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1: 3.2 Linear Algebra

… ►With $𝐲={[y_1,y_2,\mathrm{\dots },y_n]}^{\mathrm{T}}$ the process of solution can then be regarded as first solving the equation $𝐋𝐲=𝐛$ for $𝐲$ (forward elimination), followed by the solution of $𝐔𝐱=𝐲$ for $𝐱$ (back substitution). … ►In solving $𝐀𝐱={[1,1,1]}^{\mathrm{T}}$, we obtain by forward elimination $𝐲={[1,-1,3]}^{\mathrm{T}}$, and by back substitution $𝐱={[\frac{1}{6},\frac{1}{6},\frac{1}{6}]}^{\mathrm{T}}$. … ►and back substitution is $x_n=y_n/d_n$, followed by …

2: About the Project

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Figure 1: The Editors and 9 of the 10 Associate Editors of the DLMF Project (photo taken at 3rd Editors Meeting, April, 2001). …The back row, from left to right: William P. …

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3: 3.6 Linear Difference Equations

… ►(This part of the process is back substitution.) …

4: Philip J. Davis

… ►Davis left NBS in 1963 to become a faculty member in the Division of Applied Mathematics at Brown University, but during the early development of the DLMF, which started in 1998, he was invited back to give a talk and speak with DLMF project members about their plans. …

5: 27.13 Functions

… ►This conjecture dates back to 1742 and was undecided in 2009, although it has been confirmed numerically up to very large numbers. …

6: 8.15 Sums

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8.15.2 $$a\sum _{k=1}^{\mathrm{\infty }}\left(\frac{{\mathrm{e}}^{2\pi \mathrm{i}k(z+h)}}{{\left(2\pi \mathrm{i}k\right)}^{a+1}}\mathrm{\Gamma }(a,2\pi \mathrm{i}kz)+\frac{{\mathrm{e}}^{-2\pi \mathrm{i}k(z+h)}}{{\left(-2\pi \mathrm{i}k\right)}^{a+1}}\mathrm{\Gamma }(a,-2\pi \mathrm{i}kz)\right)=\zeta (-a,z+h)+\frac{z^{a+1}}{a+1}+\left(h-\frac{1}{2}\right)z^a,$$ $h\in [0,1]$.

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Symbols:

$\zeta (s,a)$: Hurwitz zeta function, $\pi$: the ratio of the circumference of a circle to its diameter, $[a,b]$: closed interval, $\in$: element of, $\mathrm{e}$: base of natural logarithm, $\mathrm{i}$: imaginary unit, $\mathrm{\Gamma }(a,z)$: incomplete gamma function, $z$: complex variable, $a$: parameter and $k$: nonnegative integer

Proof sketch:

Start with and $z>0$. For $\mathrm{\Gamma }(a,\pm 2\pi \mathrm{i}kz)$ take integral representation (8.2.2) and use the substitution $t=\pm 2\pi \mathrm{i}kz(1+\tau )$. The sum and the integral can be interchanged, and the sum can be evaluated via (4.6.1). Use integration by parts. This will result in $\left(h-\frac{1}{2}\right)z^a$ plus two integrals with infinitely many poles. The residue theorem (§1.10(iv)) will give us an infinite series which can be identified via (25.11.1). For other values of $a$ and $z$ use analytic continuation.

Referenced by:

§25.11(x), §25.11(x), §8.15, Erratum (V1.1.3) for Additions

Permalink:

http://dlmf.nist.gov/8.15.E2

Encodings:

pMML, png, TeX

Addition (effective with 1.1.3):

This equation was added.

See also:

Annotations for §8.15 and Ch.8

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7: 29.10 Lamé Functions with Imaginary Periods

… ►The substitutions …

8: 10.12 Generating Function and Associated Series

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9: 10.35 Generating Function and Associated Series

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10: 27.5 Inversion Formulas

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