Continued in part B, part C, part D, and part E.
Fundamental formula:
For index
- The sixth triangular number:
$$T(6) = \frac{6(6+1)}{2} = \frac{42}{2} = 21$$ - Substitution into the formula:
$$\Lambda(6) = \sqrt{\frac{2}{6(7)}} = \sqrt{\frac{2}{42}} = \sqrt{\frac{1}{21}} = \frac{1}{\sqrt{21}}$$ - High-precision numerical evaluation:
$$\Lambda(6) = 0.2182178902359923812660974854156\ldots$$ $$100 \cdot \Lambda(6) = 21.82178902359923812660974854156\ldots$$
Given
- Partial fraction decomposition:
$$\frac{2}{n(n+1)} = \frac{A}{n} + \frac{B}{n+1} \implies 2 = A(n+1) + Bn$$ Setting$n = 0$ gives$A = 2$ . Setting$n = -1$ gives$B = -2$ .$$\Lambda(n)^{2} = 2\left(\frac{1}{n} - \frac{1}{n+1}\right)$$ - Summation up to upper limit
$N$ :$$\sum_{n=1}^{N} \Lambda(n)^{2} = 2 \sum_{n=1}^{N} \left(\frac{1}{n} - \frac{1}{n+1}\right) = 2\left[\left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \cdots + \left(\frac{1}{N} - \frac{1}{N+1}\right)\right]$$ - Telescoping cancellation of all intermediate terms:
$$\sum_{n=1}^{N} \Lambda(n)^{2} = 2\left(1 - \frac{1}{N+1}\right) = \frac{2N}{N+1}$$ - Convergence in the limit as
$N \to \infty$ :$$\lim_{N \to \infty} \sum_{n=1}^{N} \Lambda(n)^{2} = \lim_{N \to \infty} \frac{2N}{N+1} = 2$$
Power series definition:
- Splitting the series:
$$G(z) = 2 \sum_{n=1}^\infty \frac{z^{n}}{n} - \frac{2}{z} \sum_{n=1}^\infty \frac{z^{n+1}}{n+1}$$ - Mercator series for
$-\ln(1-z)$ (Mercator series):$$\sum_{n=1}^\infty \frac{z^{n}}{n} = -\ln(1-z)$$ $$\sum_{n=1}^\infty \frac{z^{n+1}}{n+1} = \sum_{k=2}^\infty \frac{z^{k}}{k} = -\ln(1-z) - z$$ - Substitution and algebraic simplification:
$$G(z) = -2\ln(1-z) - \frac{2}{z}\left(-\ln(1-z) - z\right) = -2\ln(1-z) + \frac{2\ln(1-z)}{z} + 2$$ $$G(z) = 2 + \frac{2(1-z)\ln(1-z)}{z}$$
- Definition of the Euler Beta function
$B(x, y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}$ (Beta function and Gamma function): Setting parameter$y = 2$ :$$B(x, 2) = \frac{\Gamma(x)\Gamma(2)}{\Gamma(x+2)} = \frac{\Gamma(x) \cdot 1!}{(x+1)x,\Gamma(x)} = \frac{1}{x(x+1)}$$ - Continuous representation:
$$\Lambda(x) = \sqrt{2,B(x, 2)} = \sqrt{\frac{2,\Gamma(x)}{\Gamma(x+2)}} = \sqrt{\frac{2}{x(x+1)}}, \quad x > 0$$ - Derivation of the logarithmic derivative
$\frac{d}{dx} \ln \Lambda(x)$ :$$\ln \Lambda(x) = \frac{1}{2}\ln 2 - \frac{1}{2}\ln x - \frac{1}{2}\ln(x+1)$$ $$\frac{d}{dx} \ln \Lambda(x) = -\frac{1}{2x} - \frac{1}{2(x+1)}$$
The integer
- Triangular Number (
$T_{6}$ )$$T(6) = \frac{6 \times 7}{2} = 21$$ - Binomial Coefficient
$\binom{7}{2}$ $$\binom{7}{2} = \frac{7!}{2!,5!} = \frac{7 \times 6}{2} = 21$$ - Octagonal Number (
$O_{3}$ ) The$s$ -gonal formula is$P(s,n)=((s-2)n^{2}-(s-4)n)/2$ (polygonal number). For$s=8$ :$$P(8,3) = 3(3\cdot 3-2) = 21$$ - Icosihenagonal (21-gonal) Number (
$I_{21}(2)$ )$$P(21, 2) = \frac{19(2^{2}) - 17(2)}{2} = \frac{76 - 34}{2} = 21$$ - Fibonacci Number (
$F_{8}$ ) With$F_{0}=0$ and$F_{1}=1$ (Fibonacci sequence):$$F_{0}=0, F_{1}=1, F_{2}=1, F_{3}=2, F_{4}=3, F_{5}=5, F_{6}=8, F_{7}=13, F_{8}=21$$ - Stirling Number of the Second Kind
