# truncated exponential series

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## 6 matching pages

##### 1: 8.11 Asymptotic Approximations and Expansions
This reference also contains explicit formulas for the coefficients in terms of Stirling numbers. …
##### 2: 2.11 Remainder Terms; Stokes Phenomenon
When a rigorous bound or reliable estimate for the remainder term is unavailable, it is unsafe to judge the accuracy of an asymptotic expansion merely from the numerical rate of decrease of the terms at the point of truncation. … If the results agree within $S$ significant figures, then it is likely—but not certain—that the truncated asymptotic series will yield at least $S$ correct significant figures for larger values of $x$. … Truncation after 5 terms yields 0. … The process just used is equivalent to re-expanding the remainder term of the original asymptotic series (2.11.24) in powers of $1/(x+5)$ and truncating the new series optimally. … Optimum truncation occurs just prior to the numerically smallest term, that is, at $s_{4}$. …
##### 3: 9.7 Asymptotic Expansions
In (9.7.5) and (9.7.6) the $n$th error term, that is, the error on truncating the expansion at $n$ terms, is bounded in magnitude by the first neglected term and has the same sign, provided that the following term is of opposite sign, that is, if $n\geq 0$ for (9.7.5) and $n\geq 1$ for (9.7.6). … In (9.7.9)–(9.7.12) the $n$th error term in each infinite series is bounded in magnitude by the first neglected term and has the same sign, provided that the following term in the series is of opposite sign. …
$\mathrm{Ai}\left(x\right)\leq\frac{e^{-\xi}}{2\sqrt{\pi}x^{1/4}}$ ,
$|\mathrm{Ai}'\left(x\right)|\leq\frac{x^{1/4}e^{-\xi}}{2\sqrt{\pi}}\left(1+% \frac{7}{72\xi}\right)$ ,
$\mathrm{Bi}\left(x\right)\leq\frac{{\mathrm{e}}^{\xi}}{\sqrt{\pi}x^{1/4}}\left% (1+\left(\chi(\tfrac{7}{6})+1\right)\frac{5}{72\xi}\right),$
##### 4: 19.5 Maclaurin and Related Expansions
###### §19.5 Maclaurin and Related Expansions
Coefficients of terms up to $\lambda^{49}$ are given in Lee (1990), along with tables of fractional errors in $K\left(k\right)$ and $E\left(k\right)$, $0.1\leq k^{2}\leq 0.9999$, obtained by using 12 different truncations of (19.5.6) in (19.5.8) and (19.5.9). … An infinite series for $\ln K\left(k\right)$ is equivalent to the infinite product …
##### 5: 11.6 Asymptotic Expansions
If the series on the right-hand side of (11.6.1) is truncated after $m(\geq 0)$ terms, then the remainder term $R_{m}(z)$ is $O\left(z^{\nu-2m-1}\right)$. …
11.6.5 $\mathbf{H}_{\nu}\left(z\right),\mathbf{L}_{\nu}\left(z\right)\sim\frac{z}{\pi% \nu\sqrt{2}}\left(\frac{ez}{2\nu}\right)^{\nu},$ $|\operatorname{ph}\nu|\leq\pi-\delta$.
More fully, the series (11.2.1) and (11.2.2) can be regarded as generalized asymptotic expansions (§2.1(v)). …
##### 6: 2.10 Sums and Sequences
$e^{C}$ is sometimes called Glaisher’s constant. … In both expansions the remainder term is bounded in absolute value by the first neglected term in the sum, and has the same sign, provided that in the case of (2.10.7), truncation takes place at $s=2m-1$, where $m$ is any positive integer satisfying $m\geq\frac{1}{2}(\alpha+1)$. … The asymptotic behavior of entire functions defined by Maclaurin series can be approached by converting the sum into a contour integral by use of the residue theorem and applying the methods of §§2.4 and 2.5. … What is the asymptotic behavior of $f_{n}$ as $n\to\infty$ or $n\to-\infty$? More specially, what is the behavior of the higher coefficients in a Taylor-series expansion? … Here the branch of $\left(e^{-i\alpha}-z\right)^{-1/2}$ is continuous in the $z$-plane cut along the outward-drawn ray through $z=e^{-i\alpha}$ and equals $e^{i\alpha/2}$ at $z=0$. …