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怎樣展開$\tan(z)$
The simplest meromorphic functions with an infinite number of poles are the non-entire trigonometric functions. As an example, $\tan(z)$ is meromorphic with poles at $(n + \frac{1}{2})\pi$, $n = 0, \pm 1, \pm 2, ...$ The contours $\Gamma_k$ will be squares with vertices at $\pm \pi k \pm \pi k i$ traversed counterclockwise, $k > 1$, which are easily seen to satisfy the necessary conditions.
On the horizontal sides of $\Gamma_k$,
\[z = t \pm \pi k i,\ \ t \in [-\pi k, \pi k],\]
so
\[\lvert\tan(z)\rvert^2 = \frac{\sin^2(t)\cosh^2(\pi k) + \cos^2(t)\sinh^2(\pi k)}{\cos^2(t)\cosh^2(\pi k) + \sin^2(t)\sinh^2(\pi k)}\]
$\sinh(x) < \cosh(x)$ for all real $x$, which yields
\[\lvert\tan(z)\rvert^2 < \coth^2(\pi k)\]
For $x > 0$, $\coth(x)$ is continuous, decreasing, and bounded below by 1, so it follows that on the horizontal sides of $\Gamma_k$, $\lvert\tan(z)\rvert < \coth(\pi)$. Similarly, it can be shown that $\lvert\tan(z)\rvert < 1$ on the vertical sides of $\Gamma_k$.
With this bound on $\lvert\tan(z)\rvert$ we can see that
\[\oint_{\Gamma_k} \left|\frac{\tan(z)}{z}\right| dz \le \operatorname{length}(\Gamma_k) \max_{z\in \Gamma_k} \left|\frac{\tan(z)}{z}\right| < 8k \pi \frac{\coth(\pi)}{k\pi} = 8\coth(\pi) < \infty.\]
That is, the maximum of $\lvert\frac{1}{z}\rvert$ on $\Gamma_k$ occurs at the minimum of $\lvert z\rvert$, which is $k\pi$.
Therefore $p = 0$, and the partial fraction expansion of $\tan(z)$ looks like
\[\tan(z) = \sum_{k=0}^{\infty} \left(\operatorname{PP}(\tan(z); z = \lambda_k) + \operatorname{Res}_{z=\lambda_k} \frac{\tan(z)}{z}\right).\]
The principal parts and residues are easy enough to calculate, as all the poles of $\tan(z)$ are simple and have residue -1:
\[\operatorname{PP}(\tan(z); z = (n + \frac{1}{2})\pi) = \frac{-1}{z - (n + \frac{1}{2})\pi}\]
\[\operatorname{Res}_{z=(n + \frac{1}{2})\pi} \frac{\tan(z)}{z} = \frac{-1}{(n + \frac{1}{2})\pi}\]
We can ignore $\lambda_0 = 0$, since both $\tan(z)$ and $\frac{\tan(z)}{z}$ are analytic at 0, so there is no contribution to the sum, and ordering the poles $\lambda_k$ so that $\lambda_1 = \frac{\pi}{2}, \lambda_2 = \frac{-\pi}{2}, \lambda_3 = \frac{3\pi}{2}$, etc., gives
\[\tan(z) = \sum_{k=0}^{\infty} \left[\left(\frac{-1}{z - (k + \frac{1}{2})\pi} - \frac{1}{(k + \frac{1}{2})\pi}\right) + \left(\frac{-1}{z + (k + \frac{1}{2})\pi} + \frac{1}{(k + \frac{1}{2})\pi}\right)\right]\]
\[\tan(z) = \sum_{k=0}^{\infty} \frac{-2z}{z^2 - (k + \frac{1}{2})^2\pi^2}\] |
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hbghlyj
Post time 2024-4-19 18:58
