# residue

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##### 1: 27.15 Chinese Remainder Theorem
By the Chinese remainder theorem each integer in the data can be uniquely represented by its residues (mod $m_{1}$), (mod $m_{2}$), (mod $m_{3}$), and (mod $m_{4}$), respectively. Because each residue has no more than five digits, the arithmetic can be performed efficiently on these residues with respect to each of the moduli, yielding answers $a_{1}\pmod{m_{1}}$, $a_{2}\pmod{m_{2}}$, $a_{3}\pmod{m_{3}}$, and $a_{4}\pmod{m_{4}}$, where each $a_{j}$ has no more than five digits. …
##### 2: 1.10 Functions of a Complex Variable
The coefficient $a_{-1}$ of $(z-z_{0})^{-1}$ in the Laurent series for $f(z)$ is called the residue of $f(z)$ at $z_{0}$, and denoted by $\Residue_{z=z_{0}}[f(z)]$, $\Residue\limits_{z=z_{0}}[f(z)]$, or (when there is no ambiguity) $\Residue[f(z)]$. …
###### §1.10(iv) Residue Theorem
Suppose that … is analytic in $\mathbb{C}$, except for simple poles at $z=z_{n}$ of residue $a_{n}$. …
##### 3: 2.5 Mellin Transform Methods
2.5.11 $\Residue_{z=n}\left[x^{-z}\mathscr{M}\mskip-3.0muf\mskip 3.0mu\left(1-z\right)% \mathscr{M}\mskip-3.0muh\mskip 3.0mu\left(z\right)\right]=(a_{n}\ln x+b_{n})x^% {-n},$
2.5.35 $I_{jk}(x)=\sum_{p_{jk}<\Re z
2.5.46 $\Residue_{z=k}\left[-\zeta^{z-1}\Gamma\left(1-z\right)\pi\csc\left(\pi z\right% )\right]=\left(-\ln\zeta+\psi\left(k\right)\right)\dfrac{\zeta^{k-1}}{(k-1)!},$
2.5.47 $\Residue_{z=1}\left[-\zeta^{z-1}\Gamma\left(1-z\right)\mathscr{M}\mskip-3.0muh% _{2}\mskip 3.0mu\left(z\right)\right]=\left(-\ln\zeta-\gamma\right)-\mathscr{M% }\mskip-3.0muh_{1}\mskip 3.0mu\left(1\right),$
##### 4: 5.2 Definitions
It is a meromorphic function with no zeros, and with simple poles of residue $(-1)^{n}/n!$ at $z=-n$. …$\psi\left(z\right)$ is meromorphic with simple poles of residue $-1$ at $z=-n$. …
##### 5: 5.19 Mathematical Applications
By translating the contour parallel to itself and summing the residues of the integrand, asymptotic expansions of $f(z)$ for large $|z|$, or small $|z|$, can be obtained complete with an integral representation of the error term. …
##### 6: 8.15 Sums
8.15.2 $a\sum_{k=1}^{\infty}\left(\frac{{\mathrm{e}}^{2\pi\mathrm{i}k(z+h)}}{\left(2% \pi\mathrm{i}k\right)^{a+1}}\Gamma\left(a,2\pi\mathrm{i}kz\right)+\frac{{% \mathrm{e}}^{-2\pi\mathrm{i}k(z+h)}}{\left(-2\pi\mathrm{i}k\right)^{a+1}}% \Gamma\left(a,-2\pi\mathrm{i}kz\right)\right)=\zeta\left(-a,z+h\right)+\frac{z% ^{a+1}}{a+1}+\left(h-\tfrac{1}{2}\right)z^{a},$ $h\in[0,1]$.
##### 7: 27.2 Functions
Such a set is a reduced residue system modulo $n$. …
##### 8: Mathematical Introduction
 $(a,b]$ or $[a,b)$ half-closed intervals. … residue. …
##### 9: Bibliography V
• J. F. Van Diejen and V. P. Spiridonov (2001) Modular hypergeometric residue sums of elliptic Selberg integrals. Lett. Math. Phys. 58 (3), pp. 223–238.