# normalizing factor

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##### 1: 3.6 Linear Difference Equations
It therefore remains to apply a normalizing factor $\Lambda$. … The normalizing factor $\Lambda$ can be the true value of $w_{0}$ divided by its trial value, or $\Lambda$ can be chosen to satisfy a known property of the wanted solution of the form … …
##### 2: 3.7 Ordinary Differential Equations
The eigenvalues $\lambda_{k}$ are simple, that is, there is only one corresponding eigenfunction (apart from a normalization factor), and when ordered increasingly the eigenvalues satisfy …
##### 3: 18.39 Applications in the Physical Sciences
All are written in the same form as the product of three factors: the square root of a weight function $w(x)$, the corresponding OP or EOP, and constant factors ensuring unit normalization. … With the normalization factor $\left(c\,h_{n}\right)^{-\ifrac{1}{2}}$ the $\psi_{n}$ are orthonormal in $L^{2}\left(\mathbb{R},\,\mathrm{d}x\right)$. …
##### 4: 28.5 Second Solutions $\operatorname{fe}_{n}$, $\operatorname{ge}_{n}$
The factors $C_{n}(q)$ and $S_{n}(q)$ in (28.5.1) and (28.5.2) are normalized so that
28.5.5 $(C_{n}(q))^{2}\int_{0}^{2\pi}(f_{n}(x,q))^{2}\,\mathrm{d}x=(S_{n}(q))^{2}\int_% {0}^{2\pi}(g_{n}(x,q))^{2}\,\mathrm{d}x=\pi.$
##### 6: Bibliography C
• B. C. Carlson and J. FitzSimons (2000) Reduction theorems for elliptic integrands with the square root of two quadratic factors. J. Comput. Appl. Math. 118 (1-2), pp. 71–85.
• B. C. Carlson (1964) Normal elliptic integrals of the first and second kinds. Duke Math. J. 31 (3), pp. 405–419.
• D. C. Cronemeyer (1991) Demagnetization factors for general ellipsoids. J. Appl. Phys. 70 (6), pp. 2911–2914.
• Cunningham Project (website)
• S. W. Cunningham (1969) Algorithm AS 24: From normal integral to deviate. Appl. Statist. 18 (3), pp. 290–293.
• ##### 7: 19.16 Definitions
19.16.1 $R_{F}\left(x,y,z\right)=\frac{1}{2}\int_{0}^{\infty}\frac{\,\mathrm{d}t}{s(t)},$
19.16.2 $R_{J}\left(x,y,z,p\right)=\frac{3}{2}\int_{0}^{\infty}\frac{\,\mathrm{d}t}{s(t% )(t+p)},$
19.16.2_5 $R_{G}\left(x,y,z\right)=\frac{1}{4}\int_{0}^{\infty}\frac{1}{s(t)}\*\left(% \frac{x}{t+x}+\frac{y}{t+y}+\frac{z}{t+z}\right)t\,\mathrm{d}t.$
19.16.4 $s(t)=\sqrt{t+x}\sqrt{t+y}\sqrt{t+z}.$
It should be noted that the integrals (19.16.1)–(19.16.2_5) have been normalized so that $R_{F}\left(1,1,1\right)=R_{J}\left(1,1,1,1\right)=R_{G}\left(1,1,1\right)=1$. …
##### 8: 18.2 General Orthogonal Polynomials
The orthogonality relations (18.2.1)–(18.2.3) each determine the polynomials $p_{n}(x)$ uniquely up to constant factors, which may be fixed by suitable standardizations. … If the polynomials $p_{n}(x)$ ($n=0,1,\ldots,N$) are orthogonal on a finite set $X$ of $N+1$ distinct points as in (18.2.3), then the polynomial $p_{N+1}(x)$ of degree $N+1$, up to a constant factor defined by (18.2.8) or (18.2.10), vanishes on $X$. … , of the form $w(x)\,\mathrm{d}x$) nor is it necessarily unique, up to a positive constant factor. However, if OP’s have an orthogonality relation on a bounded interval, then their orthogonality measure is unique, up to a positive constant factor. …
##### 9: 28.12 Definitions and Basic Properties
In consequence, for the Floquet solutions $w(z)$ the factor $e^{\pi\mathrm{i}\nu}$ in (28.2.14) is no longer $\pm 1$. …
28.12.2 $\lambda_{\nu}\left(-q\right)=\lambda_{\nu}\left(q\right)=\lambda_{-\nu}\left(q% \right).$
As in §28.7 values of $q$ for which (28.2.16) has simple roots $\lambda$ are called normal values with respect to $\nu$. For real values of $\nu$ and $q$ all the $\lambda_{\nu}\left(q\right)$ are real, and $q$ is normal. … If $q$ is a normal value of the corresponding equation (28.2.16), then these functions are uniquely determined as analytic functions of $z$ and $q$ by the normalization
##### 10: 10.22 Integrals
10.22.72 $\int_{0}^{\infty}J_{\mu}\left(at\right)J_{\nu}\left(bt\right)J_{\nu}\left(ct% \right)t^{1-\mu}\,\mathrm{d}t=\frac{(bc)^{\mu-1}\sin\left((\mu-\nu)\pi\right)(% \sinh\chi)^{\mu-\frac{1}{2}}}{(\frac{1}{2}\pi^{3})^{\frac{1}{2}}a^{\mu}}{% \mathrm{e}}^{(\mu-\frac{1}{2})\mathrm{i}\pi}Q^{\frac{1}{2}-\mu}_{\nu-\frac{1}{% 2}}\left(\cosh\chi\right),$ $\Re\mu>-\tfrac{1}{2},\Re\nu>-1,a>b+c,\cosh\chi=(a^{2}-b^{2}-c^{2})/(2bc)$.