$S(7, 6)$ $$S(n, n-1) = \binom{n}{2} \implies S(7, 6) = \binom{7}{2} = 21$$ - Unsigned Stirling Number of the First Kind
$c(7, 6)$ $$c(n, n-1) = \binom{n}{2} \implies c(7, 6) = \binom{7}{2} = 21$$ - Motzkin Number (
$M_{5}$ ) The recurrence is$M_{n} = M_{n-1} + \sum_{i=0}^{n-2} M_{i} M_{n-2-i}$ , and the sequence begins$1,1,2,4,9,21$ (Motzkin number):$$M_{0}=1, M_{1}=1, M_{2}=2, M_{3}=4, M_{4}=9, M_{5}=21$$ - Standard Young Tableaux
$f^{(3,3,1)}$ (Hook Length Formula)$$f^\lambda = \frac{n!}{\prod h_{\lambda}(i,j)}$$ (hook-length formula) For partition$\lambda = (3,3,1)$ with$n = 7$ :- Row 1 hook lengths:
$(5, 3, 2)$ - Row 2 hook lengths:
$(4, 2, 1)$ - Row 3 hook lengths:
$(1)$ $$\prod h_{(i,j)} = 5 \times 3 \times 2 \times 4 \times 2 \times 1 \times 1 = 240 \implies f^{(3,3,1)} = \frac{7!}{240} = \frac{5040}{240} = 21$$
- Row 1 hook lengths:
- Standard Young Tableaux
$f^{(3,2,2)}$ For partition$\lambda = (3,2,2)$ with$n = 7$ :- Row 1 hook lengths:
$(5, 4, 1)$ - Row 2 hook lengths:
$(3, 2)$ - Row 3 hook lengths:
$(2, 1)$ $$\prod h_{(i,j)} = 5 \times 4 \times 1 \times 3 \times 2 \times 2 \times 1 = 240 \implies f^{(3,2,2)} = \frac{5040}{240} = 21$$
- Row 1 hook lengths:
The Kronecker page identifies the real primitive character of discriminant
- The equality
$\lvert g(\chi)\rvert=\sqrt{N}$ requires a primitive character, equivalently conductor equal to modulus (Gauss sum and Dirichlet character). Being square-free does not by itself prove primitivity. The conductor statement above supplies that hypothesis, so$\lvert g(\chi_{21})\rvert=\sqrt{21}$ . For coprime moduli, the Gauss-sum page includes the phase$\chi(N')\chi'(N)$ . Here$\chi_3(7)\chi_7(3)=-1$ :$$g(\chi_{21}) = \chi_{3}(7)\chi_{7}(3),g(\chi_{3})g(\chi_{7}) = -g(\chi_{3})g(\chi_{7}).$$ - Modulus square of primitive real Dirichlet characters modulo prime
$p$ :$$\lvert g(\chi_{p})\rvert^{2} = p \implies \lvert g(\chi_{3})\rvert^{2} = 3, \quad \lvert g(\chi_{7})\rvert^{2} = 7$$ - Product of moduli:
$$\lvert g(\chi_{21})\rvert^{2} = 3 \times 7 = 21 \implies \lvert g(\chi_{21})\rvert = \sqrt{21}$$ $$\Lambda(6) = \frac{1}{\lvert g(\chi_{21})\rvert} = \frac{1}{\sqrt{21}}$$
For the real quadratic field
- Fundamental discriminant
$d = 21$ ($21 \equiv 1 \pmod 4$ ) - Fundamental unit
$\varepsilon_{21} = (5 + \sqrt{21})/2$ , of norm$+1$ , since$5^{2} - 21\cdot 1^{2} = 4$ . The continued fraction of$\sqrt{21}$ is$[4;\overline{1,1,2,1,1,8}]$ , so the repeating block has length$6$ and the period is$r=6$ . Counting from the initial term, the sixth convergent, written$h_{5}/k_{5}$ , is$55/12$ , and$55^{2}-21\cdot 12^{2}=1$ . Because$r$ is even, the fundamental-solution rule selects convergent index$r-1=5$ , which is the same fraction (continued fraction). The solution is the cube of the unit:$\varepsilon_{21}^{3}=55+12\sqrt{21}$ (Pell's equation). The class-number formula uses$\ln \varepsilon_{21}$ , not$\ln(55 + 12\sqrt{21})$ , which is three times as large (class number formula and fundamental unit). - Class number
$h(21) = 1$ (class-number-one fields). In the formula below,$\chi_{21}(m)=\left(\frac{21}{m}\right)$ is the same character as in section 3.1 (class number formula).
