi}$

and so the expansion holds for $x = \pi$.

Also note that because $\lambda$ is not an integer, $\sin \lambda \pi \ne 0$ and so $\cot \pi \lambda$ is defined.


So, setting $x = \pi$:

\(\ds \cos \lambda \pi\) \(=\) \(\ds \frac {2 \lambda \sin \lambda \pi} \pi \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n \pi} {\lambda^2 - n^2} }\)
\(\ds \leadsto \ \ \) \(\ds \pi \cot \pi \lambda\) \(=\) \(\ds 2 \lambda \paren {\frac 1 {2 \lambda^2} + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {\cos n \pi} {\lambda^2 - n^2} }\) Definition of Real Cotangent Function
\(\ds \) \(=\) \(\ds \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {2 \lambda \cos n \pi} {\lambda^2 - n^2}\) simplification
\(\ds \) \(=\) \(\ds \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \paren {-1}^n \frac {2 \lambda \paren {-1}^n} {\lambda^2 - n^2}\) Cosine of Multiple of Pi
\(\ds \) \(=\) \(\ds \frac 1 \lambda + \sum_{n \mathop = 1}^\infty \frac {2 \lambda} {\lambda^2 - n^2}\) simplification

$\blacksquare$


Sources