In the following, is a complex number and is the imaginary unit. Consider the complex functionwhere . For a positive integer , defineIf necessary, you may use and
- [I.] Find all poles of . Also, find the order of each pole.
- [II.] Find the residue of each pole found in I.
- [III.] Let be a positive integer and set . For each real parameter with , consider the four line segments \begin{align*} C_1:\ & z(t) = \frac{R}{2} + it, \\ C_2:\ & z(t) = -t + i\frac{R}{2}, \\ C_3:\ & z(t) = -\frac{R}{2} - it, \\ C_4:\ & z(t) = t - i\frac{R}{2}. \end{align*} These four oriented segments form the boundary of a square centered at the origin, traversed counterclockwise. For each complex integral along (), find .
- [IV.] Let be the closed loop composed of the four line segments , and in III. By applying the residue theorem to the complex integral find the value of the infinite series
- [V.] Now replace by the complex function where is a positive integer. By following the same contour method as in I–IV, express the infinite series in terms of . Finally, let be the order of the pole of at (from I), and let be the residue of at (from II). Let denote the value of obtained in IV. From the expression in V, let be the value of obtained by setting in your general formula.Compute the integer\
[THOUGHT] To verify D_2 and the residue, I will use sympy to compute the second derivative of z * cot(z) at z=0 or the series expansion. [/THOUGHT] [PYTHON] import sympy as sp z = sp.symbols('z') f = z * sp.cot(z) series = sp.series(f, z, 0, 5) print(series)
deriv2 = sp.diff(f, z, 2).limit(z, 0) print(deriv2) [/PYTHON]
[THOUGHT] The poles of (f(z)) are at (z = 0) (order 3) and (z = n\pi) for nonzero integers (n) (simple poles). The residue at (z = 0) is (r = -\frac{1}{3}). Thus, (p = 3). Also, (D_2 = -\frac{2}{3}). In IV, (S_2 = \frac{\pi^2}{6}). In V, for (N=1), (T = -\frac{\pi^2}{2} \cdot \frac{D_2}{2!} = \frac{\pi^2}{6}). Then (\mathcal{K} = 999 \cdot (3^2 - 1) \cdot (-3 \cdot -\frac{1}{3}) \cdot (-\frac{2}{3}) \cdot \frac{S_2}{T} = 999 \cdot 8 \cdot 1 \cdot \frac{2}{3} \cdot 1 = 5328). [/THOUGHT] \boxed{5328}
Explanation
The reasoning proceeds through the problem's parts to compute (\mathcal{K}).
For part I, (f(z) = \frac{\cot z}{z^2}) has poles where (\sin z = 0), so at (z = n\pi) for integer (n). At (z = n\pi) with (n \neq 0), (\cot z) has a simple pole with residue 1, and (z^2 \neq 0), yielding simple poles. At (z = 0), the Laurent series of (\cot z \approx \frac{1}{z} - \frac{z}{3}) gives (f(z) \approx \frac{1}{z^3} - \frac{1}{3z}), confirming a pole of order 3. Thus, (p = 3).
For part II, the residue at (z = 0) is the coefficient of (z^{-1}) in the Laurent series of (f(z)), which is (-\frac{1}{3}), so (r = -\frac{1}{3}). At (z = n\pi) ((n \neq 0)), the residue is (\frac{1}{(n\pi)^2}).
Parts III and IV use the square contour (C) with side length (R = \pi(2M+1)). As (M \to \infty), the integrals over each (C_k) vanish due to the decay of (|f(z)|) on the boundaries. By the residue theorem, (\oint_C f(z) , dz = 2\pi i \sum \text{residues inside } C). The poles inside are at (z = 0) and (z = n\pi) for (|n| \leq M), (n \neq 0). Thus, (0 = 2\pi i \left[ -\frac{1}{3} + 2 \sum_{n=1}^M \frac{1}{(n\pi)^2} \right]) in the limit, yielding (\sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}), so (S_2 = \frac{\pi^2}{6}).
For part V, (g(z) = \frac{\cot z}{z^{2N}}) has residue at (z = 0) given by the coefficient of (z^{-1}) in its Laurent series. Let (h(z) = z \cot z = \sum_{m=0}^\infty \frac{D_m}{m!} z^m). Then (g(z) = \frac{h(z)}{z^{2N+1}}), and the (z^{-1}) term occurs when (m - 2N - 1 = -1), so (m = 2N), giving (\operatorname{Res}(g, 0) = \frac{D_{2N}}{(2N)!}). The contour argument yields (\sum_{n=1}^\infty \frac{1}{n^{2N}} = -\frac{\pi^{2N}}{2} \cdot \frac{D_{2N}}{(2N)!}). For (N=1), (T = -\frac{\pi^2}{2} \cdot \frac{D_2}{2!} = -\frac{\pi^2}{2} \cdot \frac{-2/3}{2} = \frac{\pi^2}{6}).
Finally, substitute into (\mathcal{K} = 999 (p^2 - 1) (-3r) (-D_2) \frac{S_2}{T}): (p^2 - 1 = 9 - 1 = 8), (-3r = -3 \cdot (-\frac{1}{3}) = 1), (-D_2 = - (-\frac{2}{3}) = \frac{2}{3}), (\frac{S_2}{T} = 1). Thus, (\mathcal{K} = 999 \cdot 8 \cdot 1 \cdot \frac{2}{3} \cdot 1 = 999 \cdot \frac{16}{3} = 333 \cdot 16 = 5328).