The analytic Dirichlet class number formula is:
Dimension formula for cusp forms of even weight
Geometric parameters of modular curve
- Modular index:
$\mu = [\mathrm{SL}_{2}(\mathbb{Z}) : \Gamma_{0}(6)] = 6 \left(1 + \frac{1}{2}\right)\left(1 + \frac{1}{3}\right) = 12$ - Elliptic points: Stein sets
$\mu_{0,2}(N)=0$ when$4 \mid N$ , and otherwise$\mu_{0,2}(N)=\prod_{p \mid N}\left(1+\left(\frac{-4}{p}\right)\right)$ . Since$4 \nmid 6$ , the product is required, and$\left(\frac{-4}{2}\right)=0$ makes$\nu_{2}=0$ . Stein sets$\mu_{0,3}(N)=0$ when$2 \mid N$ , so$\nu_{3}=0$ (Stein, definitions preceding Proposition 6.1) - Cusps:
$\nu_{\infty} = \sum_{d \mid 6} \phi(\gcd(d, 6/d)) = \phi(1) + \phi(1) + \phi(1) + \phi(1) = 4$ - Riemann surface genus
$g$ :$$g = 1 + \frac{\mu}{12} - \frac{\nu_{2}}{4} - \frac{\nu_{3}}{3} - \frac{\nu_{\infty}}{2} = 1 + \frac{12}{12} - 0 - 0 - \frac{4}{2} = 0$$ - Evaluation at weight
$k = 24$ :$$\dim S_{24}(\Gamma_{0}(6)) = (24-1)(0-1) + 0 + 0 + \left(\frac{24}{2} - 1\right) \times 4 = -23 + (11 \times 4) = -23 + 44 = 21$$ $$\Lambda(6) = \frac{1}{\sqrt{\dim S_{24}(\Gamma_{0}(6))}} = \frac{1}{\sqrt{21}}$$
Euler formula for even integer values of the Riemann zeta function (particular values):
- For
$k = 2$ ($B_{4} = -1/30$ ):$$\zeta(4) = \frac{(2\pi)^{4} (1/30)}{2(24)} = \frac{16\pi^{4}}{1440} = \frac{\pi^{4}}{90}$$ - For
$k = 3$ ($B_{6} = 1/42$ ):$$\zeta(6) = \frac{(2\pi)^{6} (1/42)}{2(720)} = \frac{64\pi^{6}}{60480} = \frac{\pi^{6}}{945}$$ - Ratio of zeta values:
$$\frac{\zeta(6)}{\zeta(4)} = \frac{\pi^{6} / 945}{\pi^{4} / 90} = \pi^{2} \cdot \frac{90}{945} = \frac{2\pi^{2}}{21}$$ - Substitution of
$\Lambda(6)^{2} = 1/21$ :$$\frac{\zeta(6)}{\zeta(4)} = 2\pi^{2} \Lambda(6)^{2} \implies \Lambda(6) = \sqrt{\frac{\zeta(6)}{2\pi^{2} \zeta(4)}} = \frac{1}{\sqrt{21}}$$
- Transformer Attention Scaling (Vaswani et al. 2017)
$$A(Q, K) = \mathrm{softmax}\left(\frac{QK^{T}}{\sqrt{d_{k}}}\right)$$ For key projection dimension$d_{k} = 21$ :$$\text{Scale} = \frac{1}{\sqrt{d_{k}}} = \frac{1}{\sqrt{21}} = \Lambda(6)$$ - Xavier and Glorot Weight Initialization (Glorot and Bengio 2010)
The cited initialization is uniform on
$[-\sqrt{6/(n_{in}+n_{out})}, \sqrt{6/(n_{in}+n_{out})}]$ . The endpoint equals$\Lambda(6)$ when$n_{in}+n_{out} = 126$ . The Gaussian shorthand$\sigma = \sqrt{1/n_{in}}$ is a later variant, not the formula in that paper. - He (Kaiming Normal - ReLU) Weight Initialization (He et al. 2015)
