# as z→0

(0.019 seconds)

## 1—10 of 610 matching pages

##### 1: 4.31 Special Values and Limits
4.31.1 $\lim_{z\to 0}\frac{\sinh z}{z}=1,$
4.31.2 $\lim_{z\to 0}\frac{\tanh z}{z}=1,$
##### 2: 10.72 Mathematical Applications
If $f(z)$ has a double zero $z_{0}$, or more generally $z_{0}$ is a zero of order $m$, $m=2,3,4,\dotsc$, then uniform asymptotic approximations (but not expansions) can be constructed in terms of Bessel functions, or modified Bessel functions, of order $1/(m+2)$. The number $m$ can also be replaced by any real constant $\lambda$ $(>-2)$ in the sense that $(z-z_{0})^{-\lambda}$ $f(z)$ is analytic and nonvanishing at $z_{0}$; moreover, $g(z)$ is permitted to have a single or double pole at $z_{0}$. … In regions in which the function $f(z)$ has a simple pole at $z=z_{0}$ and $(z-z_{0})^{2}g(z)$ is analytic at $z=z_{0}$ (the case $\lambda=-1$ in §10.72(i)), asymptotic expansions of the solutions $w$ of (10.72.1) for large $u$ can be constructed in terms of Bessel functions and modified Bessel functions of order $\pm\sqrt{1+4\rho}$, where $\rho$ is the limiting value of $(z-z_{0})^{2}g(z)$ as $z\to z_{0}$. … In (10.72.1) assume $f(z)=f(z,\alpha)$ and $g(z)=g(z,\alpha)$ depend continuously on a real parameter $\alpha$, $f(z,\alpha)$ has a simple zero $z=z_{0}(\alpha)$ and a double pole $z=0$, except for a critical value $\alpha=a$, where $z_{0}(a)=0$. …These approximations are uniform with respect to both $z$ and $\alpha$, including $z=z_{0}(a)$, the cut neighborhood of $z=0$, and $\alpha=a$. …
##### 3: 15.15 Sums
15.15.1 $\mathbf{F}\left({a,b\atop c};\frac{1}{z}\right)=\left(1-\frac{z_{0}}{z}\right)% ^{-a}\sum_{s=0}^{\infty}\frac{{\left(a\right)_{s}}}{s!}\*\mathbf{F}\left({-s,b% \atop c};\frac{1}{z_{0}}\right)\left(1-\frac{z}{z_{0}}\right)^{-s}.$
Here $z_{0}$ (${\neq 0}$) is an arbitrary complex constant and the expansion converges when $|z-z_{0}|>\max(|z_{0}|,|z_{0}-1|)$. …
##### 4: 1.10 Functions of a Complex Variable
Then $z=z_{0}$ is an isolated singularity of $f(z)$. … For example, $z=0$ is a branch point of $\sqrt{z}$. … Furthermore, if $g(z)$ is analytic at $z_{0}$, then … Let $w_{0}=f(z_{0})$. … Also, if in addition $g(z)$ is analytic at $z_{0}$, then …
##### 5: 9.4 Maclaurin Series
9.4.1 $\mathrm{Ai}\left(z\right)=\mathrm{Ai}\left(0\right)\left(1+\frac{1}{3!}z^{3}+% \frac{1\cdot 4}{6!}z^{6}+\frac{1\cdot 4\cdot 7}{9!}z^{9}+\cdots\right)+\mathrm% {Ai}'\left(0\right)\left(z+\frac{2}{4!}z^{4}+\frac{2\cdot 5}{7!}z^{7}+\frac{2% \cdot 5\cdot 8}{10!}z^{10}+\cdots\right),$
9.4.2 $\mathrm{Ai}'\left(z\right)=\mathrm{Ai}'\left(0\right)\left(1+\frac{2}{3!}z^{3}% +\frac{2\cdot 5}{6!}z^{6}+\frac{2\cdot 5\cdot 8}{9!}z^{9}+\cdots\right)+% \mathrm{Ai}\left(0\right)\left(\frac{1}{2!}z^{2}+\frac{1\cdot 4}{5!}z^{5}+% \frac{1\cdot 4\cdot 7}{8!}z^{8}+\cdots\right),$
9.4.3 $\mathrm{Bi}\left(z\right)=\mathrm{Bi}\left(0\right)\left(1+\frac{1}{3!}z^{3}+% \frac{1\cdot 4}{6!}z^{6}+\frac{1\cdot 4\cdot 7}{9!}z^{9}+\cdots\right)+\mathrm% {Bi}'\left(0\right)\left(z+\frac{2}{4!}z^{4}+\frac{2\cdot 5}{7!}z^{7}+\frac{2% \cdot 5\cdot 8}{10!}z^{10}+\cdots\right),$