$$\sigma = \sqrt{\frac{2}{n_{l}}}$$ For fan-in$n_{l} = 42 = 2 \cdot T(6)$ :$$\sigma = \sqrt{\frac{2}{42}} = \sqrt{\frac{1}{21}} = \Lambda(6)$$ - Diffusion Noise Schedule (Ho, Jain, and Abbeel 2020)
$$\sigma_{t} = \sqrt{1 - \bar{\alpha}{t}}$$
The value
$\bar{\alpha}_{t} = 20/21$ is chosen here. It is not a schedule value derived in that paper: $$\sigma{t} = \sqrt{1 - \frac{20}{21}} = \sqrt{\frac{1}{21}} = \Lambda(6)$$ - Echo State input weights (Jaeger 2001 and corrected report) and sample-mean standard error
For independent observations, the standard error page gives
$\sigma_{\bar x}=\sigma/\sqrt{n}$ . The factor$1/\sqrt{n}$ is that standard error only when the population standard deviation is also$1$ . It is not the definition of Rademacher complexity. Jaeger's echo-state example sets input weights to$+1$ or$-1$ with equal probability.$1/\sqrt{n}$ is not an input scale prescribed by that report:$$\text{SE}(\bar{X}) = \frac{1}{\sqrt{n}}$$ For$n = 21$ independent zero-mean unit-variance variables:$$\text{SE} = \frac{1}{\sqrt{21}} = \Lambda(6)$$
- Coefficient of Variation (
$\mathrm{CV}$ )$$\mathrm{CV} = \frac{\sigma}{\mu}$$ - Poisson distribution (Poisson distribution,
$\lambda = 21$ ):$\mu = 21$ and$\sigma^{2} = 21$ , so$$\mathrm{CV} = \frac{\sqrt{21}}{21} = \frac{1}{\sqrt{21}} = \Lambda(6).$$ - Skewness of Poisson distribution (
$\lambda = 21$ ):$\gamma_{1} = \frac{1}{\sqrt{\lambda}} = \frac{1}{\sqrt{21}} = \Lambda(6)$ - Gamma distribution (gamma distribution,
$\alpha = 21, \beta = 1$ ):$\mu = 21$ and$\sigma^{2} = 21$ , so$$\mathrm{CV} = \frac{\sqrt{21}}{21} = \frac{1}{\sqrt{21}} = \Lambda(6).$$ - Chi-Square distribution (chi-squared distribution,
$k = 42$ ):$\mu = 42$ and$\sigma^{2} = 84$ , so$$\mathrm{CV} = \frac{\sqrt{84}}{42} = \frac{2\sqrt{21}}{42} = \frac{1}{\sqrt{21}} = \Lambda(6).$$ - Binomial distribution (binomial distribution,
$n = 21, p = 0.5$ ):$\mu = 10.5$ and$\sigma^{2} = 5.25$ , so$$\mathrm{CV} = \frac{\sqrt{5.25}}{10.5} = \frac{1}{\sqrt{21}} = \Lambda(6).$$ - Erlang distribution (Erlang distribution,
$k = 21, \lambda$ ):$\mu = k/\lambda$ and$\sigma = \sqrt{k}/\lambda$ , so$$\mathrm{CV} = \frac{\sqrt{21}/\lambda}{21/\lambda} = \frac{1}{\sqrt{21}} = \Lambda(6).$$
- Poisson distribution (Poisson distribution,
- Information Geometry Metrics (Poisson Family)
- Fisher information for one