9.4.4 $\mathrm{Bi}'\left(z\right)=\mathrm{Bi}'\left(0\right)\left(1+\frac{2}{3!}z^{3}% +\frac{2\cdot 5}{6!}z^{6}+\frac{2\cdot 5\cdot 8}{9!}z^{9}+\cdots\right)+% \mathrm{Bi}\left(0\right)\left(\frac{1}{2!}z^{2}+\frac{1\cdot 4}{5!}z^{5}+% \frac{1\cdot 4\cdot 7}{8!}z^{8}+\cdots\right).$
##### 6: 20.4 Values at $z$ = 0
###### §20.4 Values at $z$ = 0
20.4.3 $\theta_{2}\left(0,q\right)=2q^{1/4}\prod\limits_{n=1}^{\infty}\left(1-q^{2n}% \right)\left(1+q^{2n}\right)^{2},$
20.4.4 $\theta_{3}\left(0,q\right)=\prod\limits_{n=1}^{\infty}\left(1-q^{2n}\right)% \left(1+q^{2n-1}\right)^{2},$
20.4.5 $\theta_{4}\left(0,q\right)=\prod\limits_{n=1}^{\infty}\left(1-q^{2n}\right)% \left(1-q^{2n-1}\right)^{2}.$
20.4.12 $\frac{\theta_{1}'''\left(0,q\right)}{\theta_{1}'\left(0,q\right)}=\frac{\theta% _{2}''\left(0,q\right)}{\theta_{2}\left(0,q\right)}+\frac{\theta_{3}''\left(0,% q\right)}{\theta_{3}\left(0,q\right)}+\frac{\theta_{4}''\left(0,q\right)}{% \theta_{4}\left(0,q\right)}.$
##### 7: 4.6 Power Series
4.6.3 $\ln z=(z-1)-\tfrac{1}{2}(z-1)^{2}+\tfrac{1}{3}(z-1)^{3}-\cdots,$ $|z-1|\leq 1$, $z\neq 0$,
4.6.4 $\ln z=2\left(\left(\frac{z-1}{z+1}\right)+\frac{1}{3}\left(\frac{z-1}{z+1}% \right)^{3}+\frac{1}{5}\left(\frac{z-1}{z+1}\right)^{5}+\cdots\right),$ $\Re z\geq 0$, $z\neq 0$,
4.6.6 $\ln\left(z+a\right)=\ln a+2\left(\left(\frac{z}{2a+z}\right)+\frac{1}{3}\left(% \frac{z}{2a+z}\right)^{3}+\frac{1}{5}\left(\frac{z}{2a+z}\right)^{5}+\cdots% \right),$ $a>0$, $\Re z\geq-a$, $z\neq-a$.
If $a=0,1,2,\dots$, then the series terminates and $z$ is unrestricted. …
##### 8: 32.4 Isomonodromy Problems
32.4.4 $\mathbf{A}(z,\lambda)=(4\lambda^{4}+2w^{2}+z)\begin{bmatrix}1&0\\ 0&-1\end{bmatrix}-i(4\lambda^{2}w+2w^{2}+z)\begin{bmatrix}0&-i\\ i&0\end{bmatrix}-\left(2\lambda w^{\prime}+\frac{1}{2\lambda}\right)\begin{% bmatrix}0&1\\ 1&0\end{bmatrix},$
32.4.6 $\mathbf{A}(z,\lambda)=-i(4\lambda^{2}+2w^{2}+z)\begin{bmatrix}1&0\\ 0&-1\end{bmatrix}-2w^{\prime}\begin{bmatrix}0&-i\\ i&0\end{bmatrix}+\left(4\lambda w-\frac{\alpha}{\lambda}\right)\begin{bmatrix}% 0&1\\ 1&0\end{bmatrix},$
32.4.8 $\mathbf{A}(z,\lambda)=\begin{bmatrix}\tfrac{1}{4}z&0\\ 0&-\tfrac{1}{4}z\end{bmatrix}+\begin{bmatrix}-\tfrac{1}{2}\theta_{\infty}&u_{0% }\\ u_{1}&\tfrac{1}{2}\theta_{\infty}\end{bmatrix}\dfrac{1}{\lambda}+\begin{% bmatrix}v_{0}-\tfrac{1}{4}z&-v_{1}v_{0}\\ \ifrac{(v_{0}-\tfrac{1}{2}z)}{v_{1}}&\tfrac{1}{4}z-v_{0}\end{bmatrix}\frac{1}{% \lambda^{2}},$
32.4.10 $zu_{0}^{\prime}=\theta_{\infty}u_{0}-zv_{0}v_{1},$
32.4.12 $zv_{0}^{\prime}=2v_{0}u_{1}v_{1}+v_{0}+(u_{0}(2v_{0}-z)/v_{1}),$
##### 9: 29.16 Asymptotic Expansions
The approximations for Lamé polynomials hold uniformly on the rectangle $0\leq\Re z\leq K$, $0\leq\Im z\leq{K^{\prime}}$, when $nk$ and $nk^{\prime}$ assume large real values. …
##### 10: 7.9 Continued Fractions
7.9.1 $\sqrt{\pi}e^{z^{2}}\operatorname{erfc}z=\cfrac{z}{z^{2}+\cfrac{\frac{1}{2}}{1+% \cfrac{1}{z^{2}+\cfrac{\frac{3}{2}}{1+\cfrac{2}{z^{2}+\cdots}}}}},$ $\Re z>0$,
7.9.2 $\sqrt{\pi}e^{z^{2}}\operatorname{erfc}z=\cfrac{2z}{2z^{2}+1-\cfrac{1\cdot 2}{2% z^{2}+5-\cfrac{3\cdot 4}{2z^{2}+9-\cdots}}},$ $\Re z>0$,
7.9.3 $w\left(z\right)=\frac{i}{\sqrt{\pi}}\cfrac{1}{z-\cfrac{\frac{1}{2}}{z-\cfrac{1% }{z-\cfrac{\frac{3}{2}}{z-\cfrac{2}{z-\cdots}}}}},$ $\Im z>0$.