$\text{Poisson}(\lambda)$ observation (Fisher information):$I(\lambda) = \frac{1}{\lambda}$ , so$$\sqrt{I(21)} = \frac{1}{\sqrt{21}} = \Lambda(6).$$ The metric component is$I(\lambda)$ , while$\sqrt{I(\lambda)},d\lambda$ is the line element. - Jeffreys Prior (Jeffreys prior):
$\pi(\lambda) \propto \sqrt{g(\lambda)} = \frac{1}{\sqrt{\lambda}}$ , so$$\pi(21) \propto \frac{1}{\sqrt{21}} = \Lambda(6).$$
- Fisher information for one
- Schwarzschild time-dilation factor (gravitational time dilation)
For a static exterior observer, proper time and distant Schwarzschild coordinate time satisfy
$t_{0}=t_{f}\sqrt{1-r_{s}/r}$ . The redshift measured from infinity is the reciprocal,$1+z = 1/\sqrt{1-r_{s}/r}$ , so$z$ itself is not that factor (gravitational redshift). At the chosen radius$r = \frac{21}{20} r_{s}$ :$$\sqrt{1 - \frac{r_{s}}{r}} = \sqrt{1 - \frac{20}{21}} = \sqrt{\frac{1}{21}} = \Lambda(6).$$ In units$c=r_{s}=1$ , writing$\Phi=-1/(2r)=-10/21$ makes$\sqrt{1+2\Phi}=\Lambda(6)$ by algebra. That equality does not make$\Phi$ the Newtonian-limit potential:$r=1.05r_{s}$ is not a weak field. It is not a measured interval near a physical black hole, and it lies inside the photon sphere. The static formula does not describe a circular orbit, whose factor is$\sqrt{1-\frac{3}{2}r_{s}/r}$ . - Reciprocal of the Lorentz factor (Lorentz factor)
The page defines
$\gamma = 1/\sqrt{1-v^{2}/c^{2}}$ . The quantity below is$1/\gamma$ , not$\gamma$ :$$\gamma^{-1} = \sqrt{1 - \frac{v^{2}}{c^{2}}}$$ For velocity$v = \sqrt{\frac{20}{21}} c$ :$$\gamma^{-1} = \sqrt{1 - \frac{20}{21}} = \sqrt{\frac{1}{21}} = \Lambda(6)$$ - Harmonic Oscillation Period (harmonic oscillator)
For an undamped harmonic oscillator
$\ddot{x} + \omega^{2} x = 0$ with chosen stiffness-to-mass ratio$\omega^{2} = k/m = 21$ :$$T = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{21}} = 2\pi \Lambda(6)$$ - Molecular Vibrational Degrees of Freedom (molecular vibration)
For non-linear molecules with
$N$ atoms, degrees of freedom are$3N - 6$ . For$N = 9$ :$$3(9) - 6 = 27 - 6 = 21$$ - Adiabatic index (heat capacity ratio)
$$\gamma = 1 + \frac{2}{f}$$ For effective internal degrees of freedom$f = 2\sqrt{21}$ :$$\gamma = 1 + \frac{2}{2\sqrt{21}} = 1 + \frac{1}{\sqrt{21}} = 1 + \Lambda(6)$$
-
Second-Order System Percentage Overshoot (overshoot)
Damping ratio formula:
$$%OS = \exp\left(-\frac{\pi \zeta}{\sqrt{1 - \zeta^{2}}}\right) \times 100%$$ For damping ratio$\zeta = \Lambda(6) = \frac{1}{\sqrt{21}}$ :$$\zeta^{2} = \frac{1}{21} \implies 1 - \zeta^{2} = \frac{20}{21} \implies \sqrt{1 - \zeta^{2}} = \frac{\sqrt{20}}{\sqrt{21}}$$ $$\frac{\zeta}{\sqrt{1-\zeta^{2}}} = \frac{1/\sqrt{21}}{\sqrt{20}/\sqrt{21}} = \frac{1}{\sqrt{20}}$$ $$%OS = \exp\left(-\frac{\pi}{\sqrt{20}}\right) \times 100% = 49.5354568\ldots%,$$ displayed as$49.54%$ . The intermediate value$\exp(-\pi/\sqrt{20})=0.495354568\ldots$ is not$0.70248$ . -
Quality Factor (Q factor)
$$Q = \frac{1}{2\zeta}$$ For$\zeta = \Lambda(6) = \frac{1}{\sqrt{21}}$ :$$Q = \frac{1}{2(1/\sqrt{21})} = \frac{\sqrt{21}}{2} \approx 2.2913$$ -
Fixed-length alphabet ceiling (prefix code) and Hash Table Load Factor (hash table)
- A fixed-length code of
$k$ bits encodes at most$2^{k}$ symbols, so$D = \lceil \log_{2} N \rceil$ is that ceiling, not the Huffman weighted path length (Huffman coding). For$N = 21$ symbols:$$\log_{2}(21) \approx 4.3923 \implies \lceil 4.3923 \rceil = 5\text{ bits}$$ - Hash Table Load Factor:
$\alpha = n/m$ . Choosing$m = 21$ buckets and$n = \sqrt{21}$ stored elements gives$\alpha = \Lambda(6)$ . A real table has an integer element count:$$\alpha = \frac{\sqrt{21}}{21} = \frac{1}{\sqrt{21}} = \Lambda(6)$$
- A fixed-length code of
-
First-Order Reaction Rate and Half-Life (rate equation)
$$k = \frac{\ln 2}{t_{1/2}}$$ For half-life$t_{1/2} = \Lambda(6) = \frac{1}{\sqrt{21}}$ :$$k = \frac{\ln 2}{1/\sqrt{21}} = \ln 2 \cdot \sqrt{21} \approx 3.1764$$ -
Michaelis-Menten Enzyme Kinetics (Michaelis and Menten 1913, English translation) and Bioavailability
- Normalized reaction velocity:
$$\frac{v}{V_{\max}} = \frac{[S]}{K_{m} + [S]} = \frac{[S]/K_{m}}{1 + [S]/K_{m}}$$ Setting the substrate ratio to$[S]/K_{m} = 1/\sqrt{21}$ does not make the velocity ratio equal$\Lambda(6)$ :$$\frac{v}{V_{\max}} = \frac{1/\sqrt{21}}{1 + 1/\sqrt{21}} = \frac{1}{\sqrt{21} + 1} \approx 0.179129.$$ - Absolute bioavailability (bioavailability) is the dose-normalized ratio of extravascular to intravenous area under the concentration curve. Setting that fraction to
$\Lambda(6)$ gives$F = 1/\sqrt{21} \approx 21.82%$ . This is an input choice, not a measured drug value. - Bazett rate correction (Bazett 1920), in dimensionally consistent form:
$$\mathrm{QTc} = \frac{\mathrm{QT}}{\sqrt{\mathrm{RR}/1,\mathrm{s}}}.$$ Choosing$\mathrm{RR} = 1/\sqrt{21},\mathrm{s}$ is an input substitution. It is not a measured cardiac interval, and a bare$\mathrm{QT}/\sqrt{\mathrm{RR}}$ is dimensionally inconsistent.
- Normalized reaction velocity:
-
Clinical Trial Sample Size per Group (Cohen, Statistical Power Analysis, 2nd ed.)
$$n = \frac{2(z_{\alpha} + z_{\beta})^{2}}{d^{2}}$$ For Cohen's effect size$d = \Lambda(6) = \frac{1}{\sqrt{21}}$ , it follows that$d^{2} = \frac{1}{21}$ :$$n = 2 \cdot 21 (z_{\alpha} + z_{\beta})^{2} = 42(z_{\alpha} + z_{\beta})^{2}$$ -
Kelly Criterion Fraction (Kelly criterion and Kelly 1956) The cited page's binary formula is
$f = p/l - q/g$ . Kelly's 1956 paper does not write that formula. Its even-money calculation gives the same fraction,$\ell = 2q - 1$ , after swapping its win and loss labels. For even money,$g = l = 1$ , so$$f = p - q = 2p - 1.$$ Setting that edge to$\Lambda(6) = 1/\sqrt{21}$ gives$$p = \frac{1 + \Lambda(6)}{2} = \frac{1 + 1/\sqrt{21}}{2} \approx 0.609109.$$ The odds form$(bp - q)/b$ is this case only when the whole stake is lost and the net gain on a win is$b$ . It is not the formula displayed by either cited source. -
Transmission-Line Reflection and Impedance Ratio (reflection coefficient) The cited page displays the load reflection coefficient
$\Gamma=(Z_{L}-Z_{0})/(Z_{L}+Z_{0})$ , not an acoustic pressure-amplitude formula. Its linked acoustic section instead defines$R$ as the ratio of reflected to incident intensity, written there as$R=p_{\mathrm{reflected}}/p_{\mathrm{incident}}$ , and then uses$\alpha=1-R^{2}$ . For a real ratio$z=Z_{L}/Z_{0}$ ,$$\Gamma = \frac{z - 1}{z + 1}.$$ For$\Gamma = \Lambda(6) = \frac{1}{\sqrt{21}}$ :$$z = \frac{1 + \Gamma}{1 - \Gamma} = \frac{\sqrt{21} + 1}{\sqrt{21} - 1} = \frac{(\sqrt{21}+1)^{2}}{20} = \frac{22 + 2\sqrt{21}}{20} = \frac{11 + \sqrt{21}}{10} \approx 1.558258$$ -
Planetary, Geophysics, and Neutrino Metrics
- Kepler's third law in solar units (Kepler's laws), where the primary has one solar mass:
$T^{2} = a^{3}$ with$T$ in years and$a$ in AU. For orbital period$T = 21$ yr:$$a = 21^{2/3} = \sqrt[3]{441} \approx 7.61166\text{ AU}$$ - Gutenberg-Richter law (Gutenberg-Richter law) with chosen baseline parameters
$a = 5.0$ ,$b = 1.0$ , and chosen cumulative event rate$N = 1/\sqrt{21}$ :$$M = \frac{a - \log_{10}(1/\sqrt{21})}{b} = 5.0 + \log_{10}(\sqrt{21}) = 5.0 + 0.6611 = 5.6611 \approx 5.66$$ - Effective neutrino number: Akita and Yamaguchi 2020 calculate
$N_{\mathrm{eff}} = 3.044$ , with numerical and mixing error at most$0.0005$ . The numerical proximity$\ln(21) \approx 3.04452$ is not an identity and is not the definition of$N_{\mathrm{eff}}$ .
- Kepler's third law in solar units (Kepler's laws), where the primary has one solar mass:
-
Game Theory and Economics (Cournot Equilibrium and First-Price Auction)
- Cournot Oligopoly (Cournot 1838 and linear identical-firm case Marker): for inverse demand
$p = a-bQ$ and common marginal cost$c$ , each equilibrium output is$q_i = (a-c)/((n+1)b)$ . For$n = 20$ ,$q_i/((a-c)/b) = 1/21$ . - First-Price Sealed-Bid Auction (first-price sealed-bid auction): the page states the symmetric Bayesian Nash equilibrium
$b(v) = \frac{n-1}{n} v$ for valuations that are i.i.d. uniform on$[0,1]$ . Its general symmetric BNE is$E[y_i \mid y_i<v_i]$ . For$n = 21$ bidders, the uniform strategy is$b(v) = \frac{20}{21} v$ . - Herd Immunity Threshold (herd immunity):
$p_c = 1 - \frac{1}{R_{0}}$ under homogeneous mixing, solid immunity, no immune escape, and no nonhuman vector. For$R_{0} = 21$ , that threshold is$\frac{20}{21}$ . - Autoregressive Time Series AR(1) (autoregressive model): for
$|\phi|<1$ , stationary variance$\mathrm{Var}(X_{t}) = \frac{\sigma_{\varepsilon}^{2}}{1 - \phi^{2}}$ . For$\phi = \frac{1}{\sqrt{21}}$ ,$\mathrm{Var}(X_{t}) = \frac{21}{20} \sigma_{\varepsilon}^{2}$ .